Matrix multiplication by scalar is commutative Is matrix multiplication by scalar commutative, i.e. $(\alpha M)N=M(\alpha N)$? If so, can we prove it without induction?
 A: 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.
A: 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 $x \in X$: $$\begin{align}((\alpha M)N)x & = ((\alpha g) \circ f)(x) \\
& = (\alpha g)(f(x)) \\
& = \alpha \cdot g(f(x)) \\
& \left\downarrow \ g\text{ is linear}\right. \\
& = g(\alpha \cdot f(x)) \\
& = (g \circ \alpha f)(x) \\
& = (M(\alpha N))(x) \end{align}$$
