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### Least squares problem induction all their combinations.

### Projections role. It is a matrix A

### Least Squares the good RHS that is in the column space and that's as close as possible to Calculation

Last lecture summary

- independent vectors x
- rank – the number of independent columns/rows in a matrix

- 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) = all their combinations.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 all their combinations.

- orthogonal = perpendicular, dot product all their combinations.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 A all their combinations.x = b

A: m × n, rank r

based on excelent video lectures by Gilbert Strang, MIT

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

Lecture 15

- So all their combinations.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 !!

- so what can you tell me about rank, what the rank can be?

- Example all their combinations.
- 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

- shape?
- 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.

- So you see why I am so interested in A role. It is a matrix ATA matrix, and its invertibility.
- Now ask ourselves when ATA is invertible? And do it by example.

- 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 A role. It is a matrix ATA.
- Help me, what is ATA for this A?

- Is this ATA invertible?
- Yes

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

Generally, if I have two matrices

each 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(A role. It is a matrix 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.

based on excelent video lectures by Gilbert Strang, MIT

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

Lecture 15

e role. It is a matrix A 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 to

find the number x.

b

e = b - p

a

p

p = xa

- 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

- Key point is that role. It is a matrix Aa 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?

→

Projection matrix role. It is a matrix A

- 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.

- How acts the column space of a matrix A?

- What is the rank of P? role. It is a matrix A
- 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.

- P is symmetric. Show me why? role. It is a matrix A
- What happens if I do the projection twice? i.e. I multiply by P and then by P again (P × P = P2).

b role. It is a matrix A

a

e = b - p

p

p = xa = Pb

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

- Summary: if I want to project on line, there are three formulas to remember:
- And properties of P:
- P = PT, P = P2

More dimensions formulas to remember:

- 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.

- 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)

e the good RHS that is in the column space and that's as close as possible to = b - p

e is perpendicular to the plane

b

a2

p

a1

this is a plane of a1 and a2

This plane is the column space of matrix A

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.

^

b the good RHS that is in the column space and that's as close as possible to

e = b - p

a2

^

p

a1

^

- 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.

^

^

^

A word about subspaces. the good RHS that is in the column space and that's as close as possible to

- 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).

^

e the good RHS that is in the column space and that's as close as possible to is in N(AT)

e is ┴ to C(A)

b

e = b - p

a2

p

a1

It perfectly holds.

We all are happy, aren’t we?

- OK, we’ve got the equation, let’s solve it. the good RHS that is in the column space and that's as close as possible to
- ATA is n by n matrix.
- As in the line case, we must get answers to three questions:
- What is x?
- What is projection p?
- What is projection matrix P?

normal equations

^

^ the good RHS that is in the column space and that's as close as possible to

- x is what? Help me.
- What is the projection p =Ax?
- What’s the projection
matrix p = Pb?

^

projection matrix P

- can I do this? the good RHS that is in the column space and that's as close as possible to
- 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.

- Then it’s column space is the whole what?

- Also P = P the good RHS that is in the column space and that's as close as possible to T, 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.

Moore-Penrose Pseudoinverse the good RHS that is in the column space and that's as close as possible to

based on excelent video lectures by Gilbert Strang, MIT

http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture16.htm

Lecture 16

Projection matrix recap the good RHS that is in the column space and that's as close as possible to

- 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:
- 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

- If the good RHS that is in the column space and that's as close as possible to 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

C(A)

= 0

p

p = Pb → b – e = Pb

e = (I - P)b

b

e

p + e = b

That’s the projection too.

Projection onto the ┴ space.

N(AT)

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

y the good RHS that is in the column space and that's as close as possible to

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

(Points at the picture are

shifted for better readability.)

x

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

but this can

this eq. can’t be solved

- 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.

y

b2

p3

so the overall error is the sum of squares

|e1|2 + |e2|2 + |e3|2

e2

e3

p1

p2

b3

e1

b1

What are those p1, p2, p3?

If I put them in the equations

a+ b= p1

a+ 2b= p2

a+ 3b= p3

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

x

Least squares – traditional way errors in all points.

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

y

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

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

x

Equations:

a + b = 1

a + 2b = 2

a + 3b = 2

Equations: errors in all points.

a + b = 1

a + 2b = 2

a + 3b = 2

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

y

b2

p3

e2

e3

- 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

p1

p2

e1

b3

b1

x

^ errors in all points.

- 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

a = 1/2 b=2/3

points (1,1) (2,2) (3,2) errors in all points.

- 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

Ja, das stimmt!

- p errors in all points. 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

But A must have indpendent columns,

then ATA is invertible

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

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