Bounded operator norms of powers implies identity This exercise comes from $\textit{Analysis on Lie Groups}$ by Faraut. It follows the first chapter concerning the linear group and various decompositions; namely, polar, gram, and orthogonal decompositions of $GL_n(\mathbb{R})$ and $O(n)$. This is a piece of a larger exercise aiming to prove the maximality of the orthogonal group as a compact subgroup of $GL_n(\mathbb{R})$.

Let $P$ be a positive definite symmetric matrix over $\mathbb{R}$ for which there exists some $C>0$ so that $$||{P^k}||\leq C\hspace{0.5in}\forall k\in\mathbb{Z}.$$Prove that $P=I$.

where $||\cdot||$ is operator norm on $M_n(\mathbb{R})$. I conjecture that the following result should be relevant:

Every $A\in GL_n(\mathbb{R})$ decomposes uniquely as the product $$A=RP$$ where $R\in O(n)$ and $P$ is a positive definite symmetric matrix. Moreover, the map $$O(n)\times \{\text{Positive definite}_n(\mathbb{R})\}\rightarrow GL_n(\mathbb{R}):(R,P)\mapsto RP$$ is a homeomorphism.

The tricky part for me is getting a rigidity statement out of the norm inequality; I am not really sure what observations on the operator norm of $P$ or its powers would force it to be the identity. I have used the decomposition to rewrite the hypothesis of the problem as a statement on merely a general linear matrix, but it's been a bit of a dead end. Any insight, thoughts would be helpful.
 A: Since $P$ is symmetric, there exists an orthogonal matrix $O$ such that
$O^TPO = D, \tag 1$
where $D$ is a diagonal matrix; the diagonal entries of $D$ are the eigenvalues of $P$, and every non-diagonal element of $D$ is $0$.  We may thus represent $D$ as
$D = \text{diag}(d_1, d_2, \ldots, d_n); \tag 2$
since $P$ is positive definite, $d_i > 0$, $1 \le i \le n$; indeed, if we choose $\vec e_i$ to be an eigenvector of $D$ corresponding to $d_i$, e.g., the components $(\vec e_i)_j$ are given by
$(\vec e_i)_j = \delta_{ij}, \tag 4$
whence
$D \vec e_i = d_i \vec e_i, \tag 5$
then
$d_i \Vert \vec e_i \Vert^2 = d_i \langle \vec e_i, \vec e_i \rangle= \langle \vec e_i, d_i \vec e_i \rangle = \langle \vec e_i, D \vec e_i \rangle = \langle \vec e_i, O^TPO \vec e_i \rangle = \langle O \vec e_i, PO \vec e_i \rangle > 0, \tag 6$
since $P$ is positive definite; this proves $d_i > 0$, $1 \le i \le n$.  Furthermore from (1) we obtain
$PO = IPO = OO^TPO = OD, \tag 7$
whence
$PO \vec e_i = OD \vec e_i = O(d_i \vec e_i) = d_i O \vec e_i, \tag 8$
and we see that $O \vec e_i$ is an eigenvector of $P$ associated with $d_i$. (8) yields
$P^2O \vec e_i = d_i PO \vec e_i = d_i^2 O \vec e_i, \tag 9$
$P^3O \vec e_i = d_i^2 PO \vec e_i = d_i^3 O \vec e_i, \tag{10}$
and in general for every integer $k > 0$,
$P^k O \vec e_i = d_i^{k - 1} PO \vec e_i = d_i^{k - 1} d_i O \vec e_i = d_i^k O \vec e_i, \tag{11}$
which follows from a simple induction the basis of which may be taken to be (8)-(10); indeed, from (11) we may write
$P^{k + 1} O \vec e_i = P d_i^k O \vec e_i = d_i^k PO \vec e_i = d_i^k d_i O \vec e_i = d_i^{k + 1} O \vec e_i, \tag{12}$
the essential step whicn establishes (11) for all $k > 0$.  Next, we obtain from (11) 
$\Vert P^k O \vec e_i \Vert^2 = \langle P^k O \vec e_i, P^k O \vec e_i \rangle =  \langle d_i^k O \vec e_i, d_i^k O \vec e_i \rangle = d_i^{2k} \langle O \vec e_i, O \vec e_i \rangle = d_i^{2k} \Vert O \vec e_i \Vert^2, \tag{13}$
whence
$\Vert P^k O \vec e_i \Vert = d_i^k \Vert O \vec e_i \Vert, \tag{14}$
from which it follows, based upon the definition of the operator norm, that
$\Vert P^k \Vert \ge d_i^k; \tag{15}$
now suppose $d_i > 1$ for some value of the index $i$; then it follows from (15) that for $k$ sufficiently large,
$\Vert P^k \Vert > C \tag{16}$
for any real $C$, clearly contradicting the hypothesis placed upon $P$.  Therefore 
$d_i \le 1, \; 1 \le i \le n. \tag{17}$
We may multiply (11) through by $P^{-k}d_i^{-k}$, $k > 0$:
$d_i^{-k} O \vec e_i = d_i^{-k}P^{-k} (P^k O \vec e_i) = d_i^{-k}P^{-k} (d_i^k O \vec e_i) = P^{-k}O \vec e_i, \tag{18}$
or
$P^{-k}O \vec e_i = d_i^{-k} O \vec e_i, \tag{19}$
and thus
$\Vert P^{-k}O \vec e_i \Vert = d_i^{-k} \Vert O \vec e_i \Vert; \tag{20}$
as (15) follows from (14), so it follows from (20) that
$\Vert P^{-k} \Vert \ge d_i^{-k}; \tag{21}$
if now $d_i < 1$, (21) forces
$\Vert P^{-k} \Vert \ge d_i^{-k} > C \tag{22}$
for any real $C$ provided $-k$ is sufficiently large negative.  This result also contradicts the assumptions imposed upon $P$, and hence we conclude that $d_i = 1$ for all $i$.  Then (2) becomes
$D = I, \tag{23}$
and then (1) yields
$P = IPI = (OO^T)P(OO^T) = O(O^TPO)O^T = ODO^T = OIO^T = I, \tag{24}$
the result that we seek.
A: I am posting a solution to my own question, not out of any perceived incorrectness of Robert Lewis' answer, but rather because a slicker approach occurred to me Re: the comment following the problem statement.
Let us use eigen-decomposition to write down the matrix $P$ as follows: $$P=O^TDO$$ for $O\in O(n)$, $D=\text{diag}(\lambda_1,...,\lambda_n)$ for eigenvalues $\lambda_i$. We show $\lambda_i\equiv 1$. Since $O$ is orthogonal, we have $$||P^k||=||O^TD^kO||=||D^k||=\max_{i}|\lambda_i|^k.$$
For $k>0$, we have $$\max_i|\lambda_i|^k=(\max_i|\lambda_i|)^k.$$Since this is bounded as $k\rightarrow\infty$ we must have $\max_i|\lambda_i|\leq 1$. For $k<0$, we have $$\max_i |\lambda_i|^k=(\min_i |\lambda_i|)^k.$$Since this is bounded as $k\rightarrow-\infty$, we must have $\min_i|\lambda_i|\geq 1$. In turn, $$1\leq\min_i|\lambda_i|\leq\max_i|\lambda_i|\leq 1$$forcing $\min_i|\lambda_i|=\max_i|\lambda_i|=1$. Since all eigenvalues of $P$ are positive, claim follows.
