TITLE: Properties of matrices $M=UDU^*$ with $UU^*=Id$
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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.