Let $S$ be any subset of $V^*$ for some finite dimensional space $V$. Define $\text{Ann}(S) = \{v \in V \mid f(v) = 0 \text{ for all } f \in S\}$. Let $W_{1}$ and $W_{2}$ be subspaces of $V^*$. Prove that $\text{Ann}(W_{1}+W_{2}) = \text{Ann}(W_{1}) \cap \text{Ann}(W_{2})$.
My attempt: ($\text{Ann}(W_{1}) \cap \text{Ann}(W_{2}) \subseteq \text{Ann}(W_{1}+W_{2})$): Let $w \in \text{Ann}(W_{1}) \cap \text{Ann}(W_{2})$. We need to show that $w \in \text{Ann}(W_{1}+W_{2})$. Since $w \in \text{Ann}(W_{1}) \cap \text{Ann}(W_{2})$, $w \in \text{Ann}(W_{1})$ and $w \in \text{Ann}(W_{2})$. By definition, $f_{1}(w) = 0$ for all $f_{1} \in W_{1}$ and $f_{2}(w) = 0$ for all $f_{2} \in W$. Every $f$ in $W_{1} +W_{2}$ can be written as $f = f_{1}+f_{2}$ for some $f_{1} \in W_{1}$ and $f_{2} \in W_{2}$. And note that $$(f_{1}+f_{2})(w) = f_{1}(w) +f_{2}(w) = 0.$$
This implies that $w \in \text{Ann}(W_{1}+W_{2})$.
($\text{Ann}(W_{1}+W_{2}) \subseteq \text{Ann}(W_{1}) \cap \text{Ann}(W_{2})$): We let $w \in \text{Ann}(W_{1}+W_{2})$. By definition, $f (w) = 0$ for all $f \in W_{1} +W_{2}$. And we write $f = f_{1}+f_{2}$ for some $f_{1} \in W_{1}$, $f_{2} \in W_{2}$. This means that $(f_{1}+f_{2})(w) = f_{1}(w) +f_{2}(w) = 0$. But I'm confused about how to go from here to show that $f_{1}(w) = f_{2}(w) = 0$. Please help! Thanks!
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$\begingroup$ Simply, $W_1 \subseteq W_1 + W_2$ and $W_2 \subseteq W_1 + W_2$, since both contain the zero vector. And if $f(v)=0$ for all $f$ in some $A$, this is clearly also true for all subsets of $A$. $\endgroup$altwoa– altwoa2025-04-13 19:03:26 +00:00Commented Apr 13 at 19:03
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$\begingroup$ You probably meant $f_2(w)$, not $f_2(2)$. $\endgroup$jjagmath– jjagmath2025-04-13 20:13:24 +00:00Commented Apr 13 at 20:13
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$\begingroup$ @jjagmath Yes! Thanks for pointing it out! $\endgroup$Miranda– Miranda2025-04-13 20:26:05 +00:00Commented Apr 13 at 20:26
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Suppose that $f\notin\operatorname{Ann}(W_1)$. Then there is some $w\in W_1$ such that $f(w)\ne0$. But $w\in W_1+W_2$; so, this shows that $f\notin\operatorname{Ann}(W_1+W_2)$. By the same argument,$$f\notin\operatorname{Ann}(W_2)\implies f\notin\operatorname{Ann}(W_1+W_2).$$We deduce that$$f\in(\operatorname{Ann}(W_1))^\complement\cup(\operatorname{Ann}(W_2))^\complement\implies f\in(\operatorname{Ann}(W_1+W_2))^\complement.$$In other words,$$f\in(\operatorname{Ann}(W_1)\cap\operatorname{Ann}(W_2))^\complement\implies f\in(\operatorname{Ann}(W_1+W_2))^\complement,$$which is the same thing as asserting that$$\operatorname{Ann}(W_1+W_2)\subset\operatorname{Ann}(W_1)\cap\operatorname{Aut}(W_2).$$
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$\begingroup$ The question is about $\text{Ann}$, not $\text{Aut}$. $\endgroup$jjagmath– jjagmath2025-04-13 20:11:13 +00:00Commented Apr 13 at 20:11
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$\begingroup$ @jjagmath It was a typo. I meant $\operatorname{Ann}$. $\endgroup$José Carlos Santos– José Carlos Santos2025-04-13 20:21:19 +00:00Commented Apr 13 at 20:21
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$\begingroup$ I'm curious. Did you really write it wrong more than 10 times or do you use some automated process to type your posts? $\endgroup$jjagmath– jjagmath2025-04-13 21:58:19 +00:00Commented Apr 13 at 21:58
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$\begingroup$ @jjagmath The first time, I wrote
\operatorname{Aut}and then I did COPY + PASTE for all other instances. $\endgroup$José Carlos Santos– José Carlos Santos2025-04-13 22:02:49 +00:00Commented Apr 13 at 22:02