I have this matrix, and I need to say where the matrix could be diagonalized. Alpha is a real parameter.
$$ \begin{matrix} 1 & \alpha \\ 3 & 1 \\ \end{matrix} $$ I found characteristic polynomial, $ det(a) = (t*I - A) $
here's the final result: $$ \begin{matrix} 0 & -\alpha \\ -3 & 0 \\ \end{matrix} $$ and by computing the determinant of this 2-by-2 matrix I got: $0*0 - (-3*(-\alpha)$, which is equal to -3*(\alpha). alpha is equal to zero, and its algebraic multiplicity is equal to 1, same for its geometric multiplicity.
I think there's an error here, because it's strange that the characteristic polynomial has this solution, because it's not possible to go forward, since according to spectral theorem a squared matrix could be diagonalized if and only if the sum of geometric multiplicities is equal to the order of the matrix. And it's not my case, because I have only 1 eigenvalue with algebraic multiplicity equal to 1, so the geometric multiplicity must be 1 as well.