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Solution

(a)

2u 3v+ w= 2(5, 1, 7, 2) 3(0, 1, 0, 4) + (0, 0, 0, 0)= (10, 2, 14, 4) (0, 3, 0, 12)= (10, 1, 14, 16).

(1)

(b)

u + 2v+ w= (5, 1, 7, 2) + 2(0, 1, 0, 4) + (0, 0, 0, 0)= (5, 1, 7, 2) + (0, 2, 0, 8)= (5, 3, 7, 6).

(2)

(c)

0u + 0v+ 1765w= (0, 0, 0, 0) (3)

(d)

(u + w) (u + 3w) = ((5, 1, 7, 2) + (0, 0, 0, 0)) ((5, 1, 7, 2) 3(0, 0, 0, 0))= (0, 0, 0, 0).

(4)

Exercise 3 Areu= (1, 0, 0), v= (1/2, 1/2, 0) andw= (3, 1, 0) coplanar?

We can immediately see that the answer is yes, since all these vectors has null

component in the direction

k, then they lie on the plane xy. However, let us

prove it using algebra, that is by computing the determinant of the three vectors.We know indeed that the vectors are coplanar if and only if their determinantis null.

u v w=

u1 u2 u3v1 v2 v3w1 w2 w3

=u1 v2 v3w2 w3

u2 v1 v3w1 w3

+ u3 v1 v2w1 w2

and v1 v2w1 w2

=v1w2 w1v2.In our case

u v w=

1 0 01/2 1/2 0

3 1 0

= 1 1/2 01 0

= 0Other way: notice that w = 2u + 2v, that isw is a linear combination ofu

and w. This property again is equivalent to coplanarity.In terms of dimensions, what does to be coplanar means? It means that u,

vand w belong to a subspace of dimension 2 in R3. A plane in R3 is a subspaceof dimension 2. Can we find a basis for this subspace? We just need two vectorsbelonging to it that are linearly independent: for example, B ={u, v}.

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2 Week 2: straight lines and planes in R3

Notation

1. Vector: u= (u1, . . . , uN)RN.2. Point: P = (p1, . . . , pN) AN, where AN is the affine space modeled

on RN. Sometimes in the exercises the notationP = (p1, . . . , pN) RNoccurs, which sounds confusing (same notation for points and vectors!),but actually the context always makes clear whether a point or a vectoris meant.

3. Let u(1), . . . , u(m) RN, m1.[u(1), . . . , u(m)] = span(u(1), . . . , u(m)) :=

{v

R

N : v = t1u(1) +

t2u(2) + tmu(m), t1, . . . , tm R}.4. Given two points P andQ, P Q= Q P.

Reminder 2 (Straight Line) A straight line passing through a point P =(p1, . . . , pN) AN and parallel to span(u), u = (u1, . . . , uN) RN is the setlP,u={P+ tu, tR}. A line is determined also by a couple of pointsP0, P1:lP0,P1 ={P0+ t(P1 P0)), tR}

Cartesian Equation of a straight line in R2: a generic point of a line hascoordinates(x, y) = (p1, p2) + (u1, u2)t. Let us eliminate the parametert: fromthe first component of the previous vector equation, we obtain t = xp1

u1; by

substitution in the second component we have y = p2+ u2xp1u1

. Notice that

this equality corresponds to det

y p2 x p1u2 u1

= 0.

What happens inRN, N >2?Let us consider N = 3: from (x,y,z) = (p1, p2, p3) + (u1, u2, u3)t, by the

same procedure, we get two equations y = p2+ u2xp1u1

and z = p3+ u3yp2u2

.

The Cartesian expression of a straight line inR3 is a system between two lin-ear equations inx,y, z (in the next, we will notice that this corresponds to theintersection between two planes).

Reminder 3 (Plane) A plane passing through a point P RN and orientedaccording to span(u, v), u, v RN is the set P,u,v ={P +u+ v, , R}. A plane is determined also by a triple of (non collinear) pointsP0, P1.P2:P0,P1,P2 =

{P0+ (P1

P0)) + (P2

P0), ,

R

}.

Cartesian Equation of a plane inR3: (x,y,z) =P+ u + v (x,y,z) P0, u v= 0.Exercise 6 Determine whetherL1 andL2 are parallel or coincident:(a) L1: (1, 2, 2) + t(3, 5, 7), L2: (7, 12, 16) + t(6, 10, 14);(b) L1: (3, 4, 2) + t(1, 0, 7), L2: (6, 8, 4) + t(2, 0, 14).

Solution(a) First, L1 and L2 are parallel since (6, 10, 14) = 2(3, 5, 7). Then, if they in-tersect in a point, they turn out to be coincident. Let us solve the vector equa-tion: (1, 2, 2) +t1(3, 5, 7) = (7, 12, 16) +t2(6, 10, 14), that is, (6, 10, 14) =

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(t1+ 2t2)(3, 5, 7). Then if 2 =t1+ 2t2 we obtain a common point (for instance,

t1= 2 and t2= 0). This is sufficient to state that L1 and L2 are coincident.(b) L1 and L2 are parallel, but not coincident, in fact:

(3, 4, 2) + t1(1, 0, 7) = 2(3, 4, 2) + 2t2(1, 0, 7)(3, 4, 2) + (2t2 t1)(1, 0, 7) = 0

corresponds to

3 + 2t2 t1= 04 + 0 = 02 + 7(2t2 t1) = 0

in which the second equality is false. Hence, the system has no solution and thelines do not intersect.

Exercise 7 Determine a vector equation of the lineL passing throughP1 and

P2 and determine whether the pointQ is onL:(a) P1= (3, 5, 1), P2= (2, 0, 7), Q= (0, 10, 19)(b) P1= (2, 0, 5), P2= (1, 0, 0), Q= (4, 2, 3)(c) P1= (4, 1, 5), P2= (6, 2, 3), Q= (0, 1, 2).

SolutionA vector equation of the (unique) straight line passing through two points P1,

P2 isOP1+ t

P1P2, Moreover, Q is on the line ifQ =

OP1+ t

P1P2 for some t.

(a)L : (3, 5, 1) + t(1, 5, 6) and (3, 5, 1) + t(1, 5, 6) = (0, 10, 19) fort = 3,henceQ lies on L;(b)L : (2, 0, 5) + t(3, 0, 5) and (2, 0, 5) + t(3, 0, 5)= (4, 2, 3) for anytR, henceQ is not on L;(c)L : (4, 1, 5) + t(2, 3, 8) and (4, 1, 5) + t(2, 3, 8)= (0, 1, 2) for anytR,henceQ is not on L.

Exercise 8 Determine whether the given lines intersect or are skew; if theyintersect, find the point of intersection.(a) L1: (3, 5, 1) + t(1, 0, 2), L2: (1, 2, 5) + t(4, 3, 2)(b)L1: line through(2, 3, 3)and(6, 1, 7), L2: line through(6, 2, 8)and(7, 2, 9).(c) L1: (2, 4, 3) + t(3, 5, 2), L2: (5, 1, 3) + t(2, 0, 1)

Solution

(a) First, L1 and L2 are not parallel since (1, 0, 2) and (4, 3, 2) are linearlyindependent (that is, there is no a R such that (1, 0, 2) =a(4, 3, 2)). To findthe possible point of intersection, we have to solve the vector equation:

(3, 5, 1) + t1(1, 0, 2) = (1, 2, 5) + t2(4, 3, 2).

Second component: t2= 1. Third component: t1t2= 2. t14t2=2. Thereis no solution.

(b) L1 : (2, 3, 3) + t(2, 1, 2) and L2 : (6, 2, 8) + t(1, 0, 1). Then, Thevector equation (2, 3, 3) +t1(2, 1, 2) = (6, 2, 8) +t2(1, 0, 1) has no solution.(2, 4, 3) + t1(3, 5, 2) = (5, 1, 3) + t2(2, 0, 1) has solutiont1= 1 andt2=2.

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Exercise 9 Determine whether the following sets of points inR3 are collinear

(that is, lie on the same straight line):(a) (2, 1, 4), (4, 4, 1), (6, 7, 6).(b) (1, 2, 3), (4, 2, 1), (1, 1, 2).(c) (1, 2, 3), (5, 4, 7), (3, 1, 5).

SolutionWe can solve this exercise by writing the straight line passing through two pointsand then checking if the third point lies on the this line.(a)L : (2, 1, 4)+t(2, 3, 5) is the line passing through the first two points; fixingt= 2 we obtain the third point, then the three points are collinear.(b) L : (1, 2, 3) + t(5, 0, 2); notice that y = 2 is constant in this line, then(1, 1, 2) cannot lie on L.

(c) L: (1, 2, 3) +t(2, 3, 2); when t= 1, we obtain (3, 1, 5), hence the pointsare collinear and lie on L.

Exercise 10 Find the linear equation for the planeHthat contains the point

Q= (4, 0, 7) and has the normal vectorn = 4j k.

SolutionLet us call P = (x,y,z) the generic point ofH, then

QP = (x 4, y , z 7) lies

onH. Furthermore,QP = (x 4, y , z 7) is orthogonal ton = (0, 4, 1), that

is (x 4, y , z 7) (0, 4 1) = 0. Thus, the equation ofH is

4y z+ 7 = 0.

Exercise 11 Find the vector equation for the plane containing P1 = (2, 1, 2),P2= (3, 4, 2) andP3= (1, 7, 2).

SolutionThe Cartesian equation is easily obtain from

P1P2 P1P3 P1P = 0. For any

u, v, wR3, we have u v w= u v w.

P1P2P1P3P1P = (1, 3, 0) (1, 6, 0) (x 2, y 1, z 2)=

1 1 x 23 6 y 20 0 z 2

= 6(z 2) 3(z 2) = 0

Hence, the Cartesian equation is z = 2 (x, y can assume any real value!). Theplane is parallel to the x, y-plane. Now let us choose (at random) a triple of noncollinear points ofz = 2, e.g., P1= (1, 0, 2),P2= (0, 1, 2) andP0= (0, 0, 2): wecan write the parametric expression as P0,P1,P2 : P0+ (P1 P0)) +(P2P0), , R, that is P0,P1,P2 : (0, 0, 2) + (1, 0, 0) + (0, 1, 0) = (,, 2).

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Exercise 12 Determine whether the planes2x + y z= 1and3x 2y + z = 4intersect in a line, if they do, find a vector equation for the line.

SolutionBy summing the equation, we obtain 5x y = 5. Then we can write, e.g., y infunction ofx and then z in function ofx. One variable (x in our case) remainsfree: the system has solution and the planes intersect. The intersection mightbe a plane (if the planes coincide: this is not the case) or a line. The fact thatit is a line relies on the fact that there is one free variable, hence we can writethe equation in function of just one parameter t, which is the typical case forlines. Letx = t, theny = 5t 5,z = 2t + 5t 5 1 = 7t 6 and the parametricequation of the line is

L: (0, 5, 6) + t(1, 5, 7).

Reminder 4 How to compute the distance between a point P and a plane :if n is the normal to andQ any point on, the distance betweenP andis equal to the component of the vectorP Q along the direction ofn. Such acomponent is computed using the scalar product:

d(P, ) =P Q, n

whereQ is any point of(recall thatn is a unit vector). If : ax +by+cz +d=0, we know thatn =

1a2+b2+c2

(a,b,c)T, therefore we easily obtain

d(P, ) =|a(x x0) + b(y y0) + c(z z0)|

a2 + b2 + c2

whereP = (x0, y0, z0) andQ= (x,y,z).

Reminder 5 How to compute the distance between a point P and a straightline l: given anyQl,

d(P, l) =P Q, nlwherenl is a unit vector orthogonal to l. LetQ+vt, t R be the parametricexpression forl. Then, we have that(v (Q P)) v is parallel to nl. Hence,we can writenl =

(v(QP))v|(v(QP))v| .

Using the fact thatu v, w=u, v w, we obtain

d(P, l) =

P

Q,

(v (Q P)) v|(v (Q P)) v|

=

|(Q

P)

v

|v| |.

Reminder 6 How to compute the distance between two skew lines l1 and l2with directional vectorsv1 andv2:

d(l1, l2) =|P2 P1, v1 v2|

|v1 v2|whereP1l1 andP2l2.

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3 Week 3: straight lines, planes, spheres and

circumferences in R3

Reminder 7 (Relationship between lines with cross/dot products) Therelationship between two lines can be studied in different ways. One way is touse only products: let us see a possible procedure.

Let l: P+ tu andr: Q + tv be two lines inR3.

u v

= 0 = 0

PQ u

PARALLEL

= 0

NOT PARALLEL

= 0

= 0= 0= 0

COINCIDENT

PARALLELINTERSECTING SKEW

DISTINCT

The proof of the different steps is easy and left to the reader.

Reminder 8 (Distance between skew lines) Let l : P +tu and r : Q+tv be two skew lines in R3. There exists only one straight line s : S+tnperpendicular to both of them. Notice thatn must be perpendicular to both uandv, then we compute it using the cross product. In particular, letn be theunit vector (this will be useful in the next)

n = u v|u v| .

The minimum distance between l and r is the length of the segment thatjoins l and r on that perpendicular line. We can compute it in the followingway. Keep l fixed and shift r, without changing its direction, along p until itmeets l. l and the shiftedr determine a plane that we name1. Analogously,keep r fixed and shift l without changing its direction, alongp until it meetsr:we obtain a plane2 which is parallel to1. Now, it is sufficient to compute the

distance between these two planes to get the minimum distance betweenl andr.Notice that we can write1: (x,y,z) n + d1= 0, 2: (x,y,z) n + d2= 0and also: P n + d1= 0 andQ n + d2= 0. In conclusion

d(l, r) =|d2 d1|=|P Q, n |.Exercise for the reader: prove that|d2 d1| is actually the distance between

1 and2. Hint: choose any point point of1 and then compute the distancebetween this point and2 using the known formula.

Exercise 13 Compute the distance between l : t(0, 2, 3) and r : 2x y+ z =3x + y 2z 1 = 0.

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Figure 1: Minimum distance between two skew lines

Solution

Let us write r in a parametric form. Summing 2x y+ z = 0 and 3x+y2z 1 = 0 we obtain 5xz = 1. Let us fix x = t, where t R is theparameter. Thenz = 5t1 and y = 2x+ z = 7t1. Hence, r is given by(x,y,z) = (0, 1, 1) + t(1, 7, 5). Do l and r have an intersection point? No,because t1(0, 2, 3) = (0, 1, 1) + t2(1, 7, 5) has no solution. Moreover, l andr are not parallel since there is no R such that (1, 7, 5) = (0, 2, 3). Inconclusionl and r are skew. Now,

u v= det

i 0 1

j 2 7k 3 5

= (11, 3, 2).

|u v|= 121 + 9 + 4 = 134Thenn = (11,3,2)

134 . Finally,

d(l, r) =|(0, 1, 1), n |= 1134

.

Reminder 9 (Sphere) A sphere in Rn is the set of all the points that havefixed distance from a given point, that is, givenC Rn and R > 0, sphere ={P Rn :|CP|= R. C is called the center of the sphere, R is the radius. InR3: C= (x0, y0, z0), P = (x,y,z), then the equation is(x x0)2 + (y y0)2 +(z z0)2 =R2: this should sound familiar.

Reminder 10 (Circumference inR3

) The intersection of a sphere and aplane in R3 is always a circumference. Any circumference in R3 can be ex-pressed as the intersection between a sphere and a plane.

Exercise 14 GivenS: x2 + 2x + y2 + z24z= 8, write the equation of a plane with normaln = (0, 3, 3)and such that S is a circumference of radius 1.

SolutionBy completion of the squares, we have: (x + 1)2 + y2 + (x2)2 = 8+ 1+ 4 = 13.Then C = (1, 0, 2) is the center of the sphere and R = 12 is its radius. Ingeneral, ifris the radius of the circumference, by Pythagoras, R2r2 =d(C, ).

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R

d(, C)

r

Figure 2: Projection of a sphere intersected by a plane

In this case, : 3y+ 3z+= 0 where the parameter Ris not known.Since the distance between a point (x0, y0, z0) and a planex + y + z + = 0

is |x0+y0+z0+|2+2+2

we have:

d(C, ) =|6 + |/

18

Therefore R2 r2 = 131 = 12 =|6 +|/18, hence =6362. Inconclusion, we have two planes that fulfill the required property: 1 : y +z

2 + 122 and 2: y + z 2 122.

4 Week 4: vector spaces

Reminder 11 (Vector space) An informal definition: a vector space is anyset which is closed with respect to two operations that have the known propertiesof sum and product.

A more formal definition: a vector space is a setVassociated with a fieldKsuch that we can define

(a) a binary1 operation, that we call sum and we indicate by +, betweentwo elements ofV with the following properties:

(a.1) closure: for anyu, v

V, u + v

V

(a.2) common properties: the sum is commutative, associative, and thereexists an additive identity (an element0V such thatv+ 0= v);

(a) a binary operation, that we call product, between an element of V andan element ofK with the following properties:

(a.1) closure: for anyuV, aK auV(a.2) common properties: the product is commutative, associative, distribu-

tive, and there exists a product identity (1 K such that 1u = u for anyuV).

1binary operation = operation between two objects

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Notice that the operations are not priorly defined. One can define them as

he prefers, provided that the required properties hold.Actually, we have been working with vector spaces since the beginning ofthe course: Rn with the usual operations is a vector space (on the field K = R).

From now on, we will call vectors the elements of a vector space. You cannow forget the physical interpretation of a vectors: in linear algebra, sequenceof numbers, polynomials, functions... are vectors!

Reminder 12 (Span set) We say that G V is a generating set (= set ofgenerators = span set, and we write[G] =V orV =spanG) forV if any vectorinVcan be written as linear combination of elements inV.

Next exercises will focus on the identification of vector spaces. Notice that inmany cases, we start by verifying the presence of a null element (= an additive

identity), since it is easy to check.Exercise 15 Determine whether of the following sets of functions is a vectorspace on the fieldR:

(a) R3[x] ={all the polynomials inx of degree3}.(b) R=3[x] ={all the polynomials inx of degree = 3}.(c) V ={all the functionsf C2[0, 1] such thatf(x) =x2f(x)}.(d) V ={all the functionsf C1[0, 1] such thatf(x) =f2(x)}.(e) V ={all the polynomials inx having a zero atx= 5}.(f ) V ={all the polynomials inx of kinda0+ a1x3, a0, a1 R}.(g) V ={all the polynomials inx of kinda0+ a1x3, a0, a1 R, a0= 0}.(h) V ={all the functionsf C[0, 1] that are monotone increasing}.

Solution

(a) A generic element ofR3[x] has the form a0 + a1x + a2x2 + a3x3,ai R,i= 0, 1, 2, 3.

Sum: we use the usual sum for polynomials. For any ai, bi R, i =0, 1, 2, 3, a0+ a1x+a2x

2 +a3x3 +b0+ b1x+b2x

2 +b3x3 = (a0+ b0) +

(a1+ b1)x + (a2+ b2)x2 + (a3+ b3)x

3 R3[x]closure w.r.t. sum; Null element: P(x)0 R3[x] is the additive identity for the sum; Product: R,(a0+a1x+a2x2+a3x3) =a0+a1x+a2x2+a3x3 R3[x]closure w.r.t. product.

This is enough to state thatR3[x] is a vector space.(b) P(x) 0 / R=3[x]: this is enough to conclude that R=3[x] is not a

vector space. If you want on other counterexample: x3 R=3[x],x3 +x2 R=3[x], but their sum / R=3[x].

(c)

f0(x)0V since trivially f0(x) =f0(x) =f0(x) = 0 If f(x), g(x) V, then f(x) +g(x) = h(x) V by the linearity of the

derivative operation: h(x) =f(x) + g(x) =x2(f(x) + g(x)) =x2h(x).

Analogously for the product: g(x) =cf(x),cR, theng (x) =cf(x) =cx2f(x) =x2g(x).

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Hence,V is a vector space.

(d)Vcontains the null elementf0(x)0, but it is not closed with respect tothe sum: given f(x), g(x)V,f(x) + g(x) =h(x) thenh(x) =f(x) + g(x) =f2(x) + g2(x)=h2(x) =f2(x) + g2(x) + 2f(x)g(x).

(e)P0(x)0V since triviallyP0(0) = 0. Then, let sum two vectors ofC,say two polynomials P1(x) and P2(x) such that P1(5) =P2(5) = 0: we obtainP1(x) + P2(x) =P3(x)V asP3(5) =P1(5)+ P2(5) = 0. Finally, let check theproduct: given cR, P1(x)V, cP1(x) =Q(x)V, as Q(5) =cP1(5) = 0. Vis a vector space.

(f) Analogous to (a): V is a vector space.(g) Analogous to (b): there is no null element, V is not a vector space.(h) Letfbe monotone increasing: if I multiply it by the scalar

1, I obtain

a decreasing function: V is not closed w.r.t. product, hence it is not a vectorspace.

Exercise 16 DoesG={1, x , x2, x3 x} spanR3[x]?AndG={x, x2, x3 x}?AndG={1, x , x2, x3 x, x + 3}?

SolutionG spans R3[x] if any vector R3[x] can be written as linear combination ofvectors inG. That is, given a generica0 + a1x + a2x

2 + a3x3, there existbi R,

i= 0, 1, 2, 3 such that I can write it as b0(1) + b1x + b2x2 + b3(x3 x)?We recall that two polynomials are equal when the coefficients of same degree

are equal, then

a0+ a1x + a2x2 + a3x

3 =b0(1) + b1x + b2x2 + b3(x3 x)corresponds to the linear system (in the unknowns bi, i = 0, 1, 2, 3)

a0=b0a1= b1 b3a2= b2a3= b3

which has solution

b0=a0b1= a1+ a3b2= a2b3= a3.

As the system has solution, G spans R3[x]. Notice that, fixed ai, i =0, 1, 2, 3, the system has a unique solution, which implies that there is only onepossible representation for each vector ofR3[x] as linear combination of vectorsofG.

Gdoes not span R3[x] asP(x)1 cannot be written as linear combinationof vectors ofG.

G spans R3[x] in a redundant way: for some v V there is morethan one representation as as linear combination of vectors ofG. For example:

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2x = 2(x) but also 2x =2(x+ 3) 6(1). This can seen also solving thelinear system as done before: we obtain a linear system with 4 equations and 5unknowns.

This idea of redundancy is very important. We now introduce the conceptof basis, which is an non redundant generating set.

Reminder 13 (Basis) Given a vector spaceV, BV is a basis forV if (a)V =spanB and (b)B is linearly independent (l.i.), that is, ifB ={b1, . . . , bn},a1b1+ a2b2+ . . . anbn= 0, a1, . . . , an K if and only ifa1= = an = 0.Exercise 17 Find a basis for R3[x], and deduce from it a basis for V ={P(x)R3[x] such thatP(5) = 0} and forW ={P(x)R[x] such thatP(5) =0}Solution

In the previous exercise we proved that G ={1, x , x2

, x3

x} and G ={1, x , x2, x3 x, x + 3}span R3[x]. Moreover, it is easy to prove that G isbasis: in facta0(1)+a1x+a2x2+a3(x3x) = 0 if and only ifa0= = a3= 0,that is,1, x , x2, x3 x are linearly independent. This is not true forG: it issufficient to notice that an element can be written as linear combination of otherelements, for examplex + 3 =(x) 3(1). Then G is linearly dependent.

The first point of the exercise is solved, but let us do some further consid-erations before finding a basis for V. Since G in the previous exercise is nota span set for R3[x], intuition suggests that the right number of elements toform a basis forR3[x] is 4. This is true, in the sense that it is necessary tohave 4 elements to be a basis ofR3[x], but it not sufficient: {x, 2x, 3x, 4x} isnot a basis!

Intuition also suggests that we need one element of each possible degree to

generate R3[x]. A good choice is then B ={1, x , x2, x3}, which is actually thesimplestbasis we can find, and is called standard or canonical basis.

Let us start from B to deduce a basis for V. Why not starting fromG? Itis the same, but starting from the canonical basis may make the computationssimpler.

A generic polynomial ofR3[x] can be written as a0+ a1x+a2x2 +a3x3.Now, if we impose the constraintP(5) = 0, we have a0+ a15 + a25

2 + a353 = 0.

One coefficient is then fixed if the others are given, for instance a0 =a15 a252a353. In conclusion, any polynomial inVcan be written asa15a252a35

3 + a1x + a2x2 + a3x

3 =a1(x 5) + a2(x2 52) + a3(x3 53); we have thenjust 3 coefficients, andV= span{x 5, x2 52, x3 53}. Moreover, it is easy tonotice that C={x 5, x2 52, x3 53} is linearly independent, then a basis.

We observe that for V the number of elements of basis is 3. Since V isR3[x] + a constraint, we may say that if we add constraints we decrease thenumber of elements a basis. This number actually is important to describe avector space (see next reminder).

Let us now find a basis for W. Here there is no fixed degree: the number ofthe elements of a basis ofR[x] is infinite and a natural basis is {1, x , x2, x3, x4, . . . }.A basis forWis the analogous ofC, but with no stop at degree 3:{x 5, x2 52, x3 53, x4 54, x5 55, . . . }.Reminder 14 (Dimension of a vector space) A vector space has infinitedifferent bases. One property is common to all the bases: the number of elementsis always the same. We call this number the dimension of the vector space.

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Exercise 18 Which are the dimensions ofV andW in the previous exercise?

SolutiondimV= 3, dimW =.

Exercise 19 Which of the following is a basis forR3?A={(1, 2, 0), (0, 0, 2)}B= A {(1, 2, )}C=A {(0, 1, 0), (1, , 0)}D= A {(1, , 0)}

SolutionWe sum up the solution in the following table (l.i. is linearly independent forshort):

span l.i. basisA NO YES NOB NO NO NOC YES NO NOD YES YES YES

5 Week 5: linear maps and matrices

Reminder 15 (Linear map) Given two vector spacesU andV on a fieldK,a map f :U

V is a linear map if for any,

K and for anyu, w

U,

f(u + w) =f(u) + f(w).

Reminder 16 (Kernel and Image of a linear map) We define

kerf={uU :f(u) = 0V} where0V is the null element ofV}; Imf ={vVsuch that there existsuU :f(u) =v}.

It can be proved that ker and Im are subspaces of U and V, respectively, andthat

dim(kerf) +dim(Imf) =dimU.

Exercise 20 Find the linear mapf : R3

R such thatf(1, 2, 1) =f(0, 0, 1) =

0 andf(0, 1, 0) = 1. Find kerf and Imf.

SolutionA linear map is determined when it is defined over the elements of a basis ofU,that is, when we know the images through fof the elements of a basis.

The mean of this statement will be clear at the end of this exercise.First of all, we notice that B ={b1, b2, b3}, where b1 = (1, 2, 1), b2 =

(0, 0, 1), b3= (0, 1, 0), is a basis for R3 .

Our goal is to write explicitly f(u) = ... for any u R3. As B is a basis,we know that there exist ,, R such that u = b1+ b2+ b3, and in

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particular it is easy to compute that = u1, =u3 u1, =u2 2u1. Nowusing the linearity:

f(u) =f(b1+ b2+ b3) =f(b1) + f(b2) + f(b3)

=u1f(b1) + (u3 u1)f(b2) + (u2 2u1)f(b3) =u2 2u1.In conclusion, f(u) =u2 2u1.

In order to find kerf and Imf, let us first discuss their dimensions. Thetheorem mentioned before states that dim(kerf) + dim(Imf) = dim(U) = 3.Now, Imf R, then its dimension can be 0 or 1. Moreover, if it were equal tozero, Imf={0}, but this is not the case since 1 belongs to it by the hypotheses ofthe exercise. Then dim(Imf) = 1, from which we conclude that dim(kerf) = 2.In this case the dimensions also determine the subspaces themselves. In fact,Imf = R (as for any U subspace of W, if dimU = dimW, then U = W),

and as{b1, b2} kerf and b1 and b2 are linearly independent, then kerf =span{b1, b2}.Quiz 1 Which of the following is not a linear map?

(a) U=V = R3,u= (u1, u2, u3)R3, f(u) = (u3, u2, u1);(b) U=C[0, 1], V = R,gU, f(g) =g(0);(c) U= R2, V =C[0, 1],uR2 f(u) =u1ex + u2e2x;(d) U= R2, V =C[0, 1],u= (u1, u2)R2 f(u) =u1u2ex.

Solution

(a)u, vR3, , R,

f(u + v) =f(u1+ v1, u2+ v2, u3+ v3)= (u3+ v3, u2+ v2, u1+ v1)

=f(u) + f(v).

This is a linear map.(b)g, hU, , R,

f(g+ h) = (g+ h)(0) =g(0) + h(0)

=f(g) + f(h).

This is a linear map.(c)u, vR2, , R,

f(u + v) = (u1+ v1)ex + (u2+ v2)e2x

=u1ex + u2e

2x + v1ex + v2e

2x

=f(u) + f(v).

This is a linear map.(d)u, vR2, , R,

f(u + v) = (u1+ v1)(u2+ v2)ex

f(u) + f(v)(u1u2v1v2)ex

This is not a linear map.

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Quiz 2 Let f : R2(x) R3 a linear map such that f(1)= 0, f(19x)= 0,f(2x

2

x)= 0. What is true?(a) f is necessarily injective;(b) f is necessarily surjective;(c) dim(Imf)1;(d) dim(kerf) = 2 necessarily.

Solution

As for all kind of functions, a linear map f : U V is injective whenf(u) = f(v), u, v U implies u = v, and is surjective if Imf = V. As adifference from other functions, for linear maps it can be proved that surjective injective.

We know that dimR2(x) = 3, then dim(ker(f)) + dim(Im(f)) = 3. The

possible combinations are: 0+3; 1+2; 2+1; 3+0.Sincef(1)= 0 we know that ker(f)= R2(x), hence dim(kerf)< 3 (remindthat ifSis a subspace ofV and dim(S) = dim(V), then S= V!).

A linear map is surjectiveinjectivekerf={0} dim(kerf) = 0.(a) and (b) are not true since we can easily find a counterexample: f(a0+

a1x+a2x2) =a0+a1. In fact, f(1) = 1, f(19x) = 19f(x) = 19, f(2x2 x) =f(x) =1;fis not injective since ker(f) ={a0+a1x+a2x2such thata0+a1=0}={a0 a0x + a2x2, a0, a2 R}, then dim(ker(f)) = 22

(c) is true since dim(ker(f)) {0, 1, 2}; then dim(Imf) {3, 2, 1}.(d) can be true, but it is not necessarily true.

Reminder 17 (Matrix) A matrix is a table of numbers. The numbers that filla matrix are called theentriesof the matrix; in these notes will mainly consider

matrices whose entries are real numbers. We writeA Rmn

to name a realmatrixA, withm rows andn columns:

a11 a12 a1na21 a22 a2na31 a32

. . . ... a3n

......

... . . .

...am1 am2 amn

aij R, i= 1, . . . , m;j = 1, . . . , n (5)

Sum of matrices and product between a scalar and a matrix are naturallydefined as for vectors inRn, element by element. It follows that two matricesmust have the same dimensionsm, n to be summed.

Reminder 18 (Product between matrices) The definition of the productbetween matrices is somehow less natural. Given A Rmn and B Rnp,the generic entry(i, j) of the productAB is defined as

(AB)ij =n

h=1

aihbhj (6)

Notice that the number of columns ofA must be equal to the number of rowsofB . Moreover, the product is not commutative!

2this is easily proved by noting that ker(f) ={a0 a0x + a2x2, a0, a2 R}is spanned by{1 + x, x2} or also using the equivalence to R3

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Exercise 23 Find a linear map f : R3 R2 such thatf(1, 5, 0) = (0, 0) andf(0, 2, 0) = (1, 0). Is it possible to express the action offas the product betweena matrix and a vector?

SolutionWe know that a linear map is well defined when defined on a basis of its domain.This is not the case, since fis defined only on two (linearly independent) vectorsofR3: we need to define it on another (linearly independent) vector, that we canarbitrarily choose. For example, (0, 0, 1) is linearly independent from (1, 5, 0)and (0, 2, 0), and we arbitrarily decide that f(0, 0, 1) = (0, 0).

Notice that kerf= span{(1, 5, 0), (0, 0, 1)}, and dim(Imf) = 1.Letb1= (1, 5, 0), b2= (0, 2, 0), b3= (0, 0, 1).Then u R3, u = b1 +b2 +b3 = (, 5+ , ), thus = u1,

= u2 5u1, = u3.f(u) =f(b1+ b2+ b3) =f(b1) + f(b2) + f(b3)

=u1f(b1) + (u2 5u1)f(b2) + u3f(b3) = (u2 5u1)(1, 0) = (u2 5u1, 0).The second question is: is there a matrix A such that fu = Au (u has to

be thought as a column vector, that is R3 = R31)? First of all A must have3 columns in order to make the product Au feasible. Second, the result ofAumust be a vector in R2, therefore A must have 2 rows.

Then, we have to fill A:

? ? ?? ? ?

u1u2u3

=

u2 5u1

0

In the first row, we have a11u1+ a12u2+ a13u3 = u2

5u1, then a11 = 0,

a12= 1,a13=5. In the second row, we have a21u1+ a22u2+ a23u3= 0, thena21 = a22 = a23 = 0. Notice that this is the only one possible choice of thecoefficientsaij, given that u is not fixed.

0 1 50 0 0

u1u2u3

=

u2 5u1

0

This example shows that f can be associated with a matrix. This can bedone for any linear map, in different ways based on different bases, as will beinvestigated in the next.

Quiz 3 Given the linear map f : R2 R4 such that (1, 1) ker(f) and(2,

1)

f1(1,

1, 1,

1) (u

f1(v) means that f(u) = v, f1 indicates

the inverse image), the matrixM such that, for alluR2, f(u) =Mu is(a)

M=

2 21 12 2

1 1

(b)

M=

1 11 11 1

1 1

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(c)

M=

1 1 1 11 1 1 1

(d)

M=

1 1 1 31 1 1 1

Quiz 4 Givenfdefined in Quiz 3,(a) ker(f) =span{(1, 1)} and Im(f) =span{(1, 1, 1, 1)}(b) dim(kerf) +dim(Imf) = 4(c) ker(f) =span{(1, 1), (0, 1)}(d) A basis for ker(f) is{(1, 1), (0, 0)}

Quiz 5 Givenfdefined in Quiz 3,(a) f is surjective(b) f is injective(c) g(x1, x2, x3, x4) = (x1, x1) is a linear map such thatg(kerf) =Imf(d) g(x1, x2, x3, x4) = (x1, x1) is a linear map such thatg(Imf) =kerf

Solutionof Quiz 3: (b)(c) and (d) are false because the dimension of the matrix should be 4 2, not2 4. The kernel condition holds for both (a) and (b), but if (a) were true,

f

21

= 2f

10 f 01 =

2

121

which contradicts the hypotheses.Solutionof Quiz 4: (a)We know that dim(kerf) + dim(Imf) = dim(R2) = 2. Then (b) is false. Imageand kernel contain at least a nonnull vector, then dim(kerf)1, dim(Imf)1,but since their sum is 2, we must have dim(kerf) = dim(Imf) = 1, and (a)holds true. (c) is then false since ker(f) cannot be generated by two linearlyindependent vectors. (d) is false because (0, 0) never belongs to a basis (whilekerf = span{(1, 1), (0, 0)} would have been correct, since not stating that

{(1,

1), (0, 0)

}is a basis).

Solutionof Quiz 5: (d)The kernel of f has not dimension 0, then (a) and (b) are false. (c) is falsesince g is defined over R4, so g(ker(f)), ker(f) being a subspace ofR2, makesno sense. (d) is true as any vector in Imf can be written as a(1, 1, 1, 1),a R, then g(a(1, 1, 1, 1)) = ag(1, 1, 1, 1) = a(1, 1) which actually isthe expression for any vector in ker(f).

6 Week 6: more on matrices

Reminder 19 Some definitions:

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The kernel of a matrixA RMN is the set ker(A) ={x RN : Ax =0R

M

}. The range of a matrix is the space generated by its columns. The rank of a matrix is the dimension of the range.

Reminder 20 Given the linear system Ax = b, A RMN, x RN, bRM, and denoting by rank(A)and rank(A|b)the ranks of A and of the enlarged

matrix(A|b), we know that:- If rank(A)< rank(A|b), then the system has no solution;- If rank(A) =rank(A|b) =, then the sytem admits solution. In particular, itadmits a unique solution whenN= and infinitely many solutions whenN > (more precisely, N is the number of free unkwnon components).

As a consequence, anyhomogeneoussystem (say whenb = 0RM) admits

at least a solution (the null solution).

Reminder 21 A matrix row-reduced if there is at least one special entry (say,an entry below which there are only zeros) in each row.

A matrix is reduced to row-echelon form if its nonnull rows (say, rows thathave at least one entry different from zero) are above the null rows, and eachleading entry (say, the first nonzero entry of a row) is striclty to the right of theleading entry of the row above it.

Notice that each leading entry is a special entry if the matrix is in row-echelon form, which implies that a matrix in row-echelon form is row-reduced(while the vice versa is not always true).

Analogous definitions hold for column-reduction; here we will work only on

row-reduction.Letrj be thejth row of the matrix. The allowed operations to row reduce a

matrix are:

rjrirjcrjrjrj+ cri

(7)

Reminder 22 The rank of a matrix is equal to the number of nonzero rows ofthe corresponding reduced matrix.

Reminder 23 Given a square matrix A RNN, its inverse (if it exists) isthe matrix B

R

NN such that AB = I, where I is the identiti matrix withsuitable dimensions. We indicate the inverse byA1.

The following statements are equivalent:

1. ker(A) ={0}.2. rank(A) =n (the rank is maximal).

3. A is invertible.

4. The columns ofA are linearly independent.

5. The rows ofA are linearly independent.

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6. det(A)= 0.7. The reduced ofA has no null rows.

8. The super-reduced ofA is the identity matrix.

We say thatA is nonsingular if anyone of these equivalent statements hold.

Exercise 24 Consider the matrix

A=

1 0 10 1 0

3 0 1

. (8)

DoesA admit an inverse? If yes, compute it.

SolutionWe can check invertibility using any of the statements in 23. For example, Ican compute det(A) =4. Or reduce the matrix: if I substitute the third rowr3 byr3+ r1, I obtain the row-reduced matrix, and it has 3 non null rows. Orcompute the kernel by AX= 0. Or...

After having verified the invertibility, let us compute the inverse.Method 1: by row-reduction

1 0 10 1 0

3 0 1

x11 x12 x13x21 x22 x23

x31 x32 x33

=

1 0 00 1 0

0 0 1

(9)

For simplicity (this is not necessary) let us split the system column by column.

First we solve 1 0 10 1 0

3 0 1

x11x21

x31

=

10

0

. (10)

Let us consider the enlarged matrix

1 0 1 10 1 0 0

3 0 1 0

(11)

and let us row-reduce A. The second row already has a special entry, whilethe first row does not have it. We can, for instance, perform r3r3+r1 (onthe enlarged matrix):

1 0 1 10 1 0 0

4 0 0 1

(12)

We stop here and we rewrite explicitly the system:

x11+ x31= 1x21= 04x11= 1

(13)

Then,x21= 0, x11= 14 , and x31=

34 .

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Now, we solve

1 0 10 1 0

3 0 1

x12x22

x32

=

01

0

. (14)

Byr3r3+ r1, 1 0 1 00 1 0 1

4 0 0 0

(15)

hence,x32= 0, x22= 1 and x12+ x32= 0, then x12= 0.Similarly, we compute the third column, and we finally obtain:

A1

=

14

0 14

0 1 034

0 14

. (16)

Method 2: by reduction to row-echelon form (Gauss elimination)This method is analogous to Method 1, but we reduce the matrix to its row-echelon form. Let us consider the enlarged matrix

1 0 1 10 1 0 0

3 0 1 0

(17)

and let us reduce it to its row-echelon form. The rule about the position ofthe leading entry is not fulfilled in the third row, but it sufficient to perform

r3r3 3r1 to have the row-echelon form: 1 0 1 10 1 0 0

0 0 4 3

(18)

Then,x31= 34

, x21= 0, and x11= 14

.Observe that, in this example, row-reduced and row-echelon form mainly

differ for the position of the zero entries. Row-echelon form always has a triangleof zeros in the left-bottom part.

Method 3: super-reduction Given that A1 exists, we have tried tocompute it by solving AX=In. As AA1 =A1A= In, let us multiply bothside ofAX=In byA

1: A1AX=A1In, then InX=A1, then X=A1.

Therefore, if super-reduce (A|In), that is, we obtain (In|B), then B = A1

.In our case:

1 0 1 1 0 00 1 0 0 1 03 0 1 0 0 1

(19)

Byr3r3 2r1 1 0 1 1 0 00 1 0 0 1 0

0 0 4 3 0 1

(20)

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In order to super-reduce, we need to nullify the entry in position (1, 3), and we

can do it byr14r1+ r3: 4 0 0 1 0 10 1 0 0 1 0

0 0 4 3 0 1.

(21)

Finally,r1r1/4, r1 r1/4: 1 0 0 1/4 0 1/40 1 0 0 1 0

0 0 1 3/4 0 1/4.

(22)

In conclusion,

A1 = 1/4 0 1/4

0 1 03/4 0 1/4.

(23)

Method 4: Algebraic complementsIf you like determinants, there is a nicemethod to compute the inverse: it has been proved that

A1 = 1

det(A)CT

whereC is aNNmatrix whose entries Cij are obtained through three steps:(a) we eliminate row i and column j in A; (b) we compute the determinant ofthe so-obtained matrix; (c) we multiply the result by (1)i+j .

We have, C11 = (1)2 det

1 00

1

=1, C12 = 0, C13 =3, C21= 0,

C22=4, C23= 0,C31=1, C32= 0, C33= 1.Moreover, det(A) =4. The reader can now write the final solution and

check whether it matches to ones obtained with the other methods.Remark: what about if det(A) = 0? This method seems to be not feasible,

because we have to divide by 0 ... this is not really true: if det(A) = 0,A is notinvertible so we simply do not compute the inverse!

SUPPLEMENTARY EXERCISES

Exercise 25 Solve the matrix equation:

1 1 12 1 01

1 1

X=

1 1 34 3 21

2 5

SolutionLet callAandBthe two matrices and super-reduce (A|B): if we obtain a matrixof form (I3|C), where I3 is the identity matrix of dimension 3, then X = C.In fact, AX = B is equivalent to I3X = C then X = C. Notice that thissuper-reduction (with the identity matrix instead ofA at the end) is possibileonly when (A) is maximal (otherwise we would obtain some null row).Let us proceed with the super-reduction: 1 1 1 1 1 32 1 0 4 3 2

1 1 1 1 2 5

r2r22r1r3r3r1

1 1 1 1 1 30 1 2 2 5 4

0 2 2 0 1 2

r3r32r2

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1 1 1 1 1 30

1 2 2 5

4

0 0 2 4 11 10 r2r2+r3

1 1 1 1 1 30

1 0

2

6 6

0 0 2 4 11 10

r2r2

r31

2r3

1 1 1 1 1 30 1 0 2 6 6

0 0 1 2 11/2 5

r1r1r2+r3

1 0 0 1 3/2 40 1 0 2 6 6

0 0 1 2 11/2 5

In conclusion,

X=

1 3/2 42 6 6

2 11/2 5

To check the correctness of the solution, compute AX and verify that it isequal toB .

Exercise 26 Use determinants to verify whether the following sets of vectorsare linearly dependent:

A ={(1, 2, 2), (2, 1, 2), (2, 2, 1)}B ={(2, 1, 2, 1), (3, 1, 1, 3), (1, 2, 2, 2), (0, 1, 2, 3)}

SolutionGiven A, let us construct the matrix with the vectors ofA on the rows:

X=

1 2 22 1 2

2 2 1

.

Then, A is linearly dependent if and only if the rows ofXare linearly depen-dent. This is equivalent to det(X) = 0 (see reminder 23). Then, let us computethe determinant:

1 2 22 1 22 2 1

= 1 1 22 1

2 2 22 1

+ 2 2 12 2

=3 2 (2) + 2 2 = 5.

The determinant is nonnull, hence A is linearly independent.One could obtain the same result by reduction: the reduced ofX turns out tohave no null rows, hence (X) = 3, which is equivalent to our result as statedin the reminder.Notice also that one could construct the matrix putting the vectors ofA on thecolumns: the procedure is analogous.

Now, let us consider B and let us construct the matrix:

Y =

2 1 2 13 1 1 31 2 2 20 1 2 3

.

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Let us compute its determinant:

2 1 2 13 1 1 31 2 2 20 1 2 3

= 21 1 32 2 21 2 3

+3 1 31 2 20 2 3

+ 23 1 31 2 20 1 3

3 1 11 2 20 1 2

= 2(12) + 9 + 2(18) (15) =36.In conclusion, det(Y)= 0, hence its rows are linearly independent and B islinearly independent.

7 Week 7: more on linear maps

Reminder 24 Given any basis of a vector space U, any vector u U canbe uniquely written as linear combination of elements of that basis. If B ={b1, b2, . . . , bm} is a basis and

u= 1b1+ 2b2+ + mbmwe denote by

(u)B= (1, 2, . . . , m)

the representation ofu with respect to B .Notice that the order of the elements of the basis is fundamental: 1 is the

coefficient ofb1, 2 ofb2, etc, and the order must be preserved.Let f : U V a linear map, B a basis for U, and Ca basis for V. The

matrixMC,B

f

associated withfwith respect to the basesB andC is the matrix

whose columns are(f(b1))C, (f(b2))C, . . . . Clearly, MC,Bf has dim(V)rows and

dim(U) columns.

Exercise 27 LetV = R3[x] be the set of all the polynomials inx, with coef-ficients inR, and with degree3.

(1) Prove thatB={1, x , x2, x3 x} is a basis forV.(2) Extract from{(1, 0), (1, 1), (4, 1), (0, 1)} a basisC ofW = R2.(3) Prove that f : V W defined by f(a0+ a1x+ a2x2 +a3x3) = (a0+

a1, a1 a2+ 3a3) is a linear map.(4) Find the matrix associated withfwith respect to the basesB andC.

Solution

(1) We know that dim(V) = 4. Then B has the right number of elementto be a basis. It is then sufficient to prove that it is linearly independent.

Is B linearly independent? We can answer using different methods.Using the definition: prove that 0(1) + 1x+ 2x2 +3(x3 x) = 0 if

and only ifi = 0 for i = 0, 1, 2, 3. This is true since0(1) + 1x+2x2 +3(x

3 x) = 0 = 0 + 0x + 0x2 + 0x3, that is,

0 =00 =1 30 =20 =3

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which has the unique solution i= 0, i = 0, 1, 2, 3.

Using the equivalence toR4

:1 =1 + 0x + 0x2 + 0x3 (1, 0, 0, 0)x(0, 1, 0, 0)x2 (0, 0, 1, 0)x3 x(0, 1, 0, 1)then we write the matrix

1 0 0 00 1 0 00 0 1 00 1 0 1

.

Now, it quite immediate to reduce it (add the second row to the fourth one)and notice that the reduced matrix has no null rows. That is, the rank is 4, andthe rows are linearly independent.

(2) We know that dim(W) = 2. Then it is sufficient to choose a couple oflinearly independent vectors to form a basis, for instance C={(1, 0), (1, 1)}.

(3) Given any two sequences of coefficients a and bR4, it is easy to provethatf(Pa+ Pmathbfb) =f((a0+ b0) + (a1+ b1)x . . . ) = (ao+ b0+ a1+ b1, a1+b1 (a2+ b2) + 3 (a3+ b3)) = (ao+ a1, a1 a2+ 3a3) + (bo+ b1, b1 b2+ 3b3) =f(Pa) + f(Pmathbfb) and f(Pa) =f(Pa) =f(Pa) for anyR.

(4) It is clear that MC,Bf R24.Now letc1= (1, 0) and c2= (1, 1):f(1) = (1, 0) =c1=c1+ 0c2 then f(1)C= (1, 0)f(x) = (1, 1) =c2 then f(x)C= (0, 1)f(x2) = (0,

1) =c1

c2 then f(x2)C= (1,

1)

f(x3 x) = (1, 1 + 3) = (1, 2) =3c1+ 2c2thenf(x3 c)C= (3, 2).How do I have calculated, (1, 2) =3c1+ 2c2 and the other coefficients?

I guess it is quite intuitive: in order to obtain 2 in the second entry, given thatthe second entry of c1 is 0, the unique contribution is from c2 and I have tomultiply it by 2; at this point I have to arrange the coefficient ofc1 in order toobtain1 in the first entry.

More formally: you have to solve the system (1, 2) =1c1+22c2 in theunknowns1 and2.

In conclusion

MC,Bf

1 0 2 30 1 1 2

.

Remark: what about ifC was the canonical basis? In this case,f(bi)C =f(bi) wherebi, i = 1, 2, 3, 4 are the vectors in B .

Quiz 6 Let f : R2(x) R3 a linear map such that f(1)= 0, f(19x)= 0,f(2x2 x)= 0. What is true?

(a) f is necessarily injective;(b) f is necessarily surjective;(c) dim(Im(f))1;(d) dim(ker(f)) = 2 necessarily.

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Solution

We know that dim(R2(x)) = 3, then dim(ker(f)) + dim(Im(f)) = 3. Thepossible combinations are: 0+3; 1+2; 2+1; 3+0.

Since f(1)= 0 we know that ker(f)= R2(x), hence dim(ker(f)) < 3(remind that ifSis a subspace ofV and dim(S) = dim(V), then S=V!).

For a linear map, we know that surjective injective, which in turn areequivalent to ker(f) ={0}, which in turn is equivalent to dim(ker(f)) = 0.

(a) and (b) are not true since we can easily find a counterexample: f(a0+a1x+a2x

2) =a0+a1. In fact, f(1) = 1, f(19x) = 19f(x) = 19, f(2x2 x) =

f(x) =1;fis not injective since ker(f) ={a0+a1x+a2x2such thata0+a1=0}={a0 a0x + a2x2, a0, a2 R}, then dim(ker(f)) = 23

(c) is true since dim(ker(f)) {0, 1, 2}; then dim(Im(f)) {3, 2, 1}.(d) can be true, but it is not necessarily true.

8 Week 8: endomorphisms, eigenvectors, eigen-

values

Reminder 25 1. A linear mapffrom a vector spaceVto itself (f :VV)is also called endomorphism

2. Given a vector spaceV of dimensionn and a basisB = (b1, . . . , bn)of it,anyvV there exists a unique sequence of scalar coefficients1, . . . , nsuch thatv=

ni=1 ibi. We write

(v)B = (1, . . . , n)

and we call(v)B the representation ofv with respect to (w.r.t.) the basisB.

3. The matrix associated withf :VV w.r.t. to B is

MB,Bf =

......

...(f(b1))B (f(b2))B (f(bn))B

......

...

that is, MB,Bf is the matrix that has on the columns the representationsoff(b1), . . . , f (bn) w.r.t. to B.

Notice thatMB,B

f turns out to be a square matrix. Moreover, ifV = Rn

andB is the canonical basis, for anyvV, f(v) =MB,Bf v, that is, theaction off onv corresponds to multiplyv by the matrixMB,Bf .

There is then an equivalence between endomorphisms and square matri-ces! This is why in the next we will define properties for both endomor-phisms/square matrices.

4. GivenA Rnn, a vectorv Rn such that is called eigenvector ofA ifthere exists a scalar such thatAv= v

3this is easily proved by noting that ker(f) ={a0 a0x + a2x2, a0, a2 R}is spanned by{1 + x, x2} or also using the equivalence to R3

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5. Such is called eigenvalue ofA.

6. (Eigenvectors and eigenvalues can be equivalently defined for endomor-phisms: f(v) =v)

7. How to compute the eigenvalues? By solving the equation det(AI) = 0.det(A I) is called thecharacteristic polynomial.

8. If you know the eigenvalues, you can find the eigenvectors by solving thesystems(A I)v= 0 for each value of. (Why eigenvalues and eigen-vectors are computed in this way? This is easy to prove, and left to thereader as an exercise).

9. We calleigenspace the vector space generated by an eigenvector.

10. Algebraic Multiplicity of = multiplicity of as zero of the characteristicpolynomial. Example: if ( 2)(3)2, 2 and 3 are the zeros, and 2 hasalgebraic multiplicity 1, while 3 has algebraic multiplicity 2. In the next,we will denote the algebraic multiplicity by.

11. Geometric Multiplicity of = dimension of the eigenspace of . In thenext, we will denote the geometric multiplicity by.

12. It holds1 for any eigenvalue. Moreover, the sum of the algebraicmultiplicities is equal to n.

13. If for each eigenvalue of A = , A is said to be diagonalizable (orequivalently we say that the endomorphismfassociated withA is simple)

Quiz 7 Letfbe an endomorphism ofR2 such thatu = (1, 2)is an eigenvectorassociated with the eigenvalue 2 and such that v = (1, 3) f1(u) (that is,f(v) =u).(a)

ME,Ef =1

5

3 38 6

.

is the matrix offwith respect to the canonical basisE={e1, e2}={(1, 0), (0, 1)}(b)

ME,Ef =1

5

1 38 6

is the matrix offwith respect to the canonical basisE={e1, e2}={(1, 0), (0, 1)}(c)

MB,Bf =

2 10 0

is the matrix off with respect to the basisB={u, v}(d)

MB,Bf =

2 10 0

is the matrix off with respect to the basisB={u, f2(v)} (f2(u) =f(f(u)))

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Quiz 8 Givenfdefined in Quiz 7,

(a) rank(f) = 2(b) null(f) = 2(c) null(f) = 1(d) a basis for ker(f) is{(1, 34)}

Quiz 9 Letfbe the endomorphism ofR3 defined asf(x,y,z) = (x + y + z, x +y+ z, x + y+ z):(a) ker(f) ={(,, ), , R}(b) ker(f) =span{(0, 0, 1), (0, 1, 1)}(c) f(x,y,z) = (0, 0, 0) if and only ifx + y z = 0(d)(1, 0, 1) and(0, 1, 1) are linearly independent vectors belonging to ker(f)

Quiz 10 Givenfdefined in Quiz 9(a) ( , , ) is an eigenvector associated with eigenvalue 3(b) ( , , ) is an eigenvector associated with eigenvalue 2(c) ( , , ) is not an eigenvector(d) a nontrivial subspace of ker(f) must have dimension 2

Solution7 (c)

Let us compute ME,Ef =

f

10

f

01

.

f

12

=f

10

+ 2f

01

2

12

f

13

=f

10

+ 3f

01

12

Combining these equations we obtain

f

10

=

1

5

48

f

01

=

1

5

36

and

ME,Ef =1

5

4 38 6

.

Then (a) and (b) are false. (d) is false since f2(u) =f(f(u)) = 2u, thenu and

f2

(u) are linearly dependent and do not form a basis.To verify that (c) is true, let us compute MB,Bf

f

12

2

12

+ 0

13

f

13

12

+ 0

13

Then the entries of the first colum are 2 and 0, and the entries of the secondcolum are 1 and 0.Solution8 (c)

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The rank of the matrix is 1, then the kernel has dimension 1. Moreover it is

easy to compute that a basis for the kernel is{(1, 4

3)}.Solution9 (d)The kernel contains all those vectors (x,y,z) s.t. x+y+z = 0, that is, of

kind (a,b, a b) a, b R. Since (a,b, a b) =a(1, 0, 1) +b(0, 1, 1), it isclear that{(1, 0, 1), (0, 1, 1)} is a basis for the kernel.Solution10 (a)

(d) is false because any nontrivial subspace (= a subspace different from{0}and from the space itself) must have dimension 1.

9 Week 9: multiple choice quizzes

Quiz 11 The matrix

1 1 10 2 0

3 0 1

(a) has no eigenvectors since it is square(b) has only one eigenvalue=2(c) has two complex eigenvalues(d) has only real eigenvectors

Quiz 12 The matrix

1 0 04 k 00 0 1

(24)

(a) ifk= 1 has null eigenvectors(b) ifk= 1 has an eigenspace which is a plane(c) ifk= 0 the endomorphism associated with is surjective(d) ifk= 0 the endomorphism associated with is not linear

Quiz 13 Letfbe a non surjective endomorphism ofR3 having 2 and 4 amongits eigenvalues. Then

(a) the algebraic multiplicity of the eigenvalue 2 is 2.(b) there exist two linearly independent eigenvectors for the eigenvalue2(c) f has only one eigenspace(d) fhas 3 distinct eigenvalues

Quiz 14 The rank of a matrix(a) corresponds to the dimension of the kernel(b) can be computed only for square matrices(c) can be computed only for endomorphisms(d) never exceeds the minimum dimension of the matrix

Quiz 15 A linear map fromR3 to R3[X](a) if the image has dimension2, the kernel is a straight line(b) can have image of dimension4(c) if the image has dimension2, the kernel has dimension2(d) can have kernel of dimension4

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Quiz 16 Let us consider the systemAx= 0 where

A=

1 3 12 6 0

0 0 1

.

Its solution set is(a) a straight line(b) does not exist since the matrix is not reduced(c) the union of two parallel planes(d) is not a vector space

Quiz 17 The linear system

x y+ z t= 1x + 2y+ 3z+ 3t= 02x y 4z 2t= 8

(25)

(a) has one free unknown as 4 unknowns - 3 equations =1

(b) has one free unknown as the determinant of

1 11 2

is nonzero

(c) has2 solutions(d) has no solution

Quiz 18 The setV ={u= (u1, u2, u3, u4)R4 :u1+ 3u2 6u4= 0}(a) is not a subspace as it does not contain the null vector(b) has bases composed by two vectors(c) is the kernel of a surjective linear map f : R4 R(d) is the image of an injective endomorphismf : R4

R4

Quiz 19 The matrix 1 1 11 2 0

0 0 1

(26)

(a) has an eigenvalue with algebraic multiplicity equal to 2(b) has an eigenvalue with geometric multiplicity equal to 2(c) all the eigenvectors have = (i.e., the matrix is diagonalizable)(d) has an eigenspace of dimension 3

Quiz 20 a 00 a

a

R

(a) is not diagonalizable(b) has all the eigenvectors null(c) has two eigenspaces respectively generated by(a, 0) and(0, a)(d) has eigenspace =R2

Quiz 21 A planeax + by+ cz+ d= 0 inR3

(a) is a vector space of dimension2 if it contains(0, 0, 0)(b) is a vector space of dimension 2 only if it is parallel to the coordinate

planez= 0(c) is never a vector space(d) is always a vector space

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Quiz 22 Givenf : R3 R3, f(x,y,z) = (x,y,y+ z),

M=

1 0 00 2 1

0 1 0

(a) M=MB,Bf whereB={(1, 1, 0), (0, 0, 1), (0, 1, 0)}(b) M=MB,Bf whereB is the canonical basis

(c) M=MB,Bf whereB ={(1, 1, 0), (0, 0, 1), (1, 1, 1)}(d) M=MB,Bf whereB={(1, 0, 0), (0, 1, 1), (0, 1, 0)}

Solutionof Quiz 11 (d)Notice that (a) is completely nonsense.

Let us compute the eigenvalues/eigenvectors of the matrix.

det

1 1 10 2 0

3 0 1

= (1 )(2 )(1 ) + 3(2 + )

= (2 + )[3 + (1 2)] = (2 + )(4 2) = (2 + )(2 + )(2 )The zeros of the characteristic polynomial are 1= 2 with 1= 1 and 2=2with 2= 2.

Let us compute also the eigenvectors:= 2:

1 1 10 4 03 0 3

v= 0v = (, 0, ) =(1, 0, 1), RNotice that the geometric multiplicity, which is the dimension of the spacegenerated by (1, 0, 1), is 1= 1

=2: 3 1 10 0 0

3 0 1

v= 0v = (, 0, 3) =(1, 0, 3), R

and 2= 1.Solutionof Quiz 12 (b)

First notice that (d) is nonsense (by definition an endomorphism is a linearmap!).

If k = 1, it is easy to see that the characteristic polynomial is (1)3,then = 1 and = 3. The associated eigenvector is given by (0, , ),, R and since (0, , ) = (0, 1, 0) +(0, 0, 1), then the eigenspace isspan{(0, 1, 0), (0, 0, 1)}. But a subspace of dimension 2 in R3 is a plane. In fact,the parametric formula of a plane is (x,y,z) =P+ t1Q + t2RwhereP,Q, Rarepoints in R3 and t1, t2 are scalar coefficients. Now, (x,y,z) = P +t1Q+t2Ris not always a vector space (it may not contain the null element!), but if itcontains the null element and then we can write it as (x,y,z) =t1Q+t2R forsuitable value ofQ and R, this is a vector space (generated by Qand R).

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In order to check (c), we first remind that the following are equivalent: 1.

an endomorphism is injective; 2. an endomorphism is surjective; 3. an endo-morphism is bijective, and finally 4. the kernel of the endomorphism containsonly the null element.

One can verify that (c) is false observing that ifk= 0 the matrix has rank2 (non maximal), while a an endomorphism is bijective only if the associatedmatrix has maximal rank (can the reader explain why?).Solution of Quiz 13 (d) If an endomorphism is non surjective, it always hasthe eigenvalue 0.

In fact, we know that (see the solution of the previous quiz), if an endomor-phism fis not surjective, then its kernel contains something more than 0. Letus suppose that v=0 is in the kernel off. Then f(v) =0, or alsof(v) = 0v.Thenv is an eigenvector for the eigenvalue 0!

In conclusion we have 3 different eigenvalues 1 = 0, 2 = 2, 3 = 4, Since

the sum of the algebraic multiplicities must be 3, then 1= 2= 3= 1 (then(a) is false). It follows also that1 = 2 = 3 = 1, which implies that (b) and(c) are false.Solutionof Quiz 14 (d)

No computation is required for this quiz, just study the theory.Solutionof Quiz 15 (a)

Use the theorem dim(ker(f)) + dim(Imf) = dim(V) where f :VW. Inthis case dim(V)=3 and dim(W)=4, then it is easy to prove that the uniquefeasible statement is (a): if the image has dimension 2, the kernel must havedimension 1. But a subspace of dimension 1 in R3 is nothing but a straight linepassing throgh zero!Solutionof Quiz 16 The reduced matrix is

1 3 10 0 10 0 0

.

Therefore the solution set is{v = (3a,a, 0) a R}, and (3a,a, 0) =a(3, 1, 0) is a straight line.Solutionof Quiz 17 (d)

The augmented matrix of the system is: 1 1 1 1 11 2 3 3 0

2 1 4 2 8

.

We reduce it by r3r3+ r1+ r2 1 1 1 1 11 2 3 3 0

0 0 0 0 9

.

In conclusion rank(A) = 2, while rank(A|B) = 3, then the system has nosolution.Solution of Quiz 18 (c) V is subspace of dimension 3. An injective endomor-phism must have ker of dimension 0, then iffwas injective, it should have kerof dimension 4 (that is, Imf = R4). Then (c) is the unique feasible statement.We can explicitly write that f(u1, u2, u3, u4) =u1+ 3u2 6u4.

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Solution of Quiz 19 (c) The eigenvalues of the matrix are 1, and 35

2 . They

all have = = 1.Solutionof Quiz 20 (d)The matrix has eigenvalue = a with = 2, and the eigenspace is R2.

Solution of Quiz 21 (a) This point has already been discussed in the solutionof Quiz 12. Examples: the planex+y+z = 0 is vector space, in particular itis generated by (1, 0, 1) and (0, 1, 1); x + y+ z= 1 is not a vector space.Solutionof Quiz 22 (d)

(c) is false because B is not a basis!(b) is false because it is easy to prove that ifB is the canonical basisf(v) =

MB,Bf v where

MB,Bf =

1 0 00 1 0

0 1 1

Let us consider (a), b1 = (1, 1, 0)T, b2 = (0, 0, 1)

T . Then (a) is false as(f(b1)) = (1, 1, 1)

T = b1 + b2, and (f(b1))B = (1, 1, 0)T, which is different

from the first column of the given matrix.(d) is true, in fact:Letb1= (1, 0, 0)

T, b2= (0, 1, 1)T, b3= (0, 1, 0)

T

(f(b1)) = (1, 0, 0)T =b1 (f(b1))B = (1, 0, 0)T

(f(b2)) = (0, 1, 2)T = 2b2 b3 (f(b1))B = (0, 2, 1)T

(f(b3)) = (0, 1, 1)T =b2 (f(b1))B = (0, 1, 0)T.

10 Week 10: other multiple choice quizzes

Reminder 26 Given a square matrixARNN,1. given B RNN, we say thatA and B are similar if there exists P

RNN such thatA= P1BP (or equivalentlyB= P AP1).

2. Similar matrices have the same eigenvalues, the same rank, the same de-terminant.

3. A is diagonalizable if it is similar to a diagonal matrix. 4

4. If a matrix A Rnn is diagonalizable, then the collection of its eigen-vectors is basis forRn, and vice versa.

5. A symmetric matrix is diagonalizable and its eigenvectors are orthogonal.

Some of these quizzes are taken or inspired by the book of Carlini 50 mul-tiple choices in geometry.

Quiz 23 LetA R33 with det(A)=0.(a) A is the matrix of an isomorphsim(b) the rank ofATA is 3

4In the previous weeks I said that a matrix is diagonalizable if the algebraic and geometricmultiplicities coincide for each eigenvalue. This is more a property or an operative definitionwhich is useful to solve the exercises (thati is, to verify the diagonalizability.) The formaldefinition is the one I have written here.

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(c) A has the null eigenvalue

(d) the row-echelon form ofA has just one null row

Quiz 24 The matrix 0 00 0

(a) is not diagonalizable(b) has the eigenvalue10(c) the only eigenspace ofA is generated by(1, 0) and(0, 1)(d) any vector ofR2 is an eigenvector ofA

Quiz 25 LetV andWsubspaces ofR5 and let dim(V) = 4. What is true?(a) it is possible to findWsuch that dim(V + W) = 5(b) if dim(W)=2, dim(V

W)

1

(c) if dim(W)=1, dim(V W) = 0(d) V W contains an infinite number of vectors

Quiz 26 Consider the following subspaces of R3: U =span{(1, 0, 1), (0, 1, 0)}andV =span{(1, 2, 3)}.

(a) any endomorphism ofR3 having eigenspacesU andV is simple(b) any matrix having eigenspace U has the vector (1, 1, 1)T on the second

column(c) there exists a symmetric real matrix with eigenspacesU andV(d) there exists a unique symmetric real matrix with eigenspacesU andV

Quiz 27 Consider the following subspaces of R3: U =span{(1, 0, 1), (0, 1, 0)}andV =span{(1, 0, 1)}.(a) all the endomorphisms ofR3 having eigenspacesU andV are simple

(b) any matrix having eigenspace U has the vector (1, 1, 1)T on the secondcolumn

(c) there exists a symmetric real matrix with eigenspacesU andV(d) there exists a unique symmetric real matrix with eigenspacesU andV

Quiz 28 LetA R44 have two distinct eigenvalues1 and2, and1 = 3.LetV denote the eigenspace associated with the eigenvalue. Then

(a) dim(V1)dim(V2)(b) dim(V1)< dim(V2)(c) V1+ V2= R4(d) V1 is a cube

Quiz 29 Letf andg be endomorphisms ofV. Then(a) if is an eigenvalue for bothf andg, then it is eigenvalue ofg(f)(b) ifv is an eigenvector for bothf andg, then it is eigenvector ofg(f)(c) if is an eigenvalue for f and if is an eigenvalue for g, + is

eigenvalue off+ g(d) 0 is an eigenvalue forfonly if it is eigenvalue off(g)

Solutionof Quiz 23 (c)An isomorphism (= bijective endomorphism) is always associated with a

matrix of maximal rank. If the determinant is null, the rank is not maximal.

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Morevoer, det(ATA) = det(AT)det(A) = 0. The row-echelon form will have

some null rows, but not necessarily one.Solutionof Quiz 24 (c)It is clear that the eigenspace is R2. Notice the difference between (c) and

(d): (d) is false just because (0, 0, 0)R2 never is an eigenvector!Solutionof Quiz 25 (a)

(b) is false as dim(VW) 2 from what we know about intersectionsof subspaces; analogously (c) is false, as dim(VW) 1; (d) is false sinceV W ={0}is not impossible.Solutionof Quiz 26 (a)

An endomorphism is simple if it admits a basis of eigenvectors, and in thiscase (1, 0, 1), (0, 1, 0), (1, 2, 3) form a basis.

(b) is false: if we consider the eigenvectorv= (0, 1, 0)T, we have, for some

R,

Uv= v a b

c

01

0

=

ab

c

=

01

0

. (27)

(c) is false since the eigenvectors of a symmetric real matrix are orthogonal;(d) is false as (c) is false.

Solutionof Quiz 27 (c)(a) and (b) are false (see previous quiz)(d) is false: as a counterexample, do the eigenvalue/eigenvector analysis of

1/2 0 1/2

0 1 01/2 0 1/2

1/2 0 1/2

0 0 0

1/2 0 1/2

Can you derive thegenericexpression of a symmetric matrix with eigenspaces

U and V?Solutionof Quiz 28 (a)

In general, we know that. Since13, 21 and2 +1 = 4,then 1 = 3, 2 = 1; moreover 2 = 1. Hence, (a) is true and (b) is false.(d) makes no sense.

Notice that such an A is diagonalizable (the multiplicities coincide for eacheigenvalue), hence the eigenvectors form a basis. As a consequence the eigenspacesV1 and V2 intersect only in 0, and their sum is the whole space R

4.Observe that ifA was not diagonalizable, the interesection could be larger

and the sum could be smaller.Solutionof Quiz 29

(This quiz is bit more difficult than the other ones)Let us start from (d). Remind that Statement A holds only if statement

B holds means that A implies B (while Statement A holds if statementB holds means that B implies A). Then, we have to check whether 0eigenvalue off implies 0 eigenvalue off(g). To see that this is not alwaystrue, letf(v) = 0 (v is the eigenvector of eigenvalue 0): ifv is not in the imageofg , then (d) is false. For instance let image(g)=span(0, 1) andv = (1, 0): thisprovides a clear counterexample.

(c) is true only if f and g have the same eigenvector for and . Coun-terexample: f(x, y) = (x, 0) andg(x, y) = (0, y) both have the eigenvalue 1, but(f+ g)(x, y) = (x, y) is the identity map, and has no eigenvalue equal to 2.

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The same example can be used to check that (a) is not generally true: fand

g have the eigenvalue 1, while g(f) maps anything in (0, 0).(b) is true: iff(v) =v andg (v) =v, then g (f(v)) =g(v) =(g(v)) =v, so v is an eigenvector for g (f) with eigenvalue .

11 Week 11: more on similar and diagonalizable

matrices, and change of basis

Reminder 27 LetvV, dim(V) =N, B ={b1, . . . , bN}, C={c1, . . . , cN}be bases ofV, and(v)B be the vector of coefficients ofv with respect to the basisB. We have

(v)B =PB,C(v)C

where PB,C

has (c1)B, . . . (cN)B on the columns. It is known that PB,C

isinvertible, and(PB,C)1 =PC,B; PB,C is called change basis matrix.

Reminder 28 If we consider a linear map f :VW,B , B bases ofV,C, Cbases ofW then

MC,B

f =PC,CMC,Bf P

B,B .

For endomorphisms, f :VV, B, Cbases ofV,

MC,Cf =PC,BMB,Bf P

B,C = (PB,C)1MB,Bf PB,C.

Reminder 29 A square matrixP is said to be orthogonal ifP1 =PT.IfA is symmetric, it also diagonalizable, andA= PP1 ( is a diagonal

matrix) whereP is orthogonal, thenA= PPT

.

Exercise 28 Let

A=

2 0 10 1 2

0 0 1

B=

2 0 10 1 0

0 0 1

C=

1 0 10 1 0

1 0 1

(a) Compute the eigenvalues and the eigenvectors;(b) Discuss the diagonalizability.

Solution

|A I|= 2

0 1

0 1 20 0 1

= (2 )(1 )2 = 0.The eigenvalues are 1 = 1 with algebraic multiplicity 1 = 2 and 2 = 2,2 = 1. Given that for any eigenvalue the geometric multiplicity is lessor equal than the algebraic one: 1, then 2 = 1.In order to compute 1 , let us solve the system (A 1I)v= (0, 0, 0)T. Since

(A I) = 1 0 10 0 2

0 0 1

.

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and reducing it one obtains

1 0 10 0 2

0 0 0

.

it is easy to find that the solution is v = (0, 1, 0)T, R. The eigenvectorsof 1 = 1 are then all the vectors v1 = (0, 1, 0)

T, R, = 0, whilethe corresponding eigenspace is V1 = span{(0, 1, 0)}. In particular, 1 =dim(V1) = 1.Since1=1 , thenA is not diagonalizable.Finally, let us compute the eigenvectors of2= 2:

(A

2I) =

0 0 10

1 2

0 0 1

.

and reducing it one obtains 0 0 10 1 2

0 0 0

.

then v2= (1, 0, 0)T.

Notice that B has the same characteristic polynomial, then the same eigen-values ofA. However, for B 1 = 2. In fact

(B I) = 1 0 1

0 0 00 0 0

.then, rank(B I) = 1 a n d 1 = 3 1 = 2. Explicitly, the solution of(B I)v = (0, 0, 0)T is the set{(,, )T, , R}. Then the eigenvectorsare all the nonnull solutions and given that (,, ) =(1, 0, 1) + (0, 1, 0),V1 = span{(1, 0, 1), (0, 1, 0)}.Given that the geometric and algebraic multiplicities coincide for all the eigen-values, B is diagonizable.In particualar, notice that v2 = (1, 0, 0), R, = 0 are the eigenvalues of2= 1. Then, we can diagonalize in this way:

= P1BP

where

=

1 0 00 1 0

0 0 2

P =

1 0 10 1 0

1 0 0

.

P1 can be easily obtained with the method of algebraic complements or bysuper-reduction (by the transformations: r1 r1+ r3; r3 r3; r3 r1 onthe matrix (P|I), so that we obtain (I|P1).

P1 =

0 0 10 1 0

1 0 1

.

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Cis diagonalizable since it is symmetric.

|C I|=

1 0 10 1 01 0 1

= (1 )3 (1 ) = (1 )( 2) = 0.

The eigenvalues are1= 0, 2= 1,3= 2.We then have V1 = span{(1, 0, 1)},V2 = span{(0, 1, 0)},V3 = span{(1, 0, 1}.We normalize the eigenvectors and we write them on the columns of the

matrixP:

P = 1

2

1 0 10 2 0

1 0 1

Notice that 1 0 10 1 0

1 0 1

= 1

2

1 0 10 2 0

1 0 1

0 0 00 1 0

0 0 2

1

2

1 0 s10 2 0

1 0 1

Exercise 29 Let

A=

1 1k 2

B={(2, 1), (1, 1)} (28)

(a) For which values of k R the endomorphism f of R2 associated to thematrixA with respect to the basisB is simple?

(b) Write the matrix of the endomorphism with respect to the standard basis.

Solution(a) The eigenvalues ofAare given byp() = (1)(2)k= 23+2k=0. The roots of the polynomial are not real (hence A is not diagonalizable) fork

14

, as in this case A has two real

distinct eigenvalues 1= 2. Remind that this is sufficient to havei =i ,i= 1, 2, since in general .In conclusion, fis simple for any k >1

4.

(b) We can explicitly write f:

f(2, 1) = 1(2, 1) + k(1, 1)f(1, 1) = 1(2, 1) + 2(1, 1)

Then, on the standard basis, f(1, 0) = f(2, 1)f(1, 1) = (k2)(1, 1) andf(0, 1) = 2f(1, 1) f(2, 1) = (2, 1) + (4 k)(1, 1) = (6 k, 5 k) and finallyf(x, y) =xf(1, 0) + yf(0, 1) =x(k 2, k 2) + y(6 k, 5 k).

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ME,Ef =

k 1 6 kk 1 5 k . (30)Other method: use ME,Ef =P

E,BMB,Bf PB,E

PE,B =

2 11 1

. (31)

PB,E =

1 11 2

. (32)

2 11 1

1 1k 2

1 11 2

. (33)

Exercise 30 Letfbe the endomorphism ofR22 defined as

f

x yz t

=

x + 3y 2x + 2y

z+ 2t 2z+ t

(34)

(a) Compute eigenvalues, eigenvectors, eigenspaces off.(b) Isf simple?(c) Write a basis of eigenvectors forR2,2.(d) Is the matrix associated to fdiagonalizable? If yes, write the change-of-basismatrixPto change to a basis of eigenvectors.

SolutionWe are not given a basis. Let us choose the standard basis:

B={b1, b2, b3, b4}=

1 00 0

,

0 10 0

,

0 01 0

,

0 00 1

Now, f(b1) = b1+ 2b2, f(b2) = 3b1+ 2b2, f(b3) = b3+ 2b4, f(b4) = 2b3+ b4,then

MB,Bf =

1 3 0 02 2 0 00 0 1 20 0 2 1

(a) Let us compute the eigenvalues:

1

3 0 02 2 0 00 0 1 20 0 2 1

= (1){(2)[(1)24]}3{(2[(1)24]}= (1)2(3)(4).

The eigenvalues ofMB,Bf are 1 =1, 2 = 3, 3 = 4 and 1 = 2, 2 = 1,3 = 1. Let us compute the eigenvectors: for1=1:

2 3 0 02 3 0 00 0 2 20 0 2 2

y11y12y13y14

=

0000

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We easily compute thaty1= (, 23,, )T, , R not both zeros.The corresponding eigenvector fo fis y11b1+y12b2+y13b3+y14b4=

2

3

.

Notice that this calculus actually corresponds to put the components ofy1 in amatrix: this is true if and only if we work with the standard basis.

Given that

2

3

=

1 2

30 0

+

0 01 1

the corresponding

eigenspace off is

span

1 2

30 0

,

0 01 1

.

Its dimension is 2, then 1 = 2.This is sufficient to conclude that f is simple: in fact,2 and3 have algebraicmultiplicity 1, then also their geometric multiplicty must be equal to 1. Inconclusion, i = i , i = 1, 2, 3, that is, f is simple (and the correspondingmatrix is diagonalizable).

Exercise 31 Letfbe the endomorphism ofR2 such that(1, 2)is an eigenvectorofffor the eigenvalue 1 and(0, 1)ker(f). Say whetherfis simple just lookingat the conditions (no computations!).

SolutionNotice that (0, 1)ker(f) is equivalent to state that (0, 1) is an eigenvector forthe eigenvalue 0. Then{(1, 2), (0, 1)}is a basis of eigenvectors (or equivalently,fhas two distinct eigenvalues) which implies that fis simple.

Exercise 32 Let

A=

0 a b0 0 c

0 0 0

witha, b, c= 0. Prove thatAis diagonalizable without computing its eigenvalues.SolutionUse the definition: A is diagonalizable if there exists an invertible matrix Psuch that P1AP= where is a diagonal matrix.Suppose that such Pexists. Now, notice that

A3 =AAA =

0 0 00 0 0

0 0 0

= 0R3,3

Since (P1AP)3 = 3 and (P1AP)3 =P1AP P1AP P1AP =P1A3P =0R3,3 then it must be

3 = 0R3,3 . Given that is diagonal, this implies that = 0R3,3 . In fact, if

=

x 0 00 y 0

0 0 z

then

3 =

x

3 0 00 y3 00 0 z3

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and 3 = 0R3,3 impliesx = y = z = 0.

Now since P1

AP = A = PP1

, if = 0R3,3

, then A = 0R3,3

whichcontradicts the hypothesis. Then, we have prove by contradiction thatPcannotexist.