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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?

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  • $\begingroup$ $P$ 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

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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.

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