When is the given matrix not diagonizable? 
Given the following matrix $$A = \begin{bmatrix} 0 & 1\\ bk+b & -bk\end{bmatrix}$$ where $b \neq 0$ and $k \neq 0$, find when the given matrix is not diagonizable.

My initial thought would be when the matrix doesn't have $2$ linearly independent eigenvectors. 
In this case it would be when the eigenvalue has a multiplicity of 2.
I.e. $\lambda_1=\lambda_2 $.
I find the eigenvalues:
$
\lambda_1=\frac{-bk+\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}\\
\lambda_2=\frac{-bk-\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}
$
I then let them equal each other:
$
(1)\qquad\frac{-bk+\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}=\frac{-bk-\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}\\
\quad\\
\quad\\
(2)\qquad\sqrt{(bk)^2-4\cdot(-bk-b)}=-\sqrt{(bk)^2-4\cdot(-bk-b)}
$
Here my initial thought would be that the only case this is true if the term under the square root is equal zero.
$
b^2k^2+4bk+4b=0\\
b(bk^2+4k+4)=0
$
Since $b\neq0$ I need the term inside the parantheses to be equal 0.
$
bk^2+4k+4=0
$
I'm a bit lost from here.
I know the solution is 
$
-bk-k=(\frac{-bk}{2})^2
$.
The solution was provided by someone else here on Stack Exchange, he used the determinant and the trace of the matrix to come to that conclusion, which I can't seem to replicate.
 A: We have that


*

*$Tr(A)=\lambda_1+\lambda_2=-bk$

*$\det(A)=\lambda_1\lambda_2=-(bk+b)$
and $\lambda_1=\lambda_2=\lambda$ implies


*

*$\lambda=-\frac{bk}2$

*$\lambda^2=-(bk+b)$
and therefore $\lambda_1=\lambda_2$ when
$$\left(-\frac{bk}2\right)^2=-(bk+b)$$
Finally recall that for $\lambda_1\neq \lambda_2$ $A$ is diagonalizable but for $\lambda_1=\lambda_2$ we need to check directly whether or not $A$ is diagonalizable.
A: $$b\neq 0, k\neq 0, A=\begin{pmatrix} 
0 & 1\\
bk+b & -bk
\end{pmatrix}\Rightarrow \det (A-xI)=\begin{vmatrix}
-x & 1\\
bk+b & -bk-x
\end{vmatrix}=0\Rightarrow$$
$$\Rightarrow x^2+bkx-(bk+b)=0\Rightarrow x=\dfrac{-bk\pm \sqrt{b^2k^2+4bk+4b}}{2}$$
$$x_1=x_2\Rightarrow b^2k^2+4bk+4b=0\Rightarrow b^2k^2=-4(bk+b)\Rightarrow \left( \dfrac{bk}{2}\right)^2=-bk-b$$
Let us now look at the eigenvectors.
$$x=-\dfrac{bk}{2}\Rightarrow A-xI=\begin{pmatrix}
\frac{bk}{2} & 1\\
-\left( \frac{bk}{2}\right)^2 & -\frac{bk}{2}
\end{pmatrix}$$
$$\begin{pmatrix}
\frac{bk}{2} & 1\\
-\left( \frac{bk}{2}\right)^2 & -\frac{bk}{2}
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}=\begin{pmatrix}
0\\
0
\end{pmatrix}\Rightarrow \left\{ \begin{array}{lcc}
             \dfrac{bk}{2}x+y=0 \\
             \\ -\left( \dfrac{bk}{2}\right)^2x-\dfrac{bk}{2}y=0
             \end{array}
   \right.\Rightarrow bkx+2y=0\Rightarrow$$
$$E_{-\frac{bk}{2}}=\langle (-2,bk)\rangle \Rightarrow \dim \left( E_{-\frac{bk}{2}}\right) =1$$
Not diagonalizable.
A: The characteristic polynomial is
$$
\det\begin{bmatrix} -X & 1\\ bk+b & -bk-X\end{bmatrix}=
X^2+bkX-(bk+b)
$$
The discriminant is
$$
(bk)^2+4(bk+b)=b^2k^2+4bk+4b
$$
If the discriminant is nonzero, then the matrix is diagonalizable (over the complex numbers), because it has two distinct eigenvalues.
If the discriminant is zero, then the matrix is not diagonalizable: a $2\times2$ matrix with a unique eigenvalue (of multiplicity $2$) is diagonalizable if and only if it is already diagonal. But it's not necessary to believe to the preceding statement: just note that the unique eigenvalue is $-bk/2$ so we need to compute the rank of $A+(bk/2)I$:
$$
A+\frac{bk}{2}I=
\begin{bmatrix} bk/2 & 1\\ bk+b & -bk/2\end{bmatrix}
$$
Such a matrix surely has rank less than $2$, but it cannot have rank $0$, which would be required for diagonalizability.
If you want diagonalizability with a real matrix, the eigenvalues must so be distinct and real; this is the same as
$$
b^2k^2+4bk+4b>0
$$
Thus the answer is:


*

*$A$ is not diagonalizable over the complex numbers if and only if $b^2k^2+4bk+4b=0$.

*$A$ is not diagonalizable over the real numbers if and only if $b^2k^2+4bk+4b\le0$.

