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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.

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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

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

Least squares problem induction

based on excelent video lectures by Gilbert Strang, MIT

Lecture 15

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


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.

  • 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.


based on excelent video lectures by Gilbert Strang, MIT

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.


e = b - 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).



e = b - 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





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.



e = b - p






  • 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.

  • 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)


e = b - p




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.

Moore-Penrose Pseudoinverse

Least SquaresCalculation

based on excelent video lectures by Gilbert Strang, MIT

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


= 0


p = Pb → b – e = Pb

e = (I - P)b



p + e = 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 are

shifted 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.


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.




so the overall error is the sum of squares

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








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


Least squares – traditional way

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


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

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



a + b = 1

a + 2b = 2

a + 3b = 2


a + b = 1

a + 2b = 2

a + 3b = 2

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







  • 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

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