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I am trying to find the eigenvalues and eigenvectors of the following 4x4 "checkerboard" matrix:

$$ \mathbf C = \begin{pmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ \end{pmatrix}$$

Here is my approach:

$$ \det(\mathbf C - \lambda \mathbf I) = \begin{vmatrix} -\lambda & 1 & 0 & 1 \\ 1 & -\lambda & 1 & 0 \\ 0 & 1 & -\lambda & 1 \\ 1 & 0 & 1 & -\lambda \end{vmatrix} = 0$$

From this I get a characteristic polynomial of:

$$\lambda^4 - 4\lambda^2 = \lambda^2(\lambda - 2)(\lambda + 2)$$

So that would yield eigenvalues of 2, -2, 0, 0.

For eigenvalue $\lambda_1 = 2$, I get eigenvector of: $$v_1 = \begin{pmatrix}1\\1\\1\\1\end{pmatrix}$$

For eigenvalue $\lambda_2 = -2$, I get eigenvector of: $$v_2 = \begin{pmatrix}-1\\1\\-1\\1\end{pmatrix}$$

This seems to check out. If I multiply $C v_1$, I get $\lambda_1 v_1 = 2 v_1$. If I multiply $C v_2$, I get $\lambda_2 v_2 = -2 v_2$.

My professor disagrees. He states that the eigenvalues should be 2, 2, 0, 0, and that the two corresponding eigenvectors associated with eigenvalue of 2 are:

$$v_1' = \begin{pmatrix}0\\1\\0\\1\end{pmatrix} \qquad v_2' = \begin{pmatrix}1\\0\\1\\0\end{pmatrix}$$

My professor further states:

Try multiplying the vectors $v'_1$ and $v'_2$ by $\mathbf C$ and you’ll see that, indeed, they are both eigenvectors with eigenvalue 2.

That still seems wrong to me. I did as he suggested and

$$ C v'_1 = \begin{pmatrix}2\\0\\2\\0\end{pmatrix} \neq 2 v'_1 \qquad C v'_2 = \begin{pmatrix}0\\2\\0\\2\end{pmatrix}$$

I believe there is a subtle mistake here. $C v'_1 = 2 v'_2$ and $C v'_2 = 2 v'_1$ but that is irrelevant, correct? That does not make the eigenvalue of $\lambda = 2$ have a multiplicity of 2, does it?

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    $\begingroup$ It seems you're right. The vectors that your professor suggested would be eigenvectors if the roles of 0 and 1 were reversed in the matrix. $\endgroup$ – Hans Lundmark Mar 30 '15 at 7:13
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Your professor is certainly wrong, provided that you haven't misquoted him. The eigenvalues of a matrix sum to the matrix trace. The trace of your matrix is zero, so it's impossible to have some positive eigenvalues, but no eigenvalues with negative real parts.

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  • $\begingroup$ My professor realized that he made the mistake @Hans Lundmark indicated: swapping all the 0's and 1's for one another. $\endgroup$ – tony_tiger Mar 31 '15 at 16:37

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