A diagonalizable $3\times3$ matrix with precisely two real eigenvalues.

Question

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?

Answer 1

The answer really depends on how you choose to count the eigenvalues:

If you simply want the number of distinct real eigenvalues, then the answer to your question is yes: take any diagonal $3\times 3$ matrix with only two distinct diagonal entries. For instance, $$ \begin{bmatrix}1&0&0\\0&1&0\\0&0&2\end{bmatrix}$$

If you want the number of real eigenvalues counted with multiplicity, then the answer is no: the characteristic polynomial of a real $3\times 3$ matrix is a real polynomial of degree $3$, and therefore has either $1$ or $3$ real roots if these roots are counted with multiplicity. In the above example, the multiplicity of $\lambda=1$ is $2$.

Answer 2

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.