Let $U$ be a proper subspace of a finite-dimensional vector space $V$ . Find a basis for $V$ containing no element of $U$.
We have no answer key for this question which i find annoying since that is the case for many questions in uni.
But if we look at a $R^2$ with basis vectors {$e_1, e_2$} and for example $U$ spans the vector $(1, 1)$. I get that this is a much simpler case but shouldn't it be the same for all other vector spaces of dimensions $n$? I struggle a lot with formally proving/showing that something is true and I'm not sure how to think when looking at arbitrary vector spaces.
Also I saw somewhere an explanation that said that: Let {$u_1, u_2, ... , u_k$} be a basis for $U$. Because $U$ is a subspace of $V$ {$u_1, u_2, ... , u_k$} are also in $V$. By adding linearly independent vectors from $V$ to the set of basis vectors of $U$ we eventually get a basis for $V$ {$u_1, u_2, ... , u_k, v_{k+1}, ... , v_n$}. And by removing all vectors that are in $U$ we get a basis {$v_{k+1}, ... , v_n$} for $V$ that contains no elements of $U$.
This I don't get however. Because how can they just remove the basis vectors of the subspace $U$? Wouldn't there be too few vectors to form the basis for $V$ after removing those? If we go back to the simpler case using the above strategy $U$ is spanned by {$u_1$} and $V$ = $R^2$. Adding one linearly independent vector from $V$ to the set gets us a basis for $V$ {$u_1, v_2$} and after removing $u_1$ we only have {$v_2$} left which is not a basis for $V$.
Am I wrong thinking this way? Can someone please explain.
What you "saw somewhere" is of course incorrect, as $v_{k+1},\ldots,v_n$ cannot be a basis for $V$ if you had a basis for $V$ with $n$ elements; this one just has $n-k$ elements.
Instead, let us consider the following: if $v_1,\ldots,v_n$ form a basis for $V$, and $1\lt i\leq n$, then $$v_1+v_i, v_2+v_i,\ldots, v_{i-1}+v_i, v_i, v_{i+1},\ldots,v_n$$ is still a basis. Why? Well, clearly it spans! If $\mathbf{x}\in V$, then we can write $$\mathbf{x} = \alpha_1v_1+\cdots +\alpha_nv_n.$$ Then $$\begin{align*} \mathbf{x} &= \alpha_1v_1+\cdots + \alpha_n v_n\\ &= \alpha_1(v_1+v_i) - \alpha_1v_i + \alpha_2(v_2+v_i) - \alpha_2v_i + \cdots \\&\qquad \mathop{+}\alpha_{i-1}(v_{i-1}+v_i) - \alpha_{i-1}v_i + \alpha_iv_i + \cdots + \alpha_n v_n\\ &= \alpha_1(v_1+v_i) + \alpha_2(v_2+v_i) + \cdots + \alpha_{i-1}(v_{i-1}+v_i)\\ &\qquad \mathop{+} (\alpha_i-(\alpha_1+\cdots+\alpha_{i-1}))v_i + \alpha_{i+1}v_{i+1} + \cdots + \alpha_n v_n. \end{align*}$$ Since it is a spanning set of vectors with exactly $n$ vectors, it is a basis.
Now, let $u_1,\ldots,u_k$ be a basis for $U$. Extend to a basis $u_1,\ldots,u_k,v_{k+1},\ldots,v_n$ of $V$. Note that $v_{k+1}\notin U$, and that $k+1\leq n$. Consider $$u_1+v_{k+1}, u_2+v_{k+1},\ldots, u_k+v_{k+1},v_{k+1},\ldots,v_n.$$ This is a basis. We know for sure that $v_{k+1},\ldots,v_n$ are not in $U$. Show that $u_i+v_{k+1}$, with $i=1,\ldots,k$, is also not in $U$.