For which numbers is the matrix diagonalizable? We have the matrix \begin{equation*}A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\end{equation*} 
I want to find for which real numbers $a,b,c,d$ the matrix is diagonalizable in $\mathbb{R}^{2\times 2}$ and for which in $\mathbb{C}^{2\times 2}$. 
The charachteristic polynomial is \begin{align*}\det (A-\lambda I)=\lambda^2-(a+d)\lambda +(ad-cb)\end{align*} 
So, the eigenvalues are
\begin{align*}\lambda^2&-(a+d)\lambda +(ad-cb)=0 \\  \Rightarrow &\lambda_{1,2}=\frac{(a+d)\pm \sqrt{(a+d)^2-4(ad-cb)}}{2}=\frac{(a+d)\pm \sqrt{a^2+2ad+d^2-4ad+4cb}}{2}\\ & =\frac{(a+d)\pm \sqrt{a^2-2ad+d^2+4cb}}{2}  =\frac{(a+d)\pm \sqrt{(a-d)^2+4cb}}{2}\end{align*} 
We have the following cases: 
As an element of $\mathbb{R}^{\times 2}$ we have the following: 


*

*Expression under the root $< 0$: no real eigenvalue, so the matrix is not diagonalizable.

*Expression under the root $> 0$: two different eigenvalues, that means that the matrix is not diagonalizable, or not? 

*Expression under the root $= 0$, we have an eigenvalue of multiplicity $2$. What do we have in this case?  


As an element of $\mathbb{C}^{\times 2}$ we have the following: 


*

*Expression under the root $\neq 0$: two different eigenvalues , that means that the matrix is not diagonalizable, right? 

*Expression under the root $= 0$, we have an eigenvalue of multiplicity $2$. What do we have in this case?  

 A: A matrix is diagonalizable iff its minimum polynomial is a product of simple linear factors. 
Also, the distinct roots of the minimum polynomial are the same as the distinct eigenvalues of the matrix.
So for a 2x2 matrix, the only case when it is NOT diagonalizable is when the matrix has 2 repeated eigenvalues.
The matrix has repeated eigenvalues when:
$$det(A-\lambda I) = \lambda^2-\lambda \cdot (a+d) + (ad-bc)$$
has repeated roots....ie: $b^2-4ac=0$, which should simplify to: $$(a-d)^2+4bc=0$$
So all values of a, b, c and d that satisfy $(a-d)^2+4bc\ne0$ make A diagonalizable.
A: The matrix can be diagonalised iff it has a complete set of eigenvectors.
To be more explicit for a $2\times2$ matrix we can distinguish a few cases:

*

*If the two eigenvalues are distinct, which occurs when $(a-d)^2+4bc\neq0$, then the eigenvectors are linearly independent, and the matrix can be diagonalised.


*If the eigenvalues are identical (degenerate), one needs to check if the eigenvectors are linearly dependent. The degenerate condition requires $(a-d)^2+4bc = 0$, and the eigenvalues are $\lambda_1 = \lambda_2 = (a-d)/2$. The equations for the eigenvector(s) $\mathbf{x} = (x,y)$ are therefore
\begin{align}
   \begin{cases}
  \frac{a-d}{2} x + b y = 0 \\
  c x - \frac{a-d}{2} y = 0
\end{cases}
\end{align}
These equations are linearly dependent because of the degeneracy condition, so there are two possibilities:

*

*We can determine no relation between $x$ and $y$, and therefore two eigenvectors exist (we can choose any linearly independent pair). This can occur only if all coefficients of the eigenvectors system of equations vanish, i.e. $(a-d) = b = c = 0$. The matrix in this cases reduces to the (already diagonal) form
\begin{equation}
A = \begin{bmatrix}
 a & 0 \\
0 & a
\end{bmatrix}
\end{equation}


*As soon as one of the coefficients of the eigenvectors equations does not vanish, we establish a relation between $x$ and $y$, thus choosing one specific direction for the (only) eigenvector. As a consequence the matrix is defective and not diagonalisable. This can manifest in two forms:

*

*If $(a-d)=0$ then we must have that either $b=0$ or $c=0$, because we still must satisfy the degenerate condition $(a-d)^2+4bc = 0$. In this case the matrix has the form
\begin{equation}
A = \begin{bmatrix}
 a & b \\
0 & a
\end{bmatrix} \mbox{ with } b\neq0 \qquad \mbox{ or } \qquad  
A = \begin{bmatrix}
 a & 0 \\
c & a
\end{bmatrix} \mbox{ with } c\neq0.
\end{equation}
which are classic non-diagonalisable Jordan blocks.


*If $(a-d)\neq0$ then, for  $(a-d)^2+4bc = 0$ to hold, we must have that both $b\neq0$ and $c\neq0$. Also in this case there exist only one eigenvector, and the matrix is non-diagonalisable. An example of such a matrix is
\begin{equation}
A = \begin{bmatrix}
 5 & 1 \\
-4 & 1
\end{bmatrix} 
\end{equation}
