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.
Best Answer
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.