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Start with the unweighted case. We have $L=BB^T$ where $B$ is (what I call) the incidence matrix of an orientation of $G$. So $B$ has one 1 and one $-1$ in each column with all other entries zero. The …

Since the eigenvalues are real, and since their sum is the trace of $A$, which is zero, we see that either all eigenvalues are zero, or there are both positive and negative eigenvalues. … For more regular graphs there are bounds on the size of cliques and cocliques that involve the least eigenvalues, and these can be rearranged to get upper bounds on the least eigenvalue. …

Let $z$ be the vector with all entries equal to 1. If $Az=0$ and $B=A+A^T$, then $z^TBz=0$. Choose an orthogonal basis for $\mathbb{R}^n$ with $z$ as its first vector. If $C$ is the matrix representin …

Amongst the polynomials that can arise as characteristic polynomials of tridiagonal matrices with zero diagonal, one finds the Hermite polynomials. Schur showed that Hermite polynomials of even degree …

$$ \begin{pmatrix}1&0\\\0&k^{-1/2}\end{pmatrix} \begin{pmatrix}A_1&A_2\\\ kA_2^T&A_3\end{pmatrix} \begin{pmatrix}1&0\\\0&k^{1/2}\end{pmatrix} = \begin{pmatrix}A_1& k^{1/2}A_2\\\ k^{1/2}A_2^T&A …

The short summary is that eigenvalues of $A$ provide no information useful towards computing the eigenvalues of $A+D$. …

Then $\pi$ is equitable and $G/\pi$ is a path, so its eigenvalues are all simple. But if $G$ is not bipartite, its least eigenvalue is not simple. …

If M is Hermitian then H + aI is positive semidefinite when a is large enough. So the Hermitian + positive semidefinite case is essentially the same as the Hermitian case