Is it possible for the basis of two distinct eigenspaces to share elements I am solving a problem of finding a diagonal matrix $A$ such that $A$
such that $T: A \rightarrow A^t $ for $A \in \text{Mat}(2x2)$ over $(\Bbb R)$. I have found two eigenvalues $1$, and $-1$ and it seems that they have the following Eigenspaces
$$E_1={(1,0,0,0)^t,(0,1,1,0)^t}$$
$$E_2 = {(1,0,0,0)^t,(0,1,-1,0)^t,(0,0,0,1)^t}$$
Is this possible or have I made some kind of error?
Edit: I clearly made an error with the eigenspace of 1, namely that there was only one vector (0,1,1,0).

If $x$ is an eigenvector for $1$ and for $-1$, you have
$$
x=Ax=-x.
$$
So $2x=0$, and then $x=0$. In summary, a non-zero vector cannot be an eigenvector for two different eigenvalues for the same matrix. 
