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?
 A: Hint:
First it depends on which field you want it to be diagonalisable, e.g. $\mathbf Q$, $\mathbf R$,  $\mathbf C$?
Supposing the characteristic  polynomial splits on the base field, the necessary and sufficient condition for a matrix to be  diagonlisable, is NOT to have distinct eigenvalues (this is only a sufficient condition), but the geometric multiplicity being equal to the algebraic multiplicity, i.e. if $\lambda$ is an eigenvalue with algebraic multiplicity $m_\lambda$ and eigenspace $E_\lambda$:
$$\dim E_\lambda=m_\lambda$$
(it is always true that $\;\dim E_\lambda \le m_\lambda$).
A: Note that to diagonalize the matrix you need a basis of eigenvectors for $\mathbb{R^2}$.
When we find two distinct eigenvalues that is given for free.
If eigenvalues are not distinct we need to check whether or not for the eigenvalues 2 (which has algebraic multiplicity 2) we can find 2 linearly independent eigenvectors, in this case you can diagonalize the matrix.
Otherwise you can only obtain a Jordan form.
In this case since $\lambda_1=8$ and $\lambda_2=4$, we can diagonalize $\forall k$.
A: The characteristic polynomial is $(t-8)(t-4)$, so the eigenvalues are $t_1 = 4$ and $t_2 = 8$.  By inspection, when $v_2 = (1,0)^T$, $\begin{pmatrix}8&k\\ 0&4\end{pmatrix} v_2 = 8v_2$.
To find another eigenvector $v_1$.
$$\begin{pmatrix}4&k\\ 0&0\end{pmatrix} \begin{pmatrix}v_{11}\\ v_{12}\end{pmatrix} = 0$$
So choose $v_1 = (k,-4)^T$, which correspond to $t_1 = 4$, and $v_2 = (1,0)^T$ corresponding to $t_2 = 8$.
A: 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.
