Is $T^T = I$ if $T^ T$ is unitary?

Question

Let $\mathcal{H}$ be a complex Hilbert space which has at most countable basis, and $T: \mathcal{H} \to \mathcal{H}$ be a bounded linear operator. I am trying to prove that $T^* T = I$ if $T^* T$ is unitary. (Here $T^*$ denotes the adjoint of $T$ and $I$ is an identity map.) Letting $U:=T^* T$, I found out this is true if we assume that $U$ is diagonalizable. Indeed let $\{ \psi_k \}_{k=1}^N$ be a basis of $\mathcal{H}$, where $N \in \mathbb{N} \cup \{\infty\}$ , such that $U\psi_k = \lambda_k \psi_k \,\, (\lambda_k \in \mathbb{C})$ for all $k$. Since $U$ is unitary, $| \lambda_k | = 1$ for all $k$. In addition,$$\lambda_k = \lambda_k ( \psi_k, \psi_k ) = ( U\psi_k, \psi_k) = (T\psi_k, T\psi_k) \geq 0$$ hence we have $\lambda_k = 1$, which shows that U is an identity. Clearly this method cannnot be applied to the case in which $U$ is not diagonalizable. I would like to have any advice for the general case. Thank you in advance. (This is my first post and I apologize for any inconvenience.)

score 2 · Answer 1 · answered Apr 20 '17 at 11:22

2

The spectrum of a unitary is contained in $S^1 \subset \mathbb C$ and since $T^*T$ is positive, its spectrum is contained in $[0,\infty)$. Hence, if $T^*T$ is unitary, it must be the identity.

answered Apr 20 '17 at 11:22

score 1 · Accepted Answer · answered Apr 20 '17 at 11:21

1

answered Apr 20 '17 at 11:21

Exodd

10,844

Is $T^*T = I$ if $T^* T$ is unitary?

2 Answers2

Is $T^T = I$ if $T^ T$ is unitary?