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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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    The line on top is notation for taking the Complex Conjugate. Have fun!2012-12-09

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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 $\mathbb{C}$.

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    Thanks. I will try doing it and come back for help if I get suck. :)2012-12-09