~~REVEAL~~

==== Row-column & matrix multiplication ====

  * The **row-column product** of $a$ and $b$ is defined by \[\!\!\!\!\!\!\!\!\!\!ab=[\begin{smallmatrix}a_1&a_2&\dots&a_n\end{smallmatrix}]\left[\begin{smallmatrix}b_1\\b_2\\\vdots\\b_n\end{smallmatrix}\right]=a_1b_1+a_2b_2+\dots+a_nb_n.\]

  * $AB=$ matrix of all "row-of-$A$ times col-of-$B$" products
  * \[\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \def\r{\left[\begin{smallmatrix}1&0&5\end{smallmatrix}\right]}\def\rr{\left[\begin{smallmatrix}2&-1&3\end{smallmatrix}\right]}\left[\begin{smallmatrix}1&0&5\\2&-1&3\end{smallmatrix}\right]\left[\begin{smallmatrix} 1&2\\3&4\\5&6\end{smallmatrix}\right]\def\s{\left[\begin{smallmatrix}1\\3\\5\end{smallmatrix}\right]}\def\ss{\left[\begin{smallmatrix}2\\4\\6\end{smallmatrix}\right]}=\left[\begin{smallmatrix}{\r\s}&{\r\ss}\\{\rr\s}&{\rr\ss}\end{smallmatrix}\right]=\left[\begin{smallmatrix}26&32\\14&18\end{smallmatrix}\right].\]

==== Matrix multiplication: the definition ====

  * 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 multiplication|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 $A=\mat{1&2\\3&4\\5&6}$, $B=\mat{2&1&1\\1&2&0\\1&0&2\\2&2&1}$ 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 ====

{{page>commute}}

  * Because it's true that $AB=BA$ for every choice of matrices $A$ and $B$, we say that **matrix multiplication is not commutative**.

==== Commuting matrices II ====

  * What can we say about a pair of 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** pairs of square matrices $A$ and $B$ of the same size do commute...
  * ....but not all! 
  * See examples above.

==== 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!

==== 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 that $I_nA=A$ for $A$: $n\times m$ ====

  * $I_nA$ is $n\times m$ (from definition of matrix multiplication)
  * So $I_nA$ has same size as $A$
  * $\text{row}_i(I_n)=[0~0~\dots~0~1~0~\dots~0]$, with $1$ in $i$th place
  * $\text{col}_j(A)=\mat{a_{1j}\\a_{2j}\\\vdots\\a_{nj}}$
  * So $(i,j)$ entry of $I_nA$ is \[\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\text{row}_i(I_n)\cdot \text{col}_j(A)= 0a_{1j}+0a_{2j}+\dots+1a_{ij}+\dots+0a_{nj} =a_{ij}\]
  * same as $(i,j)$ entry of $A$.
  * So $I_nA=A$

==== More proofs====

  * Proof that $A=AI_m$ for $A$: $n\times m$ is very similar (exercise)
  * Now if $B$ is $n\times n$, take $n=m$ and $A=B$ above:
    * $I_nB=B$ and $BI_n=B$
    * So $I_nB=B=BI_n$
    * So $I_n$ commutes with $B$, for any $n\times n$ matrix $B$.