Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $L$ be a bounded linear operator on a Hilbert space. Without assuming finite dimensions, can we express the operator norm of $L$ in terms of the spectrum of the positive operator $L^{\dagger}L$?

More precisely, does the following hold.

$\sup \big\{ \| L\phi \| \,\big|\, \phi \in \mathcal{H} \land \|\phi\| = 1 \big\} = \sqrt{ \max \sigma(L^{\dagger} L)}$

share|cite|improve this question
For $L^*L$ we have that $\|L^*L\| = r(L^*L)$ - spectral radious of $L^*L$, i.e. $r(L^*L) = \sup_{\lambda \in \sigma(L^*L)}\lambda$. Since $B(H)$ is a C*-algebra $\|L^*L\| = \|L\|^2$. – Frank Tessla Mar 3 '13 at 10:02
up vote 3 down vote accepted

I will convert Frank's comment into the answer.

As it was showed in this answer (theorem 3.4) for every normal element $a$ of a unital $C^*$-algebra $A$ holds $$ \Vert a\Vert=r(a) $$ For your particular case $A=\mathcal{B(H)}$ and $a=L^*L$. Using $C^*$-identity and the fact that $L^*L$ is normal we get $$ \Vert L\Vert^2=\Vert L^*L\Vert=r(L^*L)=\max\sigma(L^*L) $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.