Basis of a vectorspace If $V$ is a vectorspace with basis $\{e_1,e_2,e_3\}$, and $W = vct\{e_1,e_2\}$. Is it in general true that there exists a basis $\{v_1,v_2,v_3\}$ of $V$ with $v_i \notin W$ for $i=1,2,3$?
I thought yes, first I considered the canonical example of $\mathbb{R}^3$ with the standard basis $\{e_1,e_2,e_3\}$. Then if you take $vct\{e_1,e_2\}$, and take a basis for $\mathbb{R}^3$, $\{v_1,v_2,v_3\}$, as
$$
\{(1,1,1),(0,1,1),(0,0,1)\}
$$
none of which is in $vct\{e_1,e_2\}$. 
So intuitively to me it seems plausible that it is true in general but how to prove this in general? One option I thought of: given $\mathbb{R}^3$ and any basis, we can make a basis transformation to $\{e_1,e_2,e_3\}$ and then apply the above. given any $V$ with dimension 3, it is isomoprhic to $\mathbb{R}^3$, so the same procedure works. 
However, firstly, I'm not sure if this is correct? 
Secondly, this is a question that appears after a chapter when isomorphisms, linear maps, and basis transformations have not yet been introduced, so there should be a way to do it without?
 A: Hint: What can be said about $\{e_1+e_3, e_2+e_3, e_3\}$?
A: It is true in general.

Let $V$ be a finite-dimensional vector space and $W$ its subspace. Then there exists a basis for $V$ not intersecting $W$.

Assume $\dim V = n$ and $\dim W = k$. Let $\{w_1, \ldots, w_k\}$ be a basis for $W$ and $\{w_1, \ldots, w_k, v_{k+1}, \ldots, v_n\}$ its extension to a basis for $V$. Let's check that $$B = \{w_1 + v_{k+1}, w_2 + v_{k+1}, \ldots, w_n + v_{k+1}, v_{k+1}, v_{k+2}, \ldots, v_n\}$$ is a basis for $V$ and that $B \cap W = \emptyset$.
Since $\mathrm{card} \,B = n$, to show that $B$ is a basis it is enough to check that it is linearly independent.
Indeed, let $\alpha_1, \ldots, \alpha_k, \beta_{k+1}, \ldots, \beta_n$ be scalars such that:
$$0 = \sum_{i=1}^k \alpha_i(w_i + v_{k+1}) + \sum_{j=k+1}^n \beta_jv_j = \sum_{i=1}^k \alpha_iw_i + \left(\sum_{i=1}^k \alpha_i + \beta_{k+1}\right)v_{k+1} + \sum_{j=k+2}^n \beta_jv_j$$
This is a linear combination of vectors from the basis $\{w_1, \ldots, w_k, v_{k+1}, \ldots, v_n\}$ so we have $\alpha_i = 0$ for $i = 1, \ldots, k$, then $\sum_{i=1}^k \alpha_i + \beta_{k+1} = 0$ and finally $\beta_j = 0$ for $j = k + 2, \ldots n$. The equation $\sum_{i=1}^k \alpha_i + \beta_{k+1} = 0$ implies $\beta_{k+1} = 0$. Therefore, all scalars are $0$.
Thus, $B$ is a basis. Now assume $w_i + v_{k+1}\in W$. Then we would also have $v_{k+1} = -w_i + (w_i+v_{k+1}) \in W$, which is a contradiction because $v_{k+1} \notin W$ (otherwise $\{w_1, \ldots, w_k, v_{k+1}, \ldots, v_n\}$ would not be linearly independent.)

In your case, we can use this result to conclude that $\{e_1 + e_3, e_2 + e_3, e_3\}$ is a basis for $V$ disjoint with $W$.
