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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.

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$2h_1h_2\leq h_1^2+h_2^2$ hence $|T(h)|\leq \sqrt{a_{11}^2+a_{21}^2+|a_{11}a_{12}|+|a_{21}a_{22}|+a_{12}^2+a_{22}^2}|h|$. For the general case, you can use Cauchy-Schwarz. –  Davide Giraudo Jul 31 '12 at 22:26
    
I hate to pry further, and I definitely see the benefit of applying that inequality, but while I cannot find a counterexample, I can't seem to work out a proof. I drew a picture and tried to justify it in terms of area, but it didn't reveal any insight to me. –  Kyle Schlitt Jul 31 '12 at 22:33
    
Perhaps I'll start by just working out the general case and seeing if I can figure out how to apply Cauchy-Schwarz. (Thank you, by the way.) –  Kyle Schlitt Jul 31 '12 at 22:34
    
I started calculating $|T(h)|$ for general $m,n\geq 1$ and I think I now see where you are applying Cauchy-Schwarz. Thanks again! –  Kyle Schlitt Jul 31 '12 at 22:42
    
@Kyle All norms on finite dimensional euclidean space are equivalent. This comes from the fact that the unit sphere in finite dimensional space is compact. –  user38268 Jul 31 '12 at 22:44
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4 Answers 4

up vote 4 down vote accepted

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.

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Thanks Norbert! –  Kyle Schlitt Jul 31 '12 at 22:43
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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.

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Thanks for the suggestion! I am familiar with the general result that all norms are equivalent on finite dimensional spaces (that's the one you are referring to I'm sure). But applying this would seem to invent (or at least use) much more machinery than was covered in the chapter or mentioned in the hint. This is the same reason that I decided to search for a more direct argument instead of using the theoretical one I gave intially. –  Kyle Schlitt Jul 31 '12 at 22:39
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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 $M = n\max\|T(e_i)\|.$

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This is a different (and much simpler) argument which also does not rely on other machinery. Thanks for pointing it out! –  Kyle Schlitt Aug 1 '12 at 20:03
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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.

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