Hey n22.
Hint: Consider the determinant of the matrix. How does this relate to the conditional of being diagonalizable?
I need to learn to read all posts in a thread! I spent some time writing out a proof then realized that Johng had already given exactly that proof!
I am not sure what Chiro's point is. Obviously the determinant is 0 but that alone does not prevent the matrix being diagonalizable. It does mean that 0 is an eigenvalue. If the other eigenvalue is NOT 0, the matrix is diagonalizable. If both eigenvalues are 0, the matrix still might be "diagonalizable" but, in fact, the matrix in that case would have to be the 0 matrix.
We can use |A| to assess the eigen-values (since the product of the eigen-values is equal to |A|) and from this, I was going to look at the results of a non full-rank (i.e. has at least one row of zeroes since the matrix is diagonal) and its effect on being able to diagonalize the matrix.
Also note that the diagonalization is a form of a rotation so if you aren't able to do the rotation, then you can also prove results regarding whether the diagonalization exists (with regards to whether the rotation provided by the eigen-vectors and eigen-values actually works).