# Is $(\mathbf{V} \cap \mathbf{W})^{\bot}=(\mathbf{V}^{\bot} \cap \mathbf{W}^{\bot})$?

Is $(\mathbf{V} \cap \mathbf{W})^{\bot}=(\mathbf{V}^{\bot} \cap \mathbf{W}^{\bot})$? I tried element-chasing, but I am getting confused when trying to determine mutual containment.

-
Try examples. Try at least a few examples to see if it works. – Mariano Suárez-Alvarez Feb 21 '11 at 3:59
And don't look back at this page before that, because someone is going to post an answer soon, thereby ruining this for you! – Mariano Suárez-Alvarez Feb 21 '11 at 3:59
I told you..... – Mariano Suárez-Alvarez Feb 21 '11 at 4:06
Guilty... Maybe I should temporarily hide it, but I suppose it's too late. Dear Jared, I hope that I have not ruined your joy of discovery. – Jonas Meyer Feb 21 '11 at 4:08
By the way, $\LaTeX$'s way to say 'perpendicular' is \perp (which gives a $\perp$) and not \bot (which results in a $\bot$) – Mariano Suárez-Alvarez Feb 21 '11 at 4:34

No. Try $W=\{0\}$, $V\neq\{0\}$.
What is true is that the perp of the span of two spaces is the intersection of the perps: $(V\vee W)^\perp=V^\perp\cap W^\perp$.
If the ambient space has dimension $n$, then taking the orthogonal complement takes a subspace of dimension $k$ to a subspace of dimension $n-k$. Intersection in general reduces the dimension. So your equation looks suspect.
A nice analogy is from propositional logic: the orthogonal complement is similar to negation, and intersection is similar to logical and. So according to de Morgan's laws, if in one side of the equation you have $\cap$, in the other side you should have the dual operator, viz. $+$ (what Jonas uses $\lor$ for).