Proof: Any subspace basis has same number of elements

[NikhilXLII]

lol, bj

[jeaniedomino]

cool clip thanks for sharing ..

[humarukh]

It is Confusing , all the swapping things , this thing has to be more clear and less confusing / contradicting

[CantBeDisplayed]

did you mispeak at 10:55 ?

nonzero should be said zero??

[dalcde]

Actually, without loss of generalization, we could assume b_1 doesn't equal to 0.

[mollierdiag]

I really enjoyed that, thank you!

[HapaLife]

bj, lulz

[yynotx]

isn't the claim B spans the space negate the claim A spans the space--yet in the discussion of "a2" you cite the linear independence of A..

.time to rewind!

[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

[VeryEvilPettingZoo]

A side note. Your comment similar to "I can rename the labels of my set B1' at will, so that bj having non-zero coefficient could've been renamed to be b1" is extremely common in math. It's typically phrased "Since we've shown that non-zero coef accompanies at least one of the b's, without loss of generality we'll say it's with the b1 vector" (or said even more concisely). It's so common it's often abbreviated "W.L.O.G." with little explanation at all.

[VeryEvilPettingZoo]

Here's a briefer form (not actually different at all) of your correct final proof of the invariance of basis cardinality, with |.| meaning set cardinality:

Let A, B both be (finite) basis for V.

Then A basis, B spans (since basis implies spans) implies |A| <= |B| (by what you just proved). But similarly, B basis, A spans, so |B| <= |A|. Therefore |A| <= |B| <= |A|. Therefore |A| = |B|.

[VeryEvilPettingZoo]

You misspoke at 18:23. You said that if X spans V and has 5 elements, then no set that spans V can have fewer than 5 elements. I think you meant to say either "..., then no basis for V can have more than 5 elements" or "if X is a basis for V and has 5 elements, then...".

[VeryEvilPettingZoo]

There's a small error. Your claim that B1' and B2' must be linearly dependent (l.d.) is false. The reason is that the "new" a1 or a2 you're including with the b's MIGHT ALREADY BE one of those b's, and the original set might be linearly independent. Ex: If B = [v1, v2] ([...] means set) is a basis for V, and I make B' = [w, v1, v2], for some w in V, it does not follow that B' is l.d.. That's because it might be that w = v1 or w = v2, in which case B' = B, and so B' is l.i., since B is.

[VeryEvilPettingZoo]

This little problem is bypassed by saying, in your example, since a1 is in V, and B spans V, have a1 = sum(0<i<m+1, di*bi) for some coefficients di (from the def of the span of a set of vectors). Then a1 not 0 (since A is l.i.), so not all di's can be zero. This gets you to the same place, so your proof is otherwise unchanged.

[KingRobbStark]

Nice, wise screen.