Show that for a given linear transformation $T:\Bbb R^m\to\Bbb R^n$ there exists $M\in\Bbb R$ such that $|T(h)|\le M|h|$ for any $h\in\Bbb R^m$

Question

I'm having a little trouble demonstrating this basic result. I was trying to proceed as follows:

If $h=(x_1,\dots,x_m)$ and $T$ is given by the matrix $(a_{ij})$, then $T(h)=\bigl(\sum_{i=1}^m a_{1i}x_i,\dots,\sum_{i=1}^m a_{ni}x_i\bigr)$. Then, $$|T(h)|^2=\bigl(\sum_{i=1}^m a_{1i}x_i\bigr)^2+\dots+\bigl(\sum_{i=1}^m a_{ni}x_i\bigr)^2=\sum_{i=1}^m a_{1i}^2x_i^2+\dots+\sum_{i=1}^m a_{ni}^2x_i^2+\sum_{1\le i,j\le m} c_{ij}x_ix_j=\sum_{k=1}^n a_{k1}^2x_1^2+\dots+\sum_{k=1}^n a_{km}^2x_m^2+\sum_{1\le i,j\le m} c_{ij}x_ix_j=A_1^2x_1^2+\dots+A_m^2x_m^2+C$$

Assuming everything is correct up till here, this is the part I have trouble with. If $C$ were not there, I could simply take $M$ to be $max\{A_i\}$ and be done with this. Is there a way to express $C$ in terms of $|h|$ somehow?

Answer 1

Cauchy-Schwarz is the only tool you need, which gives $$ \left(\sum_{i=1}^m a_i x_i\right)^2 \le \sum_{i=1}^ma_i^2 \sum_{i=1}^mx_i^2 $$ plug this into your $ (\sum_{i=1}^m a_{1i} x_i)^2 + \dots + (\sum_{i=1}^m a_{ni} x_i)^2$ line and it should come out cleanly

Answer 2

There's a far less computational way to do this. This uses a bit of topology: namely the notion of compactness. Indeed, the unit ball $S^{m - 1} = \{ (x_1, \dots, x_m) \in \mathbb R^m : \sum x_i^2 = 1 \}$ is compact. Hence, its image under $T$, $T[S^{m - 1}] \subseteq \mathbb R^n$, is compact as well, as linear maps are continuous. Hence, $T[S^{m - 1}]$ has an element of maximal norm. Call that maximal norm $C$. By definition, $C$ satisfies $|T(x)| \leq C$ for all $|x| = 1$. Then for $x \in \mathbb R^m - 0$, $\left|T\left(\frac{x}{|x|}\right) \right|\leq C$. Hence, $|T(x)| \leq C |x|$ for all $x \neq 0$. Of course, this obviously holds for $0$ as well and we are done.

While this approach doesn't yield as explicit of a form for $C$, I think it's conceptually clearer and an easier proof to work with.