I have a quick question on finding eigenvectors for a linear transformation. I'm given:
$T(A) = A^t$ where $A = M_2$ i.e. a $2 \times 2$ matrix consisting of real numbers.
So the general approach is to solve $T(v) = \lambda v$, and from there I set up:
$ \left\{\begin{align*} a &= \lambda a \\ d &= \lambda d \\ b &= \lambda c \\ c &= \lambda b \end{align*}\right. $
which suggests that that $\lambda = 1$ is the only eigenvalue. Given that, I can verify that $T(v) = \lambda v$ holds when $\lambda = 1$ and $b = c$.
But I'm not sure how to describe the eigenvector. How do I put it into terms? I was thinking that a basis for the solution set would be:
$a \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\\ 1 & 0 \end{pmatrix} $
but that's not an eigenvector. How exactly would I describe the eigenvector for $\lambda = 1$?
Thanks in advance.
Edit: So if you have a matrix with two eigenvectors, both of which are 2 x 2 matrices, how would you go about diagonalizing it? i.e. if you have to solve $\Lambda = SAS^{-1}$, how do you determine S?