Question

Prove that A=[a b, c d] is diagonalizable if -4bc<(a-d)^2 and is not diagonalizable if -4bc>(a-d)^2.

Answer 1

\[A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\]

Answer 2

\[A-xI=\left[\begin{matrix}a-x & b \\ c & d-x\end{matrix}\right]\\charactristic- function -is \\(a-x)(d-x)-bc\\so\\x^2-(a+d)x+ad-bc=0\\x=\frac{ (a+d)\pm \sqrt{(a+d)^2-4(ad-bc)} }{ 2(a+d)\\so \\if\\ }\]

Answer 3

\[(a+d)^2-4(ad-bc) <0 \\ roots \\are \\imaginery \\a^2+d^2+2ad-4ad+4bc<0\\(a-d)^2<-4bc\]

Answer 4

Thanks @amoodarya! There's only one small error: the denominator of the quadratic should just be 2 I believe ;).

Answer 5

yes you are right