If $B = UD{U^*}$ is real matrix and $D$ is diagonal matrix whit ${d_i} \ne {d_j}$. then $U=VW$ Let $B\in M_n$


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*$B = UD{U^*}$ is real matrix and $D$ is diagonal matrix whit $d_i$ is real positive and ${d_i} \ne {d_j}$.for $i,j=1,2, ...., n$.

*$U$ is unitary matrix


Can we prove that $U=VW$ where $V$ is real orthogonal and $W$ is a diagonal unitary matrix?
 A: Yes. Let $u_1, \ldots, u_n$ denote the columns of $U$ so that $Bu_i = d_i u_i$ and $u_i \in \mathbb{C}^n$. Since the eigenvalues are distinct, we have $1 = \dim \mathrm{ker}_{\mathbb{C}} (B - d_iI) = n - \mathrm{rank}_{\mathbb{C}} (B - d_iI)$. Since $B$ and $d_i$ are real, we can compute the rank of $B - d_iI$ over $\mathbb{R}$ and get the same answer, so $\mathrm{rank}_{\mathbb{R}} (B - d_iI) = n - 1$ and $\dim \mathrm{ker}_{\mathbb{R}} (B - d_iI) = 1$. Choose unit-length elements $v_i \in \mathrm{ker}_{\mathbb{R}} (B - d_iI)$. They will be real eigenvectors of $B$ with eigenvalues $d_i$ and in particular complex eigenvectors of $B$. Since $1 = \dim \mathrm{ker}_{\mathbb{C}} (B - d_iI)$, we must have $u_i = w_i \cdot v_i$ for some $0 \neq w_i \in \mathbb{C}$. Finally,
$$ 1 = \left<u_i, u_i\right> = u_i^{*} u_i = \overline{w}_i v_i^t w_i v_i = |w_i|^2 \left<v_i, v_i\right> = |w_i|^2 $$
so $|w_i| = 1$. Letting $W = \mathrm{diag}(w_1, \ldots, w_n)$ you have $U = VW$.
A: The answer to this one is yes!
In particular, note by the real spectral theorem that there is a (real) orthogonal $V$ such that
$$
UDU^* = B = VDV^*
$$
Now, we have
$$
UDU^* = VDV^* \implies\\
(V^*U)D(V^*U)^* = D \implies\\
(V^*U)D = D(V^*U)
$$
Which is to say that $V^*U$ commutes with $D$.  Verify that this can only occur if $V^*U$ is diagonal (of particular importance is the fact that the $d_i$ are distinct). 
So, setting $W = V^*U$, we have $U = VW$, as desired.
