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. $$ This matrix is already in (complex) Schur form, and the obvious procedure to make the eigenvalues real and similar in spirit to what you propose would be to make the diagonal entries real (that is … (So the general message is that even eigenvalues that aren't close to the real axis might need only a small perturbation to get them there.) …

In view of Terry's comment on Michael's answer, it's perhaps worth pointing out that this monotonicity also fails if both domains are required to be convex. We can take $M_2=[0,L]^2$ as a square of si …

The question of whether we have finitely or infinitely many eigenvalues below zero is probably answerable, but seems a bit tricky. … EDIT 2: I'm reasonably confident now that there are only finitely many eigenvalues, though to show it properly would probably require some work. …

Write $P$ for the projection onto the $v_j$, and let $N$ denote the number of eigenvalues $\ge \delta/2$ of the matrix $PSP$. …

As a general rule of thumb, it's usually most convenient in one-dimensional problems to work with solutions of the ODE $-y''+Vy=Ey$ rather than operator theoretic methods. Here, everything follows fro …

There is a more general reason why any such statement will fail: If you consider a $P$ consisting of $N$ separate copies of the same basic region $P_0$, then $|P|=N|P_0|$, while everything else in you …

Update: I intended this to be a complete answer originally, but my "counterexample" was based on a miscalculation. So this is now more a collection of remarks on what I think the question is about. W …

To put this into context, the limit $$ \lim_{L\to\infty} \frac{\# \textrm{ eigenvalues in }I \textrm{ of the problem on } \{0,\ldots, L\} }{L} $$ (assuming it exists) is one way of defining the density …

We cannot have strict inequalities in all cases since you could have $B=0$. After this adjustment, we can obtain the claim as follows. Let me slightly change notations and consider $A(s)=A-sB$ (so $s …

The eigenvalues of $A$ are $\lambda(z)=T(z)/2 \pm (1/2)\sqrt{T^2(z)-4}$, $T(z)=\textrm{tr}\: A(z)$. …

Next attempt: By min-max, it's clear that the largest eigenvalue satisfies $\lambda_1=M_{11}n^{2(h_3-h_1)}(1+o(1))$ (test on $e_1$; nothing else in the matrix is as large as the $11$ entry, so this pr …

Since $B$ has rank $2$ and trace zero, this implies that the eigenvalues of $B$ are $\pm \sqrt{k(n-k)}$. Finally, we maximize this over $k$ to obtain the claim. …

Here's a slightly different (direct) method that gives similar conclusions: Since we have a double periodic eigenvalue at $\widehat{E}$ (let me in fact assume that $\widehat{E}=0$, for convenience), t …

No, this isn't working at all. Since $N,P$ commute (you would have to be more specific what exactly you mean by this since the operators are unbounded, but let's just assume we have the right version) …

Here, the $\mu_j$ are the eigenvalues of $CA+AC$, compressed to $L(v_1,\ldots, v_k)$. … In other words, they are the eigenvalues of the $k\times k$ matrix with entries $$ \langle v_j, (CA+AC) v_m\rangle = 2\lambda \langle v_j, Cv_m\rangle . $$ Now we only need to make sure that not all the …