- #1
e(ho0n3
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Let B be a basis of a vector space V. If U is a subspace of V, is it true that a subset of B may serve as a basis for U?
Basis of Vector Space V: Subset of B = Basis of U?
- Thread starter e(ho0n3
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In summary: This isn't true for infinite-dimensional spaces. So this is a finite-dimensional phenomenon. But in this case, it is true that there are infinitely many different subspaces of dimension k which are not isomorphic to each other.
- #2
LukeD
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Let V be R^2. One basis of R^2 is {(1,0),(0,1)} (we'll call this B). Let U be {(x,y): y = 2x}. This is a subspace of R^2. One basis of U is {(1,2)}, and every basis of this space contains a multiple of the vector (1,2). Since no multiple of (1,2) is in B, there is no subset of B that is a basis of U.
However, it is possible to have a subset of B be a basis for U. For instance, if B were {(1,2),(2,1)} (this is indeed a basis of V), then {(1,2)} is a basis for U and a subset of B.
However, it is possible to have a subset of B be a basis for U. For instance, if B were {(1,2),(2,1)} (this is indeed a basis of V), then {(1,2)} is a basis for U and a subset of B.
- #3
n_bourbaki
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e(ho0n3 said:
This question is so trivially false that it tells me that you didn't bother to think about any concrete examples at all. The first thing you do if you're not sure about something is you see what happens in a case you can work out by hand, if at all possible. A trivial consequence of this statement would be that a vector space only had a finite number of subspaces, for heaven's sake. I apologise for sounding harsh.
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tiny-tim
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e(ho0n3 said:
Hi e(ho0n3!
Hint: take a really easy example to visualise: let U be ordinary space (R3), and let B be (1,0,0) (0,1,0) and (0,0,1).
Can you draw a two-dimensional subspace which doesn't include any of B?
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HallsofIvy
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Be careful exactly what you are saying. If U is a proper subspace of V, then a basis for V cannot be a basis for U because it will span a space, V, larger than U.
What is true is that there always exists a basis for V that contains a basis for U.
What is true is that there always exists a basis for V that contains a basis for U.
- #6
tiny-tim
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HallsofIvy said:
Hi HallsofIvy!
A basis, yes, but not necessarily the given basis.
- #7
e(ho0n3
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I have a hard time coming up with examples or counterexamples. Thanks guys.
- #8
n_bourbaki
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Does that mean you didn't understand LukeD's example? Or you don't understand that in a vector space of dimension 2 with basis {u,v} then not evey subspace is spanned by either u or v, which is surely the most obvious and trivial example you should have first thought of. Or you do, but just didn't think of them yourself and you're explaining why?
Surely it is clear that if I have n (linearly independent) vectors, then subsets of these span exactly 2^n possible vector subspaces? And that almost all vector spaces have a lot more subspaces than that?
Surely it is clear that if I have n (linearly independent) vectors, then subsets of these span exactly 2^n possible vector subspaces? And that almost all vector spaces have a lot more subspaces than that?
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- #9
e(ho0n3
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This one.n_bourbaki said:
Are you asking? The first one is clear to me. The second one isn't.
- #10
n_bourbaki
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In one thread you're asking about how to distribute operators over tensor products in relation to quantum computing, and in another you're do not know that a vector space of dimension at least 2 (over something like the field of complex numbers) has infinitely many distinct subspaces? This bothers me.
Consider R^2. How many lines through the origin are there?
Consider R^2. How many lines through the origin are there?
- #11
e(ho0n3
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It bothers me a lot more.n_bourbaki said:
Countless. I understand now. Thus, for any n-dimensional space V, since it is isomorphic to Rn, it contains a subspace isomorphic to R2, and since R2 contains infinitely many subspaces, V has infinitely many subspaces. Right?
- #12
n_bourbaki
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e(ho0n3 said:
any n-dimensional real vector space. But a similar analysis for any field will tell you something. But the 2 wasn't important: R^n contains infinitely many proper subspaces of any dimension.
FAQ: Basis of Vector Space V: Subset of B = Basis of U?
What is a vector space?
A vector space is a mathematical structure that consists of a set of vectors and two operations, addition and scalar multiplication, that satisfy specific axioms. It is a fundamental concept in linear algebra and is used to model various physical and abstract systems.
What is a basis of a vector space?
A basis of a vector space is a set of linearly independent vectors that span the entire space. It is the smallest set of vectors that can be used to express any vector in the space through linear combinations.
How is a subset of a basis related to the basis of a vector space?
A subset of a basis is a smaller set of vectors that is still linearly independent and can be used to express a subset of the vectors in the basis of the vector space. It is a smaller basis for a subspace of the original vector space.
Why is it important to understand the basis of a vector space?
Understanding the basis of a vector space is important because it allows us to represent vectors in a simple and concise way. It also helps us to understand the structure of a vector space and its subspaces, which is crucial in solving problems in linear algebra and other related fields.
How do you determine if a set of vectors is a basis of a given vector space?
To determine if a set of vectors is a basis of a given vector space, we need to check two things: linear independence and spanning. If the vectors are linearly independent and span the entire space, then they form a basis. This can be checked through various methods, such as solving a system of equations or using the rank-nullity theorem.