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matrix subspace - dimension?

Started by DyDx Dec 9, 2002 at 4:31 PM 0 replies 550+ views
Original Post
DyDx
DyDx
I have a final wednesday in Linear Algebra and, looking through my notes, I found a question on an old exam asking what the dimension of a particular subspace of matrices is - 3x3 matrices with 3 basis matrices. (this particular subspace is all 3x3 skew-symmetric matrices) I know that the dimension of a subspace of vectors is the # of basis vectors, but what is it for a matrix subspace? I got the question wrong (entirely!) so now I''m curious, just in case something like this pops-up on my final. At the bottom of my paper I do have (n(n-1))/2 written, which I think I may have written when he was going over the exam, but I don''t remember the reasoning behind it. Any help would be appreciated!
alvaro
alvaro
Forget about matrices for a second. In this context, you are just considering them as vectors. As a vector space, 3x3 matrice is the same as R^9. Only how you arrange the numbers change. The dimension is always the cardinal of a basis, so the answer is 3. I guess skew symmetric matrices means these:

[0 +a +b]
[-a 0 +c]
[-b -c 0]

So it makes sense that it has dimension 3.

(n(n-1))/2 is probably just the number of entries in a triangle like the one formed by a, b and c in the matrix above.

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