$U(n)$ is a Maximal Compact Subgroup

Question

I am practicing for my Lie algebra exam and came across this problem which is split into several parts: I have done 1

1. The first is to show that $U(n)$ is compact which i did.
2. Let $Q \in End(n,\mathbb{C})$ be positively defined hermitian and $C$ a constant such that $\forall k \in \mathbb{Z} , ||Q^k|| \leq C$. Show that $Q=I$.
3. Show that $U(n)$ is a maximal compact subgroup.

I have done 1, but I have no idea how to start 2 and I am not sure about my proof for 3.

Some useful information:

We are using the usual subordinate norm of $GL(n,\mathbb{C})$ of $\mathbb{C^n}$ with the hermitian product and its accompanying norm:

$<(z_1,...,z_n),(z^{'}_{1},...,z^{'}_{n})> = \sum_{i=1}^{n} z_{i}\overline{z}_{i}^{'}$

$<Q^{k-1}(x),Q^k(x)> > 0$ and $||Q^k|| = supp\{\frac{||Q^k(x)||}{||x||}| x\in\mathbb{C^n}\}\leq C$

I am having trouble with this question any suggestions are appreciated.

My idea of a proof for 3:

We are doing this to prove that $U(n)$ in $GL(n,\mathbb{C})$ is a maximal compact subgroup. The idea is that if $K$ is a compact subgroup of $GL(n,\mathbb{C})$ containing $U(n)$, then $K = U(n)$ by using its polar decomposition and the fact that $GL(n,\mathbb{C})$ it is a Banach space and

$K \simeq U(n)\times D$ where $D\subset HDP(n)$. Since $K$ is compact it must be bounded. The product norm is defined as $||(x,y)|| = max\{||x||,||y||\}$, which implies that every element of $D$ must be bounded like we have above, hence, $D = \{I_n\}$ and $K \simeq U(n)\times \{I_n\} \simeq U(n)$.

Does this proof work?

Answer 1

For 2) remark that such a matrix $A$ is diagonalizable and its eigenvalues are positive real, if $c$ is an eigenvalue distinct of $1$ such that $A(x)=cx, \|x\|=1$, if $c>1$, $lim_{k\in \mathbb{N},k\rightarrow+\infty}\|A^k(x)\|=lim_kc^k=+\infty$. If $c<1$, $lim_{k\in\mathbb{N},k\rightarrow+\infty}\|A^{-k}(x)\|=lim_kc^{-k}=+\infty$, we deduce that the eigenvalues are $1$ and $A=I$.

For $3$, if $K$ is a compact subgroup, take an Hermitian product $b$ and a Haar measure on $K$, $\langle x,y\rangle =\int_Kb(kx,ky)$ is an Hermitian product invariant by $K$, this implies that $K$ is contained in a subgroup conjugated of $U(n)$.