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Let $V$ be a finite dimensional vector space and $U^0$ and $W^0$ denote the annihilators of subspaces $U$ and $W$, respectively. Then I need to show that $(U\cap W)^0 = U^0 + W^0$.

I can show that $U^0 + W^0 \subseteq (U\cap W)^0$ like so: Let $f\in U^0$ and $g\in W^0$. Then, because $U^0, W^0 \subseteq (U\cap W)^0$, we see that if $f\in U^0$ and $g\in W^0$, then for any $v\in U\cap W$ and $a,b \in \Bbb F$, $(af+bg)(v) = af(v) + bg(v) = 0$. Thus $af+bg\in (U\cap W)^0$. But because $f$ and $g$ were arbitrary, this implies that any linear combination of elements of $U^0$ and $W^0$ is an element of $(U\cap W)^0$. I.e. $U^0+W^0 \subseteq (U\cap W)^0$.

But I haven't had any luck proving the other direction. Obviously I should start with $f\in (U\cap W)^0$. But then it looks like I need to somehow break that function into two to show its in $U^0 + W^0$. That's throwing me off.

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    $\begingroup$ If you have one inclusion you can also try to show that the dimensions are equal. $\endgroup$ – Arnaud D. Feb 25 '17 at 18:28
  • $\begingroup$ That might work. Good thinking! $\endgroup$ – Bobbie D Feb 25 '17 at 18:29
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    $\begingroup$ You should prouve that $U^0\cap W^0=(U+W)^0$ and use duality. $\endgroup$ – Adren Feb 25 '17 at 18:43
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    $\begingroup$ @TheBirdistheWord $U^0$ is the subset of $V'$ that annihilates all of $U$. $(U\cap W)^0$ is the subset of $V'$ while only annihilates the part of $U$ that's shared with $W$. So intuitively, $U^0$ has strictly stronger conditions on it so it's a subset of $(U\cap W)^0$ (analogously the condition of being a green apple is strictly stronger than just being an apple). That's just the intuitive argument, though. I suggest you try to prove it. $\endgroup$ – Bobbie D Mar 4 '17 at 15:55
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    $\begingroup$ @TheBirdistheWord More generally, you can prove that if $U\subset W$, then $W^0\subset U^0$. The argument is basically the one that Bobbie D gives in his comment above. $\endgroup$ – Arnaud D. Mar 4 '17 at 23:16
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Since you have already proved that $U^0 + W^0 \subseteq (U\cap W)^0$, it is enough to prove that their dimensions are equal. Now we have $$\dim(U\cap W)^0=n-\dim (U\cap W),$$ and \begin{align}\dim(U^0+W^0) & =\dim (U^0)+\dim (W^0)-\dim (U^0\cap W^0)\\ & =2n-\dim(U)-\dim(W)-\dim (U+W)^0 \\ & = 2n-\dim(U)-\dim(W)-n+\dim(U+W)\\ &= n-\dim (U\cap W).\end{align}

Note that to obtain the second line, we used the equality $U^0\cap W^0=(U+W)^0$, which simply says that a linear functional $f$ vanishes on $U+W$ if and only if it vanishes on $U$ and $W$.

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  • $\begingroup$ Sorry to trouble you. I was going through your answer again as a study review. I wonder if there are possibly a few typos in the lower section. I think ultimately in the last line the $V$ should be $U$. Also maybe you would clarify how you get the second line as I would thing the third term should be $\dim (W)$ and (my lack of math skills) I wonder how you get the fourth term. Then maybe the third line will become evident. Thanks for your effort. With regards, $\endgroup$ – user12802 Apr 1 '17 at 12:44
  • $\begingroup$ @Andrew There were indeed a couple errors, thank you for pointing that out. I've also added some explanation for the second line, I hope that helps. $\endgroup$ – Arnaud D. Apr 1 '17 at 13:19

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