$f \in \mathrm{End} (\mathbb{C^2})$ $f(e_1)=e_1+e_2$ $f(e_2)=e_2-e_1$. Eigenvalues of f and the bases of the associated eigenspaces Let $f \in \mathrm{End} (\mathbb{C^2})$ be defined by its image on the standard basis $(e_1,e_2)$: 
$f(e_1)=e_1+e_2$
$f(e_2)=e_2-e_1$
I want to determine all eigenvalues of f and the bases of the associated eigenspaces.
First of all how does the transformation matrix of $f$ look like?
Is it 
$\begin{pmatrix}1 &-1 \\1 &1 \end{pmatrix}$?

Yes, the matrix of $f$ is
$$
A=\begin{pmatrix}
1 & -1 \\
1 & 1
\end{pmatrix}
$$
because $f(e_1)=1e_1+1e_2$ and $f(e_2)=(-1)e_1+1e_2$ and you fill the columns with the coefficients of the linear combinations.
The characteristic polynomial is $p(X)=(1-X)^2+1=X^2-2X+2$, so the eigenvalues are $\lambda_1=1+i$ and $\lambda_2=1-i$.
You need to find a vector in the null space of $A-(1+i)I$: Gaussian elimination yields
$$
\begin{pmatrix}
-i & -1 \\
1 & -i
\end{pmatrix}
\xrightarrow{R_1\gets i R_1}
\begin{pmatrix}
1 & -i \\
1 & -i
\end{pmatrix}
\xrightarrow{R_2\gets R_2-R_1}
\begin{pmatrix}
1 & -i \\
0 & 0
\end{pmatrix}
$$
Since you want to solve the system $(A-(1+i)I)x=0$, a nonzero solution is
$$
\begin{pmatrix} i \\ 1 \end{pmatrix}
$$
Do similarly for the other eigenvalue.
