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