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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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    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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    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