I am trying to prove that
I have managed to prove one inclusion
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
then and
since is a linear transformation:
thus and
I dont think this helps me. I can't necessarily say and
What should I do from here? Is this the right path and I'm just not catching something?
Sorry, I have another question having to do with this particular equality.
I've read you can use this to prove that
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
but I have found no equalities that I could apply to.
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 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.
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 ?)
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"
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".