# Algebraic operations on linear maps

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
• Matrix of linear map with respect to bases for $V$ and $W$
• Basis for $V$: $\{v_1,\ldots,v_n\}$
• Column $j$: coordinates of $T(v_j)$ with respect to basis of $W$
• null $T=\{v\in V:Tv=0\}$, range $T=\{Tv:v\in V\}$
• Fundamental theorem: dim null $T$ + dim range $T$ = dim $V$
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• colspace(A) = range T, null(A) = null(T)

## Addition and scalar multiplication of linear maps

$S,T:V\to W$ are linear maps, $\lambda\in F$. Sum of $S$ and $T$, denoted $S+T$, is defined as $(S+T)v = Sv + Tv.$ The scalar product of $\lambda$ and $T$, denoted $\lambda T$, is defined as $(\lambda T)v = \lambda (Tv).$

Results

• $S+T$ and $\lambda T$ are linear maps from $V\to W$.

• $\mathcal{L}(V,W)$: set of all linear maps from $V$ to $W$

• $\mathcal{L}(V,W)$ is a vector space over $F$

• under addition and scalar multiplication defined above

## Product of linear maps

$U,V,W$: vector spaces over $F$ and $T:U\to V$, $S:V\to W$ are linear maps. Product of $S$ and $T$, denoted $ST$, is defined as $(ST)u = S(Tu).$ $ST$: linear map from $U$ to $W$; composition of the two maps $T$ and $S$

Algebraic properties of linear maps

• Multiplication is associative

• $(T_1T_2)T_3=T_1(T_2T_3)$
• Multiplication need not be commutative

• If $ST$ is defined, $TS$ may not even be defined

• Even if $ST$ and $TS$ are defined, they may not be the same map

• Distributive property

• $(S_1+S_2)T = S_1T + S_2T$ and $S(T_1+T_2) = ST_1 + ST_2$

## Addition and scalar multiplication of matrices

\begin{align} \begin{bmatrix} A_{11}&\cdots&A_{1n}\\ \vdots&\ddots&\vdots\\ A_{m1}&\cdots&A_{mn} \end{bmatrix}+\begin{bmatrix} B_{11}&\cdots&B_{1n}\\ \vdots&\ddots&\vdots\\ B_{m1}&\cdots&B_{mn} \end{bmatrix}&=\begin{bmatrix} A_{11}{+}B_{11}&\cdots&A_{1n}{+}B_{1n}\\ \vdots&\ddots&\vdots\\ A_{m1}{+}B_{m1}&\cdots&A_{mn}{+}B_{mn} \end{bmatrix}\\[5pt] \lambda\begin{bmatrix} A_{11}&\cdots&A_{1n}\\ \vdots&\ddots&\vdots\\ A_{m1}&\cdots&A_{mn} \end{bmatrix}&=\begin{bmatrix} \lambda A_{11}&\cdots&\lambda A_{1n}\\ \vdots&\ddots&\vdots\\ \lambda A_{m1}&\cdots&\lambda A_{mn} \end{bmatrix} \end{align}

$V,W$: finite-dimensional and $S,T:V\to W$. Fix bases for $V$ and for $W$.

$M(S), M(T), M(S+T), M(\lambda T)$ are matrices for $S$, $T$, $S+T$ and $\lambda T$ with respect to the chosen bases. Then, \begin{align} M(S+T)&=M(S)+M(T)\\ M(\lambda T)&=\lambda M(T) \end{align}

## Matrix multiplication

$\begin{bmatrix} A_{11}&\cdots&A_{1n}\\ \vdots&\ddots&\vdots\\ A_{m1}&\cdots&A_{mn} \end{bmatrix}\begin{bmatrix} B_{11}&\cdots&B_{1k}\\ \vdots&\ddots&\vdots\\ B_{n1}&\cdots&B_{nk} \end{bmatrix}=\begin{bmatrix} C_{11}&\cdots&C_{1k}\\ \vdots&\ddots&\vdots\\ C_{m1}&\cdots&C_{mk} \end{bmatrix}$

$C_{ij}=\sum_{l=1}^n A_{il}B_{lj}$

(dot product of $i$-th row of $A$ and $j$-th column of $B$)

$U,V,W$: finite-dimensional and $T:U\to V$, $S:V\to W$. Fix bases for $U$, $V$ and $W$.

$M(S), M(T), M(ST)$ are matrices for $S$, $T$ and $ST$ with respect to the chosen bases. Then, $M(ST)=M(S)M(T)$

## Interesting matrix multiplications

$\begin{bmatrix} b_1&b_2&\cdots&b_n \end{bmatrix}\begin{bmatrix} a_1\\a_2\\\vdots\\a_n \end{bmatrix}=a_1b_1+\cdots+a_nb_n$

• $S: F^n\to F$, $T:F\to F^n$

• $ST: F\to F$

$\begin{bmatrix} a_1\\a_2\\\vdots\\a_m \end{bmatrix}\begin{bmatrix} b_1&b_2&\cdots&b_n \end{bmatrix}=\begin{bmatrix} a_1b_1&a_1b_2&\cdots&a_1b_n\\ a_2b_1&a_2b_2&\cdots&a_2b_n\\ \vdots&\vdots&\vdots&\vdots\\ a_mb_1&a_mb_2&\cdots&a_mb_n \end{bmatrix}$

• $S:F\to F^m$, $T:F^n\to F$

• $ST: F^n\to F^m$

## More on matrix multiplication

$\begin{bmatrix} A_{11}&\cdots&A_{1n}\\ \vdots&\ddots&\vdots\\ A_{m1}&\cdots&A_{mn} \end{bmatrix}\begin{bmatrix} B_{11}&\cdots&B_{1k}\\ \vdots&\ddots&\vdots\\ B_{n1}&\cdots&B_{nk} \end{bmatrix}=\begin{bmatrix} C_{11}&\cdots&C_{1k}\\ \vdots&\ddots&\vdots\\ C_{m1}&\cdots&C_{mk} \end{bmatrix}$

$C_{ij}=\sum_{l=1}^n A_{il}B_{lj}$

• $C_{ij}$: ($i$-th row of $A$) $\times$ ($j$-th column of $B$)

• $i$-th row of $C$: ($i$-th row of $A$) $\times$ $B$

• $j$-th column of $C$: $A$ $\times$ ($j$-th column of $B$)

• $C=\sum_{l=1}^n (l$-th column of $A) \times (l$-th row of $B)$