Annihilators and subspaces

**Mollier** · March 10th 2010, 07:44 PM

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.

**josipive** · March 11th 2010, 07:07 AM

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.

**tonio** · March 11th 2010, 06:39 PM

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"

**Mollier** · March 14th 2010, 12:44 AM

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!

**tonio** · March 14th 2010, 03:14 AM

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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