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$$ A=\begin{bmatrix} a & b & b\\ b & a & b\\ b & b & a\\ \end{bmatrix} and \space B=\begin{bmatrix} 0 & A\\ A & 0\\ \end{bmatrix} $$

What are the eigen values and eigen vectors of B? Any help is appreciated.

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2 Answers 2

For each eigenvalue $\lambda_n$ and eigenvector $x_n$ of a matrix $A$, the matrix $\pmatrix{0&A\\A&0}$ has eigenvalues $\pm\lambda_n$ with eigenvectors $\pmatrix{x_n\\x_n}$ and $\pmatrix{x_n\\-x_n}$. Your matrix $A$ becomes a constant matrix if we subtract $a-b$ on the diagonal, so it has eigenvalue $a-b$ with eigenspace the orthogonal complement of a constant vector (i.e. all vectors that sum to zero); since it's symmetric, the remaining eigenvector must be orthogonal to this eigenspace, so it's the constant vector, and the eigenvalue is $a+2b$ by inspection.

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Thanks. But my B matrix is a little different from what you assumed it to be. –  saroj Aug 23 '12 at 5:14
    
@saroj: No, it's just a little different from how I accidentally wrote it :-) I've corrected that; the rest of the answer should be OK. –  joriki Aug 23 '12 at 6:05

As an addition to the below answer, you may also want to notice that $A$ is a circulant matrix. That is, each row/column is a circularly shifted version of the row/column next to it. Hence, its eigenvectors are same as the $3$-point discrete Fourier transform (DFT) basis. The corresponding eigenvalues can be found out by taking the inner product of the eigenvector with the first row of $A$.

You can read more on circulants here.

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