Properties of matrices $M=UDU^*$ with $UU^*=Id$ I recently came across some matrices of the form $M=UDU^*$ (the superscript $*$ denotes the conjugate transpose), where $U \in \mathbb{C}^{r\times n}$ with $r<n$, $D \in \mathbb{C}^{n \times n}$ a diagonal matrix, and $UU^*=\text{Id} \in \mathbb{C}^{r \times r}$ the identity matrix.
(Note that $U$ is rectangular, so the last condition is not that $U$ is unitary).
I am interested in the eigenvalues of $M$, in particular how they are related to the eigenvalues of the matrix $D$. 
Any hints?
 A: Let $U=PSQ^\ast$ be a singular value decomposition. Since $U$ has orthonormal rows, $S=(I_r,0_{r\times(n-r)})$. Therefore $UDU^\ast$ is unitarily equivalent to the leading principal $r\times r$ submatrix of $Q^\ast DQ$. In other words, you are essentially asking about the relationship between the eigenvalues of an arbitrary normal matrix and the eigenvalues of its principal submatrix.
It is well-known (pls consult any reference book on advanced linear algebra) that


*

*The spectrum of any complex square matrix lies inside the matrix's numerical range.

*The numerical range of any complex square is a superset of all numerical ranges of its principal submatrices.

*The numerical range of a normal matrix is precisely the convex hull of the eigenvalues.


Therefore, the eigenvalues of $U^\ast DU$ are convex combinations of the eigenvalues of $D$.
If $D$ happens to be real, there is a more refined relationship between the eigenvalues of a Hermitian matrix with the eigenvalues of its principal submatrix, namely, Cauchy's interlacing inequality.
A: The eigenvalues of $M$ are convex combinations of the eigenvalues (i.e. the diagonal entries) of $D$. 
To see this, note that $P=U^*U$ is a projection, since $P^2=U^*UU^*U=U^*U=P$. Concretely, it is the projection onto the subspace spanned by the rows of $U$. 
Since the eigenvalues of $AB$ are the same as the eigenvalues of $BA$, the eigenvalues of $M=UDU^*$ are the same as the eigenvalues of $DU^*U=DP$, which are also the eigenvalues of $(DP)P$ and so are also the eigenvalues of $PDP$. 
The case of eigenvalue zero is trivial: if $M$ has zero as eigenvalue, then at least one $d_k=0$, and $0=1\,d_k$ is a convex combination. 
Now suppose that $w$ is a nonzero unit eigenvector of $PDP$ with nonzero eigenvalue $\lambda$. So $PDPw=\lambda w$; multiplying on the left by $P$, we get $PDP=\lambda Pw$. Note that $$Pw=\lambda^{-1}PDPw=w.$$ Now
$$
\lambda=\langle\lambda w,w\rangle=\langle PDPw,w\rangle=\langle DPw,Pw\rangle=\langle Dw,w\rangle.
$$
If we write $d_1,\ldots,d_n$ for the diagonal elements of $d$, we have
$$\tag{1}
\lambda=\langle Dw,w\rangle=\sum_{j=1}^n\,|w_j|^2\,d_j.
$$
As $w $ is a unit vector, the numbers $|w_1|^2,\ldots,|w_n|^2$ are convex coefficients. 
Finally, in $(1)$ we can be a little more specific about the numbers $w_j$. We saw above that $Pw=w$; this means that $w$ is in the span of the rows of $U$, say $v_1,\ldots,v_r$. As these are orthonormal and $w$ is a unit vector, we have $w=\sum_{k=1}^r\alpha_k\,v_k$, with $|\alpha_1|^2,\ldots,|\alpha_r|^2$ convex coefficients. Then, writing $e_1,\ldots,e_n$ for the canonical basis, 
$$
w_j=\langle w,e_j\rangle=\sum_{k=1}^r\alpha_k\langle v_k,e_j\rangle=\sum_{k=1}^r\alpha_k\,U_{k,j}.
$$
