#Linear Algebra Direct sum help D:

21 messages · Page 1 of 1 (latest)

wise monolith
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Hm.
Well, we can treat it the following way:
Imagine W as a hyperplane of dimension 3 in ℝ^4(x). For now, let's consider just ℝ^4 instead of ℝ^4(x).
So, it will have an equation:
ax + by + cz + dw + e = 0
So, this hyperplane will have a normal line with vector {a, b, c, d}, which should be a satisfactory T for your problem.

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At least, if I understood the problem correctly...

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I am not too good with sums/intersections of subspaces...

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Oh, the dim is 2. Hmm...

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Yeah, then my approach won't work.

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Nah, seems good.

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Hm.
How about we search for a couple of vectors orthogonal to v1 and v2?

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So, we can try picking a couple of random vectors and performing Gram-Schmidt on them.

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Actually, I don't think it's even necessary for them to be orthogonal.

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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.

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How about u1 = x, u2 = 1? Maybe that will work.

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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.

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If it's 4, all good.

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If it's less than 4, we'll look for other vectors.

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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.

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Great!
So, I think we can take T to be ℒ{x, 1}.

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Well, probably...
I'm still not 100% sure my thoughts are correct.

plain solar
wise monolith
thin lintelBOT
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