linear algebra - Prove that the matrix A is diagonalizable if $-4bc < (a-d)^2$

I'm trying to prove that matrix A is diagonalizable
if $-4bc < (a - d)^2$ and is not diagonalizable if $-4bc > (a - d)^2$

$A = \begin{bmatrix}
a & b\\

c & d
\end{bmatrix}$

I'm not sure how to start. I was thinking of taking the inverse of A but I'm sure if that's the right approach.

Answer

I am assuming that you want to work in the field of real numbers. In this case you have the following

1) If the matrix is diagonalizable, then it must have both the roots real

2) If the matrix has eigenvalues that are not repeated, then it can be diagonalized with the following additional comment: If the eigenvalues are not real, the diagonalization would require complex numbers.

3) The only real matrices with repeated eigenvalues that is also diagonalizable are multiples of the identity matrix, i.e. diagonal matrices with equal values on the diagonal.

To OP:
In one of your comments you have
$$
\lambda^2-(a+b)\lambda+ad+bc$$
It should be
$$
\lambda^2-(a+d)\lambda+ad-bc$$

So for real roots that are distinct, you need
$$
(a+d)^2 -4(ad - bc) = a^2 + 2 a d + d^2 - 4 a d + 4bc = (a-d)^2 + 4 bc \gt 0$$
or
$$ (a-d)^2 \gt -4bc$$

Note $a=d, b=1, c=0$ is not diagonalizable. So you cannot change the $\gt$ to $\ge$.
Also, $a=d, b=c=0$ is diagonalizable. So
$$
\text{If }(a-d)^2 = -4bc, \text{ the matrix is diagonalizable if and only if $b=0$ and $c=0$}$$

** Added in reply to OP **

Let $\Delta$ be the discriminant

There are 3 cases

1. $\Delta < 0$. Both the eigenvalues are complex so matrix can not be diagonalized.


2. $\Delta >0$. Both eigenvalues are real and distinct (different). Matrix can be diagonalized



3. $\Delta=0$. Both eigenvalues are equal. Then there are two cases: Matrix is diagonal, i.e. $b=c=0$. Matrix is not diagonal $b \ne 0$ or $c \ne 0$. In the first case the matrix is diagonalizable (it is already diagonal!). In the second case the matrix cannot be diagonalized.