If $A\in M_n$ is Hermitian, $a\in \mathbf{R}$, and $y\in \mathbf{C}^n$. Let $\hat{A}=\left[\begin{array}{cc}A&y\\y^\ast&a\end{array}\right]$. Then what's the possible value of rank $\hat{A}$-rank $A$?
It's easy to see that the eigenvalues of $A$ and $\hat{A}$ are interlacing. But what's the relationship between the interlacing eigenvalues and rank?
My approach: WLOG, let eigenvalues of $A$ to be $\{0,\lambda_2,\ldots,\lambda_{n-1}\}$, $B$ to be $\{\mu_1,\mu_2,\ldots,\mu_{n}\}$, and $\lambda_i\neq 0$ ,then we have
$$\begin{array}{cccccc}\text{The eigenvalues of }A:&&0&\lambda_1&\lambda_2&\cdots\\ \text{The eigenvalues of }\hat{A}: &\mu_1&\mu_2&\mu_3&\mu_4&\cdots \end{array}$$ Since the eigenvalues are interlacing, $\mu_1\leq 0$ and $\mu_2\geq 0$, if $\mu_1=0$ and $\mu_2=0$, rank $\hat{A}$-rank $A=0$. If $\mu_1<0$ and $\mu_2=0$, rank $\hat{A}$-rank $A=1$, if $\mu_1=0$ and $\mu_2>0$, rank $\hat{A}$-rank $A=1$ and finally, if $\mu_1<0$ and $\mu_2>0$, rank $\hat{A}$-rank $A=2$. Is this a valid solution?
Adding a row or column to a matrix either leaves the rank the same or inreases the rank by 1. So the difference of ranks cannot exceed 2. $$\text {first example }\begin{bmatrix}1&2&0\\2&4&0\\0&0&0\end{bmatrix} \text{2x2 and 3x3 matrices both have rank 1}$$ $$\text {second example }\begin{bmatrix}1&2&1\\2&5&1\\1&1&1\end{bmatrix} \text{2x2 has matrix rank2, 3x3 matrix has rank 3} $$ $$\text {third example }\begin{bmatrix}1&2&1\\2&4&1\\1&1&1\end{bmatrix} \text{2x2 matrix has rank 1, 3x3 matrix has rank 3}$$ Thus the difference of ranks can be 0, 1 or 2.