Proof: Any subspace basis has same number of elements | Linear Algebra | Khan Academy

Proof: Any subspace basis has same number of elements

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[stgr]

Thanks! This is a much longer proof than I would have thought. The following minor corrections might help some viewers:
At 10:53 you say, "cuz if all of these guys (pointing at 'c2b2' and greater) were non-zero, then you wouldn't be able to say that these gu-- ... that this vector and that vector (pointing at 'a2' and 'c1a1') are lineraly independent because it would... they would be scalar multiples of each other."
I suspect you meant to say, "cuz if all of these guys were ZERO, then 'a2' and 'c1a1' would be scalar multiples of each other and thus not lineraly independent." At 13:48 you say it correctly in the same context.
At 16:53 you say, "Then A becomes lineraly independent." I suspect you mean to say, "Then A becomes lineraly DEPENDENT. You say/write it correctly at 17:07.
At 19:27, you say, "... we know that if Y is a basis for V and X is a basis, X also spans V, so we know that X has to have fewer elements than Y." Frankly, it just goes downhill from there for about a minute. I think you meant to say, "... we know that Y is a basis for V and X is a basis for V, so we know that X and Y must have the same number of elements." You get it back on track at about 20:20.

[rupjyotidas]

Thank you for such detailed explanation

[ivoriankoua]

That was a good demonstration, thanks!

[bullzeyet]

its likely that most of the ppl on this page are not even here to learn there job is to try and correct this havard and mit graduate

[eswyatt]

These are spanning vectors, the little b's. B (big B) is not a vector. Maddening!

[samlee]

this is a college professor, shown by the handwriting

[lovepeace-jo]

i think korean subtitle have some errors... meaning is slightly different...