= Solution
For a <linear map> $A:\mathbb R^m\to\mathbb R^n$, define the <operator norm> by
$$
\boxed{\|A\|=\sup_{x\ne0}\frac{\|Ax\|}{\|x\|}=\sup_{\|x\|=1}\|Ax\|.}
$$
It is finite. To see this without assuming continuity of $A$, write $x=\sum x_j e_j$: the <triangle inequality> bounds $\|Ax\|$ by $\sum|x_j|\|Ae_j\|$. Every <norm> in finite dimensions is equivalent to the <Euclidean norm>. One proof first uses this same coordinate bound to establish continuity of the <norm> on the Euclidean unit sphere; <compactness> then gives a strictly positive minimum there. Thus the coordinate bound above is at most a fixed multiple of $\|x\|$.
Absolute homogeneity and the <triangle inequality> for this <operator norm> follow from those for $\|Ax\|$ and taking suprema. If $\|A\|=0$, then $Ax=0$ for every $x$, hence $A=0$, and the converse is immediate. For any $B:\mathbb R^\ell\to\mathbb R^m$, the defining bound $\|Ax\|\le\|A\|\|x\|$ gives
$$
\|ABx\|\le\|A\|\|Bx\|\le\|A\|\|B\|\|x\|,
\qquad \boxed{\|AB\|\le\|A\|\|B\|.}
$$
This is <submultiplicativity of the operator norm>.
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