TITLE: Two eigenvalues and an eigenvector walk into a bar...
QUESTION [8 upvotes]: Suppose I have the transformation $T(v) = Av = \lambda v$. If two of the eigenvalues are $\lambda_1$ and $\lambda_2$ where $\lambda_1=-\lambda_2$, is there a way to quickly find the eigenvector(s) for $\lambda_2$ if I know the eigenvector(s) for $\lambda_1$? I ask this because attempting to find the kernel of larger matrices with irrational numbers and writing down each step is time consuming. I noticed a subtle relationship between the eigenvectors for eigenvalues of opposite signs in that the entries are the same with the exception of a negative sign somewhere.
Any assistance is much appreciated!

REPLY [12 votes]: Sadly, the answer to this question is likely "can't be done easily" in general: the eigenvectors can be pretty much anything.
To see why, pick your favourite pair of linearly independent vectors $v, w \in \mathbb{R}^2$ (this generalizes to higher dimensions and different base fields), and some $\lambda \neq 0$. Let $$T := \left( \begin{array}{cc}v_1 & w_1\\v_2 & w_2 \end{array} \right), D := \left( \begin{array}{cc} \lambda & 0 \\ 0 & -\lambda \end{array} \right).$$
Then $T D T^{-1}$ is a $2 \times 2$ matrix with Eigenvectors $v$ and $w$, and Eigenvalues $\lambda$ (for $v$) and $-\lambda$ (for $w$).
That being said, there may be numeric tricks to help you - but these will not be straightforward.