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How do I find a complex number $\lambda $ such that $\pmatrix{ 3&-2\\2&3}\vec v$ = $\lambda\vec v $ where $ \vec v $ is non-zero. Yes, this is a homework problem, I didn't learn complex number in my previous linear alg course.

add....

So far I found $\lambda = 3-2i$ is this the only solution?

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What about the conjugate of your $\lambda$? –  André Nicolas Sep 16 '12 at 21:48

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I don't know if I understand this notation (I edited and don't know if I edited correctly =p).

To find $\lambda$, calculate the characteristic polynomial (here, $I$ denote the identity matrix, and $A=\pmatrix{3&-2\\2&3}$) $$ p(x)=\det(A-xI)=\det\pmatrix{3-x&-2\\2&3-x}=x^2-6x+13 $$ and the roots will be exactly the $\lambda$'s you want. In this case, $\lambda=3\pm 2i$.

(See this for more details, the idea of "why this works", and how to calculate $\vec v$ such that $A\vec v=\lambda\vec v$)

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Followed your procedure and it's definitely correct. I thought of solving it using eiganvectors but I don't know to solve polynomial for complex numbers. –  xiamx Sep 16 '12 at 22:07
    
@xiamx: if the polynomial $p(x)=ax^2+bx+c$, you can solve by the classical formula (independent if this is complex or no) $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ if the degree is greater, it can be hard solving... =p –  Yuki Sep 16 '12 at 22:23

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