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if space X can be described by basis b1,b2,b3,b4,b5, then if i can find some subspace of X which can using linear

combination of basis b10,b20,b30,then can I found out b40,b50 and prove they(b40,b50) must exists?

It's a question came out of looking https://www.youtube.com/watch?v=2IdtqGM6KWU&index=11&list=PLE7DDD91010BC51F8,

(Lecture 11 of MIT 18.06 Linear Algebra, Spring 2005) the professor says dim(S+U)-dim(S and U) = dim(S) + dim(U), I want to prove that as

transforming S to b1,b2,b3,b4,b5... U to b1,b2,b3,(c4,c5)... then S+U would be combining two basis and removing those common basis, to prove that.

femto
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  • The question seems to be off-topic because it is about linear algebra (which is pure mathematics). Although not totally clear, the question seems to be about the following phenomenon http://en.wikipedia.org/wiki/Basis_%28linear_algebra%29#Extending_to_a_basis – Codor Jan 16 '15 at 07:58

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ok, I find my answer, you can pick one arbitrary point from

X-{b10+b20+b30} (those point in X but not in the spaces of b10,b20,b30 forms) to use that as b40, and then picking one arbitrary point

from X-{b10+b20+b30+b40} as b50, according to basis theorem the length of

of basises of a space is same, so from that above process, the new b would generate same length of basis as original ones, neither more or less. that proved.

femto
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