Ch.10 Matrix Operations

Fix domain $V$ and codomain $W$ with bases $B=\langle\vec{\beta}_1,...,\vec{\beta}_n\rangle$ and $D=\langle\vec{\delta}_1,...,\vec{\delta}_n\rangle$.
Let $h:V\to W$ be a linear map and consider the map $r\cdot h:V\to W$ given by $\vec{v}\xmapsto{r\cdot h}r\cdot h(\vec{v})$.
If the representation of $h(\vec{v})$ is
$\text{Rep}_D(h(\vec{v}))=\begin{pmatrix}w_1\\\vdots\\w_n\end{pmatrix}$
Then $r\cdot h(\vec{v})=r\cdot(w_1\vec{\delta}_1+\cdots+w_n\vec{\delta}_n)=(r\cdot w_1)\vec{\delta}_1+\cdots+(r\cdot w_n)\vec{\delta}_n$, so
$\text{Rep}_D(r\cdot h(\vec{v}))=\begin{pmatrix}r\cdot w_1\\\vdots\\r\cdot w_n\end{pmatrix}=r\cdot\begin{pmatrix}w_1\\\vdots\\w_n\end{pmatrix}$
The output is $r$ times bigger.
So if $H$ is the matrix representing $h$, $rH$ represents $rh$.
Similarly, if $f$ is represented by $H$ and $g$ is represented by $G$, then $f+g$ is represented by $H+G$.

For two matrices with different sizes, their domains or codomains are different, so the function sum would not be possible. Therefore, the matrix sum is not defined.

Let $h:V\to W$ and $g:W\to U$ be linear.
Then $\begin{array}{rcl}g\circ h(c_1\cdot\vec{v}_1+c_2\cdot\vec{v}_2)&=&g(h(c_1\cdot\vec{v}_1+c_2\cdot\vec{v}_2))\\&=&g(c_1\cdot h(\vec{v}_1)+c_2\cdot h(\vec{v}_2))\\&=&c_1\cdot g(h(\vec{v}_1))+c_2\cdot g(h(\vec{v}_2))\\&=&c_1\cdot g\circ h(\vec{v}_1)+c_2\cdot g\circ h(\vec{v}_2)\end{array}$
shows $g\circ h$ is linear

Compute
$\begin{pmatrix}2&3&1\\1&4&2\end{pmatrix}\begin{pmatrix}0&1&1\\2&5&1\\1&2&3\end{pmatrix}$

This will result in a $2\times3$ matrix
$\begin{pmatrix}a&b&c\\d&e&f\end{pmatrix}$
$a=2(0)+3(2)+1(1)=7\\ b=2(1)+3(5)+1(2)=19\\ c=2(1)+3(1)+1(3)=8\\ d=1(0)+4(2)+2(1)=10\\ e=1(1)+4(5)+2(2)=25\\ f=1(1)+4(1)+2(3)=11$

Let $h:V\to W$ and $g:W\to X$ be represented by matrices $H$ and $G$ with respect to bases $B\subset V$, $C\subset W$, $D\subset X$ of sizes $n,r,m$.

For any $\vec{v}\in V$, the $k$-th component ($1\le k\le r$) of $\text{Rep}_C(h(\vec{v}))$ is $h_{k,1}v_1+\cdots+h_{k,n}v_n$,
so the $i$-th component of $\text{Rep}_D(g\circ h(\vec{v}))$ is
$g_{i,1}(h_{1,1}v_1+\cdots+h_{1,n}v_n)+\cdots+g_{i,r}(h_{r,1}v_1+\cdots+h_{r,n}v_n)$
Distributing and grouping by the $v$'s
$(g_{i,1}h_{1,1}+g_{i,2}h_{2,1}+\cdots+g_{i,r}h_{r,n})v_1+\cdots+(g_{i,1}h_{1,n}+g_{i,2}h_{2,n}+\cdots+g_{i,r}h_{r,n})v_n$
Each coefficient of $v$'s matches the definition for the $i,j$-th entry of the product $GH$

Let $V=\mathbb{R}^2$, $W=\mathcal{P}_2$, $X=\mathcal{M}_{2\times2}$ with bases
$\begin{array}{cc}B=\langle\begin{pmatrix}1\\1\end{pmatrix},\begin{pmatrix}1\\-1\end{pmatrix}\rangle&C=\langle x^2,x^2+x,x^2+x+1\rangle\end{array}$
$D=\langle\begin{pmatrix}1&0\\0&0\end{pmatrix},\begin{pmatrix}0&2\\0&0\end{pmatrix},\begin{pmatrix}0&0\\3&0\end{pmatrix},\begin{pmatrix}0&0\\0&4\end{pmatrix}\rangle$
Suppose $h:\mathbb{R}^2\to\mathcal{P}_2$ and $g:\mathcal{P}_2\to\mathcal{M}_{2\times2}$ have these actions
$\begin{array}{cc}\begin{pmatrix}a\\b\end{pmatrix}\xmapsto{h}ax^2+(a+b)&px^2+qx+r\xmapsto{g}\begin{pmatrix}p&p+q\\0&r\end{pmatrix}\end{array}$
and the matrices $G$ and $H$ represent $g$ and $h$.
We will check that the matrix product $GH$ represents $g\circ h$

The composition has the action
$\begin{pmatrix}a\\b\end{pmatrix}\xmapsto{g\circ h}\begin{pmatrix}a&a\\0&a+b\end{pmatrix}$

Now, we find the matrix $H$
$\begin{pmatrix}1\\1\end{pmatrix}=\text{Rep}_B(\begin{pmatrix}1\\0\end{pmatrix})\xmapsto{h}x^2+2=\text{Rep}_C(\begin{pmatrix}1\\-2\\2\end{pmatrix})$
$\begin{pmatrix}1\\-1\end{pmatrix}=\text{Rep}_B(\begin{pmatrix}0\\1\end{pmatrix})\xmapsto{h}x^2=\text{Rep}_C(\begin{pmatrix}1\\0\\0\end{pmatrix})$
So
$H=\begin{pmatrix}1&1\\-2&0\\2&0\end{pmatrix}$

Next, to find $G$
$x^2=\text{Rep}_C(\begin{pmatrix}1\\0\\0\end{pmatrix})\xmapsto{g}\begin{pmatrix}1&1\\0&0\end{pmatrix}=\text{Rep}_D(\begin{pmatrix}1\\1/2\\0\\0\end{pmatrix})$
$x^2+x=\text{Rep}_C(\begin{pmatrix}1\\1\\0\end{pmatrix})\xmapsto{g}\begin{pmatrix}1&2\\0&0\end{pmatrix}=\text{Rep}_D(\begin{pmatrix}1\\1\\0\\0\end{pmatrix})$
$x^2+x+1=\text{Rep}_C(\begin{pmatrix}1\\1\\1\end{pmatrix})\xmapsto{g}\begin{pmatrix}1&2\\0&1\end{pmatrix}=\text{Rep}_D(\begin{pmatrix}1\\1\\0\\1/4\end{pmatrix})$
So
$G=\begin{pmatrix}1&1&1\\1/2&1&1\\0&0&0\\0&0&1/4\end{pmatrix}$
Now, we compute $GH$
$GH=\begin{pmatrix}1&1&1\\1/2&1&1\\0&0&0\\0&0&1/4\end{pmatrix}\begin{pmatrix}1&1\\-2&0\\2&0\end{pmatrix}=\begin{pmatrix}1(1)+1(-2)+1(2)&1\\1/2(1)+1(-2)+1(2)&1/2\\0&0\\1/4(2)&0\end{pmatrix}=\begin{pmatrix}1&1\\1/2&1/2\\0&0\\1/2&0\end{pmatrix}$

We check this by finding the matrix representation of $g\circ h$
$\begin{pmatrix}1\\1\end{pmatrix}=\text{Rep}_B(\begin{pmatrix}1\\0\end{pmatrix})\xmapsto{g\circ h}\begin{pmatrix}1&1\\0&2\end{pmatrix}=\text{Rep}_D(\begin{pmatrix}1\\1/2\\0\\1/2\end{pmatrix})$
$\begin{pmatrix}1\\-1\end{pmatrix}=\text{Rep}_B(\begin{pmatrix}0\\1\end{pmatrix})\xmapsto{g\circ h}\begin{pmatrix}1&1\\0&0\end{pmatrix}=\text{Rep}_D(\begin{pmatrix}1\\1/2\\0\\0\end{pmatrix})$
This yields the matrix
$GH=\begin{pmatrix}1&1\\1/2&1/2\\0&0\\1/2&0\end{pmatrix}$
which is equal to what was computed before.

Are thse two matrix products equal?
$\begin{array}{cc}\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}2&4\\0&1\end{pmatrix}&\begin{pmatrix}2&4\\0&1\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix}\end{array}$

The first yields
$\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}2&4\\0&1\end{pmatrix}=\begin{pmatrix}1(2)+2(0)&1(4)+2(1)\\3(2)+4(0)&3(4)+4(1)\end{pmatrix}=\begin{pmatrix}2&6\\6&16\end{pmatrix}$

The second yields
$\begin{pmatrix}2&4\\0&1\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix}=\begin{pmatrix}2(1)+4(3)&2(2)+4(4)\\0(1)+1(3)&0(2)+1(4)\end{pmatrix}=\begin{pmatrix}14&20\\3&4\end{pmatrix}$

Clearly, they are not equal.
$\begin{pmatrix}2&6\\6&16\end{pmatrix}\ne\begin{pmatrix}14&20\\3&4\end{pmatrix}$

Matrix multiplication represents function composition, so $(FG)H$ is $(f\circ g)\circ h$ and $F(GH)$ is $f\circ(g\circ h)$, both of which are equivalent to $f\circ g\circ h$

Proofs for both sides are similar, so we only prove $F(G+H)=FG+FH$
Converting to function composition, we must prove that $f\circ(g+h)(\vec{v})=f\circ g(\vec{v})+f\circ h(\vec{v})$
$f\circ(g+h)(\vec{v})=f(g(\vec{v})+h(\vec{v}))=f(g(\vec{v}))+f(h(\vec{v}))=f\circ g(\vec{v})+f\circ h(\vec{v})$

We can apply the operation $(1/2)\rho_2$ on the matrix
$\begin{pmatrix}1&2&3\\0&2&5\\0&0&0\end{pmatrix}$
or we can apply the operator to the identity matrix and multiply with the matrix
$\begin{pmatrix}1&0&0\\0&1/2&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&2&3\\0&2&5\\0&0&0\end{pmatrix}$
The same can be done with the other operations

Gaussian reduction can be performed with row operators
$\begin{pmatrix}1&2\\3&4\end{pmatrix}\xrightarrow{-3\rho_1+\rho_2}\begin{pmatrix}1&2\\0&-2\end{pmatrix}\xrightarrow{\rho_2+\rho_1}\begin{pmatrix}1&0\\0&-2\end{pmatrix}\xrightarrow{-1/2\rho_2}\begin{pmatrix}1&0\\0&1\end{pmatrix}$
Or the elementary matrices can be multiplied instead
$\begin{pmatrix}1&0\\0&-1/2\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&0\\-3&1\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$
Each of the matrices (from the right) are the elementary matrices for each row operator step

$H$ is invertible iff it is nonsingular, so it Gaus-Jordan reduces to the identity through a series of operations
$R_r\cdot R_{r-1}\cdots R_1 \cdot H=I$
Each row operation is invertible, and the inverses are elementary.
Apply $R_r^{-1}$, then $R_{r-1}^{-1}$ etc to both sides to get
$H=R_r^{-1}\cdot R_{r-1}^{-1}\cdots R_1^{-1}\cdot I$
This represents $H$ as the product of elementary operations.

Group as
$(R_r\cdot R_{r-1}\cdots R_1) \cdot H=I$
Then we have
$H^{-1}=R_r\cdot R_{r-1}\cdots R_1\cdot I$

Find the inverse of
$A=\begin{pmatrix}1&3&1\\2&0&-1\\1&2&0\end{pmatrix}$

$\begin{array}{cccccc}&\left(\begin{array}{ccc|ccc}1&3&1&1&0&0\\2&0&-1&0&1&0\\1&2&0&0&0&1\end{array}\right)&\xrightarrow[-\rho_1+\rho_2]{-2\rho_1+\rho_2}&\left(\begin{array}{ccc|ccc}1&3&1&1&0&0\\0&-6&-3&-2&1&0\\0&-1&-1&-1&0&1\end{array}\right)&\xrightarrow[3\rho_3+\rho_1]{-6\rho_3+\rho_2}&\left(\begin{array}{ccc|ccc}1&0&-2&-2&0&3\\0&0&3&4&1&-6\\0&-1&-1&-1&0&1\end{array}\right)\\\xrightarrow{-\rho_3\leftrightarrow1/3\rho_2}&\left(\begin{array}{ccc|ccc}1&0&-2&-2&0&3\\0&1&1&1&0&-1\\0&0&1&4/3&1/3&-2\end{array}\right)&\xrightarrow[2\rho_3+\rho_1]{-\rho_3+\rho_2}&\left(\begin{array}{ccc|ccc}1&0&0&2/3&2/3&-1\\0&1&0&-1/3&-1/3&1\\0&0&1&4/3&1/3&-2\end{array}\right)\end{array}$
Thus the inverse is
$A^{-1}=\begin{pmatrix}2/3&2/3&-1\\-1/3&-1/3&1\\4/3&1/3&-2\end{pmatrix}$