TITLE: A diagonalizable $3\times3$ matrix with precisely two real eigenvalues.
QUESTION [1 upvotes]: I'm wondering, does there exist a diagonalizable $3\times3$ matrix that has precisely two real eigenvalues? I thought that there doesn't because in order for a matrix to be diagonalizable, the number of eigenvalues has to equal to the dimension of the matrix. I know that it might work with complex numbers but if no complex numbers are allowed in the matrix, then would a matrix like that exist using only real numbers?

REPLY [0 votes]: I think the bit you are missing is that the characteristic equation has to have three real roots, and there must be 3 linearly independent eigenvectors.
So the characteristic equation could be in the form $(1-λ)^{2}(2-λ)$ which technically has only two roots but still has three linearly independent eigenvectors.
{"set_name": "stack_exchange", "score": 1, "question_id": 2036944}