#Linear Algebra Direct sum help D:
21 messages · Page 1 of 1 (latest)
At least, if I understood the problem correctly...
I am not too good with sums/intersections of subspaces...
Oh, the dim is 2. Hmm...
Yeah, then my approach won't work.
Nah, seems good.
Hm.
How about we search for a couple of vectors orthogonal to v1 and v2?
So, we can try picking a couple of random vectors and performing Gram-Schmidt on them.
Actually, I don't think it's even necessary for them to be orthogonal.
We just need to pick two vectors u1, u2 such that for W = ℒ{v1, v2} and T = ℒ{u1, u2} we get dim(T) = 2, dim(W ⋂ T) = 0. Then by Grassmann we will get dim(W + T) = 4.
How about u1 = x, u2 = 1? Maybe that will work.
No, wait, hold on.
What I meant is that you need to show that dim(W + T) = 4.
So, take all four vectors and find the rank of their matrix like before.
If it's 4, all good.
If it's less than 4, we'll look for other vectors.
Yeah, that will solve the problem. Note that there's no need to check the intersection, since by Grassmann we have dim(W ⋂ T) = dim(W) + dim(T) - dim(W + T), so in case of the latter being 4 we will instantly get the required dim(W ⋂ T) = 0.
Great!
So, I think we can take T to be ℒ{x, 1}.
Well, probably...
I'm still not 100% sure my thoughts are correct.
Sorry to 'revive' the post but this question have piqued my interest because it approach a lot of Linear Algebra content.
Why to choose u1 = x, u2 = 1? I understood the problem resolution until this point.
When Fumo transformed the system using Gauss, we got vectors from rows 1 and 2, so corresponding to x^3 and x^2.
Thus, I thought of taking x and 1 to be the rest so that the system would have full rank.
I got it, thanks.
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