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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 $$C=\begin{bmatrix}A &B\\B^T &0\end{bmatrix}$$ and $$D=\begin{bmatrix}A &v\\v^T &0\end{bmatrix}$$ Let $P$ be any $n$ order principal submatrix of $D$. Suppose $\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_{d}$ be first non zero eigenvalues of $C$ 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 $i=1,2,\ldots,d$? (By interlacing property first inequality is obvious.)

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?

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 $$C=\begin{bmatrix}A &B\\B^T &0\end{bmatrix}$$ and $$D=\begin{bmatrix}A &v\\v^T &0\end{bmatrix}$$ Let $P$ be any $n$ order principal submatrix of $D$. Suppose $\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_{n+4}$ be eigenvalues of $C$ and $\mu_1\ge \mu_2 \ge \ldots \ge \mu_p$ be first non zero eigenvalues of $P$.

Is it true that $\lambda_i\ge\mu_i\ge\lambda_{i+1}$ for $i=1,2,\ldots,p$? (By interlacing property first inequality is obvious.)

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 $$C=\begin{bmatrix}A &B\\B^T &0\end{bmatrix}$$ and $$D=\begin{bmatrix}A &v\\v^T &0\end{bmatrix}$$ Let $P$ be any $n$ order principal submatrix of $D$. Suppose $\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_{d}$ be first non zero eigenvalues of $C$ 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 $i=1,2,\ldots,d$? (By interlacing property first inequality is obvious.)

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 column vector of dimension $n$. Consider $$C=\begin{bmatrix}A &B\\B^T &0\end{bmatrix}$$ and $$D=\begin{bmatrix}A &v\\v^T &0\end{bmatrix}$$ Let $P$ be any $n$ order principal submatrix of $D$. Suppose $\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_{n+4}$ be eigenvalues of $C$ and $\mu_1\ge \mu_2 \ge \ldots \ge \mu_p$ be first non zero eigenvalues of $P$.

Is it true that $\lambda_i\ge\mu_i\ge\lambda_{i+1}$ for $i=1,2,\ldots,p$? (By interlacing property first inequality is obvious.)

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 $$C=\begin{bmatrix}A &B\\B^T &0\end{bmatrix}$$ and $$D=\begin{bmatrix}A &v\\v^T &0\end{bmatrix}$$ Let $P$ be any $n$ order principal submatrix of $D$. Suppose $\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_{n+4}$ be eigenvalues of $C$ and $\mu_1\ge \mu_2 \ge \ldots \ge \mu_p$ be first non zero eigenvalues of $P$.

Is it true that $\lambda_i\ge\mu_i\ge\lambda_{i+1}$ for $i=1,2,\ldots,p$? (By interlacing property first inequality is obvious.)