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I'm taking basic linear algebra now. During a lecture on orthogonal complements, a professor said that there's a thing that is more 'general' than orthogonal complements and it's called the annihilator. He didn't provide further details, so I hopped on Google and found some explanations of this concept. Now I think I understand what the annihilator is, but I can't seem to prove a basic fact about it..

Theorem 3.14, part (1) in Advanced Linear Algebra by Steven Roman (Google Books) says that

$M \subseteq N \Rightarrow N^0 \subseteq M^0$

I don't know how to prove this (the author leaves proof of it for the reader). Intuitively I understand why $M^0$ can't be smaller than $N^0$, for if we extend the subset $M$ by adding some $n \in N$ to it, then, in case (a), some functionals from the 'original' $M^0$ won't be able to send the new element of $M$ to zero and $M^0$ will get smaller. In case (b), every functional from the 'original' $M^0$ will send $n$ to zero, and $M^0$ will stay the same. But $M^0$ can't get bigger, because the functionals that annihilate $n$ AND all the vectors in $M$ must be in $M^0$ already.

So can you please help me convert my vision of the problem into a formal proof?

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up vote 3 down vote accepted

Well, as you mention, the definition of the annihilator of a subset $S\subset V$ of your vector space is the set $S^0$ of all elements $f\in V^*$ in the dual of $V$ which satisfy $f(s)=0$ for all $s\in S$.

So, $f\in N^0$ by definition means $f(n)=0$ for all $n\in N$. Since $M\subset N$, we also have $f(n)=0$ for all $n\in M$. By definition, this means $f\in M^0$. This shows $N^0\subset M^0$.

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Many thanks. I guess the converse $N^0 \subseteq M^0 \Rightarrow M \subseteq N$ is also true: pick $m \in M$, it's annihilated by all functionals in $M^0$. Since $N^0 \subseteq M^0$, all functionals in $N^0$ also map $m$ to zero, so $m \in N$ and $M \subseteq N$. Is it right? – Vadecris Feb 13 '11 at 19:08
No. The problem is that $f(m)=0$ for all $f\in N^0$ does not imply $m\in N$. For instance, if your vector space in just $\mathbb R$. Then $\mathbb R^0=\{1\}^0(=\{0\})$ but $\mathbb R\supsetneq\{1\}$. – Rasmus Feb 13 '11 at 19:58

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