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!

 

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.

 

My quadratic equation came to:
\( \frac{a+d \pm \sqrt(a^2 +d^2 -2(ad -2bc))}{2}\)

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?

 

For tables in LaTeX, check this tutorial out:
http://forum.allaboutcircuits.com/showthread.php?t=60409

It's a bit tricky, but rewarding.

 

@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?

 

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.

 

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?

 

Yes, practically your eigenvalues are:

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

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.

 

Cool, thanks Georacer!