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Matrix multiplication

Let $A,B$ be matrices, with sizes
- $A$: $n\times m$
- $B$: $m\times k$
The product $AB$ is: the $n\times k$ matrix whose $(i,j)$ entry is \[ (AB)_{i,j} = \text{row}_i(A)\cdot \text{col}_j(B)\]
So the entries of $AB$ are all possible row-column products of a row of $A$ with a column of $B$

"Compatible" sizes for $AB$ to be defined

We need the sizes of $A$ and $B$ to be “compatible” for $AB$ to be defined
Need $A$: $n\times m$ and $B$: $m\times k$ (same numbers “in the middle”)
If $A,B$ are matrices, with sizes
- $A$: $n\times m$
- $B$: $\ell\times k$ with $\ell\ne m$,
then the matrix product $AB$ is undefined.

Example 1

If $\newcommand{\mat}[1]{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]} A=\mat{1&0&5\\2&-1&3}$ and $B=\mat{1&2\\3&4\\5&6}$,

$AB=\mat{26&32\\14&18}$
$BA=\mat{5&-2&11\\11&-4&27\\17&-6&43}$.
Note that $AB$ and $BA$ are both defined, but $AB\ne BA$
- $AB$ and $BA$ don't even have the same size.

Example 2

If $A=\mat{1&2\\3&4\\5&6}$, $B=\mat{2&1&1\\1&2&0\\1&0&2\\1&0&2\\2&2&1}$ and $C=\mat{1&3&0&7\\0&4&6&8}$,

$A$ is $3\times 2$, $B$ is $4\times 3$ and $C$ is $2\times 4$, so
$AB$, $CA$ and $BC$ don't exist (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 $BC\ne CB$ (undefined vs defined!)

Example 3

If $A=\mat{0&1\\0&0}$ and $B=\mat{0&0\\1&0}$, then

$AB=\mat{1&0\\0&0}$
$BA=\mat{0&0\\0&1}$.
So $AB$ and $BA$ are both defined and have the same size, but they are not equal matrices: $AB\ne BA$.

Example 4

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 $AB=BA$.

Example 5

If $A=\mat{1&2\\3&4}$ and $B=\mat{7&10\\15&22}$, then

$AB=\mat{37&54\\81&118}$
$BA=\mat{37&54\\81&118}$
So $AB=BA$ for these particular matrices $A$ and $B$.

Example 6

If $A=\mat{1&2\\3&4}$ and $B=\mat{6&10\\15&22}$, then

$AB=\mat{36&54\\78&118}$
$BA= \mat{36&52\\81&118}$
So $AB\ne BA$.

Commuting matrices I

We say that matrices $A$ and $B$ commute if $AB=BA$.

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

Commuting matrices II

What can we say about commuting matrices?
Suppose $AB=BA$ and think about sizes.
- $A$: $n\times m$
- $B$: $\ell\times k$
$AB$ is defined, so $m=\ell$.
$BA$ is defined, so $k=n$.
$AB$ is $n\times k$ and $BA$ is $\ell\times m$, so $n=\ell$ and $k=m$. So $n=\ell=m=k$!
$A$ and $B$ must both be $n\times n$: they're square matrices of the same size.

Commuting matrices III

If $A$ and $B$ commute, they must be square matrices of the same size.
Some square matrices $A$ and $B$ of the same size commute…
….but not all!
See examples above.

The $n\times n$ identity matrix

The $n\times n$ identity matrix is the $n\times n$ matrix $I_n$ with $1$s in every diagonal entry (that is, in the $(i,i)$ entry for every $i$ between $1$ and $n$), and $0$s in every other entry. So \[ I_n=\begin{bmatrix} 1&0&0&\dots&0\\0&1&0&\dots&0\\0&0&1&\dots&0\\\vdots & & &\ddots & \vdots\\0&0&0&\dots&1\end{bmatrix}.\]

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!

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 $B$.

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