I have the matrix $$B=\begin{pmatrix}-6&4&-17\\0&-3&-4\\0&2&3\\\end{pmatrix}$$ which I found to have eigenvalues $\lambda_1=-6, \lambda_2=-1, \lambda_3=1$.
When attempting to find the eigenvectors for $\lambda_2$ and $\lambda_3$ I found the solution to contain free variables and I was unsure whether an eigenvector can contain free variables or not.
Thanks.
Best Answer
Notice that if $v$ is an eigenvector, then for any non-zero number $t$, $t\cdot v$ is also an eigenvector.
If this is the free variable that you refer to, then yes.
Edit:
In general, if $v_i\neq 0$ satisfies $Av_i = \lambda v_i$, then $$A\left( \sum_{i=1}^k \alpha_i v_i\right) = \lambda \left( \sum_{i=1}^k \alpha_i v_i\right)$$
That is if $ \sum_{i=1}^k \alpha_i v_i \ne 0$, then it is an eigenvector with respect to the same eigenvalue.