Prove that linear operator is unitary This is my question:
Let $T: V\rightarrow V$ be a linear operator, in unitary space.
Assume that:
a) $\left|\lambda\right|=1$ for each eigenvaule of $T$.
b) $\left|Ta\right|\leq\left|a\right|\,\,\forall a\in V$.
Prove that $T$ is unitary operator.
My thoughts:
Its clear to me that if I will be able to prove that $T$ is diagonalization, then $T$ is unitary.
I also tried to prove by contradiction that there is $a\in V\,$such that $\left|Ta\right|<\left|a\right|$, but It didn't work.
 A: As @Ben Grossmann mentioned in the comment, by Schur decomposition theorem, there exists a unitary matrix $P$ such that (here for simplicity, I am just treating $T$ as a matrix, so our goal is to show $T$ is a unitary matrix):
\begin{align*}
T = P\begin{pmatrix}
e^{i\theta_1} & t_{12} & \cdots & t_{1n} \\
0 & e^{i\theta_2} & \cdots & t_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & e^{i\theta_n}
\end{pmatrix}P^*,
\end{align*}
that is, $T$ is unitary similar to an upper triangular matrix whose diagonals are all of modulus $1$ (by assumption a)). For $2 \leq j \leq n$, take $a = Pe_j$, where $e_j$ is the column vector with $j$-th position $1$ and all other entries $0$, then by assumption b) and the unitarity of $P$, we have
\begin{align*}
|Ta| = |P(t_{1j}, \ldots, t_{j - 1, j}, e^{i\theta_j}, 0, \ldots, 0)^T| = |t_{1j}|^2 + \cdots + |t_{j - 1, j}|^2 + 1 \leq |a| = 1,
\end{align*}
which implies $t_{1j} = \cdots = t_{j - 1, j} = 0$. Therefore, the upper triangular matrix is in fact diagonal. This completes the proof.
A: supposing this is finite dimensional and OP knows either Polar Decomposition or SVD, we have
$\Big\vert\det\big(T\big)\Big\vert = 1$ by (a)  and
$\Big\Vert T\Big\Vert_2 \leq 1$ by (b), which implies all singular values $\sigma_k\leq 1$, so
$1 = \Big\vert\det\big(T\big)\Big\vert=\sigma_1\cdot\sigma_2\cdot ...\cdot \sigma_n\leq 1  $
$\implies \sigma_k = 1$ for all $k$
thus the matrix representation $A$ is written as
$A=UP=UI = U$ which is unitary (Polar Decomposition)
$A=U'\Sigma V^*=U'IV^* = U'V^*$ which is unitary (SVD)
