I am trying to find the eigenvalues and eigenvectors of the following 4×4 "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?
Best Answer
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.