anhilators of vector spaces.

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October 13th 2011, 03:58 PM
Jame

anhilators of vector spaces.

I am trying to prove that

$Ann(W_1+W_2)= Ann(W_1) \cap Ann (W_2)$

I have managed to prove one inclusion

$Ann(W_1) \cap Ann (W_2) \subset Ann(W_1+W_2)$

The fact that any function in these spaces is a linear transformation was really helpful however if doesn't seem to really help me on the reverse inclusion.

Let $f \in Ann(W_1+W_2)$

then $f(w_1+w_2) = 0 \; \forall w_1 \in W_1$ and $w_2 \in W_2$

since $f$ is a linear transformation: $f(w_1+w_2) = f(w_1)+f(w_2)$

thus $f(w_1)+f(w_2) = 0 \; \forall w_1 \in W_1$ and $w_2 \in W_2$

I dont think this helps me. I can't necessarily say $f(w_1) = 0 \; \forall w_1$ and $f(w_2) = 0 \; \forall w_2$

What should I do from here? Is this the right path and I'm just not catching something?
October 13th 2011, 04:01 PM
Drexel28

Re: anhilators of vector spaces.

Quote:

Originally Posted by Jame

I am trying to prove that

$Ann(W_1+W_2)= Ann(W_1) \cap Ann (W_2)$

I have managed to prove one inclusion

$Ann(W_1) \cap Ann (W_2) \subset Ann(W_1+W_2)$

The fact that any function in these spaces is a linear transformation was really helpful however if doesn't seem to really help me on the reverse inclusion.

Let $f \in Ann(W_1+W_2)$

then $f(w_1+w_2) = 0 \; \forall w_1 \in W_1$ and $w_2 \in W_2$

since $f$ is a linear transformation: $f(w_1+w_2) = f(w_1)+f(w_2)$

thus $f(w_1)+f(w_2) = 0 \; \forall w_1 \in W_1$ and $w_2 \in W_2$

I dont think this helps me. I can't necessarily say $f(w_1) = 0 \; \forall w_1$ and $f(w_2) = 0 \; \forall w_2$

What should I do from here? Is this the right path and I'm just not catching something?

Ok, so let $\varphi\in\text{Ann}(W_1+W_2)$ then we know that $\varphi(v)=0$ for all $v\in W_1+W_2$ , right? But, isn't $W_1,W_2\leqslant W_1+W_2$ ?
October 13th 2011, 04:17 PM
Jame

Re: anhilators of vector spaces.

Oh i see

so

$f(w_1)+f(w_2) = 0 \; \forall w_1 \in W_1$ and $w_2 \in W_2$

Particularly when $w_2 = 0$ , and since f is an L.T., $f(0) = 0$

therefore $f(w_1) = 0 \; \forall w_1 \in W_1$

implies $f \in Ann(W_1)$

similarly for $W_2$

so $f \in Ann(W_1 \cap W_2)$

Thanks a lot!
October 13th 2011, 06:03 PM
Jame

Re: anhilators of vector spaces.

Sorry, I have another question having to do with this particular equality.

I've read you can use this to prove that

$Ann(W_1 \cap W_2) = Ann(W_1)+Ann(W_2)$

by asserting that they have the same dimension.

I have been playing around with this for the past hour or so. And haven't really gotten anywhere.

I've got

$dim(Ann(W_1 \cap W_2)) = dim V - dim(W_1 \cap W_2)$

but I have found no equalities that I could apply $Ann(W_1 + W_2) = Ann(W_1) \cap Ann(W_2)$ to.
October 13th 2011, 06:08 PM
Drexel28

Re: anhilators of vector spaces.

Quote:

Originally Posted by Jame

Sorry, I have another question having to do with this particular equality.

I've read you can use this to prove that

$Ann(W_1 \cap W_2) = Ann(W_1)+Ann(W_2)$

by asserting that they have the same dimension.

I have been playing around with this for the past hour or so. And haven't really gotten anywhere.

I've got

$dim(Ann(W_1 \cap W_2)) = dim V - dim(W_1 \cap W_2)$

but I have found no equalities that I could apply $Ann(W_1 + W_2) = Ann(W_1) \cap Ann(W_2)$ to.

Which part are you trying to prove. I think you've proven the dimension formula and are trying to see how to use it. Well, as I mentioned earlier it's trivial to prove that $\mathsf{Ann}(W_1)+\mathsf{Ann}(W_2)\subseteq \mathsf{Ann}(W_1\cap W_2)$ but since they have the same dimension, they must be equal. Is that what you;re after?
October 13th 2011, 06:14 PM
Jame

Re: anhilators of vector spaces.

Proving $Ann(W_1 \cap W_2) = Ann(W_1)+Ann(W_2)$

This is just an application if the equality in the very first post? How does this work?
October 13th 2011, 06:36 PM
Drexel28

Re: anhilators of vector spaces.

Quote:

Originally Posted by Jame

Proving $Ann(W_1 \cap W_2) = Ann(W_1)+Ann(W_2)$

This is just an application if the equality in the very first post? How does this work?

Ok, so $\dim\mathsf{Ann}(W_1\cap W_2)=n-\dim (W_1\cap W_2)$ where $W_1,W_2\leqslant V$ with $\dim V=n$ . Then,

$\begin{aligned}\dim (\mathsf{Ann}(W_1)+\mathsf{Ann}(W_2))&=n-\dim(\mathsf{Ann}(\mathsf{Ann}(W_1)+\mathsf{Ann}(W _2)))\\ &=n-\dim(\mathsf{Ann}(\mathsf{Ann}(W_1))\cap\mathsf{An n}(\mathsf{Ann}(W_2))\\ &=n-\dim(W_1\cap W_2)\end{aligned}$

where I have used a slightly non-obvious fact (where, and how do you prove it). So, $\mathsf{Ann}(W_1\cap W_2)\cong\mathsf{Ann}(W_1)+\mathsf{Ann}(W_2)$ and since $\mathsf{Ann}(W_1)+\mathsf{Ann}(W_2)\subseteq\maths f{Ann}(W_1\cap W_2)$ since if [tex]\varphi,\psi[/atex] are in the left hand side and $v\in W_1\cap W_2$ then $(\varphi+\psi)(v)=\varphi(v)+\psi(v)=0$ etc. But, since they are isomorphic, containment is equivalent to equality.
October 13th 2011, 09:10 PM
Deveno

Re: anhilators of vector spaces.

i'm trying to follow your reasoning, and it looks circular.

how does $\dim(\mathsf{Ann}(\mathsf{Ann}(W_1) + \mathsf{Ann}(W_2))) = \dim(\mathsf{Ann}(\mathsf{Ann}(W_1)) \cap \mathsf{Ann}(\mathsf{Ann}(W_2)))$

without first showing what we are trying to prove, namely:

$\dim(\mathsf{Ann}(W_1+W_2)) = \dim(\mathsf{Ann}(W_1)) \cap \mathsf{Ann}(W_2))$ ?
October 13th 2011, 09:25 PM
Drexel28

Re: anhilators of vector spaces.

Quote:

Originally Posted by Deveno

i'm trying to follow your reasoning, and it looks circular.

how does $\dim(\mathsf{Ann}(\mathsf{Ann}(W_1) + \mathsf{Ann}(W_2))) = \dim(\mathsf{Ann}(\mathsf{Ann}(W_1)) \cap \mathsf{Ann}(\mathsf{Ann}(W_2)))$

without first showing what we are trying to prove, namely:

$\dim(\mathsf{Ann}(W_1+W_2)) = \dim(\mathsf{Ann}(W_1)) \cap \mathsf{Ann}(W_2))$ ?

Perhaps I mixed up what we are trying to prove with what we're proving. But, we first proved (through the first three posts) that $\mathsf{Ann}(W_1+W_2)=\mathsf{Ann}(W_1)\cap\mathsf {Ann}(W_2)$ which is what I used. I'm tired and don't feel like making sure I didn't mix them up, which I did initially. So, if I did, which I don't think I did, feel free to do it up man.
October 13th 2011, 09:38 PM
Deveno

Re: anhilators of vector spaces.

from the first 3 posts, we know that they are equal (the annihilator of the sum, and the intersection of the annihilators). so they have to have the same dimension.

(assuming, of course, that their dimension is finite).

i may be mistaken, but i am wondering if what the thread starter is asking is, can we show equality from a dimensional argument?

i think we can, but we're going to have to invoke a basis, at some point.

(was your "non-obvious" assertion $\mathsf{Ann}(\mathsf{Ann}(W)) = W$ ?)
October 13th 2011, 09:40 PM
Drexel28

Re: anhilators of vector spaces.

Quote:

Originally Posted by Deveno

from the first 3 posts, we know that they are equal (the annihilator of the sum, and the intersection of the annihilators). so they have to have the same dimension.

(assuming, of course, that their dimension is finite).

i may be mistaken, but i am wondering if what the thread starter is asking is, can we show equality from a dimensional argument?

i think we can, but we're going to have to invoke a basis, at some point.

Ok, now I wonder if you realize that he asked to prove a DIFFERENT equality than the one initially asked...I originally proved "the annihilator of a sum is the interesction of annihilators" and the post that you thought I had circular reasoning on I was proving "the sum of annihilators is the annihilator of the intersection" for which I used "the annihilator of a sum is the interesction of annihilators"
October 13th 2011, 09:49 PM
Deveno

Re: anhilators of vector spaces.

ok, i got it. yes, the similarity DID confuse me. and in the case we're talking about:

Ann(W1+W2) = Ann(W1)∩Ann(W2) DOES imply Ann(W1∩W2) = Ann(W1)+Ann(W2)

for a moment, there, we were talking about our conversation, and i got "meta-lost".