Question on finding eigenvalues of another linear transformation 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? 

Arturo and joriki gave you good advices. You should find the matrix of your endomorphism $T(A) = A^t$ associated to the standard basis of $M_2(\mathbb{R})$:
$$
E_1 = 
\begin{pmatrix}
1    &   0  \\
0    &   0
\end{pmatrix}
\ , \qquad 
E_2 =
\begin{pmatrix}
0    &   1  \\
0    &   0
\end{pmatrix}
\ , \qquad 
E_3 =
\begin{pmatrix}
0    &   0  \\
1    &   0
\end{pmatrix}
\ , \qquad 
E_4 =
\begin{pmatrix}
0    &   0  \\
0    &   1
\end{pmatrix}
\ . 
$$
That is,
$$
T(E_1) = 
\begin{pmatrix}
1    &   0  \\
0    &   0
\end{pmatrix}
\ , \qquad 
T(E_2) =
\begin{pmatrix}
0    &   0  \\
1    &   0
\end{pmatrix}
\ , \qquad 
T(E_3) =
\begin{pmatrix}
0    &   1  \\
0    &   0
\end{pmatrix}
\ , \qquad 
T(E_4) =
\begin{pmatrix}
0    &   0  \\
0    &   1
\end{pmatrix}
\ . 
$$
From which you immediatly realise that there is at least one eigenvalue, $1$, with at least a $2$-dimensional subspace of eigenvectors -the span of $E_1$ and $E_4$. To get everything, you just put this data into the associated matrix of $T$:
$$
M =
\begin{pmatrix}
1  &  0  &  0  &  0  \\
0  &  0  &  1  &  0  \\
0  &  1  &  0  &  0  \\
0  &  0  &  0  &  1
\end{pmatrix}
$$
compute its characteristic polynomial:
$$
Q_M(t) = (t-1)^3 (t+1)
$$
and find out that there was an eigenvalue you didn't see: $-1$.
If you compute now the respective eigenspaces, you'll find that
$$
\ker (M - I ) = \mathrm{span} (E_1, E_2 + E_3, E_4)
$$
is the subspace of symmetric matrices and 
$$
\ker (M + I) = \mathrm{span}(E_2 - E_3)
$$
is the subspace of skew-symmetric matrices.
