For which values of $a$, $b$, and $c$ is this matrix diagonalizable? For which values of $a$, $b$, and $c$  is the following matrix diagonalizable? 
$$\begin{bmatrix}
    1 & 0 & c  \\
    1 & a & b  \\
    0 & 0 & 1 
\end{bmatrix}$$
As I understand it, in order for matrix to be diagonalizable, it be written in the form:
$$A = P \Lambda P^{-1}$$ 
where $\Lambda$ contains the eigenvalues and the columns of $P$ are the eigenvectors.
I have calculated the eigenvalues which are : $\lambda_1 = 1, \lambda_2 = a$, but I got stuck after that. Any help?
 A: Case $1$: $a=1$.
Consider the matrix $A-I$,
$$A-I=\begin{bmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 &0 &0\end{bmatrix}$$
The nullity of the matrix is less than $3$,the algebraic multiplicity corresponding to the eigenvalue $1$ , it cannot be diagonalizable.
Case $2$: $a \neq 1$:
Again, consider the matrix $$A-I=\begin{bmatrix} 0 & 0 & c \\ 1 & a-1 & b \\ 0 &0 &0\end{bmatrix}$$
How does $a,b,c$ affects the nullity of the matrices? 
To be diagonalizable, it has to be has nullity of $2$ (the algebraic multiplicity of eigenvalue $1$), i.e. the matrix $A-I$ has to be of rank $1$.
Are you able to complete the rest?
A: The characteristic polynomial of the given matrix is $(X-a)(X-1)^2$. Now, if $a=1$ for it to be diagonalizable the minimal polynomial should be $X-1$ which is not possible. So, for $a=1$ the matrix is not diagonalizable. Let's assume $ a \neq 1$ then for it to be diagonalizable the minimal polynomial should be $(X-1)(X-a)$.  Which gives $(A-I)(A-a)=\begin{pmatrix}0 & 0 & c(1-a) \\0 & 0 & c \\0  & 0 & 0 &\\ \end{pmatrix}$ for it to be diagonalizable $c=0$. So, the given matrix is diagonalizable when $ a \neq 1,  b \in \mathbb{R}, c=0$.
