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.