(Concept checking)Testing for diagonalizability For the matrix $A \in M_{n \times n}(R)$, test A for diagonalizability.
$A= \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 1 \end{bmatrix}$

Question: To test for diagonalizability, is it true that as long as we can find distinct eigenvalues, which in this case, 3, then we look for three eigenspaces that together form a eigenbasis. If in this case, we can find three linearly independent eigenvectors that that such an eigenbasis, then we can say A is diagonalizable?

 A: The matrix is diagonalizable if and only if there exist three linearly independent eigenvectors. However, you may not have to find them in order to guarantee that the matrix is diagonalizable, and you may not need to prove there are no such three vectors to prove it is not diagonalizable.
For example, a $3\times 3$ matrix that has three distinct eigenvalues is always diagonalizable (more generally, an $n\times n$ matrix with $n$ distinct eigenvalues is always diagonalizable). Because each eigenvalue has at least one eigenvector, and eigenvectors corresponding to distinct eigenvalues are linearly independent. 
If the characteristic polynomial of the matrix does not split, then the matrix is not diagonalizable.
If the matrix has repeated eigenvalues, then you just need to check the eigenvalues with multiplicity greater than $1$: you need to verify there are at as many linearly independent eigenvectors corresponding to that eigenvalue as the multiplicity (there’s always at least one).
For the matrix at hand, the characteristic polynomial is $-(t-1)(t^2+1)$. If you are working over $\mathbb{R}$, there aren’t enough eigenvalues, so it’s not diagonalizable. If you are working over $\mathbb{C}$, then the matrix has three distinct eigenvalues ($1$, $i$, and $-i$), so it is diagonalizable. No need to find any eigenvectors in this case. 
