MST10030 notes wiki

Recap: matrix multiplication and the identity matrix

• $AB$ defined if $A:n\times m$ and $B:m\times k$
  • then $AB$ is $n\times k$ …
  • with $(i,j)$ entry is $\text{row}_i(A).\text{col}_j(B)$
• Sometimes $AB=BA$ (need $A,B$ to both be $n\times n$)
  • say $A$ and $B$ commute
• But often $AB\ne BA$ (even if $A,B$ both $n\times n$)
• $n\times n$ identity matrix $I_n$: $1$s on diagonal, zeros elsewhere
• $I_n$ commutes with every $n\times n$ matrix, in a nice way…

1. Last time: proved that $I_nA=A$ for any $n\times m$ matrix A.
2. Proof that $AI_m=A$ for any $n\times m$ matrix $A$ is similar (exercise!)
3. If $B$ is any $n\times n$ matrix, then
   • $I_nB=B$ by part 1
   • and $BI_n=B$ by part 2
   • so $I_nB=B=BI_n$
   • In particular, $I_nB=BI_n$
   • So $I_n$ commutes with $B$, for every square $n\times n$ matrix $B$. ■

Algebraic properties of matrix multiplication

The associative law

• Matrix multiplication is associative.
• This means that \[(AB)C=A(BC)\] whenever $A,B,C$ are matrices which can be multiplied together in this order.
• Proof isn't too difficult but we skip it
  • it uses the known fact that if $a,b,c$ are real numbers, then $(ab)c=a(bc)$

An example using the associative law $(AB)C=A(BC)$

• $\newcommand{\m}[1]{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}A=\m{1&2\\3&4}$ commutes with $B=\m{7&10\\15&22}$
• Why?
  • Can check it (calculation) but this doesn't give a “reason”
  • We can explain it using associativity…

1. $B=AA$ (usually write as $B=A^2$).
2. Using associativity, we get $AB=A(AA)\stackrel*=(AA)A=BA.$

The same argument for any square matrix $A$ gives a proof of:

Proposition

For any square matrix $A$,

$A$ commutes with $A^2$.■

Powers of a square matrix $A$

• Define $A^1=A$
• and $A^2=AA$
• and $A^3=AAA=A(A^2)$
• and $A^4=AAAA=A(A^3)$
• ….
• $A^{k+1}=A(A^k)$ for $k\in \mathbb{N}$

Proposition: a square matrix commutes with its powers

For any square matrix $A$, and any $k\in\mathbb{N}$,

$A$ commutes with $A^k$.■

• Proof is by induction on $k$ (exercise).

The distributive laws

Proposition: the distributive laws

If $A$ is an $n\times m$ matrix and $k\in\mathbb{N}$, then:

1. $A(B+C)=AB+AC$ for any $m\times k$ matrices $B$ and $C$; and
2. $(B+C)A=BA=CA$ for any $k\times n$ matrices $B$ and $C$.

• In other words, $A(B+C)=AB+AC$ whenever the matrix products make sense, and similarly $(B+C)A=BA+CA$ whenever this makes sense.

Proof that $A(B+C)=AB+AC$

• Have $A$: $n\times m$, $B$ and $C$: $m\times k$
• So $B+C$: $m\times k$
• So $A(B+C)$: $n\times k$
• and $AB$: $n\times k$ and $AC$: $n\times k$
• So $AB+AC$: $n\times k$

• Conclusion so far: $A(B+C)$ and $AB+AC$ have the same sizes!

$A(B+C)=AB+AC$ continued

• In tutorial 4: $a\cdot (b+c)=a\cdot b+a\cdot c$ (row-col product)
• (for $a$: $1\times m$ and $b,c$: $m\times 1$)
• So the $(i,j)$ entry of $A(B+C)$ is\begin{align*}\def\xx{\!\!\!\!}\def\xxx{\xx\xx\xx\xx}\xxx\xxx\def\row{\text{row}}\def\col{\text{col}}\text{row}_i(A)\cdot \col_j(B+C) &= \text{row}_i(A)\cdot \big(\col_j(B)+\col_j(C)\big)\\ &= \text{row}_i(A)\cdot \col_j(B)+\row_i(A)\cdot\col_j(C).\end{align*}
• the $(i,j)$ entry of $AB$ is $\text{row}_i(A)\cdot \col_j(B)$; and
• the $(i,j)$ entry of $AC$ is $\row_i(A)\cdot\col_j(C)$;
• so the $(i,j)$ entry of $AB+AC$ is also $\text{row}_i(A)\cdot \col_j(B)+\row_i(A)\cdot\col_j(C)$.
• Same entries; so $A(B+C)=AB+AC$.

Proof that $(B+C)A=BA+CA$

• This is very similar, and is left as an exercise.■

Matrix equations

• A linear equation can be written using row-column multiplication.
• e.g. $ 2x-3y+z=8$ is same as $ \m{2&-3&1}\m{x\\y\\z}=8$
• or $ a\vec x=8$ where $a=\m{2&-3&1}$ and $\vec x=\m{x\\y\\z}$.

• We can write a whole system of linear equations in a similar way, as a matrix equation using matrix multiplication.
• e.g. the linear system $\begin{align*} 2x-3y+z&=8\\ y-z&=4\\x+y+z&=0\end{align*}$

• is same as $\m{2&-3&1\\0&1&-1\\1&1&1}\m{x\\y\\z}=\m{8\\4\\0}$
• or $ A\vec x=\vec b$ where $A=\m{2&-3&1\\0&1&-1\\1&1&1}$, $\vec x=\m{x\\y\\z}$ and $\vec b=\m{8\\4\\0}$.

In a similar way, any linear system \begin{align*} a_{11}x_1+a_{12}x_2+\dots+a_{1m}x_m&=b_1\\ a_{21}x_1+a_{22}x_2+\dots+a_{2m}x_m&=b_2\\ \hphantom{a_{11}}\vdots \hphantom{x_1+a_{22}}\vdots\hphantom{x_2+\dots+{}a_{nn}} \vdots\ & \hphantom{{}={}\!} \vdots\\ a_{n1}x_1+a_{n2}x_2+\dots+a_{nm}x_m&=b_n \end{align*} can be written in the form \[ A\vec x=\vec b\] where $A$ is the $n\times m $ matrix, called the coefficient matrix of the linear system, whose $(i,j)$ entry is $a_{ij}$ (the number in front of $x_j$ in the $i$th equation of the system).

Solutions of matrix equations

• More generally, might want to solve a matrix equation like \[AX=B\] where $A$, $X$ and $B$ are matrices of any size, with $A$ and $B$ fixed matrices and $X$ a matrix of unknown variables.
• If $A$ is $n\times m$, we need $B$ to be $n\times k$ for some $k$, and then $X$ must be $m\times k$
  • so we know the size of any solution $X$.
• But which $m\times k$ matrices $X$ are solutions?

Example

If $A=\m{1&0\\0&0}$ and $B=0_{2\times 3}$, then any solution $X$ to $AX=B$ must be $2\times 3$.

• One solution is $X=0_{2\times 3}$
  • because then we have $AX=A0_{2\times 3}=0_{2\times 3}$.
• This is not the only solution!
• For example, $X=\m{0&0&0\\1&2&3}$ is another solution
  • because then we have $AX=\m{1&0\\0&0}\m{0&0&0\\1&2&3}=\m{0&0&0\\0&0&0}=0_{2\times 3}.$

• So a matrix equation can have more than one solution.

Example

• Let $A=\m{2&4\\0&1}$
• and $B=\m{3&4\\5&6}$
• Solve $AX=B$ for $X$

• $X$ must be $2\times 2$
• $X=\m{x_{11}&x_{12}\\x_{21}&x_{22}}$
• Do some algebra to solve for $X$
• …
• Quicker way? Next time.