Unit four

Section 4.5

Suppose we have a matrix A_{m,n}. This acts as a function mapping \mathbb{R}^n to \mathbb{R}^m.

Fact

A matrix A maps \text{RowSp}(A) bijectivly to \text{ColSp}(A).

Proposition

Suppose x_1,\ldots,x_r is a basis of \text{RowSp}(A). Then Ax_1,\ldots,Ax_r is a basis of \text{ColSp}(A).

Proof

Suppose c_1Ax_1,\ldots,c_rAx_r = 0. then

A(c_1x_1,\ldots,c_rx_r) = 0

All c_i must be zero as x_1,\ldots,x_r is a basis. Thus Ax is independent.

Definition

The pseudoinverser of A, A^+ is basically

A|_{\text{RowSp}(A)} \colon \text{RowSp}(A) \to \text{ColSp}(A)

Note that when we multiply A^+A, we get an n\times n matrix. Lets look at what happens to each subspace under A^+A.

The Null space disapears, as A maps \text{NullSp}(A) \to 0. The row space is restored. Thus we have an interesting map

A^+A|_{\text{Row(A)}} = P_{\text{Row}(A)}

We also have

AA^A|_{\text{Col(A)}} = P_{\text{Col}(A)}

Somewhat suprisingly, (A^+)^+ = A. Every subspace of A is the same under A^T and A^+.

Example

If A_{(n\times n)} is invertible, Then

A^+ = A^{-1}.

Example

Let

\begin{align*} A = \begin{pmatrix} 0 & 2\\ 0 & 0 \end{pmatrix} \end{align*}

Then

\begin{align*} A = \begin{pmatrix} 0 & 0\\ \frac{1}{2} & 0 \end{pmatrix} \end{align*}

Note that

\begin{align*} AA^+ = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}= P_{\text{Col}(A)} \end{align*}

Example

Let

\begin{align*} A = \begin{pmatrix} a & 0\\ 0 & 0\\ 0 & b \end{pmatrix} \end{align*}

Section 5.1

Permutations matrices are matrices with the same columns as I. For an n\times n permutation matrix, there exists n! permutation matrices.

Half of these matrices are even. Meaning we can get to P via an even number of row exchanges on I. Half are odd.

Definition

We will define the operation \text{Prod}(A\colon P) as the product of all entries of A matching a non zero entry of P.

Thus we define the determinate \det A as

\det A = \sum\limits_{\text{Even perms }P} \text{Prod}(A\colon P) - \sum\limits_{\text{Odd perms }P} \text{Prod}(A\colon P)

This is actually just the cofactor expansion formula wrote in a different way. If you multiply a single row of A by a constant \mu, then your determinate gets multiplied by \mu. If you exchange two rows of A, multiply the determinate by negative 1.

And if you add a row to another, the determinate has no change. The determinate of A^T is equal to the determinate of A. This is all the same for rows and columnss.

If A is upper triangular, then \text{Prod}(A\colon P) = 0 if P \not = I. Thus \det A = \text{Prod}(A \colon I) = \Pi of the diagonal entries.

The determinate of A is equal to

\det A = (-1)^{\# \text{ of row exhchanges}} \cdot \text{Product of pivots}

Corollary

\det A \not = 0

If and only if A is invertible

Theorem

\det ZA = \det Z \det A

Thus the determinate of ZA is simply the product of the operations of Z on A.