Proof: Any subspace basis has same number of elements | Linear Algebra | Khan Academy
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Proof: Any subspace basis has same number of elements Watch the next lesson: https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/null_column_space/v/dimension-of-the-null-space-or-nullity?utm_source=YT&utm_medium=Desc&utm_campaign=LinearAlgebra Missed the previous lesson? https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/null_column_space/v/visualizing-a-column-space-as-a-plane-in-r3?utm_source=YT&utm_medium=Desc&utm_campaign=LinearAlgebra Linear Algebra on Khan Academy: Have you ever wondered what the difference is between speed and velocity? Ever try to visualize in four dimensions or six or seven? Linear algebra describes things in two dimensions, but many of the concepts can be extended into three, four or more. Linear algebra implies two dimensional reasoning, however, the concepts covered in linear algebra provide the basis for multi-dimensional representations of mathematical reasoning. Matrices, vectors, vector spaces, transformations, eigenvectors/values all help us to visualize and understand multi dimensional concepts. This is an advanced course normally taken by science or engineering majors after taking at least two semesters of calculus (although calculus really isn't a prereq) so don't confuse this with regular high school algebra. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to KhanAcademy’s Linear Algebra channel:: https://www.youtube.com/channel/UCGYSKl6e3HM0PP7QR35Crug?sub_confirmation=1 Subscribe to KhanAcademy: https://www.youtube.com/subscription_center?add_user=khanacademy
Comments
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Instead of all this gibberish, why can't you simply state that if the dimension of m is less than the dimension of n, m cannot span n and if the dimension of m is greater than the dimension of n, the vectors in m cannot be linearly independent?
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lol he couldnt handle saying bj anymore
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lol, bj
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Actually, without loss of generalization, we could assume b_1 doesn't equal to 0.
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I really enjoyed that, thank you!
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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
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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.
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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|.
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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...".
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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.
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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.
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Nice, wise screen.
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