$f \in \mathrm{End} (\mathbb{C^2})$ $f(e_1)=e_1+e_2$ $f(e_2)=e_2-e_1$. 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 
$\begin{pmatrix}1 &-1 \\1 &1 \end{pmatrix}$?
 A: If one represents the standard basis $e_1$, $e_2$ in the usual form
$e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \tag 1$
$e_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \tag 2$
and writes the matrix of $f$ as
$[f] = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}, \tag 3$
then we have, since
$f(e_1) = e_1 + e_2, \; f(e_2) = e_2 - e_1, \tag 4$
$\begin{pmatrix} 1 \\ 1 \end{pmatrix} = e_1 + e_2 = [f]e_1 = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \alpha \\ \gamma \end{pmatrix}, \tag 5$
and
$\begin{pmatrix} -1 \\ 1 \end{pmatrix} = e_2 - e_1 = [f]e_2 = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \beta \\ \delta \end{pmatrix}, \tag 6$
from which it immediately follows that
$\alpha = \gamma = \delta = 1, \tag 7$
$\beta = -1; \tag 8$
thus
$[f] = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}, \tag 9$
as anticipated by our OP user567319.  Once we have (9), it is an easy matter to find the eigenvalues of $f$, sincd they must satisfy
$0 = \det([f] - \lambda I) = \det \left ( \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} - \lambda I \right ) = \det \left (\begin{bmatrix} 1 - \lambda & -1 \\ 1 & 1 - \lambda \end{bmatrix} \right )$
$= (1 - \lambda)^2 + 1 = \lambda^2 - 2\lambda + 2; \tag{10}$
it follows from (10), using the quadratic formula, that
$\lambda = \dfrac{1}{2}(2 \pm \sqrt{-4}) = \dfrac{1}{2}(2 \pm 2 i) = 1 \pm i; \tag{11}$
it is now an easy matter to find the eigenvectors, satisfying as they do 
$\begin{pmatrix} \lambda \mu \\ \lambda \nu \end{pmatrix} = \lambda \begin{pmatrix} \mu \\ \nu \end{pmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{pmatrix} \mu \\ \nu \end{pmatrix} = \begin{pmatrix} \mu - \nu \\ \mu + \nu \end{pmatrix}, \tag{12}$
whence
$\lambda \mu = \mu - \nu, \tag{13}$
$\lambda \nu = \mu + \nu; \tag{14}$
from (13),
$(1 - \lambda) \mu = \nu; \tag{15}$
it follows that, taking $\mu = 1$, the eigenvectors are
$\begin{pmatrix} 1 \\ \mp i \end{pmatrix} = \begin{pmatrix} 1 \\ 1 - \lambda \end{pmatrix}, \; \lambda = 1 \pm i; \tag{16}$
we check the case $\lambda = 1 + i$:
$\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{pmatrix} 1 \\ -i \end{pmatrix} = \begin{pmatrix} 1 + i \\ 1 - i \end{pmatrix} = (1 + i)\begin{pmatrix} 1 \\ -i \end{pmatrix}; \tag{17}$
a check of the case $\lambda = 1 - i$ follows from this by complex conjugation, since $[f]$ is a real matrix.
A: 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.
