last lecture summary independent vectors x rank – the number of independent columns/rows in a...

Last lecture summary

• independent vectors x

• rank – the number of independent columns/rows in a matrix

0alln,combinatiozeroexcept

02211

i

nn

c

xcxcxc

552

873

321 • Rank of this matrix is 2!• Thus, this matrix is noninvertible (singular).• It’s because both column and row spaces have the same rank.• And row2 = row1 + row3 are identical, thus rank is 2.

• Column space – space given by columns of the matrix and all their combinations.

• Columns of a matrix span the column space.

• We’re highly interested in a set of vectors that spans a space and is independent. Such a bunch of vector is called a basis for a vector space.

• Basis is not unique.• Every basis has the same number of

vectors – dimension.• Rank is dimension of the column space.

• dim C(A) = r, dim N(A) = n - r (A is m x n)• row space

– C(AT), dim C(AT) = r• left null space

– N(AT), dim N(AT) = m – r• C(A) ┴ N(AT)• C(AT) ┴ N(A), row space and null space

are orthogonal complements

G. Strang, Introduction to linear algebra

• orthogonal = perpendicular, dot product aTb = a1b1+a2b2+… = 0

• length of the vector |a| = √|a|2 = √aTa• If subspace S is orthogonal to subspace T

then every vector in S is orthogonal to every vector in T.

Four possibilities for Ax = b

A: m × n, rank r

r = m & r = n square & invertible 1 solution

r = m & r < n short & wide ∞ solutions

r < m & r = n tall & thin 0 or 1 solution

r < m & r < n not full rank 0 or ∞ solutions

Least squares problem induction

based on excelent video lectures by Gilbert Strang, MIThttp://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture15.htmLecture 15

http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture15.htm

I want to solve Ax = b when there is no solution.

WHAT ??WAS ??

• So b is not in a column space.• This problem is not rare, it’s actually quite

typical.• It appears when the number of equations

is bigger than the number of unknowns (i.e. m > n for m x n matrix A)– so what can you tell me about rank, what the

rank can be?• it can’t be m, it can be n or even less

– so there will be a lot of RHS with no solution !!

• Example– You measure a position of sattelite buzzing around– There are six parameters giving the position– You measure the position 1000-times– And you want to solve Ax = b, where A is 1000 x 6

• In many problems we’ve got too many equations with noisy RHSs (b).

• So I can't expect to solve Ax = b exactly right, because there's a measurement mistake in b. But there's information too. There's a lot of information about x in there.

• So I’d like to separate the noise from the information.

• One way to solve the problem is throw away some measurements till we get nice square, non-singular matrix.

• That’s not satisfactory, there's no reason in these measurements to say these measurements are perfect and these measurements are useless.

• We want to use all the measurements to get the best information.

• But how?

• Now I want you jump ahead to the matrix that will play a key role. It is a matrix ATA.

• What you can tell me about the matrix?– shape?

• square – dimension?

• n x n– symmetric or not?

• symmetric• Now we can ask more about the matrix. The answers will

come later in the lecture– Is it invertible?– If not, what’s its null space?

• Now let me to tell you in advance what equation to solve when you can’t solve Ax = b:– multiply both sides by AT from left, and you get ATAx = ATb, but

this x is not the same as x in Ax = b, so lets call it , because I am hoping this one will have a solution.

– And I will say it’s my best solution. This is going to be my plan.

x̂

• So you see why I am so interested in ATA matrix, and its invertibility.

• Now ask ourselves when ATA is invertible? And do it by example.

51

21

11

A

• 3 x 2 matrix, i.e. 3 equations on 2 unknowns• rank = 2• Does Ax equal b? When can we solve it?

• Only if b is in the column space of A.• It is a combination of columns of A. • The combinations just fill up the plane,

but most vectors b will not be on that plane.

• So I am saying I will work with matrix ATA.• Help me, what is ATA for this A?

51

21

11

A

308

83AAT • Is this ATA invertible?

• Yes

• However, ATA is not always invertible !• Propose such A so that ATA is not invertible ?

31

31

31

A

279

93AAT

Generally, if I have two matriceseach with rank r, their product can’t have rank higher than r.

And in our case rank(A)=1, so rank(AT) can’t be more than 1.

• This happens always, rank(ATA) = rank(A).• If rank(ATA) = rank(A), then N(ATA)=N(A).

• So ATA is invertible exactly if N(A)=0. Which means when columns of A are independent.

Projections



• I want to find a point on line a that is closest to b.• My space is what?

– 2D plane• Is line a a subspace?

– Yes, it is, one dimensional.• So where is such a point?• So we say we projected vector b on line a, we projected b into

subspace. And how did we get it?• Orthogonality

ab

p

e = b - p

e is the error, i.e. how much I am wrong by, and it is perpendicular to a

And we know, that the projection p is some multiple of a, p = xa. And we want tofind the number x.

p = xa

• Key point is that a is perpendicular to e.• So I have aTe = aT(b-p) = aT(b -xa) = 0• So after some simple math we get

• I may look at the problem from another point of view.

• The projection from b to p is carried out by some matrix called projection matrix P.

• p = Pb• What is the P for our case?

xapaa

bax

T

T

aa

baap

T

T

aa

aaP

T

T

→

Projection matrix

• What’s its column space?– How acts the column space of a matrix A?

• If you multiply the matrix A by anything you always get in the column space. That’s what column space is.

– So where am I if I do Pb? • I am on the line a. The column space of P is the

line through a.

• What is the rank of P?– one

• Column times row is a rank one matrix, the columns of the matrix are row-wise-multiples of the column vector, so the column vector is a basis for its column space.

1510

128

96

,32,

5

4

3TT baba

• P is symmetric. Show me why?

• What happens if I do the projection twice? i.e. I multiply by P and then by P again (P × P = P2).

P

aa

aa

aa

aa

aa

aaP

T

T

TTT

TTT

TT

TTT

aa

aaP

T

T

• So if I project b, and then do projection again I what?– stay put– So P2 = P … Projection matrix is idempotent.

ab

p

e = b - p

p = xa = Pb

• Summary: if I want to project on line, there are three formulas to remember:

• And properties of P:– P = PT, P = P2

aa

aaPxap

aa

bax

T

T

T

T

More dimensions• Three formulas again, but different, we won’t

have single line, but plane, 3D or nD subspace.• You may be asking why I actually project?

– Because Ax = b may have no solution– I am given a problem with more equations than

unknowns, I can’t solve it.– The problem is that Ax is in the column space, but b

does not have to be. – So I change vector b into closest vector in the column

space of A. – So I solve Ax = p instead !!– p is a projection of b onto the column space– I should indicate somehow, that I am not looking for x

from Ax = b (x, which actually does not exist), but for x that’s the best possible.

px ˆA

• I must figure out what’s the good projection here. What's the good RHS that is in the column space and that's as close as possible to b.

• Let’s move into 3D space, where I have a vector b I want to project into a plane (i.e. subspace of 3D space)

b

pa1

a2

this is a plane of a1 and a2

This plane is the column space of matrix A 21 | aaA

e = b - p e is perpendicular to the plane

Apparently, projection p is some multiple of basis vectors.

p = x1a1 + x2a2 = Ax , and I am looking for x^^ ^ ^

So now I've got hold of the problem. The problem is to find the right combination of the columns so that the error vector (b – Ax) is perpendicular to the plane.

^

• I write again the main point– Projection is p = Ax– Problem is to find x– Key is that e = b – Ax is perpendicular to the

plane• So I am looking for two equations,

because I have x1 and x2.

• And e is perpendicular to the plane, so it means it must be perpendicular to each vector in the plane. It must be perpendicular to a1 and a2 !!

• So which two eqs. do I have? Help me.

^^^

b

pa1

a2

e = b - p

^ ^

A word about subspaces.• In what subspace lies (b – Ax)?

– Well, this is actually vector e, so I have ATe=0. Thus in which space is e?

– In N(AT)!• And from the last lecture, what do we

know about N(AT)?– It is perpendicular to C(A).

0ˆ0ˆ 21 xAbaxAba TT

0ˆ

xAbAT

^

e is in N(AT)e is ┴ to C(A)

It perfectly holds.We all are happy, aren’t we?

b

pa1

a2

e = b - p

• OK, we’ve got the equation, let’s solve it.• ATA is n by n matrix.• As in the line case, we must get answers

to three questions:1. What is x?

2. What is projection p?

3. What is projection matrix P?

bAxAAxAbA TTT

ˆ0ˆ

^

normal equations

• x is what? Help me.• What is the projection p = Ax?

• What’s the projection

matrix p = Pb?

bAxAA TT

ˆ

bAAAx TT1

ˆ

bAAAAp TT 1

projection matrix P

^

^

• can I do this?

• Apparently not, but why not? What did I do wrong?

• A is not square matrix, it does not have an inverse.

• Of course, this formula works well also if A was square invertible n x n matrix.– Then it’s column space is the whole what?

• Rn

– Then b is already in the whole Rn space, I am projecting b there, so the P = I.

?what111

TTTT AAAAAAAAP

• Also P = PT, and P = P2 holds. Prove P2!• So we have all the formulas

• And when will I use these equations. If I have more equations (measurements) than unknowns.

• Least squares, fitting by a line.

TTTTTT AAAAPbAAAApbAAAx111

ˆ

Moore-Penrose Pseudoinverse

bAx 1

bAAAx TT1

ˆ

TT AAAA1

Least SquaresCalculation



Projection matrix recap

• Projection matrix P = A(ATA)-1AT projects vector b to the nearest point in the column space (i.e. Pb).

• Let’s have a look at two extreme cases:

1. If b is in the column space, then Pb = b. Why?– What does it mean that b is in the column

space of A?– b is linear combination of columns of A, i.e. b

is in the form Ax.– so Pb = PAx = A(ATA)-1ATAx = Ax = b

2. If b is ┴ to the column space of A then Pb = 0. Why?

– What vectors are perpendicular to the column space?

– Vectors in N(AT)– Pb = A(ATA)-1ATb = 0

= 0C(A)

N(AT)

b

p

ep + e = b

p = Pb → b – e = Pbe = (I - P)b

That’s the projection too.Projection onto the ┴ space.

When P projects onto one subspace, I – P projects onto the perpendicular subspace

points (1,1) (2,2) (3,2)

(Points at the picture areshifted for better readability.)

OK, I want to find a matrix A, once we have A, we can do all we need.

I am looking for the best line (smallest overall error) y = a + bx, meaning I am looking for a, b.

Equations: a + b = 1 a + 2b = 2 a + 3b = 2

1 1 1

1 2 2

1 3 2

aAx b

b

bAxAA TT

ˆ

this eq. can’t be solved

but this can

x

y

• In other words, the best solution is the line with smallest errors in all points.

• So I want to minimize length |Ax – b|, which is the error |e|, actually I want to minimize the never-zero quantity |Ax – b|2.

x

y

e1

e2

e3

so the overall error is the sum of squares|e1|2 + |e2|2 + |e3|2 p1 p2

p3

b1

b3

b2

What are those p1, p2, p3?If I put them in the equationsa + b = p1

a + 2b = p2

a + 3b = p3

I can solve them. Vector [p1,p2,p3] is in the column space

Least squares – traditional way

• least squares problem – “metoda nejmenších čtverců” … the sum of square of errors is minimized

x

y

points (x,y) : (1,1) (2,2) (3,2)

I am looking for a line: a + bx = y


x

y

e1

e2

e3p1 p2

p3

b1

b3

b2


points (x,y) : (1,1) (2,2) (3,2)

- So if there is a solution, each point lies on that line: a + b = 1, a + 2b = 2, a + 3b = 2- However, there is apparently no solution, no line at which all three

points lie.- The optimal line a+bx will go somewhere between the points. Thus

for each point, there will be some error (i.e. b value of the point on that line will differ from the required b value)

- Therefore, the errors are: e1 = a + b - 1, e2 = a + 2b - 2, e3 = a + 3b - 2

Least squares – linear algebra way

C(A)

N(AT)

b

p

e

3

2

1

31

21

11

2

2

1

p

p

p

pAb

• And now computation• Task: find p and x = [a b]

• Let’s solve that equation for

• Help me, what is ATA?

• And what is ATb?

• So I have to solve (Gauss elimination) a system of linear equations 3a + 6b =5, 6a + 14b = 11

bAxAA TT

ˆ

146

63

11

5

a = 1/2 b=2/3

^1 1 1

1 2 2

1 3 2

aAx b

b

• best line: 2/3 + 1/2x• What is p1?

– A value for x = 1 … 7/6

• And e1? – 1 - p1 = -1/6

• p2 = 5/3, e2 = +2/6, p3 = 13/6, e3 = -1/6

• So we have projection vector p, and error vector e

points (1,1) (2,2) (3,2)

616261

6133567

2

2

1

epb

Ja, das stimmt!

• p and e should be perpendicular. Verify that.

• However, e is not perpendicular not only to p. Give me another vector e is perpendicular to?– Well, e is perpendicular to column space, so?– It must be perpendicular to columns of matrix

A, i.e. to [1 1 1] and [1 2 3]• Just again, fitting by straight line

means solving the key equation

xApbAxAA TT ˆˆ But A must have indpendent columns,

then ATA is invertible

If not, oops, sorry, I am out of luck

last lecture summary independent vectors x rank – the number of independent columns/rows in a...

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