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Given a Real square matrix $A$, it's easy to show that if $v$ is an eigenvector relative to the eigenvalue $\lambda$, that is $A v=\lambda v$, then $\bar v$ is another eigenvector for $A$ relative to the eigenvalue $\bar\lambda$.

Can we say something similar about opposite eigenvalues/vectors? Specifically, if i know that $\lambda$ and $-\lambda$ are both eigenvalues of $A$, is there any relationship between the relative eigenvectors?

I've played around a bit with this but came to no interesting conclusion.

blundered_bishop
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    think about it geometrically, if the eigenvector is that direction that is only scaled by the eigenvalue, then doesn't mean any thing that some other direction is scaled by the negative of that value, the latter direction can be any other one. – Physor Jun 20 '21 at 13:45

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If $(x,y)$ and $(u,v)$ are two linearly independent vectors and $\lambda \neq 0$, you can see that $$\pmatrix{x & u \\ y & v}\pmatrix{\lambda & 0 \\ 0 & -\lambda}\pmatrix{x & u \\ y & v}^{-1}$$ has eigenvector $(x,y)$ with eigenvalue $\lambda$ and eigenvector $(u,v)$ with eigenvalue $-\lambda$. The choice of two linearly independent vectors was arbitrary, so having eigenvalue $\lambda, -\lambda$ doesn't imply anything smart sounding about the eigenvectors. Other than that there exists at least two of them that are linearly independent as $\lambda \neq -\lambda$ whenever $\lambda \neq 0$ and distinct eigenvalues correspond to linearly indepent eigenvectors.

Steve
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