MST10030 notes wiki

This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong.
=== Definition of matrix multiplication ===

{{page>matrix multiplication}}
=== Examples ===

  - If $\newcommand{\mat}[1]{\begin{bmatrix}#1\end{bmatrix}} A=\mat{1&0&5\\2&-1&3}$ and  $B=\mat{1&2\\3&4\\5&6}$, then $AB=\mat{26&32\\14&18}$ and $BA=\mat{5&-2&11\\11&-4&27\\17&-6&43}$. Note that $AB$ and $BA$ are both defined, but $AB\ne BA$ since $AB$ and $BA$ don't even have the [[same size]].
  - If $A=\mat{1&2\\3&4\\5&6}$, $B=\mat{2&1&1\\1&2&0\\1&0&2\\2&2&1}$ and $C=\mat{1&3&0&7\\0&4&6&8}$, then $A$ is $3\times 2$, $B$ is $4\times 3$ and $C$ is $2\times 4$, so 
    * $AB$, $CA$ and $BC$ don't exist (i.e., they are undefined);
    * $AC$ exists and is $3\times 4$;
    * $BA$ exists and is $4\times 2$; and
    * $CB$ exists and is $2\times 2$.
    * In particular, $AB\ne BA$ and $AC\ne CA$ and $BC\ne CB$, since in each case one of the matrices doesn't exist.
  - If $A=\mat{0&1\\0&0}$ and $B=\mat{0&0\\1&0}$, then $AB=\mat{1&0\\0&0}$ and $BA=\mat{0&0\\0&1}$. So $AB$ and $BA$ are both defined and have the [[same size]], but they are not [[equal matrices]]: $AB\ne BA$.
  - If $A=\mat{1&0\\0&0}$ and $B=\mat{0&0\\0&1}$, then $AB=\mat{0&0\\0&0}$ and $BA=\mat{0&0\\0&0}$. So $AB=BA$ in this case.
  - If $A=0_{n\times n}$ is the $n\times n$ zero matrix and $B$ is any $n\times n$ matrix, then $AB=0_{n\times n}$ and $BA=0_{n\times n}$. So in this case, we do have $AB=BA$.
  - If $A=\mat{1&2\\3&4}$ and $B=\mat{7&10\\15&22}$, then $AB=\mat{37&54\\81&118}=BA$, so $AB=BA$ for these particular matrices $A$ and $B$.
  - If $A=\mat{1&2\\3&4}$ and $B=\mat{6&10\\15&22}$, then $AB=\mat{36&54\\78&118}$ and $BA= \mat{36&52\\81&118}$, so $AB\ne BA$.

=== Commuting matrices ===

{{page>commute}}

Which matrices commute? Suppose $A$ is an $n\times m$ matrix and $B$ is an $\ell\times k$ matrix, and $A$ and $B$ commute, i.e., $AB=BA$.

  * $AB$ must be defined, so $m=\ell$
  * $BA$ must be defined, so $k=n$
  * $AB$ is an $n\times k$ matrix and $BA$ is an $\ell\times n$ matrix. Since $AB$ has the same size as $BA$, we must have $n=\ell$ and $k=m$.
Putting this together: we see that if $A$ and $B$ commute, then $A$ and $B$ must both be $n\times n$ matrices for some number $n$. In other words, they must be //square matrices of the same size//.

Examples 4 and 5 above show that for some square matrices $A$ and $B$ of the same size, it is true that $A$ and $B$ commute. On the other hand, examples 3 and 6 show that it's not true that square matrices of the same size must always commute.

Because it's not true in general that $AB=BA$, we say that **matrix multiplication is not commutative**.

=== Definition of the $n\times n$ identity matrix ===

{{page>identity matrix}}

=== Examples ===

  - $I_1=[1]$
  - $I_2=\mat{1&0\\0&1}$
  - $I_3=\mat{1&0&0\\0&1&0\\0&0&1}$
  - $I_4=\mat{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$, and so on!

=== Proposition: properties of $I_n$ ===
  - $I_nA=A$ for any $n\times m$ matrix $A$;
  - $AI_m=A$ for any $n\times m$ matrix $A$; and
  - $I_nB=B=BI_n$ for any $n\times n$ matrix $B$. In particular, $I_n$ commutes with every other square $n\times n$ matrix $B$.

=== Proof of the proposition ===

1. We want to show that $I_nA=A$ for any $n\times m$ matrix $A$. These matrices the [[same size]], since $I_n$ has size $n\times n$ and $A$ has size $n\times m$, so $I_n A$ has size $n\times m$ by the definition of [[matrix multiplication]], which is the same as the size of $A$.

 Note that $\text{row}_i(I_n)=[0~0~\dots~0~1~0~\dots~0]$, where the $1$ is in the $i$th place, by definition of the [[identity matrix]] $I_n$; and the $j$th column of $A$ is $\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\end{bmatrix}$. The (i,j) entry of $I_nA$ is 
$\text{row}_i(I_n)\cdot \text{col}_j(A)$, by the definition of [[matrix multiplication]], which is therefore
\begin{align*} [0~0~\dots~0~1~0~\dots~0]\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\end{bmatrix} &= 0a_{1j}+0a_{2j}+\dots+0a_{i-1,j}+1a_{ij}+0a_{i+1,j}+\dots+0a_{nj} \\&= a_{ij}.\end{align*}
So the matrices $I_nA$ and $A$ have the same size, and the same $(i,j)$ entries, for any $(i,j)$. So $I_nA=A$.