How to find a multiple eigenvalue of a square matrix?

Question

$$ \ A= \begin{bmatrix} 2 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 3 & 5 & -2 & -2 \\ -2 & 3 & 3 & 5 \\ \end{bmatrix} $$

For this matrix, I am supposed to figure out which eigenvalue is a multiple eigenvalue. The answer to the problem is -1, but I am struggling to understand how to get to that conclusion.

I know that for upper or lower triangular matrices, you can find the eigenvalues just by looking at the diagonal values, but that can't happen in this case since there's a single value preventing it from being triangular at (3,4).

I also know that this can be done by finding $ det(A-\lambda I) $ and then using cofactor expansion, but since this question is taken from a past linear algebra exam at my university, I feel like there has to be a less tedious way of going about it.

Is there a simpler way to come to an answer of -1?

Answer 1

Using $\det(A-\lambda I)$ is not that tedious as you thought.

$$ \begin{align*} \det(A-\lambda I) & = \begin{vmatrix} 2-\lambda & 0 & 0 & 0 \\ 1 & -1-\lambda & 0 & 0 \\ 3 & 5 & -2-\lambda & -2 \\ -2 & 3 & 3 & 5-\lambda \end{vmatrix} \\ & =-(-2) \begin{vmatrix} 2-\lambda & 0 & 0 \\ 1 & -1-\lambda & 0 \\ -2 & 3 & 3 \end{vmatrix} +(5-\lambda) \begin{vmatrix} 2-\lambda & 0 & 0 \\ 1 & -1-\lambda & 0 \\ 3 & 5 & -2-\lambda \end{vmatrix} \\ & =-(-2)(2-\lambda)(-1-\lambda)(3)+(5-\lambda)(2-\lambda)(-1-\lambda)(-2-\lambda) \\ & =(2-\lambda)(-1-\lambda)[-(-2)(3)+(5-\lambda)(-2-\lambda)] \\ & =(\lambda-2)(\lambda+1)(\lambda^2-3\lambda-4) \\ & =(\lambda-2)(\lambda+1)^2(\lambda-4) \end{align*} $$

Answer 2

The key point is the square of zeroes in the upper right. That says that eigenvectors are $(1,0,0,0)^T$ and $(0,1,0,0)^T$ and then two vectors of type $(0,0,p,q)^T$ and $(0,0,r,s)^T$ where the nonzero portions are eigenvectors of the lower right two by two, as written in comment by Dave. The characteristic polynomial for Dave's square is $\lambda^2 - 3 \lambda - 4= ( \lambda + 1)(\lambda - 4)$

$$ \left( \begin{array}{cc} -2 & -2 \\ 3 & 5 \end{array} \right) $$