Decomposition of a normal endomorphism into an unitary and a self-adjoint endomorphism $\phi : V \to V$ is an endomorphism and V is an unitary vector space.
I have to show:
$\phi $ is a normal endomorphism $\iff$ $\phi = \alpha \circ \beta$ and $ \beta \circ \alpha = \alpha \circ \beta$ with $\alpha$ is an unitary endomorphism and $\beta$ is a self-adjoint endomorphism.
I have already proofed one direction ("$\Leftarrow$") but I am stuck on the other direction. 
How do I proof that this decomposition exists?
 A: Suppose that $\phi$ is normal. Then, by the finite-dimensional spectral theorem, you can decompose $V$ as a direct sum
$$
 V = \oplus_{j=1}^N \ker(\phi - \lambda_j I)
$$
of pairwise orthogonal eigenspaces, where $\{\lambda_1,\dotsc,\lambda_N\}$ are the distinct eigenvalues of $\phi$, and hence you can decompose $\phi$ itself as
$$
 \phi = \sum_{j=1}^N \lambda_j P_j,
$$
where for each $j$, $P_j = P_j^2 = P_j^\ast$ is the orthogonal projection onto the eigenspace $\ker(A-\lambda_j I)$ corresponding to the eigenvalue $\lambda_j$, and where the orthogonal projections $\{P_1,\dotsc,P_N\}$ satisfy
$$
 \sum_{j=1}^N P_j = I, \quad P_j P_k = \begin{cases} P_j &\text{if $j=k$,}\\ 0 &\text{if $j\neq k$}.\end{cases}
$$
At last, let
$$
 \alpha = \sum_{j=1}^N \mu_j P_j, \quad \mu_j := \begin{cases} \frac{\lambda_j}{|\lambda_j|} &\text{if $\lambda_j \neq 0$,}\\ 1 &\text{if $\lambda_j=0$,} \end{cases}
$$
and let
$$
 \beta = \sum_{j=1}^N |\lambda_j| P_j.
$$
You can now use the properties of the orthogonal projections $\{P_1,\dotsc,P_N\}$ to conclude that this choice of $\alpha$ and $\beta$ does the job. By the way, this decomposition is well-known in functional analysis as the polar decomposition. In fact, you can even set up polar decompositions when $\phi$ isn't normal, but now you'll need a bit more care, using the spectral theorem as applied to $\phi^\ast \phi$ or $\phi \phi^\ast$.
