Let $A$ be a real symmetric matrix of order $n$ and $B=\begin{bmatrix}v &v &v &v\end{bmatrix}$ where $v$ is a non zero real column vector of dimension $n$. Consider $$C=\begin{bmatrix}A &B\\B^T &0\end{bmatrix}$$ We know that $C$ will have atleast $3$ zero eigenvalues. Removing those $3$ zeros, let $\lambda_1\ge \lambda_2\ge\ldots\ge \lambda_{n+1}$ are remaining eigenvalues of $C$. $$D=\begin{bmatrix}A &v\\v^T &0\end{bmatrix}$$ Let $P$ be any $n$ order principal submatrix of $D$ and $\mu_1 \ge \mu_2 \ge \ldots \ge \mu_n$ be eigenvalues of $P$.

Is it true that $\lambda_i\ge\mu_i\ge\lambda_{i+1}$ for $i=1,2,\ldots,n$? (By interlacing property first inequality is obvious). Or if I hope for a counter example then on which lines I should start thinking?

  • $\begingroup$ $P$ is any order $n$ principal submatrix of $D$ $\ldots$ so if $n \leq 4$, could $P$ be the zero matrix? $\endgroup$
    – Nik Weaver
    Apr 26, 2016 at 16:09
  • $\begingroup$ Yes, It can be zero. In particular, take $A=0$ and $A$ is a $n$ order principal submatrix of $D$. I don't understand why you took $n\le4$ $\endgroup$
    – Sry
    Apr 27, 2016 at 7:07
  • $\begingroup$ I have edited the question to be more elaborated. Yesterday also I put those details in a comment but that comment is no more visible and I don't know why :( $\endgroup$
    – Sry
    Apr 27, 2016 at 7:24
  • $\begingroup$ Okay, I misread the problem. I don't know the answer, but did you notice that you could put $B = 2v$ without changing nonzero eigenvalues? So the question is about interlacing between $\left[\matrix{A& v\cr v^T&0}\right]$ and $\left[\matrix{A& 2v\cr 2v^T& 0}\right]$. $\endgroup$
    – Nik Weaver
    Apr 27, 2016 at 15:17

1 Answer 1


As I explained in the comments, the nonzero eigenvalues of $\left[\matrix{A&B\cr B^T&0}\right]$ and $\left[\matrix{A&2v\cr 2v^T&0}\right]$ are the same. So there's no reason you should expect interlacing between the eigenvalues of $C$ and the eigenvalues of a principal submatrix of $D$.

I went to an eigenvalue calculator and threw in a few values and quickly found a counterexample. Take $$A = \left[\matrix{10&0\cr 0&20}\right]\qquad v = \left[\matrix{1\cr 1}\right]$$ so that the nonzero eigenvalues of $C$ are the eigenvalues of $\left[\matrix{10&0&2\cr 0&20&2\cr 2&2&0}\right]$, which came out to $-.573$, $10.371$, and $20.201$. Whereas the eigenvalues of the principal submatrix $\left[\matrix{10&1\cr 1&0}\right]$ of $D$ are $-.099$ and $10.099$. So, no interlacing.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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