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Consider a block matrix of the type

$ M = \begin{bmatrix}A & B \\B & A \end{bmatrix} $

where $A$ and $B$ are square matrices of equal dimensions. We know that

$\det(M) = \det(A-B) \det(A+B)$.

This property is very useful for computing the eigenvalues of $M$, because it implies that

$\det(M-\lambda I) = \det(A-B -\lambda I) \det(A+B -\lambda I)$,

(where $\lambda$ is a scalar and $I$ the identity matrix). Therefore, the eigenvalues of $M$ will be given by joining the sets of the eigenvalues of the matrices $A+B$ and $A-B$. This simplifies the problem because it reduces the dimensions of the matrices.

Is there a similar "trick" that we can exploit to simplify the calculations of the eigenvectors? That is, given a matrix $M$ in the block form shown above, is there a way to obtain its eigenvectors by calculating the eigenvectors of the smaller matrices $A$ and $B$ (or a linear combination of them) ?

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