Prove eigen values of matrix A are real numbers

Let A be $\ \begin{bmatrix} a & c \\ c & b\end{bmatrix}$ where $\ a,b,c, \in \mathbf R$

Prove $\ A$ eigen values are real numbers.

I guess it should be pretty straight forward so I just need to see what are solutions of characteristic polynomial which will be $\ |A - \lambda I| = (a-\lambda)(b - \lambda) - c^2 = 0$ but Im not sure how do I prove the only possible values are in $\ \mathbf R$ .

$\ \lambda^2 - \lambda a - \lambda b + ab - c^2 = 0$

• What's the discriminant of this quadratic equation? – Lord Shark the Unknown Jul 7 '18 at 16:41
• Maybe this will be of help:math.stackexchange.com/questions/354115/…. – StubbornAtom Jul 7 '18 at 16:43
• Use the general formla for $\lambda^2-\lambda (a+b)+(ab-c^2)=0$. Why the solution(s) always real? – RLC Jul 7 '18 at 16:47

Just solve the quadratic equation that you wrote and it will simplify to $λ = \frac{a+b\pm \sqrt{\left(a-b\right)^2+\left(2c\right)^2}}{2}$ from which you can see both roots are real numbers.
Edit: Write your quadratic equation as $\lambda ^2-\lambda \left(a+b\right)+ab-c^2=0$ . Now find roots using quadratic formula $\frac{-B\pm \sqrt{B^2-4AC}}{2A}$. Here $B=-(a+b)$ , $A=1$ and $C= ab-c^2$. Just substitute the values and simplify.
The roots are$$\frac{a+b\pm\sqrt{(a+b)^2-4(ab-c^2)}}2=\frac{a+b\pm\sqrt{(a-b)^2+4c^2}}2\in\mathbb R.$$