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## Return to Answer

2 added 105 characters in body

A partial answer: Let $\Sigma$ denote the manifold $x^n+y^n=0$. Away from $\Sigma$, the equation and the fact that the roots of the polynomial $X^n-x^n-y^n$ are simple, tell you that $xA+yB$ is diagonalisable, with eigenvalues that differ from each other from a multiplicative $n$th root of unity. In addition, the multiplicity of an eigenvalue is locally constant.

The complement of $\Sigma$ in ${\mathbb C}^2$ is path-connected (true for every algebraic curve), but it is not simply connected. There are many non-trivial loops around $\Sigma$. When you follow an eigenvalue $\lambda(xA+yB)$ along a loop, you end with possibly different eigenvalue. Actually, there are so many loops that you find the following: if $\lambda$ is an eigenvalue of $xA+yB$, then $\omega\lambda$ is another one for every $n$th rooth of unity $\omega$. Eeven more: because the multiplicity remains constant when you follow the loop, $\lambda$ and $\omega\lambda$ have the same multiplicity.

In conclusion: for $(x,y)\not\in\Sigma$, $xA+yB$ is diagonalisable with eigenvalues at the vertices of a regular $n$-agon, and the eigenspaces have equal dimensions.

Since in your question you seem to take $n$ equal to the size of matrices, this means that the eigenvalues are simple away from $\Sigma$. Anyway, the existence of such a pair $(A,B)$ implies that $n$ (the power) divides $N$ (the matrix size).

1

A partial answer: Let $\Sigma$ denote the manifold $x^n+y^n=0$. Away from $\Sigma$, the equation and the fact that the roots of the polynomial $X^n-x^n-y^n$ are simple, tell you that $xA+yB$ is diagonalisable, with eigenvalues that differ from each other from a multiplicative $n$th root of unity. In addition, the multiplicity of an eigenvalue is locally constant.

The complement of $\Sigma$ in ${\mathbb C}^2$ is path-connected (true for every algebraic curve), but it is not simply connected. There are many non-trivial loops around $\Sigma$. When you follow an eigenvalue $\lambda(xA+yB)$ along a loop, you end with possibly different eigenvalue. Actually, there are so many loops that you find the following: if $\lambda$ is an eigenvalue of $xA+yB$, then $\omega\lambda$ is another one for every $n$th rooth of unity $\omega$. Eeven more: because the multiplicity remains constant when you follow the loop, $\lambda$ and $\omega\lambda$ have the same multiplicity.

In conclusion: for $(x,y)\not\in\Sigma$, $xA+yB$ is diagonalisable with eigenvalues at the vertices of a regular $n$-agon, and the eigenspaces have equal dimensions.

Since in your question you seem to take $n$ equal to the size of matrices, this means that the eigenvalues are simple away from $\Sigma$.