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I got a question to solve for complex eigen value and eigenvector of the following matrix: A= $ \begin{pmatrix} -7 & 4 \\ -9 & 5 \\ \end{pmatrix} $ . I worked it out and obtained eigen value $-1$ which I wrote as $ i^2 $. but then it gives 2 eigen vectors but is it possible to have $2$ eigen vectors for $1$ eigen value?

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    $\begingroup$ It's possible, but that eigenvalue has only one linearly independent eigenvector. $\endgroup$ – Gerry Myerson May 10 '15 at 13:10
  • $\begingroup$ $-1$ is indeed an eigenvalue, but why do you write that eigenvalue as a square? There seems to be no obvious reason to go that way. $\endgroup$ – Henning Makholm May 10 '15 at 13:11
  • $\begingroup$ but it asked to solve for complex eigen values of A,and hence calculate the complex eigenvectors of A. $\endgroup$ – Serena May 10 '15 at 13:37
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the characteristic polynomial of $A=\pmatrix{-7&4\\-9&5}$ is $det(A+\lambda I) = (\lambda+1)^2.$ so $A$ has one one eigenvalue $-1$ repeating. the null space of $A+I$ has dimension $1$ and an eigenvector corresponding to the eigenvalue is $u = \pmatrix{2\\3}.$ you can find a generalized eigenvector $v = \pmatrix{-1\\-1}$ satisfying $(A+ I)v = u.$

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Yes. It simply scales the plane $span(v_1, v_2)$ by a $-1$ factor.

Or, differently, if im not wrong, the plane $span(v_1, v_2)$ is being reflected to the line given by: $span(v_1+v_2)=\alpha (v_1+v_2) = \alpha w$

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