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?



Modified interlacing of eigenvalues
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.


