Definition of matrix multiplication

Examples

1. If $\newcommand{\mat}[1]{\begin{bmatrix}#1\end{bmatrix}} A=\mat{1&0&5\\2&-1&3}$ and $B=\mat{1&2\\3&4\\5&6}$, then $AB=\mat{26&32\\14&18}$ and $BA=\mat{5&-2&11\\11&-4&27\\17&-6&43}$. Note that $AB$ and $BA$ are both defined, but $AB\ne BA$ since $AB$ and $BA$ don't even have the same size.
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 $C=\mat{1&3&0&7\\0&4&6&8}$, then $A$ is $3\times 2$, $B$ is $4\times 3$ and $C$ is $2\times 4$, so
   • $AB$, $CA$ and $BC$ don't exist (i.e., they are 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 $BC\ne CB$, since in each case one of the matrices doesn't exist.
3. If $A=\mat{0&1\\0&0}$ and $B=\mat{0&0\\1&0}$, then $AB=\mat{1&0\\0&0}$ and $BA=\mat{0&0\\0&1}$. So $AB$ and $BA$ are both defined and have the same size, but they are not equal matrices: $AB\ne BA$.
4. If $A=\mat{1&0\\0&0}$ and $B=\mat{0&0\\0&1}$, then $AB=\mat{0&0\\0&0}$ and $BA=\mat{0&0\\0&0}$. So $AB=BA$ in this case.
5. 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 $AB=BA$.
6. If $A=\mat{1&2\\3&4}$ and $B=\mat{7&10\\15&22}$, then $AB=\mat{37&54\\81&118}=BA$, so $AB=BA$ for these particular matrices $A$ and $B$.
7. If $A=\mat{1&2\\3&4}$ and $B=\mat{6&10\\15&22}$, then $AB=\mat{36&54\\78&118}$ and $BA= \mat{36&52\\81&118}$, so $AB\ne BA$.

Commuting matrices

Which matrices commute? Suppose $A$ is an $n\times m$ matrix and $B$ is an $\ell\times k$ matrix, and $A$ and $B$ commute, i.e., $AB=BA$.

• $AB$ must be defined, so $m=\ell$
• $BA$ must be defined, so $k=n$
• $AB$ is an $n\times k$ matrix and $BA$ is an $\ell\times n$ matrix. Since $AB$ has the same size as $BA$, we must have $n=\ell$ and $k=m$.

Putting this together: we see that if $A$ and $B$ commute, then $A$ and $B$ must both be $n\times n$ matrices for some number $n$. In other words, they must be square matrices of the same size.

Examples 4 and 5 above show that for some square matrices $A$ and $B$ of the same size, it is true that $A$ and $B$ commute. On the other hand, examples 3 and 6 show that it's not true that square matrices of the same size must always commute.

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

Definition of 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!

Proposition: 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 of the proposition

1. We want to show that $I_nA=A$ for any $n\times m$ matrix $A$. These matrices the same size, since $I_n$ has size $n\times n$ and $A$ has size $n\times m$, so $I_n A$ has size $n\times m$ by the definition of matrix multiplication, which is the same as the size of $A$.

Note that $\text{row}_i(I_n)=[0~0~\dots~0~1~0~\dots~0]$, where the $1$ is in the $i$th place, by definition of the identity matrix $I_n$; and the $j$th column of $A$ is $\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\end{bmatrix}$. The (i,j) entry of $I_nA$ is $\text{row}_i(I_n)\cdot \text{col}_j(A)$, by the definition of matrix multiplication, which is therefore \begin{align*} [0~0~\dots~0~1~0~\dots~0]\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\end{bmatrix} &= 0a_{1j}+0a_{2j}+\dots+0a_{i-1,j}+1a_{ij}+0a_{i+1,j}+\dots+0a_{nj} \\&= a_{ij}.\end{align*} So the matrices $I_nA$ and $A$ have the same size, and the same $(i,j)$ entries, for any $(i,j)$. So $I_nA=A$.

Proof of the proposition, continued

2. To show that $AI_m=A$ for any $n\times m$ matrix $A$ is similar to the first part of the proof; the details are left as an 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

Proposition: associativity of matrix multiplication

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.

We omit the proof, but this is not terribly difficult; it is a calculation in which you write down two formulae for the $(i,j)$ entries of $(AB)C$ and $A(BC)$, and carefully check they are equal using the fact that if $a,b,c$ are real numbers, then $(ab)c=a(bc)$.

Example

We saw above that $\newcommand{\m}[1]{\begin{bmatrix}#1\end{bmatrix}}A=\m{1&2\\3&4}$ commutes with $B=\m{7&10\\15&22}$. We can explain why this is so using associativity. You can check that $B=AA$ (which we usually write as $B=A^2$). Hence, using associativity at $\stackrel*=$, \[ AB=A(AA)\stackrel*=(AA)A=BA.\] The same argument for any square matrix $A$ gives a proof of:

Proposition

If $A$ is any square matrix, then $A$ commutes with $A^2$.■

The powers of a square matrix $A$ are defined by $A^1=A$, and $A^{k+1}=A(A^k)$ for $k\in \mathbb{N}$. Using mathematical induction, you can prove the following more general proposition.

Proposition: a square matrix commutes with its powers

If $A$ is any square matrix and $k\in\mathbb{N}$, then $A$ commutes with $A^k$.■

The distributive laws

Lemma: the distributive laws for row-column multiplication

1. If $a$ is a $1\times m$ row vector and $b$ and $c$ are $m\times 1$ column vectors, then $a\cdot (b+c)=a\cdot b+a\cdot c$.
2. If $b$ and $c$ are $1\times m$ row vectors and $a$ is an $m\times 1$ column vector, then $(b+c)\cdot a=b\cdot a+c\cdot a$.

The proof is an exercise (see tutorial worksheet 5).

Proposition: the distributive laws for matrix multiplication

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

1. First note that

• $B$ and $C$ are both $m\times k$, so $B+C$ is $m\times k$ by the definition of matrix addition;
• $A$ is $n\times m$ and $B+C$ is $m\times k$, so $A(B+C)$ is $m\times k$ by the definition of matrix multiplication;
• $AB$ and $AC$ are both $n\times k$ by the definition of matrix multiplication
• so $AB+AC$ is $n\times k$ by the definition of matrix addition.

So we have (rather long-windedly) checked that $A(B+C)$ and $AB+AC$ have the same size.

By the Lemma above, the row-column product has the property that \[a\cdot (b+c)=a\cdot b+a\cdot c.\] So the $(i,j)$ entry of $A(B+C)$ is \begin{align*}\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*} On the other hand,

• 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 $\text{row}_i(A)\cdot \col_j(B)+\row_i(A)\cdot\col_j(C)$.

So the entries of $A(B+C)$ and $AB+AC$ are all equal, so $A(B+C)=AB+AC$.

2. The proof is similar, and is left as an exercise.■