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**Jame** Proving $\displaystyle Ann(W_1 \cap W_2) = Ann(W_1)+Ann(W_2)$

This is just an application if the equality in the very first post? How does this work?

Ok, so $\displaystyle \dim\mathsf{Ann}(W_1\cap W_2)=n-\dim (W_1\cap W_2)$ where $\displaystyle W_1,W_2\leqslant V$ with $\displaystyle \dim V=n$. Then,

$\displaystyle \begin{aligned}\dim (\mathsf{Ann}(W_1)+\mathsf{Ann}(W_2))&=n-\dim(\mathsf{Ann}(\mathsf{Ann}(W_1)+\mathsf{Ann}(W _2)))\\ &=n-\dim(\mathsf{Ann}(\mathsf{Ann}(W_1))\cap\mathsf{An n}(\mathsf{Ann}(W_2))\\ &=n-\dim(W_1\cap W_2)\end{aligned}$

where I have used a slightly non-obvious fact (where, and how do you prove it). So, $\displaystyle \mathsf{Ann}(W_1\cap W_2)\cong\mathsf{Ann}(W_1)+\mathsf{Ann}(W_2)$ and since $\displaystyle \mathsf{Ann}(W_1)+\mathsf{Ann}(W_2)\subseteq\maths f{Ann}(W_1\cap W_2)$ since if [tex]\varphi,\psi[/atex] are in the left hand side and $\displaystyle v\in W_1\cap W_2$ then $\displaystyle (\varphi+\psi)(v)=\varphi(v)+\psi(v)=0$ etc. But, since they are isomorphic, containment is equivalent to equality.