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?

  • $\begingroup$ Please see this question and its answer. It can help you! math.stackexchange.com/questions/16609/… $\endgroup$ Jul 31, 2015 at 17:24
  • $\begingroup$ @KhadijaMbarki it is about hermitian and unitary matrices ... how does that fit in here? $\endgroup$
    – user251257
    Jul 31, 2015 at 17:34
  • $\begingroup$ The matrix $U$ introduced in the question is normal! $\endgroup$ Jul 31, 2015 at 18:09
  • 1
    $\begingroup$ Basically, your question is about what can be said about eigenvalues of a square principal submatrix of a normal matrix. I do not think it has a simple answer, at least it has been spent efforts on that during last half a century, and the best I could find is estimations of the Gershgorin type. $\endgroup$
    – A.Γ.
    Jul 31, 2015 at 19:25
  • 2
    $\begingroup$ @BrianFitzpatrick $U^*U=I$ gives isometry, or partial isometry, depending on the choice of acting spaces. $UU^*=I$ gives coisometry I guess. $\endgroup$
    – A.Γ.
    Jul 31, 2015 at 22:06

2 Answers 2


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.


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, $$ w_j=\langle w,e_j\rangle=\sum_{k=1}^r\alpha_k\langle v_k,e_j\rangle=\sum_{k=1}^r\alpha_k\,U_{k,j}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.