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Yes. Write $D(\lambda_1, \ldots, \lambda_n)$ as the diagonal matrix with diagonal entries $(\lambda_1, \ldots, \lambda_n)$. Let $A$ be any complex $n \times n$ matrix. We can upper-triangularize $A$ a …

If you want a doubly stochastic matrix, $$\begin{pmatrix} 0&1 & 0&0&0 \\ 1&0 & 0&0&0 \\ 0&0 & 1/3&1/3&1/3 \\ 0&0 & 1/3&1/3&1/3 \\ 0&0 & 1/3&1/3&1/3 \\ \end{pmatrix}$$ has eigenvalue $-1$.

Yes. Your matrix $H$ is symmetric, so the spectral radius is the same as the maximal eigenvalue is the same as the maximal singular value. From the singular value description, for any unit vectors $\v …

Consider the matrix $$\begin{pmatrix} 7 & 8 & 4 & 4 & 0 \\ 0 & 7 & 4 & 4 & 0 \\ 4 & 4 & 7 & 8 & 0 \\ 4 & 4 & 0 & 7 & 0 \\ 0 & 0 & 0 & 0 & 5 \\ \end{pmatrix}$$ The eigenvalues are $3 \pm 4i$, $11 \pm 4 … Let $M$ be the above matrix and let $N$ be the $10 \times 10$ matrix with block form $$\begin{pmatrix} 0 & M^2 \\ \mathrm{Id} & 0 \end{pmatrix}.$$ The eigenvalues of $N$ are $\pm3 \pm 4 i$, $\pm 11 \pm …

These eigenvalues are real numbers, since they are eigenvalues of a Hermitian matrix. … So these eigenvalues are the reciprocals of the points where a line through the origin meets $F=0$. …

A simple estimate which is often useful is that, if $A$ and $B$ are Hermitian matrices with eigenvalues $a_1 > a_2 > \ldots > a_n$ and $b_1 > b_2 > \ldots > b_n$ and the eigenvalues of the sum are $c_ … More specifically, the $3n$-tuples $(a_1, \ldots, a_n, b_1, \ldots, b_n, c_1, \ldots, c_n)$ which occur as eigenvalues of $(A,B,C)$ with $A+B=C$ are dense in the hyperplane $\sum a_i + \sum b_i = \sum …