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I’m studying dual spaces in linear algebra.

I have proved the following two results. Note: I say $U^0$ for the annihilation of $U$.

For subspaces $U,W$ of a vector space $V$,

$(U+W)^0=U^0\cap W^0$

And

$U^0+W^0\subseteq (U\cap W)^0$

With the inequality becoming equality when $V$ is finite-dimensional.

I’m struggling to see the intuition for these results/visualise them/remember them.

Is there a good way to remember these results? Or sense check any other rearranging of these symbols would be false?

Thanks

Shaun
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jet
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    The first one is very intuitive. If a linear functional annihilates $U+V$ then in particular it annihilates its subspaces $U$ and $V$. And if it annihilates $U$ and $V$ then clearly it annihilates their sum, because $f(u+v)=f(u)+f(v)=0+0=0$ for $u\in U, v\in V$. You don't need to remember that, both inclusions just follow logically. – Mark May 06 '23 at 11:03
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    Please use more descriptive titles. – Shaun May 06 '23 at 11:08
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    @Shaun Apologies- changed. – jet May 06 '23 at 11:32
  • @Mark Wow when you put it in words it does seem more enlightening. I should probably think more as I crunch the logic. What about the second result? How would you phrase that? – jet May 06 '23 at 11:33
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    @jet Again, if $f\in U^0$ and $g\in W^0$ then $f,g$ in particular annihilate the subspace $U\cap W$ (which is a subspace of both $U$ and $W$), and so does their sum $f+g$. As you see, the proofs are almost immediate. So if you don't remember if an inclusion is correct or not, just try to prove it logically. You will see for example that the reverse inclusion has no good reasons to be true. (it is true in finite dimensions though, because of dimension equality) – Mark May 06 '23 at 11:43

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