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Have a square matrix problem that involves complex numbers and am at a loss. $M$ is a square matrix with real entries. $\lambda = a + ib$ is a complex eigenvalue of $M$, show that the complex conjugate $\bar{\lambda} = a - ib$ of $\lambda$ is also an eigenvalue of $M$. Does solving this relate to a matrix having a characteristic polynomial of $(t^2-4)$? Could the community please explain?

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    $\begingroup$ Where id $t^2-4$ come from? $\endgroup$ – Marc van Leeuwen Feb 18 '15 at 9:57
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let the eigenvector corresponding to the eigenvalue $a+ib$ be $u+iv$ with both $u$ and $v$ real and $|u|^2 + |v|^2 \neq 0$. then we have $M(u+iv) = (a+ib)(u+iv)$ can be separated to give $Mu = au - bv, Mv = av + bu.$

now, $M(u -iv) = Mu - iMv = (au-bv) -i(av+bu) = a(u-iv) -ib(u-iv) = (a-ib)(u-iv)$ this shows that $a-ib$ is an eigenvalue of $M.$

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  • $\begingroup$ And you get the eigenvector for $a - bi$ to boot! Well done, Plus One!!! $\endgroup$ – Robert Lewis Feb 18 '15 at 6:05
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$M$ having real entries, its characteristic polynomial will have real coefficients. Say this polynomial is $P(X)=a_nX^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$ One has by definition of eigenvalues that $P(\lambda)=0$ taking the complex conjugate of that identity and considering that all $a_i$ are real one gets $P(\bar\lambda)=0$ and this means $\bar\lambda$ is as well an eigenvalue.

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If $Mv=\lambda v$ where $v=(v_1,...,v_n)$ is a vector of complex numbers, then $$M\bar v=\overline{Mv}=\overline{\lambda v}=\bar\lambda \bar v$$ Here by writing a line over a matrix (or vector) I mean to take the complex conjugate of every entry; the first equality holds because $M$ is real, so $\bar M=M$, and because as you can check directly $\bar M\bar v=\overline{Mv}$ for any complex matrix $M$ and vector $v$. So $\bar v$ is a $\bar\lambda$-eigenvector of $M$.

This does relate to the characteristic polynomial $\chi M$: every eigenvalue is a root of $\chi M$, and if $M$ is real then $\chi M$ has real coefficients, and if a real polynomial $p$ has a complex root $\lambda$ then $\bar \lambda$ is a root as well.

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If you consider complex eigenvalues of a matrix $M$, then you are implicitly interpreting it as the matrix of a complex-linear operator $\phi$ acting on a complex vector space$~V$, even if $M$ has only real entries. The fact that $M$ has real entries then is basis dependent (it will probably not hold for the matrix of$~\phi$ with respect to any basis of$~V$); if $V_\Re$ denotes the real span of the chosen basis and $V_\Im$ the real span of the basis vectors multiplied by $\def\ii{\mathbf i}\ii$, then the fact that $M$ has real entries means that $\phi$ commutes, not only with multiplication by$~\ii$ as any complex-linear operator must, but also with "complex conjugation": the $\Bbb R$-linear operator $J:V\to V$ that acts by a scalar$~1$ on$~V_\Re$ and by a scalar$~{-}1$ on$~V_\Im$. Note that $J$ anti-commutes with multiplication by$~\ii$, so it is not complex-linear, but conjugate-linear: (it is $\Bbb R$-linear and) $J(\lambda v)=\overline\lambda J(v)$ for all $\lambda\in\Bbb C$ and $v\in V$.

Now if $v\in V$ is an eigenvector of$~\phi$ for an eigenvalue$~\lambda\in\Bbb C$, this means that $\phi(v)=\lambda v$. Now from the above $\phi(J(v))=J(\phi(v))=J(\lambda v)=\overline\lambda J(v)$, so $J(v)$ is an eigenvector for $\overline\lambda$.

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