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

Diagonalizable means the minimal polynomial is squarefree. The full characteristic polynomial is $(\lambda - 1)^2 (\lambda - 2).$
Diagonalizable if and only if
$$ (A - I)(A - 2I) = 0.  $$
This gives a restriction on the triple $x,y,z.$ And, when we do have $y = xz,$ we have
$$
\left(
\begin{array}{ccc}
1&-x&-xz \\
0&1&z \\
0&0&1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&x&xz \\
0&2&z \\
0&0&1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&x&0 \\
0&1&-z \\
0&0&1 \\
\end{array}
\right) =
\left(
\begin{array}{ccc}
1&0&0 \\
0&2&0 \\
0&0&1 \\
\end{array}
\right)
$$
On the other hand, if $y =  xz + t$ with $t \neq 0$ we have
$$
\frac{1}{t}
\left(
\begin{array}{ccc}
0&t&zt \\
1&-x&-xz \\
0&0&t \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1&x&xz +t \\
0&2&z \\
0&0&1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
x&t&0 \\
1&0&-z \\
0&0&1 \\
\end{array}
\right) =
\left(
\begin{array}{ccc}
2&0&0 \\
0&1&1 \\
0&0&1 \\
\end{array}
\right)
$$
