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If $T$ is an operator on a finite-dimensional complex inner product space, each eigenvalue $|\lambda|=1$ and $\|Tv\|\le\|v\|$, show that $T$ is unitary.

Here's what I had in mind and where I was stuck:

$$\|Tv\|\le\|v\| \to \langle v,(I-T^*T)v\rangle \ge0$$

Therefore $I-T^*T$ is self-adjoint hence there exists an orthonormal basis of eigenvectors of $I-T^*T$. I'm able to deduce that each one of its eigenvalues is real and satisfies $\lambda\ge 0$.

A possible approach I thought I might take is calculate $\operatorname{trace}(I-T^*T)$ and maybe show that it's $0$. That would complete the proof since then we would get $\lambda=0$ for all of its eigenvalues which would imply $I-T^*T=0 \to T^*T=I$ and hence $T$ is unitary.

However: $$\operatorname{trace}(I-T^*T)=\operatorname{trace}(I)-\operatorname{trace}(T^*T)=n -\operatorname{trace}(T^*T)$$

I don't know how to evaluate the second term. I realize that $T^*T$ is positive and all of its eigenvalues are nonnegative, but I don't know how to go on from here. Any help would be greatly appreciated (or otherwise, I'd love to see other ways to go about the problem).

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    $\begingroup$ Unsure about this, but how about: Let's assume $||Tv|| < ||v||$. Then, for the eigenvectors, we have: $||\lambda v|| < ||v|| \Rightarrow |\lambda| < 1$, which isn't true… $\endgroup$
    – kumanna
    Aug 3, 2013 at 0:48
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    $\begingroup$ You can show indeed that there is equality for any eigenvector of $T$. It's not enough unfortunately since I don't know how many independent eigenvectors there are. $\endgroup$
    – Adar Hefer
    Aug 3, 2013 at 0:55

3 Answers 3

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I assume you are talking of a finite-dimensional complex inner-product space, since you took the trace and only mentioned eigenvalues. But for people who are interested in the infinite-dimensional case: it is a result of Bernard Russo, Pacific JM 1968, that there exist contractions in $B(H)$ with precsribed spectrum contained in the unit circle which are not unitary, called nonunitary unimodular contractions. Actually, he proves in Theorem 1 that a von Neumann algebra contains a nonunitary unimodular contraction if and only if it infinite. The only if goes much like below, with the Kadison-Fuglede determinant and the trace.

Proof: let $\{e_j\;;\;1\leq j\leq n\}$ be an orthonormal basis of diagonalization for $T^*T$, with $T^*Te_j=t_je_j$. Note that by assumption, $0\leq t_j\leq 1$ for every $j$ and $|\det T|=1$. Then we have, by AM-GM inequality $$ 1=|\det T|^2=\det (T^*T)=\prod_{j=1}^nt_j\leq \left(\frac{1}{n}\sum_{j=1}^nt_j\right)^n\leq \left(\frac{1}{n}\sum_{j=1}^n1\right)^n=1. $$ By the equality case of AM-GM, we deduce that $t_j=1$ for every $j$, that is $T^*T$ is the identity operator, i.e. $T$ is unitary, since we are in finite dimension. $\Box$


Note: actually, the first proof I thought about is the following, which is more interesting. It uses Hadamard's inequality. The only thing I did above is that I reproduced a standard proof of that famous inequality in this special case.

Take $\{e_j\;;\;1\leq j\leq n\}$ an orthonormal basis of diagonalization for $T^*T$, with $T^*Te_j=t_je_j$. Now let $N$ be the matrix of $T$ in this basis. Then Hadamard's inequality says that $$ |\det N|\leq \prod_{j=1}^n\|v_j\| $$ where $v_j=Te_j$ is the $j$th column of $N$. By assumptions on $T$, we have $|\det N|=|\det T|=1$ and $\|v_j\|=\|Te_j\|\leq 1$ for every $j$. Therefore the inequality forces $\|Te_j\|=1$ for every $j$. Hence $t_j=(e_j,T^*Te_j)=\|Te_j\|^2=1$ for every $j$. Thus $T^*T$ is the identity operator.

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  • $\begingroup$ You're right, I should have made it clear that I was referring to a finite-dimensional inner-product space. Thank you so much! One minor point though, shouldn't there be an absolute value on $\det{T}$ in order for it to be equal to 1? (Just makes me wonder if we need to carry it on the inequality chain you just wrote. Also, thank you for introducing Hadamard's inequality, I didn't know about it! Seems useful. $\endgroup$
    – Adar Hefer
    Aug 3, 2013 at 8:59
  • $\begingroup$ Last question would be: do you think there's an way to discern that $trace(T^*T)=n$ without resorting to those measures? (I realize my way is very roundabout...) $\endgroup$
    – Adar Hefer
    Aug 3, 2013 at 9:01
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    $\begingroup$ @AdarHefer Yes, I meant $|\det T|=1$. By the AM-GM inequality, in general, we have $|\det T|^\frac{2}{n}\leq \frac{1}{n}\mbox{tr}(T^*T)$. I could not think of a more direct way to prove that $\mbox{tr}(T^*T)=n$. If you want to avoid this convexity argument, this could also be done, like Hadamard's inequality, with Gram-Schmidt. $\endgroup$
    – Julien
    Aug 3, 2013 at 14:11
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Is this a result for finite-dimensional spaces? If $\sigma$ is the left-shift on $l^2(\mathbb{Z})$, then $T = \frac{1}{2}\sigma$ has no eigenvalues, trivially satisfying the condition that all eigenvalues are on the unit circle. The norm of $T$ is 1/2, so it also satisfies the other requirement of $T$. But $T$ here is not unitary.

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The Schur decomposition gives that there exists an orthonormal basis $\{e_1, \ldots, e_n\}$ such that the matrix of $T$ is upper triangular: $$ \begin{bmatrix}t_{11} & t_{12} & \cdots & t_{1n}\\ 0 & t_{22} & \cdots & t_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & t_{nn} \end{bmatrix}$$

where $t_{11}, \ldots, t_{nn}$ are eigenvalues of $T$.

We have $$1 = \|e_2\|^2 \ge \|Te_2\|^2 = \|t_{12}e_1 + t_{22}e_2\|^2 = |t_{12}|^2 + |t_{22}|^2 = |t_{12}|^2 + 1$$ so $t_{12} = 0$.

We have $$1 = \|e_3\|^2 \ge \|Te_3\|^2 = \|t_{13}e_1 + t_{23}e_2 + t_{33}e_3\|^2 = |t_{13}|^2 + |t_{23}|^2 + |t_{33}|^2 = |t_{13}|^2 + |t_{23}|^2 + 1$$ so $t_{13} = t_{23} = 0$.

Continuing this we get that the matrix is diagonal, hence $T$ is unitary.

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