Finding for every parameter $\lambda$ if matrix is diagonalizable Given:
$$A = \begin{pmatrix} 1 & i & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & i \end{pmatrix} \; , \; \lambda \in \mathbb C$$
For every value of $\lambda$ I have to know if the matrix $A$ is diagonalizable, and if so, I need to find an invertable matrix $C$ and a diagonalizable matrix $D$ such that $A = CDC^{-1}$.
Now I did not know how to approach such a question.
Do I start calculating the polynomial, calculating the eigenvalues and eigenvectors? But even if so, what am I looking for?
 A: Hints:
If you proceed with the calculations of eigenvalues and eigenvectors, you will see which value of $\lambda$ to avoid.
The eigenvalues, from the characteristic polynomial, are:
$$\lambda_1 = i, \lambda_2 = 1, \lambda_3 = \lambda$$
Give it a go and see if you can find the eigenvectors.
Also, can you see things in the special form of the matrix that make things easier?
A: Amzoti's tip about spotting the eigenvalues is excellent!
Hint:
Hopefully you know that a matrix is diagonalizable iff it has a full set of linearly independent eigenvectors. Given such a set of eigenvectors, you can form a matrix $B$ by putting them in the columns of $B$, and then $AB=BD$ where $D$ is a diagonal matrix with eigenvalues on the diagonal. Since the eigenvectors were LI, $B$ is invertible, so $B^{-1}AB=D$.
There is a well-known exercise that if all of the eigenvalues are distinct, $A$ has a full set of linearly independent eigenvectors. (You might take the time to prove it.)  Using that, we can cut right down to the problem cases: $\lambda\in \{1,i\}$.
