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# Proof: Any subspace basis has same number of elements

Proof: Any subspace basis has same number of elements. Created by Sal Khan.

## Want to join the conversation?

• At when Sal says/writes "set X spans subspace V and X has 5 elements you now know that no set than spans the subspace V can have fewer than 5 elements", isn't this statement incorrect? The set X didn't claim to be a BASIS for V, just span V. (I know Sal continues with "even better if X is a basis..." but it sounds like he's going on to make a separate statement). • I thought that a basis was the minimum spanning set of vectors; does it not follow directly from the definition of span and basis that any other spanning set will have at least that many vectors? • Hi msrtra, the precise definition of a basis is simply a linearly independent set than spans a vector space V. The idea that a basis is the smallest spanning set (or minimum spanning set in your words) is actually a consequence of this theorem. Your claim is actually a property of a basis (but one that must be proven nonetheless).
• I am baffled as to how the existence of a minimum number of basis elements meaningfully defines "dimensionality." Doesn't the structure of the matrix itself define a deeper concept of dimensionality that pervades all of linear algebra, with certain data selections merely restricting that dimensionality in various subsets? • I think you may be over-thinking things:

When dealing with vector spaces, the “dimension” of a vector space V is LITERALLY the number of vectors that make up a basis of V.

In fact, the point of this video is to show that even though there may be an infinite number of different bases of V, one thing they ALL have in common is that they have EXACTLY the same number of elements. For example, if X is a basis of V and X has 10 elements then any other basis of V will ALSO have exactly 10 elements (and we would say “V has dimension 10” or, equivalently, “dim(V) = 10”).

Does this make sense?

[NOTE: a vector space does not necessarily have to have a finite dimension, though this is probably something Sal will address much later in the playlist.]
• I am confused now !
If the number of column vectors in some matrix represent its dimensionality, then what do the number of elements in each vector represent?
Before introducing the concept matrices, Sal said (and correct me if I am wrong) that the number of elements in each vector represents the dimensionality of the space thatis vector lives in. For example, a vector with three components is living in a three dimensional vector space..now what :/ ? • If the components of your vector can be anything, then your dimensionality is the number of components. So he's saying that for <x,y,z>, if x y and z can vary independently, then you have 3 dimensions.

However, there are many subspaces in R^3 that have less than 3 dimensions. Remember that dimensionality is a property of vector spaces, not vectors. Take for example the subspace defined by the span of {<1,0,0>,<0,1,0>} -- the XY plane. This vector space only has two dimensions...because every element can be represented as a combination of those two spanning vectors. So <3,4,0> is a part of the space, etc.; the vector space is 2-dimensional. But that's because the three components aren't allowed to vary independently. The third one "always has to be zero"...you're restraining the dimensionality.
• Starting at , I don't understand how he resolves the contradiction. I see that there is one, but I don't know how he can draw the conclusion he does. If he can devise a subset that spans V and is smaller than the originally conceived span, why wouldn't he simply have been wrong that A is linearly independent? • The information we're originally given is as follows:
`A is a subspace created as the span of n basis vectors`
`B is a subspace created as the span of m basis vectors`
`m>n`
`B spans A`
Since there's a contradiction once given these things, we can say that all of the above statements can't be true at the same time. Since the first two statements are just definitions, it's the bottom 2 statements that are incompatible.
• How would one prove an n-dimensional linear subspace is a manifold? • What was the point of bringing up y is also a basis for V towards the end of the video? We knew x was a basis for V, too. Just looks an overcomplicated and unnecessary way of saying they have the same number of elements. Or what? • Stupid question here I know, but how come you know that b_j is redundant just because you've swapped it with -a_1? how come it is NECESSARILY redundant (in other words could it not be a coincidence that it is can be expressed as a linear combination of the others)? • I have exactly the same question as Msrtra. Sal started with A as a basis for V, so by definition (of a basis) the number of vectors in A is the minimum! Sal then goes through the elaborate process to arrive at B_m. It seems to me the justification for B_m not being a basis for V relies upon the definition of basis (around ), so why not make the video a lot shorter and just use the definition?
(1 vote) • This question misunderstands the definition of "basis." Let {a_1, ..., a_n} be a basis for a space V. The definition of a basis says that {a_1, ..., a_n} is "minimal" only in the following sense: no proper subset of {a_1, ..., a_n} can span V. In other words, if you're only allowed to include a_1, ..., a_n in your basis, then you need all n of them to span V. But in this video, we're considering a different situation: you ARE allowed to put different elements in your basis. It is conceivable that you could be clever and find some other b_1, ..., b_m which span V, with m < n; nothing in the definition of a basis automatically forbids this. Thus there is something interesting here that needs to be proven.

It seems like a lot of people are confused by this point. Hopefully the video will be redone to make this clearer. Sal can also use that as an opportunity to clean up the part where he tried to rename b_j as b_1 (at ). Contrary to what he claims, most textbooks would have done the proof that way, using the phrase "without loss of generality" to rename b_j as b_1 from the start. I'm not sure if this was bad planning or a foolish attempt to avoid having to explain the meaning of "without loss of generality," but the result is a mess. 