Interlacing for "Almost Hermitian" matrices I am wondering if there is something known about the interlacing properties of an "Almost Hermitian" matrix, in the following sense: let A be a nxn matrix so that it has a Hermitian principal minor of order (n-1)x(n-1). What interlacing properties does A possess?
Thanks!
 A: Assuming that the Hermitian minor has distinct eigenvalues, there is nothing we can say about the eigenvalues of $A$. Let the eigenvalues of the Hermitian minor be $\lambda_1$, ..., $\lambda_{n-1}$, so we can choose a matrix where your matrix looks like
$$A=\begin{pmatrix} 
\lambda_1 & & & & a_1 \\
& \lambda_2 & & & a_2 \\
& & \lambda_3 & & a_3 \\
& & & \ddots & & \\
b_1 & b_2 & b_3 & & c
\end{pmatrix}$$
The characteristic polynomial of $A$ is
$$f(x):=(x-c) \prod_{i} (x-\lambda_i) - \sum_j a_j b_j \prod_{i \neq j} (x-\lambda_i). \quad (\ast)$$
I claim that we can choose $c$ and $a_j b_j$ to make $f(x)$ be any monic degree $n$ polynomial. 
Proof:  We have $f(\lambda_j) = - a_j b_j \prod_{i \neq j} (\lambda_i - \lambda_j)$. So, assuming that the $\lambda$'s are distinct, we can choose $a_j b_j$ to make $f(\lambda_j)$ have any value. Also, we can use $c$ to fix the value of $f$ at any $x$ other than the $\lambda_i$. A monic polynomial of degree $n$ is determined by its values at $n$ points, so we can arrange for $f$ to be any degree $n$ monic polynomial. $\square$
Equation $(\ast)$ also shows that, if $\lambda_i$ occurs with multiplicity $d$ in the Hermitian minor, then it occurs with multiplicity $\geq d-1$ in $A$.
A: Assume that $a_j\neq 0$ for $j=1,\ldots,n-1$.  If the vectors $a$ and $b$ are multiples of each other, say $b = k a$, and all the $\lambda$'s are distinct then we get strict interlacing.  As before we have $f(\lambda_j) = -k a_j^2 \prod_{i\neq j}(\lambda_j-\lambda_i) \neq 0$.  Assuming that $\lambda_1 < \lambda_2 < \cdots < \lambda_{n-1}$ then $sign(f(\lambda_1)) = (-1)^{n-1}$, $sign(f(\lambda_2)) = (-1)^{n-2}$, $\ldots$, $sign(f(\lambda_{n-2})) = 1$, and finally $sign(f(\lambda_{n-1}))=-1$.  Hence, $f(\lambda_j)f(\lambda_{j+1})< 0$ for $j=1,\ldots,n-2$.  By the intermediate value theorem, $f$ has $(n-2)$ roots inside the intervals $(\lambda_j, \lambda_{j+1})$ for $j=1,\ldots,n-2$.  Because $f$ is a monic polynomial, it is unbounded as $x\rightarrow \pm\infty$ and thus $f$ has one root in $(-\infty, \lambda_1)$ and one in $(\lambda_{n-1},\infty)$.  Hence, if $\mu_1, \ldots, \mu_n$ denotes the roots of $A$ then 
\begin{equation*}
\mu_1 < \lambda_1 < \mu_2 < \cdots < \lambda_{n-1} < \mu_{n-1} < \lambda_n
\end{equation*}
Cases when some $a_j=0$ and/or some of the $\mu$'s are repeated can be dealt with similarly.
