On ``deleting a row and column'' from a differential operator (Q1307312)

Let \(A\) be a real symmetric matrix mapping \({\mathbb R}^n\) into itself and suppose that its eigenvalues are distinct. We enumerate them as \(\lambda_1<\lambda_2<\dots<\lambda_n\). Let \(P\) be the \(n\times n\) projection matrix got from the identity matrix \(I\) by replacing its \((n,n)^{\text{th}}\) element by zero and then let us form the matrix \(PAP\). If we write \(\mathbb R^n\) as the direct sum \({\mathbb R}^n=V_0\oplus E\) in which \(E\) is the one-dimensional subspace generated by the vector \([0,0,\dots,0,1]^T\) and \(V_0\) is its orthogonal complement, then the eigenvalues of the matrix operator \(PAP\), restricted as follows: \[ PAP:V_0\to V_0 \tag{1} \] are the same as those which we would have obtained by deleting the last row and column of the original matrix \(A\). We conclude then, by Cauchy's Interlace Theorem that if the eigenvalues of \(PAP\) operating as in (1) are denoted by \(\mu_k\) then \(\mu_k\in [\lambda_k,\lambda_{k+1}]\) \((k=1,2,\dots,n-1)\). We now expect this interlacing property to continue to hold when \(P\) is any projection matrix provided that \(PAP\) is taken to have, as its domain, the subspace on which \(P\) acts as the identity. The purpose of the present paper is to show that this result can even be generalized to the context in which \(A\) is an unbounded linear operator acting in a Hilbert space. Furthermore, a feature appears in this more general situation which goes unnoticed in the finite dimensional case because of its triviality there.