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Last lecture summary. Fundamental system in linear algebra : system of linear equations A x = b . nice case – n equations, n unknowns matrix notation row picture column picture linear combinations. For our matrix, can I solve A x = b for every b ? Yes

Last lecture summary

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- Fundamental system in linear algebra : system of linear equations Ax = b. nice case – n equations, n unknowns
- matrix notation
- row picture
- column picture
- linear combinations

- For our matrix, can I solve
Ax = b for every b?

- Yes
- And what it means geometrically?
- RHS fills the whole 2D space.
- But when this can go wrong?
- If two columns are on the same line, then their combination is also on this line.
- singular matrix, not invertible
- invertible matrix, columns are independent

- row times column
- shape of matrices?
- m x n . n x p = m x p

- column picture
- AB = C
- Columns of C are linear combinations of columns of A

- row picture
- Rows of C are linear combinations of rows of B

- column times row
- set of full size matrices

- I = AA-1 = A-1A
- When is square matrix invertible (i.e. nonsingular)?
- If it does not have dependent columns
- If you can’t find non-zero x such that Ax = 0

Rules

- CT = (AB)T = BTAT
- (A-1)T = (AT)-1

- What is it?
- What is symmetric matrix?
- RTR is symmetric, RRT is also symmetric.

- contains vectors – objects I can
- add together
- multiply by number

- zero must belong to the space, otherwise it can’t be called space !

linear combination

- Something smaller within the space.
- This smaller bit is also space.
- Subspaces in R2:
- all of R2
- all lines through zero
- Is this line the same as R1?

- zero vector

- subspaces come out of matrices
- column space C(A)
- take columns of A
- all their linear combinations
- They all together form a space.
- Original columns as well as zero vector are all covered by the term “all linear combinations”.

- We’ll be interested in the size of the column space now.
- What do you think, is that space the whole four dimensional space? Just use your feeling, if we start with three vectors and take their combinations, can we get the whole four dimensional space?

NO

- So somehow we get a smaller space, but how much smaller? That’s not immediate.
- Let’s make me first the critical connection with linear equations.

- Column space is a subspace of R what?
- R4

- What’s in the column space of A?

- The columns (vectors) and all their linear combinations.

Two questions

- Does Ax = b always have a solution for every b? I guess that's going to be a yes or no question.
- And then I'm going to ask which right-hand sides are okay?
- OK, let’s write Ax = b for our A.
- What is the answer to the 1st question?
- Apparently the answer for 1. is No. Why?
- the combinations of the columns don't fill the whole four dimensional space

- However, for some RHSs I can solve that Ax = b.

- Which RHSs allow me to solve this? Which vectors b allow this system to be solved? This is the critical question.
- Tell me one RHS I can solve this system for?
- All zeros.
- Tell me another RHS I can solve for?
- [1 2 3 4]

- And another?
- [6 7 8 9]

- So apparently I can solve Ax = bwhen b is a linear combination of the columns.
- In other words?

- I can solve Ax=b exactly when bis in the column space.
- If b is not a combination of the columns, then there is no x. There's no way to solve Ax = b.
- Now the question is are all the columns independent? Do they all contribute something new, or can I drop one of them and still have the same column space?

- So, would you throw some column away without changing a column space?
- Col3 = Col1 + 4
- The column space of this matrix is a two dimensional subspace of R4.

- Describe the column spaces for

I … whole space R2

A … line, the equation Ax is solvable only if b is on that line

B … whole space R2

- That’s a completely different space.
- What's in it? It contains not right-hand sides b. It contains x’s.
- All solutions (x) to the equation Ax = 0.
- So where is the null space for this example?
- All x’s form a subspace of what?
- of R3

- Column space is in R4, while null space is in different R (R3).
- OK, we’ll try to find a null space N(A) of our matrix A. Help me, find at least three x’s.

- [0 0 0] Now try to find another?
- [1 4 -1] Now try to find yet another?
- [2 8 -2]
- In other words, c-multiples of [1 4 -1]

- So geometrically, how would you describe a null space?

- It’s a line. Line in R3, through the origin.
- We have to show, that this line is a subspace.

- Show, that if I have two solutions v and w, their sum v + wis also a solution. So if Av = 0 and Aw = 0, then A(v + w) = 0
- This is actually one of the matrix laws (distributive law), that I can split A(v + w) into two pieces: Av + Aw = 0

- And similarly I have to show that if Av = 0then A times any multiple v is zero.
- A(12.v) = 0because 12 x Av = 12 x 0= 0

- Now, to fully understand the vector space, let’s change the RHS
- What’s the solution?
- [1 0 0]
- Are there any other solutions?
- [0 -4 1], etc.
- Do the solutions form a subspace?
- They do NOT, why?
- zero vector is not a solution

- What are the solutions geometrically?
- They form a plane, not going through the origin.

Independence, basis, dimension

based on excelent video lectures by Gilbert Strang, MIT

http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture09.htm

Lecture 9

- When x1, x2, …, xn are independent?
- No linear combination gives zero vector, except 0.

- I have three nonzero vectors in 2D space (plane).
- I can arrange such vectors in a 2 x 3 matrix A. Then, the vectors are dependent, if Ax = 0 (except x = 0), i.e. if there is something in the null space of A.
- In othe words, columns are independent, if there is only 0 in its null space.
- The number of independent columns in matrix A is called the rank (hodnost) of matrix A.

- When we had a columns in a matrix, we took all their combinations and that gave us the column space.
- Those vectors that we started with span (definují) that column space.
- So now I can say in shorthand the columns of a matrix span the column space.

- Columns of matrix A span the column space.
- Are the columns independent?
- It depends on that particular columns.

- But obviously we’re highly interested in a set of vectors that spans a space and is independent.
- If we didn’t have them all, we wouldn’t have our whole space
- If we had more, they wouldn’t be independent.

- Such a bunch of vector is called a basis for a vector space.

- Basis is a set of vectors they have two properties:
- I’ve got enough vectors.
- And not too many.

- Well, mathematician way of saying the same:
- they span the space
- they are independent

- From now, whenever I look at a subspace, if you give me a basis for that subspace, you've told me everything I need to know about that subspace.
- Basis is
- minimum spanning set
- maximum independent set

- Examples – R3 space
- What would be a basis for a 3D space:
- [1,0,0]T, [0,1,0]T, [0,0,1]T, identity matrix, null space is 0

- n vectors give a basis if the n x n matrix with those columns is invertible
- why?
- Matrix is invertible, if Ax = 0 only for x = 0

- are these vectors independent?
- yes

- which space they span?
- R3 (3D)

- do they form a basis?
- yes, for R3

- Are these two vectors a
- basis for any space?
- yes, and for what space?
- the one they span, i.e. their combination
- what is their combination?
- plane inside 3D space

- basis is not unique
- there are zilions of bases
- columns of any invertible 3x3 matrix form bases for 3D

- however, all bases for a given space have the same number of vectors (3 for R3, n for Rn)
- If we're talking about some other space, the column space of some matrix, or the null space of some matrix, or some other space that we haven't even thought of, then that still is true that there're lots of bases but every basis has the same number of vectors.

- this number is called dimension (how big is the space?)
- Let me repeat the four terms we’ve got now defined
- Independence - looks at combinations not being zero
- Spanning - looks at all the combinations
- Basis - combines independence and spanning
- Dimension - the number of vectors in any basis, because all bases have the same number.

- do the columns span the column space of this matrix?
- yes, by definition what the column space is

- form they a basis for the column space?
- no, they are not independent, there’s something
- in the null space

- Look at the null space N(A)
- Tell me some vector in the nullspace(solution of Ax= 0)
- [-1 -1 1 0]T

- Tell me the basis for that column space. There are many
- answers, give me the most natural answer.
- first two columns

- And the rank of the matrix is?
- two

- Great theorem comes !!!!
- rank is the number of independent columns
- rank is the number of vectors in the basis
- rank is WHAT?
- dimension!

- The rank of A is the dimension of the column space.
dim C(A) = r

- About words
- I am talking about the rank of the matrix
- and I am talking about the dimension of the vector space/subspace
- I am not talking about the dimension of a matrix, I am not talking about a rank of space
- And there is a link between the rank of matrix and the dimension of its column space.

- Null space
- we already have one vector there: [-1 -1 1 0]T
- Are there other vectors in the null space?
- Yes. So our only vector is not a basis, because it does not span.

- Tell me one vector more
- [-1 0 0 1]T

- The vectors in the null space are telling me in what way the columns are dependent. That's what the null space is doing.
- Now, what is the nullspace?
- These are two vectors in the null space. They're independent. Are they a basis for the null space? What's the dimension of the null space?
- they are independent, they form a basis, the null space is two-dimensional

dim N(A) = n – r

n is the number of columns

Four fundamental subspaces

based on excelent video lectures by Gilbert Strang, MIT

http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture10.htm

Lecture 10

- column space C(A)
- null space N(A)
- row space
- all combinations of rows of A
- rows span a row space
- they are/they are not a basis for a row space
- But I don’t like to work with row vectors, I’d like to stay with column vectors. How to get column vectors out of the rows?
- transpose

- So row space is all combinations of columns of AT - C(AT)

- null space of AT... N(AT) (called the left null space of A)

- Where are these spaces? A is m x n
- N(A) – vectors with n components (it is in Rn), solutions to Ax = 0
- C(A) – columns of A, each column has m components, it is in Rm
- C(AT) – the row space is in Rn
- N(AT) – the left null space is in Rm
- And now we want to understand these spaces, i.e. we’d like to know a basis for those spaces. And what’s their dimension?

- A = m x n
- dim C(A) = r
- dim C(AT) = r
- dim N(A) = n – r
- dim N(AT) = m – r
- dim C(A) + dim N(A) = n
- dim C(AT) + dim N(AT) = m
- The row space and the null space are in Rn. Their dimensions add to n.
- The column space and the left null space are in Rm, and their dimensions add to m.

- Please, pay attention to the fact that dimension (i.e. rank) of the column space and row space is the same.
- E. g. is this matrix singular?
- Yes, it is, because two rows are the same. Thus rank both of column and row space is two.

- All our vector spaces have been subspaces of some real n dimensional space.
- A new vector space – all 3 x 3 matrices.
- i.e. my matrices are vectors.
- They don’t look like vectors, they’re matrices, but they are vectors in my vector space because they obey the rules.

- How about subspace of this matrix space M?
- upper triangular matrices U
- symmetric matrices
- intersection of two subspaces is also a subspace
- what is the intersection of U and symmetric?
- diagonal – this is smaller subspace

- Now we will intuitively investigate a bases (and dimensions) of these three matrix spaces: all 3x3 matrices, 3x3 upper triangular, 3x3 symmetric and 3x3 diagonal matrices
- Dimension = number of members in the basis
- 3x3 matrices
- basis?
- matrices, each with 1 at different positions and the rest zeros

- dimension?
- nine

- basis?

- upper triangular matrices
- dimension?
- six

- dimension?
- symmetric
- dimension is again six

- diagonal
- dimension is three

Orthogonality

based on excelent video lectures by Gilbert Strang, MIT

http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture14.htm

Lecture 14

- Welcome to the world of orthoganility. This is a ninety degree lecture.
- What it means for vectors to be orthogonal?
- What it means for subspaces to be orthogonal?

- orthogonal = perpendicular, the angle between two vectors is 90o
- How to find whether two vectors are orthogonal? Use Pythagoras.

what is this vector in terms of a and b?

length of the vector

- So
- This leads to the inner product rule about orthogonality. Here is why:
aTa + bTb = (a+b)T(a+b) = aTa + bTb + aTb + bTa

aTb + bTa = 0 2 aTb = 0 aTb = 0

- Two vectors are orthogonal if aTb= bTa = 0

- Zero vector is orthogonal to any vector.

- Subspace S is orthogonal to subspace T. What it means?
- It means that every vector in S is orthogonal to every vector in T.

- Wall is one subspace in 3D, floor is another subspace in 3D.
- Are they orthogonal?
- No

- And why not?
- if two subspaces meet at some vector, well then for sure they're not orthogonal, because that vector is in one and it's in the other, and it's not orthogonal to itself unless it's zero.

- row space is orthogonal to the null space (in Rn)
- Why?
- if x is in the null space, then Ax = 0
- So I'm saying that a vector in the row space is perpendicular to the x in the null space.
- row 1 is orthogonal to x, their inner product is zero

- OK, so far we have shown that rows of A are orthogonal to x, but what else is in the row space?
- all their linear combinations

- To show that the linear combinations of rows are also zero is pretty easy. Help me
- row1. x = 0, so c . row1 . x = 0

- Similarly, column space is orthogonal to the left null space (in Rm)

- Next comes another definition, without proving.
- Row space and null space are orthogonal complements (in Rn).
- The orthogonal complement of a row space contains not just some vectors that are orthogonal to it, but all.
- That means that the null space contains all, not just some but all, vectors that are perpendicular to the row space.

G. Strang, Introduction to linear algebra