1. ## Characteristic equation proof

Hi

Im confused on where to start in proving this question

Let A = $\displaystyle \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$ where a, b, c and d are real numbers. Show that the characteristic equation of A is $\displaystyle \lambda^2 - tr(A)\lambda + det(A) = 0$
where tr(A) = trace of A.

Thanks

2. [quote=chuckienz;312692]Hi

Im confused on where to start in proving this question

Let A = $\displaystyle \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$ where a, b, c and d are real numbers. Show that the characteristic equation of A is $\displaystyle \lambda^2 - tr(A)\lambda + det(A) = 0$
where tr(A) = trace of A.

$\displaystyle \det (A)=ac-bd$

$\displaystyle tr(a)=a+d$

to compute the char poly we evaluate the determinate of

$\displaystyle \left(\begin{array}{cc}a -\lambda &b\\c&d -\lambda \end{array}\right)=(a-\lambda)(d-\lambda)-cb$

Expanding we get

$\displaystyle ad -a\lambda -b\lambda +\lambda^2-cb$

$\displaystyle \lambda^2-(a+b)\lambda +ad-cb=$

Now just sub we what we know from above and we are done