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Let $A = \begin{pmatrix} A_{1}&B_{1}\\ 0&A_{2} \end{pmatrix}$ where $A_2 = A_1\otimes I_n$, $B_1 \in R^{n\times n^2}$ and $A_1\in {R}^{n\times n}$ is symmetric such that the related eigenvalue problem is real valued. Because of the structure, it is clear that the eigenvalues of $A_2$ and $A_1$ coincide.

How are the eigenvectors of both matrices related to the original matrix? Numerical results show that the eigenvectors of $A$ are the eigenvectors of $A_1$ padded with zeroes, but I don't know how to prove this.

I have looked into this related question but the answer assumes different eigenvalues for the different blocks.

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If $v$ is an eigenvector of $A_1$, then $\pmatrix{v\\ 0}$ is an eigenvector of $A$.

There might not be more eigenvectors: Consider the following example with $n=1$, $A_1=A_2=B=1$. then $$ A = \pmatrix{1 & 1 \\ 0 & 1} $$ only has a one-dimensional eigenspace to the eigenvector $1$.

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