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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.

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    No. You should have union on the right hand side2014-04-23

2 Answers 2

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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: $(V\vee W)^\perp=V^\perp\cap W^\perp$.

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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).