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?

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    $\begingroup$ If you zero out some rows of $A$, you get (up to permutation) $\begin{bmatrix}A_{11} & A_{12} \\ 0 & 0\end{bmatrix}$, which has the same nonzero eigenvalues as the principal submatrix $A_{11}$. This is a classical setup; there are interlacing inequalities between a matrix and its principal submatrix; see for instance math.stackexchange.com/questions/1670000/… . $\endgroup$ – Federico Poloni Dec 31 '19 at 16:36
  • $\begingroup$ @FedericoPoloni Thanks for your comment. Perhaps I'm not aware of the relevant results: could you please provide some relevant references? P.S. I'm aware of Wey's interlacing inequality - en.wikipedia.org/wiki/Weyl%27s_inequality, which bounds the eigenvalues of the sum of two matrices, but not sure how that applies to my case? Also, I'm not taking principal minor when I'm building $A_k$ out of $A$, I'm dropping a few rows (not the corresponding columns as well). How to prove that the principal minor and $A_k$ have same egenvalues? I'd appreciate a more detailed explanation. $\endgroup$ – Learning math Dec 31 '19 at 16:43
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    $\begingroup$ There is a reference in my link, which should explain the main result. "Interlacing inequalities" in my comment referred to the theorem in the answer there, which applies to submatrices, not to Weyl's theorem on $A+B$. The fact that those two matrices have the same eigenvalues is easy to prove; just use math.stackexchange.com/questions/522385/… on $A-xI$. $\endgroup$ – Federico Poloni Dec 31 '19 at 17:23
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    $\begingroup$ as you drop rows, you might consider left eigenvectors rather than right ones. eigenvalues are the same for the left and for the right ones. $\endgroup$ – Dima Pasechnik Jan 1 at 4:27

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