Show that there exists some vector subspace $F$ with $\textrm{dim}(F)=n-m$ and $F\cap F_i=\left\{0\right\}$ Let $E$ be a $K$-vector space of dimension $n$.
Moreover, let $F_1,\ldots,F_k$ be vector subspaces of $E$ with $\textrm{dim}(F_i)\leq m$, where $0<m<n$.
Show that there exists some vector subspace $F$ of $E$ with $\textrm{dim}(F)=n-m$ and $F\cap F_i=\{0\}$ for all $i=1,2,\ldots,k$.

My first idea is to use that
$$
E\neq \bigcup_{i=1}^k F_i\tag{1}
$$
since all the vector subspaces $F_i$ are proper due to $m<n$. Hence, there must exist some vector subspace $F$ of $E$ such that
$$
E=\bigcup_{i=1}^kF_i\cup F.\tag{2}
$$
But from (1) and (2) it already follows that $F\cap F_i=\{0\}$ for all $i=1,\ldots,k$.
So, I think, it remains to show that $\textrm{dim}(F)=n-m$.
From $F\cap F_i=\{0\}$ it follows that all the sums $F+F_i$ are direct sums, i.e., $F\oplus F_i$.
I am not sure how to continue. One idea would be to use that there always exist vector subspaces $H_i$ of $E$ such that $E$ can be expressed as the direct sums
$$
E=F_i\oplus H_i, \qquad i=1,\ldots,k
$$
Maybe, one needs to compare this to $F\oplus F_i$.
That is, using the dimension formula for direct sums,
$$
n=\textrm{dim}(E)=\textrm{dim}(F_i)+\textrm{dim}(H_i).
$$
Since $\textrm{dim}(F_i)\leq m$ and $m>0$, we have
$$n-m\leq \textrm{dim}(H_i)<n.$$
Since for $x\in E$, the decomposition $x=f_i+h_i$ with $f_i\in F_i$ and $h_i\in H_i$ is unique, doesn't this imply $F=H_i$ for all $i=1,2,\ldots,k$, so that $\textrm{dim}(F)=\textrm{dim}(H_i)$, meaning
$$
n-m\leq\textrm{dim}(F)<n?
$$
Assuming this is correct, I do not see how to deduce $\textrm{dim}(F)=n-m$.
 A: As pointed out by MarktMeister in the comments, we assume $char(K)=0$.
Suppose $n=m+1$. Since $E\neq \cup_k F_k$ we can take a vector in $E-\cup_k F_k$, and the subspace $F$ generated by it satisfies the conditions.
Suppose $n=m+2$. As before I can take a vector $v$ in $E-\cup_k F_k$. If we substitute every $F_k$ by $F_k\oplus\langle v \rangle$ we are in the situation of the previous case. Take $w$ as the vector satisfying the condition for the codimension 1 case. The space $F=\langle v,w \rangle$ has dimension 2 and
$$
F\cap F_k=0.
$$
If not, $\alpha v+\beta w \in F_k$ and then $ w \in F_k \oplus \langle v \rangle$, which is a contradiction.
With this idea in mind, I think you can construct a proof by induction.
Added
I will rewrite the whole answer from the beginning, since it is not so easy as I thought at a first glance. Nevertheless, the first answer is better to understand the idea, so I will leave it there.
Suppose $m=n-1$. As before.
Assume now that the result is true for $m=n-j$. That is, if $dim(F_k)\leq n-j$ then there exists a subspace $F$ such that $dim(F)=n-m=j$ and $F\cap F_k=0$.
Now we are going to prove that the result is true for $m=n-(j+1)$. Suppose $dim(F_k)\leq n-j-1$, and let $v\in E-\cup_k F_k$. We can define $\tilde{F}_k=F_k\oplus \langle v\rangle$. These spaces satisfy the conditions of the induction hypothesis, since they are of dimension lesser or equal to $n-(j+1)+1=n-j$. Therefore there exists a space $\tilde{F}$ such that $dim(\tilde{F})=j$ and
$$
\tilde{F}\cap \tilde{F}_k=0. \tag{1}
$$
We can define $F=\tilde{F}\oplus \langle{v}\rangle$ and then:

*

*$dim(F)=j+1$

*$F\cap F_k=0$. If not, there exist $f\in \tilde{F}$ such that $f+\alpha v \in F_k$. But then $f \in \tilde{F}_k$, which is a contradiction.

So $F$ is the desired space and the result is proven.
