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 $b,c$: $m\times 1$)
• Write $\def\row{\text{row}}\def\col{\text{col}}a_i=\row_i(A)$, $b_j=\col_j(B)$, $c_j=\col_j(C)$.
• $(i,j)$ entry of $A(B+C)$ is:\begin{align*}\def\xx{\!\!\!\!}\def\xxx{\xx\xx\xx\xx}\xxx\xxx \row_i(A)\cdot \col_j(B+C) &= a_i\cdot \big(b_j+c_j\big)\\ &= a_i\cdot b_j+a_i\cdot c_j.\end{align*}
• $(i,j)$ entry of $AB$ is $a_i\cdot b_j$; and
• $(i,j)$ entry of $AC$ is $a_i\cdot c_j$;
• so $(i,j)$ entry of $AB+AC$ is also $a_i\cdot b_j+a_i\cdot c_j$
• Same sizes and 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.■