If $$M = {\begin{pmatrix} a&b\\ c&d\\ \end{pmatrix}}$$ Prove that if $\operatorname{tr}(M)^2 > 4\det(M)$, then $M$ is diagonalisable.
So the $\det(M)= ad - bc$ and the trace of $M$, $\operatorname{tr}(M)= a+d$.
Therefore, we should prove that $$ (a+d)^2 > 4(ad-bc) $$ Am I allowed to use the formula for quadratic polynomial, can I expess it like that: $$ a^2 -4(ad + bc) + d^2 $$ but what would that gain me? Could you please give me a hint? Thanks.
The characteristic polynomial of $M$ is$$(x-a)(x-d)-bc=x^2-(a+d)x+ad-bc.$$It has two distinct real roots if and only if$$(a+d)^2-4(ad-bc)>0,$$that is, if and only if $\operatorname{tr}^2M>4\det M$.