Let $A$ be an $n \times n$ matrix. Let $A_k$ be the matrix obtained by keeping the first $k$ rows of $A$ fixed and substituting $0$ for the rows $k+1$ to $n$. To be precise, we write $A= [R_1...R_k, R_{k+1}...R_n]'$ where each $R_i$ is an $1 \times n$ row vector, and $A_k=[R_1...R_k, 0, ...0]'$. Here ' denotes transpose.
I'd like to know what we can say about the non-zero eigenvalues of $A_k$, from the non-zero eigenvalues of $A$? I stated non-zero, because it's clear that $A_k$ will have zero as an eigenvalues with the corresponding eigenspace at least of dimension $n-k$. When $A$ is diagonal, we see that, when $A=diag(\lambda_1,...\lambda_k, \lambda_{k+1},...\lambda_n), A_k=diag(\lambda_1,...\lambda_k,0,...0)$. So in this case, its pretty clear, but what about the general case for $A$?
P.S. You can assume, if necessary, that $A$ is symmetric and non-negative definite, because that's all what concerns me for applications, where my $A$ is a covariance matrix.
P.P.S. I think expecting some kind of equalities would be too much to expect, so I'm hoping that with the assumption of $A$ to be symmetric and non-negative definite, we can have:
$$ trace(A_k)\leq {{\sum}}_{i=1}^{k} \lambda_i(A) $$,
where the non-negative eigenvalues of $A$ satisfy or are arranged in decreasing order: $\lambda_1(A) \geq \lambda_1(A) \geq \lambda_2(A)...\geq \lambda_n(A)\geq 0$. Is this true?
