If $X$ is a finite dimensional Hilbert Space, $A$ a bounded operator on $X$, then is $||A||_{op} \leq \rho(A)$?

Question

The question is the same as the title. Is this true or false? There are a few cases where an answer is clear.

1. The trivial case where $A = 0$. Then $||A||_{op} = \rho(A) = 0$.
2. The almost trivial case where $A = \text{Id}$. Then $||A||_{op} = \rho(A) = 1$.
3. $A$ is diagonalizable. Then there is an orthonormal basis of $X$ consisting entirely of eigenvectors of $A$. Hence any $x\in B_X:=\{x\in X: |x| = 1|\}$ has an expansion $x = \sum_\nu a_\nu e_\nu$, where $a \in \mathbb{C}$ and the $e_\nu$ are the basis eigenvectors. We have

$$||A||_{op} :=\sup_{x\in B_X}|Ax| = \sup_{x\in B_X} \sqrt{(Ax,Ax)} =\sup_{x\in B_X}\sqrt{\sum_{\nu}\sum_{\mu}a^*_\nu(x) a_\mu(x) \lambda^*_\nu\lambda_\mu\delta_{\nu\mu} }= \rho(A),$$ the last equivalence following because $\sum_\nu|a_\nu|^2=1$ for any $x \in B_X$.

What follows is my reasoning so far:

Suppose $A$ is not diagonalizable. Since every finite dimensional operator can be represented by a matrix in $\mathbb{C}^{\dim(X)\times\dim(X)}$, and from elementary linear algebra we know any such matrix has at least one eigenvector, the decomposition $R(A) = \Lambda(A)\oplus P_{\Lambda(A)}[R(A)]^{\perp}$, where $\Lambda(A) = \{\vee_\nu e_\nu:A e_\nu = \lambda_\nu e_\nu\}$ and $P$ is the standard projection operator, is always meaningful. Thus any $y \in R(A)$ can be written $y = l+r$, where $l\in\Lambda(A)$ and $r \in P_{\Lambda(A)}[R(A)]^{\perp}$. Now, since $y\in R(A)$ there is some $x\in X$ such that

$$|Ax| = |y(x)| = |l(x)+r(x)|\leq |l(x)| + |r(x)|$$

Expanding $l(x)$ in the eigenvector basis, $l(x) = \sum_{\nu<\dim(X)}a_\nu(x)e_\nu$. Thus $$ |l(x)| = \left|\sum_{\nu<\dim(X)} a_\nu(x)e_\nu\right|\leq \sum_{\nu<\dim(x)}|a_\nu\lambda_\nu|\leq\left(\sum_{\nu<\dim(x)}|a_\nu(x)|\right)\rho(A). $$ So we end up with $$ ||A||_{op} \leq \sup_{x\in B_X}\left(\sum_{\nu<\dim(x)}|a_\nu(x)|\right)\rho(A) + \sup_{x\in B_X}|Ar(x)|, $$ which is really to say, the conjecture is proven if somehow the argument (or sequence of arguments) leading to the supremum are in (converge in) the eigenspace. The main problem is there is no reason I should think this is necessarily the case.

Ideas?

Answer 1

The opposite inequality is true. The operator norm dominates the spectral radius, which is proven on the wikipedia page for spectral radius. That is, $\rho(A) \le \|A\|_{op}$. For an example where the inequality is strict, consider $$A = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}.$$ Then the spectrum of $A$ is $\{0\}$, so $\rho(A) = 0$, but $\|A\| = 1$.