Proving that $\|A\|$ is finite.

Question

Let $|v|$ be the Euclidean norm on $\mathbb{R^n} $. For $A\in \mathrm{Mat}_{n\times n}(\mathbb{R})$ we define $\displaystyle \|A\|:= \sup_{\large v\in \mathbb{R^n},\,v \neq 0}\frac{|Av|}{|v|}$. How to show that $\|A\|$ is finite for every $A$? It would be very helpful if someone could give hints. I think I should show that $\|-\|$ is bounded,but I don't know how...

This is not completely straightforward. Yes every linear operator on a finnite-dimensional real/complex normed vector space is bounded (ie continuous). This can be shown using the equivalence of all norms in finite dimension. — Julien, Feb 17 '13 at 14:15
See this thread for a more general case: http://math.stackexchange.com/questions/112985/every-linear-mapping-on-a-finite-dimensional-space-is-continuous — Julien, Feb 17 '13 at 14:23
As opposed to what I first said, it is actually straightforward with the Euclidiean norm... — Julien, Feb 17 '13 at 15:28

score 6 · Answer 1 · answered Feb 17 '13 at 15:19

6

Assume $A\in \Bbb R^{m\times n}$ and write $Ax=: y$, where $x\in\Bbb R^n$, $\ y\in\Bbb R^m$. Then by Schwarz' inequality $$y_i^2=\left(\sum_{k=1}^n a_{ik}x_k\right)^2\leq \sum_{k=1}^n a_{ik}^2\ \sum_{k=1}^n x_k^2=|a_{i\cdot}|^2\ |x|^2\qquad(1\leq i\leq m)$$ and therefore $$|y|^2=\sum_{i=0}^m y_i^2\leq C|x|^2$$ with $$C:=\sum_{i=1}^m |a_{i\cdot}|^2=\sum_{i,k} a_{ik}^2\ .$$ It follows that $$\|A\|\ \leq\ \left(\sum_{i,k} a_{ik}^2\right)^{1/2}\ .$$

answered Feb 17 '13 at 15:19

Christian Blatter

226,825

Yes, that's the best possible answer here, +1. – Julien Feb 17 '13 at 15:31
I can't agree more, the most understandable answer, +1 – dineshdileep Feb 17 '13 at 15:47

score 3 · Answer 2 · edited Feb 21 '16 at 23:25

3

Ok seems like I should make it a bit more explicit: At first we can scale the problem using the homogeneity of norms: $$ \sup_{v\in \mathbb{R}^n, v\neq 0} \frac{|Av|}{|v|}=\sup_{v\in \mathbb{R}^n, v\neq 0} |A\frac{v}{|v|}| = \sup_{|v|=1, v\in \mathbb{R}^n} |Av|$$ Now we write the Matrix $A$ like this one $$\begin{pmatrix} a & b \\ c & d\\ \end{pmatrix} = \begin{pmatrix} a& b \\0 & 0 \\ \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ c & d \\ \end{pmatrix}$$ We decompose the Matrix $A$ in a sum of matrices $A_i$ where only 1 column has not only $0$ entries. So we get $$\sup_{|v|=1,v\in \mathbb{R}^n } |Av|=\sup_{|v|=1,v\in \mathbb{R}^n} |\sum_{i=1}^n A_i v|$$ Using the triangle inequality we get $$\sup_{|v|=1,v\in \mathbb{R}^n} |Av|\leq \sup_{|v|=1,v\in \mathbb{R}^n} \sum_{i=1}^n |A_i v|$$ We can identify $A_i\in \mathbb{R}^{n\times n}$ with a vector $b_i\in \mathbb{R}^n$, $b_i$ does have the non zero entries of $A_i$. Note that $A_i v$ is a vector with only 1 non $0$ entry. So $$A_i \cdot v=\langle b_i,v \rangle \cdot e_i$$ where $e_i=(0,\dots,1,0,\dots,0)$

Let $\langle \cdot , \cdot \rangle$ be the euclidean scalar product we get: $$\sup_{|v|=1,v\in \mathbb{R}^n} \sum_{i=1}^n |A_i v| =\sup_{|v|=1,v\in \mathbb{R}^n} \sum_{i=1}^n \operatorname{abs}(\langle b_i, v\rangle) \cdot |e_i| $$ Using Cauchy Schwarz we get
$$\sup_{|v|=1,v\in \mathbb{R}^n}\sum_{i=1}^n \operatorname{abs}(\langle b_i, v\rangle) \cdot |e_i| \leq \sup_{|v|=1, v\in \mathbb{R}^n} \sum_{i=1}^n |b_i|\cdot |e_i|$$ Since this is (finally) independent of $v$ $$ \sup_{|v|=1, v\in \mathbb{R}^n} \sum_{i=1}^n |b_i|\cdot |e_i|=\sum_{i=1}^n |b_i|\cdot |e_i|$$

edited Feb 21 '16 at 23:25

John B

16,854

answered Feb 17 '13 at 14:21

Dominic Michaelis

19,935

Your $b_i$ is actually $A^tAv$ and it should be $\sqrt{\langle b_i,v\rangle}$. So your $b_i$ depend on $v$ and your reasoning is kinf od circular. You basically brought the boundedness of $A$ back to the boundedness of $A^tA$. – Julien Feb 17 '13 at 15:03
Yes, $b_i$ is a vector in $\mathbb{R}^n$, I never said otherwise. But it is $A_i^tA_iv$, so it depends on $v$. – Julien Feb 17 '13 at 15:06
1

I don't think I misunderstood anything. But note that $|A_iv|=\sqrt{\langle A_iv,A_iv\rangle}=\sqrt{\langle A_i^tA_iv,v\rangle}$. So, again $b_i=A_i^tA_iv$ depends on $v$. And it is not, in general, a column of $A$ like you claim it is. – Julien Feb 17 '13 at 15:12
could the guys who downvoted read again, and tell me if there is still something wrong? – Dominic Michaelis Feb 17 '13 at 15:23
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For instance, the sentence: "We can identify $A_i$ ..." does not justify the following equality since, again, $b_i$ is actually equal to $A_i^tA_iv$. – Julien Feb 17 '13 at 15:27
@julien thanks a lot, I now changed it, could you check if it is right now ? – Dominic Michaelis Feb 17 '13 at 16:19
1

You're welcome. Now it makes sense! – Julien Feb 17 '13 at 16:22
why there are still 2 downvotes ? i think i corrected everything ... – Dominic Michaelis Feb 21 '13 at 06:38

score 2 · Answer 3 · answered Feb 17 '13 at 14:56

Let $||\cdot ||$ be any norm on a finite dimensional vector space $X$. Then define the norm $N$ on the space of endomorphisms $\mathcal{L}(X)$ by $N(\varphi)= \sup\limits_{x \in X \backslash \{0\}} ||\varphi(x)||/||x||$. It is straightforward that $N(\varphi)= \sup\limits_{x \in X, ||x||=1} ||\varphi(x)||$. You can deduce that $N(\varphi)<+ \infty$ from the continuity of $\varphi$ (any endomorphism of a finite dimensional normed spaces is continuous) and the compacity of the sphere $\{x \in X : ||x||=1\}$ (a finite dimensional normed space is locally compact).

score 2 · Answer 4 · answered Feb 17 '13 at 15:19

Hint: Let $S=\{v\in\mathbb{R}^n\;|\;|v| = 1\}, N = \{\frac{|Av|}{|v|}\;|\;v\in\mathbb{R}^n.\;v\ne 0\}, N' = \{|Av|\;|\;v\in\mathbb{R}^n.\;|v|=1\}$

Step 1: $||A|| = \sup N = \sup N'$

Step 2: Show that $x\to|Ax|$ is a continuous map. $S$ is closed and bounded in $\mathbb{R}^n$ therefore compact, |Ax| attains max on $S$.

Done

Do you happen to be in Linear Algebra II class?

Proving that $\|A\|$ is finite.

4 Answers4