Lecture 10

A. Agarwal
April 10, 2012

Recall

Previously, we saw the inverse 𝐌 of a matrix 𝐀:

π€πŒ=πŒπ€=𝐈n

where 𝐀 is an nΓ—n matrix; that is, 𝐀 must be square to have an inverse.

In a 3Γ—3 matrix, this equates to solving each of

𝐀⁒xβ†’ =[100] 𝐀⁒xβ†’ =[010] 𝐀⁒xβ†’ =[001]

which can be formulated as the process of transforming an extended augmented matrix through row reductions:

[π€πˆ]⟢[𝐈𝐌]

Fundamental Theorem of Invertible Matrices

Let 𝐀nΓ—n be a matrix over a field 𝔽. Then the following are equivalent (TFAE): 1. 𝐀 is invertible. 2. The reduced row echelon form (rref) of 𝐀 is 𝐈. 3. The rank⁒(A)=n. 4. The columns of 𝐀 are linearly independent. 5. The rows of 𝐀 are linearly independent. 6. The homogeneous system 𝐀⁒xβ†’=0β†’ has only the zero solution xβ†’=0β†’. 7. 𝐀⁒xβ†’=bβ†’ will have a unique solution for all bβ†’βˆˆβ„n. 8. The span of the columns of 𝐀=ℝn. 9. The columns and rows of 𝐀 form a basis of ℝn. 10. The det⁑(𝐀)β‰ 0. 11. 𝐀 is the product of elementary matrices.

Proof of (6).

Consider 𝐀⁒xβ†’=0β†’. If 𝐀 has an inverse 𝐌, then

𝐀⁒xβ†’ =0β†’
πŒπ€β’xβ†’ =𝐌⁒0β†’
𝐈⁒xβ†’ =0β†’
x→ =0→

Proof of (7).

Suppose bβ†’βˆˆβ„n is an arbitrary vector. Next consider the system

𝐀⁒xβ†’ =bβ†’
πŒπ€β’xβ†’ =𝐌⁒bβ†’
xβ†’ =𝐌⁒bβ†’

so the system has a unique solution for any b→.

Let us take a closer look at some of the implications of this theorem. Suppose 𝐀 is a 3Γ—3 matrix. Suppose each of the following systems

𝐀⁒xβ†’ =[100] 𝐀⁒xβ†’ =[010] 𝐀⁒xβ†’ =[001]

has a solution, called x→1,x→2,x→3, respectively.

Suppose we want to find the solution of 𝐀⁒xβ†’=[1eΟ€]. Note that

[1eΟ€] =1⁒[100]+e⁒[010]+π⁒[001]
=1⁒𝐀⁒xβ†’1+e⁒𝐀⁒xβ†’2+π⁒𝐀⁒xβ†’3
=A⁒(1⁒xβ†’1+e⁒xβ†’2+π⁒xβ†’3)

So the solution is given by xβ†’1+e⁒xβ†’2+π⁒xβ†’3.

Elementary Matrices

An elementary matrix, 𝐄, is a matrix we obtain by performing exactly one row operation on an identity matrix.

For example, in the 2Γ—2 case,

[1001]β†’2⁒R2[1002] [1001]β†’-3⁒R1+R2[10-31] [1001]β†’R2↻R1[0110]

The matrices [1002],[10-31],[0110] are all elementary matrices.

Consider multiplying a 2Γ—2 matrix 𝐀 by the second of these elementary matrices:

[10-31]⁒𝐀 =[10-31]⁒[abcd]
=[ab-3⁒a+c-3⁒b+d]
[abcd]β†’-3⁒R1+R2 [ab-3⁒a+c-3⁒b+d]

We find that multiplying 𝐀 by an elementary matrix has the same effect as performing the same single row operation on 𝐀 as was performed to obtain the elementary matrix. This shows us that to perform one row operation on a matrix 𝐀 we just need to multiply 𝐀 on the left by an appropriate elementary matrix.

Properties of Inverses

For a matrix 𝐀, the following are true of its inverse 𝐀-1:

  1. 1.

    (𝐀-1)-1=𝐀

  2. 2.

    (k⁒𝐀)-1=1k⁒𝐀-1

  3. 3.

    𝐀-1 is unique

  4. 4.

    (𝐀𝐁)-1=𝐁-1⁒𝐀-1

Proof.

Suppose 𝐌1 and 𝐌2 act as inverses of 𝐀. That is,

π€πŒ1 =𝐌1⁒𝐀 =𝐈
π€πŒ2 =𝐌2⁒𝐀 =𝐈

Note then that

𝐌1 =𝐌1⁒𝐈
=𝐌1⁒(π€πŒ2)
=(𝐌1⁒𝐀)⁒𝐌2
=𝐈𝐌2

So 𝐌1=𝐌2. ∎

Proof.

Assume that 𝐀 and 𝐁 are both invertible.

(𝐀𝐁)⁒𝐁-1⁒𝐀-1 =𝐀⁒(𝐁𝐁-1)⁒𝐀-1
=π€πˆπ€-1
=𝐀𝐀-1
=I

Similarly we can show that 𝐁-1⁒𝐀-1⁒(𝐀𝐁)=𝐈. ∎

Question:

Can (𝐀𝐁) be invertible while one of 𝐀 or 𝐁 is not invertible? Must one be invertible, must both be invertible?

What if 𝐁 is not invertible? By the fundamental theorem of invertible matrices, we have that 𝐁⁒xβ†’=0β†’ must have a non-trivial solution, xβ†’0. Then

(𝐀𝐁)⁒xβ†’0 =𝐀⁒(𝐁⁒xβ†’0)
=𝐀⁒0β†’
=0β†’

Then the equation (𝐀𝐛)⁒xβ†’=0β†’ also has a non-trivial solution. Thus (𝐀𝐁) cannot be invertible.

Elementary Matrices and Invertible Matrices

Consider the following matrix 𝐀 and its row reduced echelon form:

𝐀=[1234]β†’-3⁒R1+R2[120-2]β†’-12⁒R2[1201]β†’-2⁒R1+R2[1001]

By the fundamental theorem of matrices, this shows us that 𝐀 is invertible.

Note though that we have said that we can represent row operations as elementary matrices. Corresponding to these row operations then, are the matrices

[1234]⁒→-3⁒R1+R2βŸπ„=[10-31]⁒[120-2]⁒→-12⁒R2βŸπ…=[100-12]⁒[1201]⁒→-2⁒R1+R2βŸπ†=[1-201]⁒[1001]

If we call the matrices between 𝐀 and 𝐈 as 𝐁 and 𝐂, then we have that

𝐁 =𝐄𝐀
𝐂 =𝐅𝐁=𝐅𝐄𝐀
𝐈 =𝐆𝐂=𝐆𝐅𝐄𝐀

and so (𝐆𝐅𝐄) is the inverse of 𝐀:

(𝐆𝐅𝐄)⁒𝐀 =𝐈

so

𝐀=(𝐆𝐅𝐄)-1=𝐄-1⁒𝐅-1⁒𝐆-1

but this assumes that the matrices 𝐄,𝐅,𝐆 are invertible. This brings us to an important question:

Question:

Are elementary matrices invertible? In essence, this equates to asking how to reverse a row operation. Every row operation can be reversed, and so every elementary matrix is invertible.

𝐄 =[10-31] 𝐅 =[100-12] 𝐆 =[1-201]
𝐄-1 =[1030] 𝐅-1 =[100-2] 𝐆-1 =[1201]

Vector Space

The term vector space shouldn’t be confused for what we know as vectors. Consider the set

β„™2 =Polynomials of degree ≀2

That is, elements of β„™2 are of the form

a⁒x2+b⁒x+c

where a,b,cβˆˆβ„.

Let u,vβˆˆβ„™2. It is apparent then that u⁒(x)+v⁒(x)βˆˆβ„™2 and for any kβˆˆβ„, k⁒u⁒(x)βˆˆβ„™2.

This is very similar to ℝ3. Consider uβ†’,vβ†’βˆˆβ„3. Then uβ†’+vβ†’βˆˆβ„3 and for any kβˆˆβ„3, k⁒uβ†’βˆˆβ„3.

In fact, many things besides standard vectors will be considered as vectors. Polynomials, functions, and matrices can all be seen as vectors in their respective vector spaces.