Last lecture summary

1 / 45

Last lecture summary - PowerPoint PPT Presentation

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

PowerPoint Slideshow about 'Last lecture summary' - tyson

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
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) = 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
• 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.

Least squares problem induction

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 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.
• 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.
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.
• 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?
• 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(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

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 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 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?

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.
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).

b

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
• 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 = 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

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.

^

^

^

• 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 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.
• 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

^

^

• 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?
• 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.
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.

Least SquaresCalculation

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
• 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 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

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

(Points at the picture are

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 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:

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

^

• 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)

• 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 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