Dimension of symmetric and skew symmetric bilinear forms

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Given the vector space consisting of all bilinear forms of a vector space V (let's call it B) it's very easy to prove that B is the direct sum of two subspaces, the subspace of symmetric and the subspace of skew symmetric bilinear forms. How would one go about determining the dimension of these spaces without any definition of bases? Or is that the only way to go about it?
 

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HallsofIvy
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Think of it in terms of matrices. If V has dimension n, so that every vector in V can be written as an n-tuple of numbers, every bilinear form on V can be associated with an n by n matrix: The vector space of all such bilinear forms is the same as the dimension of all such matrices: n2. Now look at the set of symmetric bilinear forms- that is equivalent to the space of all symmetric n by n matrices. What dimension does that have? Well, we can divide a n by n matrix into 3 parts: the main diagonal, all entries above the main diagonal and all entries below the main diagonal. Obviously, there are n entries on the main diagonal. That leaves n2- n entries not on the main diagaon, (n2- n)/2 are above the main diagonal and (n2- n)/2 are below the main diagonal. Now for symmetric matrices, we can choose any numbers we want for the main diagonal: n choices. If we then choose numbers for the region above the main diagonal, that's another (n2-n)/2 choices but, because the matrix is symmetric, the entries below the main diagonal must be the same: no more choices. We are free to choose entries for n+ (n2- n)/2= (2n+ n2- n)/2= (n2+ n)/2 places independently. That's the dimension of the space of all symmetric n by n matrices and so for all symmetric bilinear forms on an n dimensional vector space.
For the space of all anti-symmetric bilinear forms, you can do either of two things:
Since the space of all bilinear forms is the direct sum of those two subspaces, the sum of the two dimensions must be n2. Or, look at the number of independent entries: now the main diagonal must be all 0s- no choices there. But again, once you have chosen numbers for the region above the main diagonal, the entries below the main diagonal are fixed: they must be the negative of those above the main diagonal. Do you see how many independent choices you have now? Those two caculations should be the same.
 
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oh ok. So this is actually quite a lot like the problem you do in elementary linear algebra where you show that the space of nxn matrices is the direct sum of symmetric and skew symmetric. Actually, it seems about exactely the same, just that you have to think of the bilinear forms as matrices... which is a little awkward in a sense, because they're not quite the same as linear transformations. But this should work. Thanks. :)
 

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