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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 $1$?

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    $\begingroup$ Orthogonality plays a very little role in this question. It's the "complement" part that is crucial. $\endgroup$ – Quang Hoang Sep 19 '14 at 11:09
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    $\begingroup$ It looks like I have an example, which shows the answer is NO. Let $L=\textrm{span}(e_1+e_4,e_2+e_5,e_3+e_6,e_1)$. Now $V\cap L = \textrm{span}(e_1)$ and $V^\bot\cap L = \textrm{span}(e_4)$. $\endgroup$ – JT1 Sep 19 '14 at 11:24
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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 $dim(L\cap V^{\perp})>1$.

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