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$V$ is a finite dimensional vector space, $U$ and $W$ are subspaces and $U^0,W^0, (U+W)^0$ are the relevant annihilators. I would please appreciate help proving:

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

I would think that anything that annihilates $(U+W)$ would annihilate $U$ and annihilate $W$. But I am even confused about that in general, given that $\dim V=\dim U+\dim U^0$, the larger the dimension of $U$, the smaller the dimension of $U^0$.

But in that $\dim (U+W)\ge \dim U$, and similarly $\ge \dim W$, I am not even sure of that.

I would appreciate help in proving the highlighted equality as an inclusion of sets in both directions, and also, if possible, using dimensions.

Thanks

2 Answers2

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Let $f \in U^0 \cap W^0$. Then $f \in U^0$ and $f \in W^0$. Therefore, $f$ is a linear functional defined on $V$ such that $f(u) = f(w) = 0$ for all $u \in U$ and for all $w \in W$.

So, for all $x \in U+W = \left\{ \ u+w \colon \ u\in U, \ w \in W \ \right\}$, we can write $x$ as $x = u+w$ for some $u \in U$ and $w \in W$ and thus $$ f(x) = f(u+w) = f(u)+f(w) = 0+0=0,$$ showing that $f \in \left( U+W \right)^0$.

Therefore, $$U^0 \cap W^0 \subset \left( U+W \right)^0.$$

Conversely, suppose that $g \in \left( U+W \right)^0$. Then $g$ is, by definition, a linear functional defined on $V$ such that $$g(x) = 0 \ \mbox{ for all } x \in U+W. \ \tag{1}$$

Now let $\theta$ denote the zero vector in $V$, and let $u \in U$ and $w \in W$ be arbitrary.

We note that, as both $U$ and $W$ are subspaces of $V$, so $\theta \in U$ and $\theta \in W$, and therefore $$u = u+\theta \in U+W, \ \mbox{ and } \ w = \theta+w \in U+W,$$ and so by (1), we can conclude that $$g(u) = 0 = g(w),$$ and since $u \in U$ and $w\in W$ are arbitrary, we can conclude that $$g \in U^0 \cap W^0,$$ and therefore $$\left( U+W \right)^0 \subset U^0 \cap W^0.$$

Hope this helps!!

Saaqib Mahmood
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  • In reading your bio, I thought you might like these links to teachings by Noam Elkies (one of the greats). The first one has notes on metric spaces at the end of the sentence "We shall use Simmons..." http://www.math.harvard.edu/~elkies/M25b.13/index.html Also he is currently teaching a course that uses Rudin for the first part. The webpage is interesting in general and specifically, the lectures are transcribed. They are found in the line with the yellow icon "NEW." It takes a little fiddling to get them to a nice format: http://www.math.harvard.edu/~elkies/M55b.16/index.html –  Apr 01 '17 at 21:25
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Suppose $\phi\in U^0\cap W^0$, and let $v = u + w\in U + W$ be arbitrary. Then $\phi(v) = \phi(u + w) = \phi(u) + \phi(w) = 0$, so $\phi\in (U + W)^0$.

Conversely, let $\phi\in(U + W)^0$. Then since $U\subseteq U + W$ and $W\subseteq U + W$, $\phi$ annihilates both $U$ and $W$, so $\phi\in U^0\cap W^0$.

Stahl
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  • Thanks. There is a required dormant period of a few minutes before I can accept your answer. While I'm waiting, can you please tell me if this can be done comparing dimensions? Or at least what the $\dim U^0\cap W^0$ would be? With regards, –  Apr 01 '17 at 20:27
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    Well, once you prove the result as above, you have $\dim V = \dim (U + W) + \dim (U + W)^0$, so that $\dim (U^0\cap W^0) = \dim V - \dim (U + W)$ (by what you've already said). As for proving the result using dimension arguments, I'm not sure. You could possibly do it, but I don't think it would be terribly enlightening, especially since the above doesn't require finite dimensionality. If I think of a nice argument, I'll update. – Stahl Apr 01 '17 at 20:37