#1

Right, Ive been working on this problem for a while.

Basically, if we take the field of n-tuples of complex numbers, the set

U=C(e1+e2+e3.......+en)

is subspace of this field, but is also invariant under any action of the group Sn (which will just permute elements in the vectors). Now Maschke's Theorem tells us that there is an invariant subspace U' say such that C(e1,e2,....,en) is equal to the direct sum of U and U' (U' is compliment to U) and also such that U' is invariant.

We have been told to find such a U'.

I thought about it for a while and found loads of compliments but they weren't invariant. Then I tried putting in the stuff for Maschke theorem for a low number and found C(e1-e2) works for n=2.

I then also found that such a compliment is the kernel of the matrix with all elements 1/n (this projects C^n onto U and is invariant).

So I'm kind of nearly there, basically it is the subspace of C^n such that all elements of the vector sum to 0. This makes sense, think about it, if you permute these elements, they will still sum to 0. This is nice, but the answer seems a bit incomplete, and there must surely be a basis for this space. But I can't figure out what it could be!!!!! It is obvious that this is correct because you can specify any of the n-1 elements and you are stuck with the last one so it is an n-1 dimensional space. The fact that it is a subspace is clear and it is linearly independent if U so it is indeed the compliment of U and satisfies everything!

I also need to then show that it is irreducible (i.e. that there are no further such invariant spaces) but I think that this will be best accomplished when I have a basis!

Maybe I've missed something obvious, and if I can't figure it out, the answer I'll have to hand in should be ok, but if anyone has any ideas it would be much appreciated!

Thanks guys.

Basically, if we take the field of n-tuples of complex numbers, the set

U=C(e1+e2+e3.......+en)

is subspace of this field, but is also invariant under any action of the group Sn (which will just permute elements in the vectors). Now Maschke's Theorem tells us that there is an invariant subspace U' say such that C(e1,e2,....,en) is equal to the direct sum of U and U' (U' is compliment to U) and also such that U' is invariant.

We have been told to find such a U'.

I thought about it for a while and found loads of compliments but they weren't invariant. Then I tried putting in the stuff for Maschke theorem for a low number and found C(e1-e2) works for n=2.

I then also found that such a compliment is the kernel of the matrix with all elements 1/n (this projects C^n onto U and is invariant).

So I'm kind of nearly there, basically it is the subspace of C^n such that all elements of the vector sum to 0. This makes sense, think about it, if you permute these elements, they will still sum to 0. This is nice, but the answer seems a bit incomplete, and there must surely be a basis for this space. But I can't figure out what it could be!!!!! It is obvious that this is correct because you can specify any of the n-1 elements and you are stuck with the last one so it is an n-1 dimensional space. The fact that it is a subspace is clear and it is linearly independent if U so it is indeed the compliment of U and satisfies everything!

I also need to then show that it is irreducible (i.e. that there are no further such invariant spaces) but I think that this will be best accomplished when I have a basis!

Maybe I've missed something obvious, and if I can't figure it out, the answer I'll have to hand in should be ok, but if anyone has any ideas it would be much appreciated!

Thanks guys.

#2

yep thats what i thought!

#3

Totally.

#4

Yeah, right!

So hopefully you will be thinking about this all night :P

So hopefully you will be thinking about this all night :P

#5

Math + Pit = Fapping

#6

Yeah, who needs the pictures of hot chicks thread, when you can just sit and think about this juicy shit

#7

uhhh...

and i thought calc was bad(trig is still blasphemy from the devil)

and i thought calc was bad(trig is still blasphemy from the devil)

#8

Nm, I got it!

Just take the basis elements to be

bi=(0,0,.....,1,0,0,.....,-1) where the 1 is in the i-th place for each i=1,....,n-1.

It is clear that any such sum of these elements will be a vector who's components sum to 0.

Just take the basis elements to be

bi=(0,0,.....,1,0,0,.....,-1) where the 1 is in the i-th place for each i=1,....,n-1.

It is clear that any such sum of these elements will be a vector who's components sum to 0.