# Thread:

1. ## Annihilators and subspaces

Hi,

Problem:
If M and N are subspaces of a finite-dimensional vector space, then $(M \cap N)^0=M^0+N^0$ and $(M+N)^0=M^0 \cap N^0$.
( $M^0$ and $N^0$ are the annihilators of M and N, respectively)

Attempt:
Not much I can do here, but I know that if $M \subset N$, then $N^0 \subset M^0$.
I also know that $M^{00}=(=(M^0)^0)=M$. I think the author wants me to use these two facts to prove the given statement.

Any hints are greatly appreciated.

Thanks.

2. dont think that you need to use those facts that you mentioned, basicly what you have is two sets, so you need to show that one set is a subset of other and opposite.

3. Originally Posted by Mollier
Hi,

Problem:
If M and N are subspaces of a finite-dimensional vector space, then $(M \cap N)^0=M^0+N^0$ and $(M+N)^0=M^0 \cap N^0$.
( $M^0$ and $N^0$ are the annihilators of M and N, respectively)

Attempt:
Not much I can do here, but I know that if $M \subset N$, then $N^0 \subset M^0$.
I also know that $M^{00}=(=(M^0)^0)=M$. I think the author wants me to use these two facts to prove the given statement.

Any hints are greatly appreciated.

Thanks.

Hints (easy): (a) $U\subset W\Longrightarrow W^0\subset U^0$ , (b) $A\subset A^{00}\,,\,\,\forall$ subset $A\subset V$ , (c) $U=U^{00}$ if $U$ is a subspace of $V$, and thus we get:

(1) $M^0+N^0\subset (M\cap N)^0$ (why?);

(2) $(M+N)^0\subset M^0\cap N^0$ (why?)

Take now the annihilator in both sides in (1) and use (a) and (c) above:

(3) $M\cap N=(M\cap N)^{00}\subset (M^0+N^0)^0$ , and now use (2) above (with $M^0\,,\,N^0$ instead of $M\,,\,N$), and you get

(4) $(M^0+N^0)^0\subset M^{00}\cap N^{00}= M\cap N$ , so from (3)-(4) you get

(5) $M\cap N\subset (M^0+N^0)^0\subset M\cap N$ , and this means the inclusions (3)-(4) must be equalities ; but this means that the annihilators of both sides in (1) are equal , and thus we get the wanted equality because of the

Hint (easy, too) (d) If $U\,,\,W$ subspaces of $V\,,\,\,s.t.\,\,\,U\neq W$ , then $U^0\neq W^0$

Try now to complete the proof of the second equality by yourself.

Tonio

Pd. I don't know any proof of these equalities that is less messy than the above, which is not hard but requires concentration in details and lots of careful "connecting the dots"

4. Hi, apologies for the late reply. The last few days have been a mess so no time for math

Originally Posted by tonio
(1) $M^0+N^0\subset (M\cap N)^0$ (why?);

(2) $(M+N)^0\subset M^0\cap N^0$ (why?)
The only thing I can come up with is that since $M \cap N \subset M+N$, then $(M+N)^0 \subset (M \cap N)^0$, but that doesn't really prove either (1) nor (2)..

Originally Posted by tonio
Try now to complete the proof of the second equality by yourself.
I figure that I need to show that $M^0 \cap N^0 \subset (M^0+N^0)$. If I assume that (2) is true, I should be able to take the annihilator of both sides of (2) and get to the wanted result.
$((M+N)^0)^0 \subset (M^0 \cap N^0)^0$
Then by (a) I have that
$(M^0 \cap N^0) \subset (M+N)^0$

Ah, I find this to be really hard!

Thanks!

5. Originally Posted by Mollier
Hi, apologies for the late reply. The last few days have been a mess so no time for math

The only thing I can come up with is that since $M \cap N \subset M+N$, then $(M+N)^0 \subset (M \cap N)^0$, but that doesn't really prove either (1) nor (2)..

Indeed, that doesn't...but $M\cap N\subset M\Longrightarrow M^0\subset (M\cap N)^0\,,\,\,and\,\,\,also\,\,\,M\cap N\subset N\Longrightarrow N^0\subset (M\cap N)^0$ and now you get what you wanted...

I figure that I need to show that $M^0 \cap N^0 \subset (M^0+N^0)$. If I assume that (2) is true, I should be able to take the annihilator of both sides of (2) and get to the wanted result.
$((M+N)^0)^0 \subset (M^0 \cap N^0)^0$
Then by (a) I have that
$(M^0 \cap N^0) \subset (M+N)^0$

Ah, I find this to be really hard!

Yes, it really requires concentration. Try again, since the last days you did no much math (or at all).

Tonio

Thanks!
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