Prove $A$ is a diagonalisable matrix

Question

Let $A\in \mathbb{R}^{2 \times 2}$ such that $\det(A)<0$. Show that $A$ is diagonalisable.

Setting

$$A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

We get $p_A(\lambda)=\lambda^2-(a+d)\lambda+ad-bc$. How can I conclude that $A$ is diagonalisable?

Answer 1

Yes, because if $\alpha$ and $\beta$ are the eigenvalues of $A$, then $\alpha\beta=\det A<0$. Therefore, $\alpha\neq\beta$.

Answer 2

Hint: Distinct eigenvalues implies diagonalizable. Note that, $detA$=Product of eigenvalues.

Answer 3

We know that $ad-bc$ is negative. Now Descartes rule of signs guarantees that $x^2+px+q=0$ has a positive and a negative real solution, provided that $q$ is negative.