Skew symmetric maps

[Erasmus00]

I'm trying to do the following proof. Let V be an n dimensional vecotr space over the complex numbers. Let w(x1,x2,x3..xn)[the xi s are vectors] be a skew symmetric n linear form. (a map w from VxVxVxV...V to the complex numbers that is linear and each slot and changes sign when any two arguments are interchanged).

 

I'm trying to show that w = 0 if and only we have a linearly dependant set of n vectors x1,x2,x3...xn.

 

I've managed to prove that if the vectors are linearly dependant, then w is 0, but I can't prove it back. (thus establishing if and only if) Any have an ideas? I must be missing something obvious, because the book that asserted the claim also said the proof was trivial.

-Will

[Qfwfq]

I think the bit you're still missing is the easiest! :naughty:

 

An n-ple of vectors being linearly dependent is very much akin to two of them being......... equal!!!!!!!!! :eek_big: At this point I'm sure you can complete the proof.

[Erasmus00]

I think the bit you're still missing is the easiest! :naughty:

 

An n-ple of vectors being linearly dependent is very much akin to two of them being......... equal!!!!!!!!! :eek_big: At this point I'm sure you can complete the proof.

 

Yes, thats true. So if vectors are linearly dependant, then w is 0. That was immediately obvious. However, that only established an if, not an if and only if.

 

To show the only if, I need to prove that either if the vectors are linearly independant w is nonzero or that if w is 0 then the vectors are linearly dependant. Niether of these has been forthcoming.

-Will

[Qfwfq]

:naughty: Sorry! I misread, I thought you said you had demonstrated the other way around! :eek_big:

 

It's less simple, I'll give it a thought when I can...

[Erasmus00]

It's less simple, I'll give it a thought when I can...

 

I figured it out about on my walk in this morning. It is pretty obvious once you thing about for awhile. A complete set form an orthonormal basis of the space. If a linear map takes a complete set to 0, it takes everything to 0.

 

I was busy trying to prove it backwards instead of proving that the converse was true.

-Will