Question about dimension of a subspace Let $K$ be a field and define the following subspaces
$$V=\textrm{span}(e_1,e_2,e_3),\;\; V^\bot = \textrm{span}(e_4,e_5,e_6)$$
inside $K^6$. Let $\dim L=4$ and assume that $\dim L\cap V\leq 1$. Can I conclude that $\dim L\cap V^\bot > 1$? I know of course that $\dim L\cap V^\bot \geq 1$, but I'm wondering if the fact that $V^\bot$ is the orthogonal complement of $V$ would force the dimension to be strictly larger than $1$?
 A: Take $V=\mathbb{R}^{6}$ and $\{e_{i}\}_{i=1}^{6}$ as the standard
basis.
Define 
$$
L=sp\{e_{1},e_{4},e_{2}+e_{5},e_{3}+e_{6}\}
$$
then since 
$$
\begin{pmatrix}1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1
\end{pmatrix}
$$
is of rank $4$ we conclude $dim(L)=4$.
Now, a general element in $L$ is of the form
$$
ae_{1}+be_{4}+c(e_{2}+e_{5})+d(e_{3}+e_{6})=(a,c,d,b,c,d)
$$
for $a,b,c,d,e,f\in\mathbb{R}$
a general element in $V$ is of the form 
$$
(x,y,z,0,0,0)
$$
and thus the intersection is 
$$
\{(a,c,d,b,c,d)\in\mathbb{R}^{6}\mid b=c=d=0\}=\{(a,0,0,0,0,0)\mid a\in\mathbb{R}\}
$$
is of dimension $1$.
But now calculate the intersection with $V^{\perp}=sp\{e_{4},e_{5},e_{6}\}$
: the intersection is 
$$
\{(a,c,d,b,c,d)\in\mathbb{R}^{6}\mid a=c=d=0\}=\{(0,0,0,b,0,0)\mid b\in\mathbb{R}\}
$$
is also of dimension $1$.
Since 
$$
6\geq dim(L\cup V^{\perp})=dim(L)+dim(V^{\perp})-dim(L\cap V^{\perp})=7-dim(L\cap V^{\perp})
$$
We conclude that in general, we can only say that 
$$
dim(L\cap V^{\perp})\geq1
$$
and not $dim(L\cap V^{\perp})>1$.
