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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    $\begingroup$ You are right, but you can reason more simply. $\endgroup$ – copper.hat Dec 18 '12 at 7:16
  • $\begingroup$ No, the proofs are incorrect. $\endgroup$ – user1551 Dec 18 '12 at 8:05
  • $\begingroup$ 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 $0<c_1<c_2$ such that $c_1 \|A\|_i \leq \|A\|\ \leq c_2 \|A\|_i$. Hopefully my ramble makes sense... $\endgroup$ – copper.hat Dec 19 '12 at 5:21
  • $\begingroup$ It makes sense, of course. $\endgroup$ – user1551 Dec 23 '12 at 17:26

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 $\|B\| \neq 0$.


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 $\|A\|\|x\|_\infty<(1+\epsilon) < 2 = \|Ax\|_\infty$.

  • $\begingroup$ Just one question, are not all a matrix norms equivalent? $\endgroup$ – Zizo Dec 18 '12 at 20:12
  • $\begingroup$ @Zizo Yes, all matrix norms are equivalent. Why the question? $\endgroup$ – user1551 Dec 18 '12 at 20:27
  • $\begingroup$ weather there are induced by a vector norm or not, yes. I think if its the case, then my proof would be right. $\endgroup$ – Zizo Dec 18 '12 at 20:28
  • $\begingroup$ @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. $\endgroup$ – user1551 Dec 18 '12 at 20:33
  • $\begingroup$ 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. $\endgroup$ – Zizo Dec 18 '12 at 20:49

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