# Do we consider complex values for eigenvalues

I have an intro Linear Algebra assignment due tomorrow and I'm unsure of what the teacher expects. (I emailed him but he hasn't replied yet) Basically we never learned this stuff in class. It's an "extension" assignment. One of the questions is

Find the eigenvalues and eigenvectors of $A=\begin{pmatrix}3&1\\-1&2\end{pmatrix}$

I did the following:

$\det(A - \lambda I)=0$

$\lambda^2 -5\lambda + 7 = 0$

This has no real solutions. Normally, for an Linear Algebra 101 class, would this mean that there are no eigenvalues and no eigenvectors. Thanks.

P.S. It doesn't allow me to post the picture. Sorry.

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See: wolframalpha.com/input/…, you may also want to read this: math.stackexchange.com/questions/15315/…, lastly, review: math.utk.edu/~freire/complex-eig2005.pdf – Amzoti Dec 9 '12 at 22:51
@Amzoti That doesn't answer my question. I know we can have complex valued eigenvalues but isn't that very advanced for an Linear Algebra 101 class? – Jey Dec 9 '12 at 22:54
It sounds like he gave you an 'extension' assignment, so it may be intended to challenge you and get you to explore, question and see if you can figure it out - after all, that is the true purpose of learning, to learn to teach yourself and to hopefully be able to contribute to the domain of knowledge. – Amzoti Dec 9 '12 at 22:56
@Amzoti Thanks for the link to the pdf file. It is helpful. However I have a quick question, what does the notation:=, lambda (or "A") with a line on top, mean? Thanks. – Jey Dec 9 '12 at 23:15
The line on top is notation for taking the Complex Conjugate. Have fun! – Amzoti Dec 9 '12 at 23:20

You allow the possibility of complex eigenvalues and eigenvectors, you treat everything exactly the same as in the real case. If the vector space you're working in is over the field $\mathbb{R}$ rather than $\mathbb{C}$ then all this means is that your matrix is not diagonalisable over $\mathbb{R}$ even though it is diagonal over $\mathbb{C}$.