Let $F$ be a field and $V$ be a $n$-dimensional vector space over $F$.
For any subspace $W\subseteq V$ we say that $W'$ is a complement of $W$ if $$ W + W' = V.$$
I have the following question. Consider $W_1,\ldots, W_l$ finitely many subspaces of the same dimension $k$, then prove that there exists a subspace $W'$ of dimension $n-k$ such that it is a complement for all $W_i$.
I have tried by considering basis $B_i$ of $W_i$ and extending to basis of $V$; but I do not really know how to "simultaneously" extend the basis so I can define the complement $W'$.
I assume that $F$ is an infinite field, oherwise the statement is false.
By (backward) induction on $k$. If $k = n-1$, then pick any vector $v$ such that $v \notin W_j$ for every $j$ and let $W' = \mathrm{Span}(v)$.
Given $W_1, \dotsc, W_l$ of dimension $k$, take any vector $v$ such that $v \notin W_j$ for every $j$, and let $W'_j = \mathrm{Span}(W_j, v)$. Then $\dim W'_j = k+1$, and we apply the induction hypothesis on $W'_j$ to get $W'$ of dimension $n-k-1$ a complement for all $W'_j$. In particular this implies that $v \notin W'$, hence $\mathrm{Span}(W', v)$ is of dimension $n-k$ and is a complement for $W_1, \dotsc, W_l$.