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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 $

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

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  • $\begingroup$ Thanks!! i'm weak with algebra.. $\endgroup$ – bm1125 Jul 7 '18 at 16:59
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    $\begingroup$ Try not to get scared by big expressions and many variables. Sometimes we reach a point which looks like a dead end but if you have a belief that you are doing things right then keep doing it until it is unsolvable by you. Also keep checking for errors in sign and when substituting value. Don't be afraid to use quadratic formula andkeep practicing it so that you get used to it. $\endgroup$ – Seth Rollins Jul 7 '18 at 17:04
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The roots are$$\frac{a+b\pm\sqrt{(a+b)^2-4(ab-c^2)}}2=\frac{a+b\pm\sqrt{(a-b)^2+4c^2}}2\in\mathbb R.$$

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