Matrix multiplication by scalar is commutative

Question

Is matrix multiplication by scalar commutative, i.e. $(\alpha M)N=M(\alpha N)$? If so, can we prove it without induction?

Answer 1

Yes. Let $L_{ij}$ be the $i,j$th component of $(\alpha M)N$, let $R_{ij}$ be the $i,j$th component of $M(\alpha N)$.

$$(\alpha M)N=\begin{pmatrix} \alpha M_{11} &...& \alpha M{1m}\\ ...&...&...\\ \alpha M_{n1} &...& \alpha M{nm} \end{pmatrix}\begin{pmatrix} N_{11}&...&N{1r}\\ ...&...&...\\ N_{n1}&...&N_{nr}\end{pmatrix}$$

$$M(\alpha N)=\begin{pmatrix} \ M_{11} &...& M{1m}\\ ...&...&...\\ M_{n1} &...& M{nm}\end{pmatrix} \begin{pmatrix} \alpha N_{11} &...& \alpha N{1m}\\ ...&...&...\\ \alpha N_{n1} &...& \alpha N{nm} \end{pmatrix}$$

Now write out $L_{ij}$ and $R_{ij}$. They are the same.

Answer 2

Yes it's true and also easy to prove. Let $f: X \rightarrow Y$ be the linear function with matrix $N$ and $g: Y \rightarrow Z$ be the linear function with matrix $M$. You have for each vector $x \in X$: $$\begin{align}((\alpha M)N)x & = ((\alpha g) \circ f)(x) \\ & = (\alpha g)(f(x)) \\ & = \alpha \cdot g(f(x)) \\ & \left\downarrow \ g\text{ is linear}\right. \\ & = g(\alpha \cdot f(x)) \\ & = (g \circ \alpha f)(x) \\ & = (M(\alpha N))(x) \end{align}$$