How the spectrum changes after removing a line from a symmetric positive definite the matrix?

Question

I have a symmetric positive definite matrix $A$ and a diagonal matrix $P$ with $1$ or $0$ on the diagonal. Knowing the spectrum of $A$, there exists some result that helps in estimation of the spectrum of $PA$ (or $AP$)? Thanks

Answer 1

If the matrices $A$ and $P$ are commutable it holds that the product of their spectra equals the spectrum of their product.

Answer 2

We note that the eigenvalues of $PA$ are given by the poles of $[z-PA]^{-1}$ . Note further that $P$ is an orthogonal projection operator. Thus we can use the Feshbach projection formula ($Q=1-P$ ) \begin{eqnarray*} \lbrack z-H]^{-1} &=&[z-QHQ]^{-1}Q+\{P+[z-QHQ]^{-1}QHPR_{P}(z)\{P+PHQ[z-QHQ]^{-1}\} \\ R_{P}(z) &=&P[z-H]^{-1}P=[z-PHP-PHQ[z-QHQ]^{-1}QHP]^{-1}P \end{eqnarray*} to arrive at \begin{eqnarray*} \lbrack z-PA]^{-1} &=&[z-QPAQ]^{-1}Q+\{P+[z-QPAQ]^{-1}QPAPR_{P}(z)\{P+PPAQ[z-QPAQ]^{-1}\} \\ &=&\frac{1}{z}Q+PR_{P}(z)\{P+\frac{1}{z}PAQ\} \\ R_{P}(z) &=&P[z-PA]^{-1}P=[z-PAP-PPAQ[z-QPAQ]^{-1}QPAP]^{-1}P=[z-PAP]^{-1} \end{eqnarray*} Hence \begin{equation*} \lbrack z-PA]^{-1}=\frac{1}{z}Q+P[z-PAP]^{-1}\{P+\frac{1}{z}PAQ\} \end{equation*} and its singularities are $z=0$ and the poles of $[z-PAP]^{-1}$, i.e., the eigenvalues of $PAP$. From this, further expressions for the eigenprojectors of $PA$ can be obtained, see Kato, "Perturbation theory for linear operators ", Chapter 1 on finite dimensional vector spaces.

Answer 3

I found the answer. First, we have some eigenvalues $\lambda=0$ since $PA$ is singular. Suppose now $\lambda\neq 0$ and $v$ its associated eigenvector. So $\lambda v=PAv$. At a line $i$ where $P_{ii}=0$ we obtain $\lambda v_i=0$, meaning $v_i=0$. So $Pv=v$ and $\lambda v=PAPv$, thus $(\lambda,v)$ is a eigen pair of $PAP$ as well. Similarly if $(\lambda,v)$ is an eigen pair of $PAP$ it is for $PA$.