Spivak Exercise involving operator norm The exercise as stated:
If $T:\mathbb{R}^{m}\to \mathbb{R}^{n}$ is a linear transformation, show that there is a number $M$ such that $|T(h)|\leq M\cdot |h|$ for all $h\in \mathbb{R}^{n}$.
Hint: Estimate $|T(h)|$ in terms of $|h|$ and the other entries in the matrix of $T$.
By the way, here $|\cdot|$ denotes the standard Euclidean norm.

It is clear that $M$ is just the operator norm of $T$.
Indeed take $M$ to be the supremum of $|T(h)|$ over all $h$ in the unit ball of $\mathbb{R}^{n}$, which exists since $T$ is continuous (it is linear and defined on a finite dimensional space) and the unit ball of $\mathbb{R}^{n}$ is compact.  Then using linearity it is easy to show that this $M$ works as desired.
However, I'm dodging the point here, and I'd like to come up with a direct proof, so I can gain the benefit of solving this problem.  So I tried using the hint:
In the simple case where $n = 2$, I tried writing the matrix of $T$ as $(a_{ij})$ and $h = (h_1, h_2)$.
Then $T(h) = (a_{11}h_{1} + a_{12}h_2, a_{21}h_{1} + a_{22}h_{2})$.
Even in the relatively simple case of $2$ dimensions, the norm of this is ugly.
After simplification, I get:
$$|T(h)| = \sqrt{(a_{11}^{2} + a_{21}^{2})h_{1}^{2} + 2(a_{11}a_{12} + a_{21}a_{22})h_{1}h_{2} + (a_{12}^{2} + a_{22}^{2})h_{2}^{2}}$$
Any suggestions on how I can estimate this in terms of $|h| = \sqrt{h_{1}^{2} + h_{2}^{2}}$?  
Thanks as always for your attention.
 A: I will leave it to you, but I think the simplest approach is likely to be the induced operator norms for worthwhile vector norms. 
If you use the $\infty$ norm on vectors, the expansion in that norm is fairly easy, although Spivak is allowing $m$ and $n$ to differ which adds annoyance. The final step is that, as vector norms, this one and the Euclidean one are "equivalent," which has been proved on MSE several times. 
A: By Cauchy Schwarz inequality
$$
\left|\sum\limits_{j=1}^m a_{ij}h_j\right|^2\leq
\left(\sum\limits_{j=1}^m a_{ij}^2\right)\left(\sum\limits_{j=1}^m h_{j}^2\right)
$$
so
$$
|T(h)|=
\left(\sum\limits_{i=1}^n|T(h)_i|^2\right)^{1/2}=
\left(\sum\limits_{i=1}^n\left|\sum\limits_{j=1}^m a_{ij}h_j\right|^2\right)^{1/2}\leq
\left(\sum\limits_{i=1}^n\left(\sum\limits_{j=1}^m a_{ij}^2\right)\left(\sum\limits_{j=1}^m h_{j}^2\right)\right)^{1/2}=
\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^m a_{ij}^2\right)^{1/2}\left(\sum\limits_{j=1}^m h_{j}^2\right)^{1/2}=
\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^m a_{ij}^2\right)^{1/2}|h|
$$
And the desired constant is
$$
M=\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^m a_{ij}^2\right)^{1/2}
$$
Note that it is not the best possible constant in this inequality.
A: Here's the simplest, cleanest proof:
$$\|T(x)\| =\bigg\|\sum_{i=1}^{n}x_iT(e_i)\bigg\| \le \sum_{i=1}^{n}|x_i|\|T(e_i)\| \le n\max|x_i|\max\|T(e_i)\| \le n\|x\|\max\|T(e_i)\|$$
So let $M = n\max\|T(e_i)\|.$
A: Note that the unit sphere $S^{n-1}$ is compact and that a linear map is continuous so we can write
$$ M = \max_{\|x\| = 1} \|T(x)\|.$$
This $M$ does the job.
