To show that a matrix is not diagonalizable, I would just have to show that there are no eigenvalues present in the matrix.
So, for example, if I want to prove that
$$A=\begin{bmatrix} 0 & -1 \\1 & 0 \end{bmatrix}$$
is not diagonalizable – would I say that it is since the tr(A) does not form any eigenvalues?
Best Answer
You can compute the charactersitic polynomial, which is in this case equal to $x^{2}+1$. Assuming you are working in $\mathbf{R}$ this polynomial has no real roots, and hence A has no eigenvalues so A is not diagonalisable.