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Here is the proof that one of the eigenvalues indeed does not exceed $1/3$; Let $X$ be the matrix with columns $\mathbf x_1$ and $\mathbf x_2$ (writing $X=(\mathbf x_1,\mathbf x_2)$); then $M=XX^\top$. … Assume now that the smallest eigenvalue $\lambda_{ij}$ of $M_{ij}$ is larger than $1/3$, for all $i,j$; then both eigenvalues of $M_{ij}$ lie on $(1/3,3d_{ij})$. …

Let $\lambda_1$, $\lambda_2$, $\lambda_3$ be the eigenvalues of $A$, and $\mu_1,\mu_2,\mu_3$ be those of $B$ (with $\lambda_1,\mu_1\in\mathbb R$). … Since the eigenvalues are distinct, $A$ and $B$ are diagonalizable; since they commute, they are simultaneously diagonalizable, i.e., an eigenbasis for $A$ is also that for $B$. …

This is a result of joint efforts of myself and Fedor Petrov. Assume, to the contrary, that the system of equations $$ G_\mu(x)=\alpha_\mu, \qquad \mu=1,2,\dots,n $$ has two different solutions $a$ …

The eigenvalues of $A$ are obvious. Choosing the $a_{i1}$, you may add to the characteristic polynomial any polynomial of degree $\leq n-2$. …

$\def\d{\mathrm{d}}\def\Vol{\mathop{\mathrm{Vol}}}$Okay, this really seems not that straigntforward to generalize the previous answer here. However, here is the method which seems to work (and which s …

Let me rephrase the answer in different terms. Since $H_i$ are idempotent and self-adjoint, they are orthogonal projectors onto some subspaces $W_i$. Set $U=W_1\cap W_2$, and let $U_i$ be the orthogo …

If $M$ had more than one Jordan block corresponding to some eigenvalue, then its minimal polynomial's degree would be smaller than $k$. This yields that all sequences satisfying your recurrence relati …