Eigenvalues of Unitary Matrix Suppose $A \in \mathbb{C}^{n \times n}$, for all $\lambda \in \text{spec}(A)$, $|\lambda| \geq 1$, and for all 
$v \in \mathbb{C}^n$, $|(Av) \cdot v| \leq ||v||^2$. Prove $A$ is unitary.
This is what I have so far. Let $\lambda \in \text{spec}(A)$ and $v$ be its corresponding eigenvector. We have that 
$$Av = \lambda v$$
Then
$$||v||^2 \geq |(Av) \cdot v| = |\lambda v \cdot v| = |\lambda| ||v||^2 \quad \Rightarrow \quad 1 \geq |\lambda|$$
Thus we conclude that $|\lambda| = 1$ for all $\lambda \in \text{spec}(A)$.
 A: As you pointed out, all eigenvalues lie on the unit circle.
We wish to verify two more things: (1) all Jordan blocks are diagonal, and (2) eigenspaces associated to different eigenvalues of $A$ are orthogonal.
We can also assume that $1$ is an eigenvalue of $A$, by rescaling $A$.
Let us assume that one of the conditions above is not satisfied.
(1) In this case, we can find two unit vectors $u$ and $w$, such that
$$
Au = u, \quad \quad Aw= w+u \ .
$$
Note that, by Cauchy-Schwartz, $|u\cdot w| < 1$. For  $v = u + w$, we have
\begin{align*}
|Av \cdot v| = |(u+ ( w +  u))\cdot (u+ w)| = | |v|^2 + (u\cdot u) + (u\cdot w)| = ||v|^2 + 1 +(u\cdot v)| > |v^2| \ .
\end{align*}
This is contradiction, so condition (1) is satisfied.
(2) In this case, there are two unit vectors $u$ and $w$, such that
$$
Au = u, \quad \quad Aw = \lambda w, \quad |\lambda| = 1, \lambda \neq 1 \ .
$$
Write $\alpha =  (u\cdot w)$. Again, by Cauchy-Schwartz and by the negation of (2), it is $0 < |\alpha| < 1$. For $v= u-\frac{1}{2} \alpha w$, we have
\begin{align*}
|Av \cdot v| &= |(u -\frac{1}{2}\lambda \alpha  w -\frac{1}{2}\alpha w +\frac{1}{2} \alpha w) \cdot  (u -\frac{1}{2} \alpha w)| = ||v|^2 + (\frac{1}{2} \alpha w -\frac{1}{2} \lambda \alpha w)\cdot (u -\frac{1}{2} \alpha w)| \\
&=||v|^2 + \frac{1}{2} \alpha(1-\lambda) [(w\cdot u) - \frac{1}{2} \overline{\alpha}(w\cdot w)]| = 
||v|^2 +\frac{1}{2} \alpha(1-\lambda ) [\overline{\alpha} -\frac{1}{2} \overline{\alpha}]| \\
&=||v|^2 +\frac{1}{4} \alpha\overline{\alpha}(1-\lambda )| >|v|^2 
\end{align*}
This is also a contradiction, so condition (2) is satisfied.
In conclusion, this proves that $A$ is diagonalizable by an orthogonal matrix and all of its eigenvalues have norm $1$, hence it must be unitary.
