A normal endomorphism that has a matrix with only real entries over a complex vector space, has pairwise always pairwise eigenvalues, meaning that we have an eigenvalue and its complex conjugate. now i was wondering whether this statement is also true for eigenvectors. therefore the question is: if we have $\lambda$ and $\bar{\lambda}$ as eigenvalues, do we also have eigenvectors $v$ and $\bar{v}$, where v belongs to $\lambda$ and $\bar{v}$ belongs to $\bar{\lambda}$?

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    $\begingroup$ Why not apply conjugation to $Av=\lambda v$? And notice that $A=\overline A$. $\endgroup$ – awllower Jun 2 '13 at 14:49
  • $\begingroup$ thank you, that was pretty helpful $\endgroup$ – user66906 Jun 2 '13 at 14:51
  • $\begingroup$ No Problem. Glad to contribute something. $\endgroup$ – awllower Jun 2 '13 at 14:53

The normality condition is irrelevant. If $\lambda$ is a real eigenvalue of the matrix $A$, take an eigenvector $\mathbf{v}$ and write it as $\mathbf{a}+i\mathbf{b}$, where $\mathbf{a}$ and $\mathbf{b}$ are vectors with real coefficients. Then $$ A\mathbf{v}=A\mathbf{a}+iA\mathbf{b} $$ so $$ \lambda\mathbf{a}+i\lambda\mathbf{b}=A\mathbf{a}+iA\mathbf{b} $$ and equating real and imaginary parts you get $$ A\mathbf{a}=\lambda\mathbf{a},\qquad A\mathbf{b}=\lambda\mathbf{b} $$ so you find a "real" eigenvector, because one among $\mathbf{a}$ and $\mathbf{b}$ must be non zero.

If $\lambda$ is not real, you can apply conjugation: if $\mathbf{v}$ is an eigenvector, then $A\mathbf{v}=\lambda\mathbf{v}$, so also $$ A\bar{\mathbf{v}}=\bar{\lambda}\bar{\mathbf{v}} $$

This shows also that the map $\mathbf{v}\mapsto\bar{\mathbf{v}}$ is a bijection between the eigenspaces relative to $\lambda$ and $\bar{\lambda}$. It's not a linear map, but easy considerations show that the two eigenspaces have the same dimension (a linear dependency relation in one space translates into a linear dependency relation in the other, with conjugate coefficients).


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