# Simplifying matrices of linear maps by choice of basis

Aug-Nov 2020

## Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• dim null $T$ + dim range $T$ = dim $V$
• Linear equation: $Ax=b$
• Solution (if it exists): $u+$ null$(A)$
• Four fundamental subspaces of a matrix
• Column space, row space, null space, left null space
• Determinant of a square matrix
• Function with many interesting properties
• Change of basis for linear map
• (change-basis in $W$) (linear map) (change-basis in $V$)

## “Simple” matrix

$n\times n$: square matrix

• Simplest is identity

• Diagonal matrices

• As many zeroes as possible off diagonal

$m\times n$ matrices

• Non-zero entries only in main diagonal

## “Simple” matrices for linear maps

$T:V\to W$, dim $V=n$, dim $W=m$

Basis for $V$

• Basis of null $T$: $\{u_1,\ldots,u_{n-r}\}$, extend to basis of $V$

• $B_V=\{v_1,\ldots,v_r,u_1,\ldots,u_{n-r}\}$, $r$: rank

Basis for $W$

• Basis of range $T$: $\{T(v_1),\ldots,T(v_r)\}$, extend to basis of $W$

• $B_W=\{T(v_1),\ldots,T(v_r),w_1,\ldots,w_{m-r}\}$

Matrix of $T$ w.r.t. $B_V$, $B_W$

$\begin{bmatrix} \vdots&\cdots&\vdots&\vdots&\cdots&\vdots\\ T(v_1)&\cdots&T(v_r)&T(u_1)&\cdots&T(u_{n-r})\\ \vdots&\cdots&\vdots&\vdots&\cdots&\vdots \end{bmatrix}=\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}$

## Elementary row-column operations and change of basis

$A:m\times n$ matrix, rank: $r$, represents $T:F^n\to F^m$ in standard basis

There exist elementary row operations $E_i$, col operations $F_j$ s.t.

$\left(\prod_i E_i\right) A \left(\prod_jF_j\right)=\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}$

$S_L=\prod_i E_i$: $m\times m$ invertible

$S_R=\prod_jF_j$: $n\times n$ invertible

$A=S^{-1}_L\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}S^{-1}_R$

Change of basis

• Basis for $F^n$: columns of $S_R$

• Basis for $F^m$: columns of $S^{-1}_L$

$T\leftrightarrow\begin{bmatrix} I_r&0\\ 0&0 \end{bmatrix}$

$A:n\times n$ matrix, rank: $r$, represents $T:F^n\to F^n$ in standard basis

Important constraint: same basis for input and Output

Goal: Find invertible $S$ s.t. $S^{-1}AS$ is “simple”

• Same vector space, same basis

• Operators are typically used multiple times on a vector

$Av$, $A^2v$, $A^3v$,$\ldots$

• Change of basis with same basis for input and output

$S^{-1}ASv$, $(S^{-1}AS)^2v=S^{-1}A^2Sv$,$\ldots$

• Well-behaved when operator is repeatedly applied

Rest of course: study operators in above scenario