linear algebra. session 11 - math.tamu.eduroquesol/math_304...dr. marco a roque sol linear algebra....
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Abstract Linear Algebra
Linear Algebra. Session 11
Dr. Marco A Roque Sol
04 / 21 / 2020
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Inner Product.
Find the distance from the point (2, 1,−2) to the plane Π, givenby:.
6(x − 1) + 2(y − 3) + 3(z + 4) = 0.
Solution
The Solution of the above equation is : xyz
= α
10−2
+ β
03−2
+
13−4
so the vectors x1 = (1, 0,−2)T and x2 = (0, 3,−2)T span theplane Π.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
The orthogonal projection,p, of v = (2, 1,−2)T − (1, 3,−4)T =(1,−2, 2)T onto the plane Π is given by:
p =< v, x1 >
< x1, x1 >x1 +
< v, x2 >
< x2, x2 >x2
where< v, x1 >=< 1,−2, 2 > · < 1, 0,−2 >= −3
< v, x2 >=< 1,−2, 2 > · < 0, 3,−2 >= −10
< x1, x1 >=< 1, 0,−2 > · < 1, 0,−2 >= 5
< x2, x2 >=< 0, 3,−2 > · < 0, 3,−2 >= 13
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
It follows that
p = −35x1 − 10
13x2 = −35(1, 0,−2)T − 10
13(0, 3,−2)T ⇒
p = (−35 ,−
1013 ,
20865 )T
Finally, the distance from (2, 1,−2) to the palne Π is :
||o = v − p|| = ||(1,−2, 2)T − (−3
5,−10
13,
208
65)T || =
||o|| = ||(8/5,−16/13,−78/65)T || = 65||(104,−80,−78)T ||
||o|| = 65√
1042 + 802 + 782 = 65√
23300 �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Orthogonal Subspaces.
(a) Let S be the subspace of R3 spanned by the vectorsx = (x1, x2, x3)T and y = (y1, y2, y3)T , and let A be the matrixdefined by:
A =
(x1 x2 x3y1 y2 y3
)(a) Show that S⊥ = N(A).
(b) Find the orthogonal complement of the subspace of R3
spanned by x = (1, 2, 1)T and y = (1,−1, 2)T .
Solution
a) A vector v = (v1, v2, v3) belongs to S⊥ if and only if
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
< v, x >= 0
< v, y >= 0
but that is true if and only if
v1x1 + v2x2 + v3x3 = 0
v1y1 + v2y2 + v3y3 = 0
if and only if, in matrix notation
(x1 x2 x3y1 y2 y3
) v1v2v3
=
(00
)
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
if and only if
Av = 0
if and only if v ∈ S⊥ �
b) Using a), we are looking for the solutions of
Av =
(x1 x2 x3y1 y2 y3
) v1v2v3
=
(1 2 11 −1 2
) v1v2v3
=
(00
)
using elementary row operations we find the row-echelon form ofthe matrix A
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
(1 2 10 −3 1
) v1v2v3
⇒ (1 2 10 1 −1/3
) v1v2v3
=
(00
)
and we obtain the system
v1 + 2v2 + v3 = 0v2 − 1
3v3 = 0
whose solution is the one-dimensional subspace:
N(A) =
v ∈ R3/ v = α
−513
, α ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Least Squares Solutions.
Find the least squares solution of each of the following system:
−x1 + x2 = 102x1 + x2 = 5x1 − 2x2 = 20
Solution
In matrix notation, the above system can be written as
Ax = b⇒
−1 12 11 −2
( x1x2
)=
105
20
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
and we know that a vector y is a lest square approximation for thesystem if, it minimizes ||r(y)|| = ||Ay − b||, and this is happeningif y satisfies the normal equations. namely,
ATAy = ATb
(−1 2 11 1 −2
) −1 12 11 −2
( y1y2
)=
(−1 2 11 1 −2
) 105
20
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
(6 −1−1 6
)(y1y2
)=
(20−25
)⇒
y =
(2.71−3.71
)�
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Inner Product.
In C [−π, π] with inner product defined by
< f , g >=1
π
∫ π
−πf (x)g(x)dx ,
show that cos(mx) and sin(nx) are orthogonal and that both areunit vectors. Determine the distance between the two vectors.
Solution
< cos(mx), sin(nx) >=1
π
∫ π
−πcos(mx) sin(nx)dx
and using the Trig. Identitiy
cos a sin b =1
2[sin(a + b)x − sin(a− b)x ]
we obtain:Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
< cos(mx), sin(nx) >=1
2π
∫ π
−π[sin(m + n)x − sin(m − n)x ]dx =
1
2π
∫ π
−πsin(m + n)x − 1
2π
∫ π
−πsin(m − n)x = 0− 0 = 0
On the other hand
< cos(mx), cos(mx) >=1
π
∫ π
−πcos2(mx)dx =
1
2π
∫ π
−π[1+cos(2mx)]dx
=1
2π
∫ π
−π1dx +
1
2π
∫ π
−πcos(2mx)dx = 1 + 0 = 1.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
< sin(nx), sin(nx) >1
π
∫ π
−πsin2(mx)dx =
1
2π
∫ π
−π[1− cos(2mx)]dx
=1
2π
∫ π
−π1dx − 1
2π
∫ π
−πcos(2mx)dx = 1− 0 = 1
Now the distance between x = cos(mx) and y = sin(nx) is givenbiy
||x− y||2 =< x− y, x− y >=1
π
∫ π
−π[cos(mx)− sin(nx)]2dx =
1
π
∫ π
−π[cos2(mx)−2 cos(mx) sin(nx) + sin2(mx)]dx = 1−0 + 1 = 2
Therefore ||x− y|| =√
2 �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Norms
Determine whether the following define norms on C [a, b]:
(a) ||f || = |f (a)|+ |f (b)|
(b) ||f || =∫ ba |f (x)|dx
(c) ||f || = maxa≤x≤b |f (x)|
Solution
Remmember that a function || · || : V→ R is called a norm definedin the vector space V, if satisfies:
||x|| ≥ 0, ||x|| = 0 only for x = 0
||rx|| = |r |||x|| for all r ∈ R.
||x + y|| ≤ ||x||+ ||y||.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
a) ||f || = |f (a)|+ |f (b)| .
||f || = |f (a)|+ |f (b)| = 0⇒ |f (a)| = 0, |f (b)| = 0, but notnecessarily that the function f (x) = 0 in the whole interval,Therefore, this function, does not define a norm ... !!!!
b) ||f || =∫ ba |f (x)|dx .
||f || =∫ ba |f (x)|dx > 0 provided f (x) 6= 0. Now, if
||f || =∫ ba |f (x)|dx = 0 this implies, from Calculus, that f (x) = 0
on [a, b]
||rf || =∫ ba |rf (x)|dx = |r |
∫ ba |f (x)|dx = |r |||f ||.
||f + g || =∫ ba |f (x) + g(x)|dx ≤
∫ ba |f (x)|dx +
∫ ba |g(x)|dx =
||f ||+ ||g ||.
Therefore, this function, defines a norm.Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
c) ||f || = maxa≤x≤b |f (x)| .
||f || = maxa≤x≤b |f (x)| > 0 provided f (x) 6= 0. Now, if||f || = maxa≤x≤b |f (x)| = 0 this implies that f (x) = 0 on [a, b]
||rf || = maxa≤x≤b |rf (x)| = maxa≤x≤b |r ||f (x)| =
|r |maxa≤x≤b |f (x)| = |r |||f ||.
||f + g || = maxa≤x≤b |f (x) + g(x)| ≤= maxa≤x≤b |f (x)|+ |g(x)| =
||f ||+ ||g ||.
Therefore, this function, defines a norm.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Orthogonal Polynomials
Let p0, p1, . . . be a sequence of orthogonal polynomials and let andenote the lead coefficient of pn. Prove that
||pn||2 = an < xn, pn >
Solution
If p0, p1, . . . is a sequence of orthogonal polynomials, then weknow that
i) deg(pn) = n
ii)< pn, xk >= 0, for 0 ≤ k ≤ n − 1
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Thus, if pn(x) is given by
pn(x) = anxn + an−1x
n−1 + · · ·+ a1x + a0
then
||pn||2 =< pn(x), pn(x) >
||pn||2 =< anxn + an−1x
n−1 + · · ·+ a1x + a0, pn >=
an < xn, pn > +an−1 < xn−1, pn > + · · ·+ a1 < x , pn > +a0 < 1, pn >
but due to ii), we have that
||pn||2 = an < xn, pn > �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Orthogonal Polynomials
Let Tn(x) denote the Chebyshev polynomial of degree n and define
Un−1(x) =1
nT ′n(x); n = 1, 2, . . .
(a) Compute U0(x),U1(x), and U2(x).
(b) Show that if x = cos(θ) , then
Un−1 =sin(nθ)
sin(θ).
Solution
a) T1 = x ⇒ U0 = T ′1(x) = 1
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
T2 = 2x2 − 1⇒ U1 = 12T′2(x) = 2x
T3 = 4x3 − 3x ⇒ U2 = 13T′3(3) = 4x2 − 1
T4 = 8x4 − 8x2 + 1⇒ U3 = 14T′4(x) = 8x3 − 4x
b)
Un−1(x) =1
nT ′n(x)
but we know that Tn(cos θ) = cos(nθ), therefore
Un−1(x) =1
n
dTn(x)
dx=
1
n
dTn(x)
dθ
dθ
dx=
1
n
dTn(cos(θ))
dθ/dx
dθ⇒
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Un−1(x) =1
n
d cos(nθ)
dθ/dx
dθ
Un−1(x) =1
n[−n sin(nθ)] /[− sin(θ)]
Un−1(x) =sin(nθ)
sin(θ)�
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Markov Chain
Let A be an n × n stochastic matrix and let e be the vector in Rn
whose entries are all equal to 1. Show that e is is an eigenvector ofAT . Explain why a stochastic matrix must have λ = 1 as aneigenvalue.
Solution
Let A be the stochastic matrix with entries given by
A =
a11 a12 · · · a1na21 a22 · · · a2n
... · · ·an1 an2 · · · ann
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
By the definition of a stochastic matrix ( the entries of eachcolumn of A are nonnegative numbers that add up to 1 ), then Asatifies:
a1j + a2j + · · ·+ anj = 1; for all 1 ≤ j ≤ n
and At = (bij) satisfies
bi1 + bi2 + · · ·+ bin = 1; for all 1 ≤ i ≤ n
Therefore
ATe =
b11 b12 · · · b1nb21 b22 · · · b2n
... · · ·bn1 bn2 · · · bnn
11...1
=
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
b11 b12 · · · b1nb21 b22 · · · b2n
... · · ·bn1 bn2 · · · bnn
11...1
=
b11 + b12 + · · ·+ bnb21 + b22 + · · ·+ b2n
...bn1 + bn2 + · · ·+ bnn
⇒
b11 b12 · · · b1nb21 b22 · · · b2n
... · · ·bn1 bn2 · · · bnn
11...1
=
11...1
Thus, e is an eigenvector of AT with eigenvalue λ = 1. �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Markov Chain
In the Dark Ages, Harvard, Dartmouth, and Yale admitted onlymale students. Assume that, at that time, 80 percent of the sonsof Harvard men went to Harvard and the rest went to Yale, 40percent of the sons of Yale men went to Yale, and the rest splitevenly between Harvard and Dartmouth; and of the sons ofDartmouth men, 70 percent went to Dartmouth, 20 percent toHarvard, and 10 percent to Yale.
(i) Find the probability that the grandson of a man from Harvardwent to Harvard.
(ii) Modify the above by assuming that the son of a Harvard manalways went to Harvard. Again, find the probability that thegrandson of a man from Harvard went to Harvard.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Solution
i)
We first form a Markov chain with state space S = {H,D,Y } andthe following transition probability matrix :
P =
.8 0 .2.2 .7 .1.3 .3 .4
Recall: the ij th entry of the matrix Pn gives the probability that theMarkov chain starting in state i will be in state j after n steps.sd
Thus, the probability that the grandson of a man from Harvardwent to Harvard is the upper-left element of the matrix P2.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
P2 =
.7 .06 .24.33 .52 .15.42 .33 .25
ii)
If all sons of men from Harvard went to Harvard, this would givethe following matrix for the new Markov chain with the same set ofstates:
P =
1 0 0.2 .7 .1.3 .3 .4
and The upper-left element of P2 is 1. Does that make sense ????
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Markov Chain
Let consider a hypothetical market with Markov properties wherehistorical data has given us the following patterns: After a weekcharacterized of a bull market trend there is a 90% chance thatanother bullish week will follow. Additionally, there is a 7.5%chance that the bull week instead will be followed by a bearish one,or a 2.5% chance that it will be a stagnant one. After a bearishweek theres an 80% chance that the upcoming week also will bebearish, and so on.
Calculate the possibilities of a bull, bear or stagnant week for, 1and 5 weeks, into the future.
Solution
By compiling these probabilities into a table, we get the followingtransition matrix M:Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
M =
0.9 0.075 0.0250.15 0.8 0.050.25 0.25 0.05
Then create an initial vector x0 which contains information aboutwhich of the three different states any current week is in. In thisexample we will choose to set the current week as bearish,therefore x0 = (0, 1, 0)
Given the state of the current week, we can then calculate thepossibilities of a bull, bear or stagnant week for any number of nweeks into the future. In this way, after one and five weeks we willhave:
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
x1 = (0, 1, 0)
0.9 0.055 0.0250.15 0.8 0.050.25 0.25 0.05
1
= (0.15, 0.8, 0.05)
and
x5 = (0, 1, 0)
0.9 0.055 0.0250.15 0.8 0.050.25 0.25 0.05
5
= (0.48, 0.45, 0.07) �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Solve, in the traditional way ( using elementary operations -row reduction method ), each of the following systems
i)
x1 + x2 + x3 + x4 + x5 = 2x1 + x2 + x3 + 2x4 + 2x5 = 3x1 + x2 + x3 + 2x4 + 3x5 = 2
ii)
x1 + 2x2 + x3 = 12x1 + 4x2 + 2x3 = 3
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Solution
i) Using row-reduction to the augmented matrix
1 1 1 1 1 21 1 1 2 2 31 1 1 2 3 2
⇒ 1 1 1 1 1 2
0 0 0 1 1 10 0 0 1 2 0
⇒ 1 1 1 1 1 2
0 0 0 1 1 10 0 0 0 1 −1
⇒ 1 1 1 0 0 1
0 0 0 1 1 10 0 0 0 1 −1
⇒ 1 1 1 0 0 1
0 0 0 1 0 20 0 0 0 1 −1
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Thus, we have that the solution is
x5 = −1, x4 = 2, x3 = α, x2 = β, x1 = 1− α− β
x =
x1x2x3x4x5
=
1− α− β
βα2−1
=
1002−1
+α
−10100
+β
−11000
Thus, the solution is 2− dimensional subspace span by vectorsv1 = (1, 0, 1, 0, 0)T and v2 = (−1, 1, 0, 0, 0, )T
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Determine whether the following are linear transformations fromP2 to P3.
a) L(p(x)) = xp(x)
(b) L(p(x)) = x2 + p(x)
(c) L(p(x)) = p(x) + xp(x) + x2p′(x)
Solution
a) For a and b real numbers, we have:
L(ap(x) + bq(x)) = x [ap(x) + bq(x)] = x [ap(x)] + x [bq(x)] =
a[xp(x)] + b[xq(x)] = aL(p(x)) + bL(q(x))
so, the Transformation is linear.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
b) For a and b real numbers, we have:
L(ap(x) + bq(x)) = x2 + [ap(x) + bq(x)] 6=
x2 + [ap(x)] + x2 + [bq(x)] = aL(p(x)) + bL(q(x))
so, the Transformation is not linear.
c) For a and b real numbers, we have:
L(ap(x) + bq(x)) = (ap(x) + bq(x)) + x [ap(x) + bq(x)]+
x2[ap(x) + bq(x)]′ = a[p(x) + xp(x) + p′(x)]+
b[q(x) + xq(x) + q′(x)] = aL(p(x)) + bL(q(x))
so, the Transformation is Linear.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Let
b1 =
110
, b1 =
101
, b3 =
011
,
and let L be the linear transformation from R2 into R3 defined by
L(x) = x1b1 + x2b2 + (x1 + x2)b3
Find the matrix A representing L with respect to the ordered bases{e1, e2} and {b1,b2,b3}.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Solution
L(e1) = b1 + b3 = 1 · b1 + 0 · b2 + 1 · b3
L(e2) = b2 + b3 = 0 · b1 + 1 · b2 + 1 · b3
Thus, the matrix AL representing the transformation w.r.t. thegiven bases is
AL =
1 00 11 1
�
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
A linear transformation L : V→W is said to be one-to-one ifL(v1) = L(v2) implies that v1 = v2 (i.e., no two
distinct vectors v1, v2 in V get mapped into the same vectorw ∈W ). Show that L is one-to-one if and only if
ker(L) = {0V}.
Solution
if L is one to one, then
L(v1) = L(v2)⇒ v1 = v2
Now, if v ∈ ker(L) then L(v) = 0W, but we already know thatL(0V) = 0W ⇒ v = 0 for all v ∈ ker(L)⇒ ker(L) = {0V}
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
On the other hand, let’s assume that ker(L) = {0V}. Suppose that
L(v1) = L(v2)⇒
L(v1)− L(v2) = 0W ⇒
L(v1 − v2) = 0W ⇒
v1 − v2 ∈ ker(L)⇒
v1 − v2 = 0V ⇒
v1 = v2 ⇒
L is a one-to-one linear transformation. �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Let V be a subspace of F(R) spanned by functions ex , e−x . Let Lbe a linear operator on V such that
AL =
(2 −1−3 2
)is the matrix of L relative to the basis ex , e−x . Find the matrix ofL relative to the basis
cosh(x) =1
2(ex + e−x), sinh(x) =
1
2(ex − e−x)
Solution
From the above we know that
L(ex) = 2ex − 3e−x ; L(e−x) = −ex + 2e−x
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
and from the equations
cosh(x) = 12(ex + e−x), sinh(x) = 1
2(ex − e−x)⇒
we have
ex = cosh(x) + sinh(x) e−x = cosh(x)− sinh(x)⇒
L(ex) = 2ex−3e−x = 2(cosh(x)+sinh(x))−3(cosh(x)−sinh(x))⇒
L(ex) = 2ex − 3e−x = − cosh(x) + 5 sinh(x))
and
L(e−x) = −ex + 2e−x =−(cosh(x) + sinh(x)) + 2(cosh(x)− sinh(x))⇒
L(e−x) = −ex + 2e−x = cosh(x)− sinh(x)⇒
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
L(cosh(x)) = L(12(ex + e−x)) = 12L(ex) + 1
2L(e−x)⇒12 [− cosh(x) + 5 sinh(x)] + 1
2 [cosh(x)− sinh(x)]
L(cosh(x)) = 2 sinh(x)
Similarly
L(sinh(x)) = L(12(ex − e−x)) = 12L(ex)− 1
2L(e−x)⇒12 [− cosh(x) + 5 sinh(x)]− 1
2 [cosh(x)− sinh(x)]
L(sinh(x)) = − cosh(x) + 3 sinh(x)
and the matrix BL, representing L in this new basis is
BL =
(0 −12 3
)�
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Let L be a linear operator on Rn. Suppose that L(x) = 0 for somex 6= 0. Let A be the matrix representing L with respect to thestandard basis {e1, e2, . . . , en}. Show that A is singular.
Solution
A be the matrix representing L w.r.t. to standard basis, then theequation
L(x) = 0⇒ Ax = 0
is valid for a vector x 6= 0, therefore system of the equations
Av = 0
has not a unique solution, namely, v = 0 and v = x , so thatimplies that A is a singular matrix, otherwise the system wouldhave a unique solution, which is not the case. �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
let
A =
1 2 10 3 10 5 −1
(i) Find all eigenvalues of the matrix A.
(ii) For each eigenvalue of A, find an associated eigenvector.
(iii) Is the matrix A diagonalizable? Explain.
(iv) Find all eigenvalues of the matrix A2.
Solution
(i) Eigenvalues are given by the characteristic equation:
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
|A− λI| =
∣∣∣∣∣∣1− λ 2 1
0 3− λ 10 5 −1− λ
∣∣∣∣∣∣ = 0⇒
(1− λ)
∣∣∣∣ 3− λ 15 −1− λ
∣∣∣∣ = (1− λ) [(3− λ)(−1− λ)− 5] =
(λ− 1)(λ− 4)(λ+ 21) = 0
Thus the eigenvalues are: λ1 = 1, λ2 = 4, λ3 = −2 .
(ii) An eigenvector corresponding to the eigenvalue λ1 = 1 is givenby
(A− λ1I)x =
1− λ1 2 10 3− λ1 10 5 −1− λ1
x1x2x3
=
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
0 2 10 2 10 5 −2
x1x2x3
=
000
Thus, an eigenvector is
v1 =
100
=
An eigenvector corresponding to the eigenvalue λ2 = 4 is given by
(A− λ2I)x =
1− λ2 2 10 3− λ2 10 5 −1− λ2
x1x2x3
=
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
−3 2 10 −1 10 5 −5
x1x2x3
=
000
Thus, an eigenvector is
v2 =
111
=
An eigenvector corresponding to the eigenvalue λ3 = −2 is givenby
(A− λ2I)x =
1− λ3 2 10 3− λ3 10 5 −1− λ3
x1x2x3
=
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
3 2 10 5 10 5 1
x1x2x3
=
000
Thus, an eigenvector is
v3 =
11−5
Thus, the matrix A has three linearly independent vectors, namely:v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
(iii) The matrix A is diagonalizable since
A = XDX−1 =
1 1 10 1 10 1 5
1 0 00 4 00 0 −2
1 1 10 1 10 1 5
−1
(iii) The matrix A2 is given by
A2 =(XDX−1
) (XDX−1
)⇒
A2 = XD2X−1
whereD2 =
1 0 00 16 00 0 4
so the eigenvalues of A2 are 1, 16, 4 �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Find the kernel and range of each of the following linear operatorson P3 :
(a) L(p(x)) = xp′(x)
(b) L(p(x)) = p(x)− p′(x)
(c) L(p(x)) = p(0)x + p(1)
Solution
ker(L) is the set of all polynomials such that L(p(x)) = 0
a)
L(p(x)) = xp′(x) = 0⇒ p′(x) = 0⇒ p(x) = a0 = cte.
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
b)
L(p(x)) = p(x)− p′(x) = 0⇒ p′(x) = p(x)⇒ p(x) = 0 is theonly polynomial satisfying that eqation.
c)
L(p(x)) = p(0)x + p(1) = 0⇒ p(0) = 0 and p(1) = 0
Thus, the kernel is the set of all quadratic polymomials satisfayingthe above condition, that is p(x) = α(x − x2). �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Find the singular value decomposition of each of the followingmatrices:
i)
A =
(1 12 2
)ii)
B =
(2 −21 2
)Solution
i)
The matrix ATA is given by :
ATA =
(5 55 5
)Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
whose eigenvalues and eigenvectors are: λ1 = 10, λ2 = 0
v1 =
(11
); v2 =
(−11
)
The matrix is AAT :
AAT =
(2 44 8
)whose eigenvalues and eigenvectiors are λ1 = 10 and λ2 = 0
u1 =
(12
); u2 =
(−21
)Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Thus the SVD decomposition of the matrix A
A = UΣVT =
(1 −22 1
)( √10 00 0
)(1 −11 0
)T
ii)
The matrix BTB is given by:
BTB =
(5 −2−2 8
)whose eigenvalues and eigenvectors are: λ1 = 9, λ2 = 4
v1 =
(−12
); v2 =
(21
)
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
The matrix is BBT :
BBT =
(8 −2−2 5
)whose eigenvalues and eigenvectiors are λ1 = 9 and λ2 = 4
u1 =
(−21
); u2 =
(12
)Thus the SVD decomposition of the matrix B
B = UΣVT =
(−2 11 2
)(3 00 2
)(−1 22 1
)T
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Given x1 = 12(1, 1, 1,−1) and x2 = 1
6(1, 1, 3, 5)verify that these vectors form an orthonormal set in R4. Extendthis set to an orthonormal basis for R4 by finding an orthonormalbasis for the null space of
A =
(1 1 1 −11 1 3 5
)[Hint: First find a basis for the null space and then use theGram-Schmidt process.]
Solution< x1, x2 >= 1
12 [1 + 1 + 3− 5] = 0
< x1, x1 >= 1 < x2, x2 >= 1
Therefore {x1, x1} form an orthonormal set.Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Now, let’s find the Null Space of the matrix A, i. e. , solve thesystem
Av = 0⇒(
1 1 1 −11 1 3 5
)v1v2v3v3
=
(00
)Using row-reduction we have
Av = 0⇒(
1 1 1 −1 01 1 3 5 0
)⇒(
1 1 1 −1 00 0 2 6 0
)⇒
(1 1 1 −1 00 0 1 3 0
)⇒(
1 1 0 −4 00 0 1 3 0
)⇒
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
(1 1 0 −4 00 0 1 3 0
)So, the solution is given by
v1 = 4α− β, v2 = β, v3 = −3α, v4 = α
Thus, a basis for the Null space is v1 = (−1, 1, 0, 0) andv2 = (4, 0,−3,−1)
and using the Gram-schmidt process we come up with
u1 = v1||v1||
u2 = v2 − <v2,u1><u1,u1>
u1/||v2 − <v2,u1><u1,u1>
u1||
u1 = 1√2
(−1, 1, 0, 0) and u2 =√26 (2, 2,−3, 1) �
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Dr. Marco A Roque Sol Linear Algebra. Session 11
Abstract Linear Algebra Orthogonal Polynomials
Examples
Dr. Marco A Roque Sol Linear Algebra. Session 11