MST10030 notes wiki

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 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

• 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

Examples

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

Properties of $I_n$

1. $I_nA=A$ for any $n\times m$ matrix $A$;
2. $AI_m=A$ for any $n\times m$ matrix $A$; and
3. $I_nB=B=BI_n$ 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 $B$.