$$ \begin{bmatrix} 1 & 1 & 0 \\ 3 & 0 & 1 \\ 1 & -1 & 2 \\ \end{bmatrix} $$
Is the $3$ by $3$ matrix above diagonalizable given the eigenvalues $-1$ and $2$? There are only $2$ eigenvalues so the one with the multiplicity two needs a full rank eigenspace for this to be diagonalizable. I think there should be a way to tell from the matrix alone whether that's possible without doing all the row operations, but I'm not sure what I should be looking for.
A square matrix is diagonalizable if it has a complete set of eigenvectors. A complete set of eigenvalues guarantees you a complete set of eigenvectors, but having repeat eigenvalues means you need to check the eigenvectors for those eigenvalues.
The characteristic polynomial of your equation is:
$$\begin{split}0&=-x^3+3x^2-4&\\ &=-x^3+4x^2-4x-x^2+4x-4\text{ add and subtract 4x, also split the $3x^2$ in two parts}\\ &=-(x^3-4x^2+4x)-(x^2-4x+4)\\ &=-x(x^2-4x+4)-(x-2)^2\\ &=-(x+1)(x-2)^2\end{split}$$
Therefore the eigenvalue with multiplicity 2 is 2, and we need to check the eigenvectors.
$$\left(\begin{matrix}-1&1&0\\3&-2&1\\1&-1&0\end{matrix}\right)\left(\begin{matrix}v_1\\v_2\\v_3\end{matrix}\right)=0\\ \begin{cases}v_1=v_2\\ 3v_1-2v_2+v_3=0\end{cases}\\ \left(\begin{matrix}v_1\\v_1\\-v_1\end{matrix}\right)\implies\left(\begin{matrix}1\\1\\-1\end{matrix}\right) \text{ is the only eigenvector for $\lambda=2$}$$
So it is not diagonalizable.
Sorry, I don't know a faster way.
