When is the following matrix diagonalizable? $\begin{pmatrix}8&k\\ 0&4\end{pmatrix}$
So I know that for a matrix to be diagonalizable, it needs to have 2 distinct real eigenvalues. 
So I calculated the characteristic polynomial to be:
$x^2-12x+32-k$
Therefore, for the discriminant to be greater than $0$, I got the inequality:
$k>-4$
Therefore, I thought that it would diagonalizable for all $k$ except $k=-4$, but this is not correct.
Any help?

It is more concrete to just go ahead and see what happens. With the eigenvalues $8,4$ we can get a matrix with eigenvectors as columns from 
$$
R = 
\left(
\begin{array}{rr}
1 & -k \\
0 & 4
\end{array}
\right).
$$ 
We can confirm this with
$$
\left(
\begin{array}{rr}
8 & k \\
0 & 4
\end{array}
\right) 
\left(
\begin{array}{rr}
1 & -k \\
0 & 4
\end{array}
\right) =
\left(
\begin{array}{rr}
8 & -4k \\
0 & 16
\end{array}
\right)
$$
Then 
$$
R^{-1} = 
\left(
\begin{array}{rr}
1 & -\frac{k}{4} \\
0 & \frac{1}{4}
\end{array}
\right)
$$ 
and
$$
\left(
\begin{array}{rr}
1 & -\frac{k}{4} \\
0 & \frac{1}{4}
\end{array}
\right)
\left(
\begin{array}{rr}
8 & k \\
0 & 4
\end{array}
\right) 
\left(
\begin{array}{rr}
1 & -k \\
0 & 4
\end{array}
\right) =
\left(
\begin{array}{rr}
8 & 0 \\
0 & 4
\end{array}
\right)
$$
Whatever the appearances, the value of $k$ does not matter. Since someone mentioned it, we are taking the characteristic of the field to not be two.
