Fan basis for $\mathbb{C}^2$ 
Find a fan basis for the linear maps of $\mathbb{C}^2$ represented by the matrices: a)$\begin{bmatrix}1&1\\1&1\end{bmatrix}$

Solution.(a) Clearly, the vector $\begin{bmatrix}1\\-1\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue 0. Let $V_1$ be the space generated by this vector. Then $\{V_1,\mathbb{C}^2\}$ is a fan for the given linear map.
Solution´s Manual for Lang´s Linear Algebra, Rami Shakarchi.
Fan basis definition: By a fan basis we shall mean a basis $\{v_1,...v_n\}$ of V such that $\{v_1,...,v_i\}$ is a basis for $V_i$.Linear Algebra, Serge Lang.
Questions:
1) I may be misunderstanding the defintions I am studying. Since $\begin{bmatrix}1\\-1\end{bmatrix}$ generates a subspace of $\mathbb{C}^2$, then $\{V_1\mathbb{C}^2\}$ is a fan basis. Can I pick an arbitrary subspace? Why does the subspace needs to be generated by a basis that is an eigenvector?
2) If I multiply $\begin{bmatrix}1\\-1\end{bmatrix}$ by a negative scalar, it proves that $\begin{bmatrix}1\\-1\end{bmatrix}$ does not generate a subspace. Is this true?
 A: 
1) I may be misunderstanding the defintions I am studying. Since $\begin{bmatrix}1\\-1\end{bmatrix}$ generates a subspace of $\mathbb{C}^2$, then $\{V_1\mathbb{C}^2\}$ is a fan basis. Can I pick an arbitrary subspace? Why does the subspace needs to be generated by a basis that is an eigenvector?

You're missing a key part of the definition of a fan basis. 
Let $~f:V\longrightarrow V~$ be a linear map, $\dim V =n$.
Definition: A basis $( v_1, \ldots, v_n)$ of $V$ such that for all $j=1, \ldots,n$ the space $\text{span}(v_1,\ldots,v_j)$ is $f$-invariant is called a fan basis of $V$ with respect to $f$. See: A matrix-free way to find a fan basis of $V$? for where I clipped that text from.
From this definition, it would appear that you're missing a vector, namely $$v_2=\left[
\begin{array}{c}
1\\
1
\end{array}
\right].$$
You should check that this is, in fact, an eigenvector with eigenvalue 2, and this should allow you to construct a fan basis for $\Bbb{C}^2,$ in agreement with the definition Leo has provided in the link.

2) If I multiply $\begin{bmatrix}1\\-1\end{bmatrix}$ by a negative scalar, it proves that $\begin{bmatrix}1\\-1\end{bmatrix}$ does not generate a subspace. Is this true?

No this is not true. Every vector generates a subspace via the "span" operation. In particular
$$\text{span}\left(\left[
\begin{array}{c}
1\\
-1
\end{array}
\right]\right)=\left\{ \alpha \left[
\begin{array}{c}
1\\
-1
\end{array}
\right], \text{ where } \alpha \in \Bbb{C}\right\}=V_1.$$
You should verify that this is a subspace, and that for any negative scalar, say for example $-1,$ that $$-1\left[
\begin{array}{c}
1\\
-1
\end{array}
\right]=\left[
\begin{array}{c}
-1\\
1
\end{array}
\right]\in \text{span}\left(\left[
\begin{array}{c}
1\\
-1
\end{array}
\right]\right).$$
