How to determine if a 3x3 matrix is diagonalizable? The matrix is given as:
$A=\begin{bmatrix} 0 & 1 & 1 \\0 & 0 & 4 \\ 0 & 0 & 3 \end{bmatrix}$
So the matrix has eigenvalues of $0$ ,$0$,and $3$.
The matrix has a free variable for $x_1$ so there are only $2$ linear independent eigenvectors. So this matrix is not diagonalizable.
What conditions would be necessary for $A$ to be diagonalizable?
Is it simply all $3$ eigenvectors must be linearly independent? Or perhaps the opposite?
 A: As you remarked correctly, the eigenvalues,  with multiplicity, are $0,0,3$. 
A matrix is diagonalizable if and only of for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue. 
For the eigenvalue $3$ this is trivially true as its multiplicity is only one and you can certainly find one nonzero eigenvector associated to it. 
For the eigenvector $0$ however you would need to find $2$ linearly indepedent eigenvectors Yet as you said, indirectly, the eigenspace associated to $0$ is the space generated by $(1,0,0)$. Its dimension is thus one and you cannot find two independent eigenvectors for the eigenvalue $0$. 
The matrix is thus not diagonalizable. 
Necessary conditions for diagonalizable include: 


*

*There exists a basis of eigenvectors, which in you case would mean there exist $3$ linearly independent eigenvectors. (This is almost what you said, but note that I said "there exists" and not "the three eigenvectors" since there are infinitely many eigenvectors.) 

*For each eigenvalue the dimension of the respective eigenspace is equal to the multiplicity of the eigenvalue. (As mentioned at the start.)
A: A matrix $A$ is diagonalisable with distinct eigenvalues $\lambda_1,\ldots,\lambda_k$ (or a subset thereof) if and only if the product $(A-\lambda_1I)\ldots(A-\lambda_kI)$ is zero. In your example clearly $0,3$ are the only eigenvalues, but
$$
  A(A-3I)=\begin{pmatrix}0&-3&4\\0&0&0\\0&0&0\end{pmatrix}
$$
is nonzero, so $A$ is not diagonalisable.
