Show for isometry $T:V \rightarrow V$, $\frac{1}{n}\sum_{i=0}^{n-1} T^i$ converges to the orthogonal projection on $ker(T-I)$. The question is: If $V$ is finite dimensional vector space with an inner product, and $T:V \rightarrow V$ is an isometry on $V$, then show that $Q_n = \frac{1}{n}\sum_{i=0}^{n-1} T^i$ converges, as $n \rightarrow \infty$, to the linear map $P:V \rightarrow V$ that is the orthognoal projection on $ker(T-I)$.
I'm struggling with this homework problem. Things I have done:
since $\forall x\in V. \ \ |T^i x| = |T^{i-1}x| = \ \ ... \ \ = |Tx|=|x|$,
$\forall x\in V. \ \ \lim_{n\rightarrow \infty} |Q_n(x)| =  \frac{1}{n}|\sum_{i=0}^{n-1} T^ix |\leq \frac{1}{n}\sum_{i=0}^{n-1} |T^ix|=\lim_{n\rightarrow \infty}\frac{1}{n}|\sum_{i=0}^{n-1} |x| =|x|$ is bounded by convergent sequence. Thus $Q_n(x)$ is convergent. And then for each $x \in V$, let $x = x_1 + x_2$ where $x_1 \in ker(T-I)$ and $x_2 \in ker(T-I)^{\perp} $. Then $Q_n(x) = Q_n(x_1) +Q_n(x_2) = x_1 + Q_n(x_2)$. 
Then I'm trying to prove that $\sum_{i=0}^{n-1} T^i(x_2)$ is bounded for all $n$ so that $\frac{1}{n}\lim_{n\rightarrow \infty}\sum_{i=0}^{n-1} T^i(x_2)$ tends to $0$ as $n\rightarrow \infty$. But right now I have no idea how to do so. Any help would be appreciated.
Thanks.
 A: For each $x \in V$, write $x = y + z$ where $y \in \ker(T-I)$ and $z \perp \ker(T-I)$. Then


*

*Since $Ty = y$, we get $Q_n(y) = y$.

*Note that $z \in \ker(T-I)^{\perp} = \operatorname{im}(T-I)^*=\operatorname{im}(T^{-1}-I)$. In the last step, we utilized the fact that $T$ is invertible and $T^* = T^{-1}$. So there exists $w \in V$ such that $z = (T^{-1}-I)w$. Then
$$ \| Q_n(z) \| = \left\| \frac{1}{n}(T^{-1}w - T^{n-1}w) \right\| \leq \frac{\|T^{-1}w\| + \|w\|}{n} \xrightarrow[n\to\infty]{} 0 $$
and so, $Q_n(z) \to 0$.
Combining altogether, it follows that $Q_n(x) \to y$ as $n\to\infty$. Since $y$ is the orthogonal projection of $x$ onto $\ker(T-I)$, we are done.
A: To complement the existing (and more elegant) answer, here's what I had in mind following the spectral theorem.
Because $T$ is an isometry, all eigenvalues of $T$ satisfy $|\lambda| = 1$.  Because $T$ is normal, it has an orthonormal basis of eigenvectors.  Let $P_n = \frac 1n \sum_{i=0}^{n-1} T^i$.
First, we confirm that for any eigenvector $x$ associated with $\lambda = 1$ (i.e. any $x$ in $\ker (T-I)$), we find that $P_n x = x$ for all $n$, so that clearly $P_n x \to x$.
Next, for any eigenvector $y$ associated with $\lambda \neq 1$, we find that
$$
P_n y = \frac 1n \sum_{i=0}^{n-1} T^i y = \left(\frac 1n \sum_{i=0}^{n-1} \lambda_i\right) y = \frac 1n \cdot \frac{1 - \lambda^n}{1 - \lambda} \cdot y.
$$
Noting that
$
\lim_{n \to \infty} \frac 1n \cdot \frac{1 - \lambda^n}{1 - \lambda} = 0,
$
we have $P_n y \to 0$ as $n \to \infty$.
Thus, $\lim_{n \to \infty}P_n$ behaves like the projection we are looking for on a basis of $V$.  Thus, $\lim_{n \to \infty}P_n$ is indeed the projection onto $\ker(T - I)$ as desired.
