Exercise 1.4.9 (Associativity of matrix product)

Question

Prove that the binary operation of matrix multiplication of $2 	imes 2$ matrices with real number entries is associative.

Richard Ganaye · Accepted Answer

ewcommand{\R}{\mathbb{R}}

\begin{proof} 
Let $M,N,P$ be $2 	imes 2$ matrices with real number entries
Let $f, g, h$ in $\mathscr{L}(\R^2)$ be linear maps associate to $M, N, P$, so that if $\mathscr{B} =(e_1, e_2)$ is the canonical basis of $\R^2$, i.e. $e_1 = (1,0), e_2 = (0,1)$, then
$$M = \mathscr{M}_{\mathscr{B}} (f),\qquad N =  \mathscr{M}_{\mathscr{B}}(g), \qquad P =  \mathscr{M}_{\mathscr{B}}(h).$$
Since $ \mathscr{M}_{\mathscr{B}} (f \circ g) = \mathscr{M}_{\mathscr{B}}(f)  \mathscr{M}_{\mathscr{B}} (g)$, we obtain, using the associativity of $\circ$,
\begin{align*}
(MN)P & = (\mathscr{M}_{\mathscr{B}} (f) \mathscr{M}_{\mathscr{B}}(g) ) \mathscr{M}_{\mathscr{B}} (h) \
&=\mathscr{M}_{\mathscr{B}} (f \circ g) \mathscr{M}_{\mathscr{B}}(h) \
&=\mathscr{M}_{\mathscr{B}} ((f \circ g) \circ h) \
&=\mathscr{M}_{\mathscr{B}} (f \circ (g \circ h)) \
&= \mathscr{M}_{\mathscr{B}}(f) \mathscr{M}_{\mathscr{B}} (g\circ h)  \
&= \mathscr{M}_{\mathscr{B}} (f) (\mathscr{M}_{\mathscr{B}}(g) \mathscr{M}_{\mathscr{B}} (h)) \
&= M (NP).
\end{align*}
Note: More patient readers than me may verify directly that
$$\left[\begin{pmatrix} a & b\ c & d\end{pmatrix}\begin{pmatrix} e & f\ g & h\end{pmatrix}ight] \begin{pmatrix} i & j \ k & l\end{pmatrix} = 
\begin{pmatrix} a & b\ c & d\end{pmatrix}\left[\begin{pmatrix} e & f\ g & h\end{pmatrix} \begin{pmatrix} i & j \ k & l\end{pmatrix} ight].
$$
\end{proof}

Exercise 1.4.9 (Associativity of matrix product)

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Exercise 1.4.9 (Associativity of matrix product)

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