If $\dim V =n$ and $W$ is a subspace of $V$ such that $T(W)\cap W \neq \{0\}$ for every isomorphism $T$ of $V$ then $\dim W \geq \frac{n}{2}$ If $\text{dim}\,V =n$ and $W$ is a subspace of $V$ such that $T(W)\cap W \neq \{0\}$ for every isomorphism $T$ of $V,$ then prove that $\text{dim}\, W \geq \frac{n}{2}.$
My attempt: By considering the restriction of $T$ to $W,$ we can see that $\text{dim}\, T(W) =\text{dim}\, W$ (since the kernel of the restriction equals $(\text{ker} \, T)\cap W = \{0\}$). By the given condition, $\dim (T(W)\cap W)\geq 1$ and since $\dim (T(W)+W) = \dim T(W) + \dim W -\dim (T(W)\cap W),$ we get  $\dim (T(W)+W)\leq 2 \dim W-1$ which, working backwards, seems to suggest that $T(W)+W$ is a hyperplane in $V,$ for every invertible operator $T$ on $V.$ But I'm not able to see this. I haven't yet used the fact that $T$ is onto either. Any hints on how to proceed, or whether this is the right track? Thanks.
 A: We will prove that for every $W$ such that $\dim W < \frac n2$, there exists an invertible transformation $T$ such that $T(W) \cap W = \{0\}$. 
To do this let $w_1,...,w_k$ be a basis of $W$, where $k = \dim W$, and extend it to a basis $w_1,...,w_k,v_1,...,v_{n-k}$ of $V$. 
Crucially, note that $k < n-k$.
Now, define $T$ as follows : $T(w_i) = v_i$ for $1 \leq i \leq k$, then $T(v_i)= w_i$ for $1 \leq i \leq k$, and amongst the rest of the vectors $v_{k+1},...,v_{n-k}$ , $T$ can be the identity i.e. $T(v_j) = v_j$ for all $k+1 \leq j \leq n-k$. 
Then, $T$ can be extended linearly (it is enough to define $T$ on a basis).
It is seen by construction, that $T$ is surjective(therefore injective) and that $T(W) \cap W = \{0\}$.  Therefore, for $\dim W < \frac n2$, we always have a linear transformation that we do not desire. Now, it follows that if no such transformation exists, then $\dim W \geq \frac n2$.
For $n$ even, this can be strengthened to $\dim W > \frac n2$ by the same argument as above. 
