# How to solve eigenvalues from a characteristic equation.

#### u-will-neva-no

Joined Mar 22, 2011
230
Hey! I have worked out the characteristic equation using det(A-λI) = 0,
where
$$A = [\frac{a b}{c d}]$$ (ignore the line, it is a matrix but don't know how to do it in LaTex so used the fraction function)

So my characteristic equation is:
$$\lambda^2 -tr(A)\lambda + det(A) = 0$$

where $$tr(A) = a + d$$ and $$det(A) = ad - bc$$

How would I solve the eigenvalues for the above function? I understand that it is the values of $$\lambda = 0$$ but not sure how to make the function to to zero in this general case... Thanks!

Joined Jul 7, 2009
1,583
Your characteristic equation in $$\lambda$$ is a quadratic -- can't you just write down the two roots (eigenvalues) using the quadratic formula? I get $$\frac{a+b \pm \sqrt{(a+d)^2 - 4(ad-bc)}}{2}$$. As always, check the geezer's algebra. • u-will-neva-no

#### u-will-neva-no

Joined Mar 22, 2011
230
$$\frac{a+d \pm \sqrt(a^2 +d^2 -2(ad -2bc))}{2}$$

My complete question is the following. I need to obtain a relationship between the trace and the determinant of A. The question also talks about a formula
A = M $$\wedge$$ (M^-1) Would it be possible if anyone could explain/ give a solid example on how to use this formula?

#### u-will-neva-no

Joined Mar 22, 2011
230
@someonesdad, I think I was wrong to expand out and your form is alot better to deal with. Has anyone come across a problem like this before?

#### Georacer

Joined Nov 25, 2009
5,182
I agree with someonesdad's answer too. It's not that unusual of a problem. The methodology to set up your equation is standard. The fact that in the end you get a quadratic equation shouldn't trouble you, it's pretty common.

#### u-will-neva-no

Joined Mar 22, 2011
230
What I don't understand Is how I can find the eigenvectors when my two eigenvalues are messy. Can the above eigenvalues be expressed in a matrix form?

#### Georacer

Joined Nov 25, 2009
5,182

$$\left[ \begin{array}{c} \frac{a+b + \sqrt{(a+d)^2 - 4(ad-bc)}}{2} \\ \frac{a+b - \sqrt{(a+d)^2 - 4(ad-bc)}}{2} \end{array} \right]$$

Which makes your eigenvectors the solutions of the equation:
$$A \cdot x_i=\lambda_i \cdot x_i$$

I know it's fuzzy, but remember that in the end λi is a constant for your use.

• u-will-neva-no