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If a matrix $A$ satisfies $x^TAx<0$ for some vector $x \neq 0.$ I wanna show that $\|A\| \neq 0$ for any matrix norm. Another one, if the spectral radius of $B, \rho(B)>1.$ I also want to show that $\|B\| \neq 0$ for any matrix norm.

I try like this: Since $x^TAx < 0$ for some $ x \neq 0,$ then $ 0 < \|x^TAx\| \leq \|x^T\|\|A\|\|x\|.$ Thus $ \|A\| \geq \frac{\|x^TAx\|}{\|x\|\|x^T\|}>0.$ Then by the equivalent of norms, there is a number $c>0$ such that $ c\|A\| \leq \|A\|_0$ for any matrix norm $\| \cdot\|_0.$ Thus $\|A\|_0 > 0$ for any matrix norm.

The second one: If $ \rho(B)>1,$ let $ \lambda$ be an eigenvalue of $B$ such that $ | \lambda| = \rho(B),$ then $ \|x\| < \rho(A) \|x\| = |\lambda| \|x\| =| |\lambda x\| = \|Bx\| \leq \|B\| \|x\|.$ Thus $\|B\| \geq 1$ for any matrix norm.

I think I'm right, yes?

Edit: After discussion with @user1551 below, I mention that the proof above assumed that $\|Ax\| \leq \|A\|\|x\|,$ that is, the matrix norm is induced by a vector norm. Otherwise this is not true in general for any matrix norm and a contradicting example has been provided by @user1551

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    You are right, but you can reason more simply.2012-12-18
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    No, the proofs are incorrect.2012-12-18
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    Well, they are not technically correct, but in terms of what the OP is proving, they are good. The only issue is that some extra constants are needed if the norm is not an induced norm. For any matrix norm, it will be equivalent to an induced matrix norm, ie, there will exist $0 such that $c_1 \|A\|_i \leq \|A\|\ \leq c_2 \|A\|_i$. Hopefully my ramble makes sense...2012-12-19
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    It makes sense, of course.2012-12-23

2 Answers 2

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Any norm satisfies $\|A\| = 0$ iff $A=0$. To show the norm is non-zero you just need to show that $A$ is non-zero.

So clearly, if $\langle x, Ax \rangle <0$ for some $x$, then $A\neq 0$, hence $\|A\| \neq 0$.

Similarly, if $\rho(B) >0$ (not just 1), it follows that $B\neq0$, from which it follows that $\|B\| \neq 0$.

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Your work is not right. Your proof is incomplete. In proving both (a) and (b), you have assumed something like $\|Ax\|\le\|A\|\|x\|$. This is true if the matrix norm is induced by the vector norm in the inequality, but untrue in general. In fact, for any matrix $A$ and any $\epsilon>0$, there is some matrix norm $\|\cdot\|$ such that $\|A\|<\rho(A)+\epsilon$. Take $A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\ x=\begin{pmatrix}1\\1\end{pmatrix}$, we have $\|A\|\|x\|_\infty<(1+\epsilon) < 2 = \|Ax\|_\infty$.

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    Just one question, are not all a matrix norms equivalent?2012-12-18
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    @Zizo Yes, all matrix norms are equivalent. Why the question?2012-12-18
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    weather there are induced by a vector norm or not, yes. I think if its the case, then my proof would be right.2012-12-18
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    @Zizo Yes. If your matrix norm is induced by a vector norm, and that vector norm is the one you use in your proof, then your proof is correct. Yet not all matrix norms are induced norms.2012-12-18
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    I understand your concern.. If the matrix norm $\|A\|$ is not induced by a vector norm, I cannot do what I did for any matrix norm, I agree seeing the contradicting example you provided.. I think @copper.hat's proof is general.2012-12-18
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    @Zizo Yes, it is.2012-12-18
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    Just one thing please, if you're still there. I got a proof for the equivalents of norms in finite dimensional space. My guess is that matrix norms induced by vector norms are also the equivalent from the result that I've. My question is, how can I see that all vector norms are equivalent weather there induced by a vector norm or not. If my question need some research you could guide me to something to see or read2012-12-18
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    I meant, how can I see that all matrix norms are equivalent weather there are induced by a vector norm or not.. Just to correct the typo2012-12-18
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    All matrix norms are equivalent *because all matrix norms are vector norms*. That is, if $\|\cdot\|$ is a matrix norm on $M_{m,n}(\mathbb{F})$ and you identify the matrix space $M_{m,n}(\mathbb{F})$ with the vector space $\mathbb{F}^{mn}$, then $\|\cdot\|$, **by definition**, is a vector norm on $\mathbb{F}^{mn}$. A matrix norm is just a vector norm (I mean a vector norm on the vector space of matrices, i.e. on $\mathbb{F}^{mn}$; don't confuse this with an induced norm) with an additional submultiplicativity propery $\|AB\|\le\|A\|\|B\|$.2012-12-18
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    I think I see what you mean, when the matrix space is $ \mathbb{R}^{n \times n},$ then its just a vector space $ \mathbb{R}^{n^2}.$ Therefore matrix norms defined in $\mathbb{R}^{n \times n}$ are just vector norms defined in $ \mathbb{R}^{n^2}.$ You did it in a general way, this is what I've understood.2012-12-18
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    Yes. As for why all vector norms on a finite dimensional vector space (over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$) are equivalent, you may look up any textbook.2012-12-19