Finding all complex entries? Find all complex triples $(x, y, z) $such that the following matrix is diagonalizable  
$A = \begin{bmatrix}1&x&y \\ 0&2 & z \\0&0&1\end{bmatrix}$
my attempts :
matrix A is an upper triangular matrix.
so the the eigenvalues of A are diagonal entries $1,2,1$
This implies that $A$ is diagonalizable.
Case 1: if $x=y= z= 0$ 
case 2 : if $x\neq y \neq z \neq 0$
Now Im confused that How can  i find all complex triples $(x, y, z) $such that the following matrix is diagonalizable

This is equivalent to the fact that 
$$
A-I_3 = 
\begin{bmatrix} 
0 &x&y \\ 
0&1 & z \\
0&0&0
\end{bmatrix}
$$
is such that $dim(ker(A-I_3))=2$ which is, in turn, equivalent to 
$$
dim(Im(A-I_3))=1
$$
and the fact that the two columns 
$$ 
\begin{bmatrix} 
x&y \\ 
1 & z \\
0&0
\end{bmatrix}
$$
are proportional. You finally find 
$$
xz=y
$$
