# Vector spaces?

hi,

I am confused about vector spaces and subspaces. I've just started a book on linear algebra, and i understood the 1st chapter which delt with gaussian reduction of systems of linear equations, and expressing the solution set as matricies, but the 2nd chapter deals with vectors and i'm perplexed.

I've learnt that if the solution of a system of equations has 3 unknowns, and 2 free variables for example, then the solution is a plane in 3 dimensional space. Now, is that 3 dimensional space a vector space? Is the plane a subspace? I really don't understand the concept of vector spaces... the book just lists 10 proofs that something is a vector space but i'd like to know what it is in geometric terms. Thanks.

matt grime
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A vector space is something that satisfies the axioms of 'being a vector space'. Unenlightening? But R^3 is a vector space, planes in R^3 that go through the origin are vector spaces, lines that go through the origin are vector spaces.

Consider all two dimensional vectors in the x-y plane of the form $$u = (u_{1},u_{2})$$ where $$u_{1}$$ is the x-component and $$u_{2}$$ is the y-component. Basically a vector space is a fancy name for saying that the collection of all those vectors have certain properties.

Those are not 10 proofs but axioms. Any set of elements which satsify those axioms forms a vector space. Have you done vector algebra? Those arrows with "head and tail" form a vector space. You can check this for yourself by making sure that under vector addition (the usual way) and multiplication of a vector by a scalar it satsifies the axioms. But this is just one example. The concept of a vector space is more general and applies to many other seemingly different parts of the mathematics.

planes in R^3 that go through the origon are vector spaces...so in the case of a homogenous system of equations with 2 unkowns, that plane is a vector space within a vector space right. But what if that plane doesn't go through the origon, what's it called then? A subspace?

No a plane that goes through the origin is a subspace. A subspace is a vectorspace (i.e. subset of a vector space).

A subspace is a vectorspace (i.e. subset of a vector space).

Just to add... a subet of a vector space which satisfies the closure axioms (with the same operations, of course).

matt grime
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planes in R^3 that go through the origon are vector spaces...so in the case of a homogenous system of equations with 2 unkowns, that plane is a vector space within a vector space right. But what if that plane doesn't go through the origon, what's it called then? A subspace?

It is just a plane. If you really want to be fancy it is an affine subspace, but this is not at all helpful for you. Part of the definition of a vector subspace is that it contains the origin, so your question is trivial to answer (but it is still a good question).

HallsofIvy
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A plane that does not go through the origin is not a "subspace", in particular, the sum of two members of the plane is not, in general, in the plane. A subspace is a subset that is, itself, a vector space. A plane that does not go through the origin is called a "linear manifold".

Note: the 10 axioms (not "proofs") that define a linear vector space include basic properties of the operations- associative, commutative, distributive. If you are given a subset of the vector space you know that those are already true since they are true for any members of the vector space. To prove the subset is a "subspace" you need only show the "closure" properties: u+ v is in the subset, av is in the subset, 0 is in the subset.

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this is so confusing

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this is so confusing

Not if you read the definitions carefully and go through some examples. Do so, and then return to this thread and read what HallsofIvy wrote. i have to 'try just a little bit harder'

as Janis Joplin would say.

to add another question on the topic: if a given plane does not go through the origin, i.e. it is a linear manifold, does that mean that it also cannot be a vector space? If it can't be a subspace, because it does not contain the zero vector (if I understand this right), then I would think that it also doesn't satisfy one of the vector space axioms that the set containing x,y must contain a vector "phi" such that x+phi = x. Am I on the right track here?

matt grime
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No. Such a thing is called an affine subspace. If you are completely ignoring the origin as the identity, then that means you're automatically ignoring the previous addition as the addition of vectors. Since you're completely redefining everything then any plane can be given the structure of a vector space in many ways. Anyw object, any space has the potential to be a vector space. In this case it is straightfoward. Pick O any point in the plane. Any other point in the plane has a unique represnentation as O+x, and we can define (O+x)+(O+y):=O+(x+y), etc. O is then the origin.

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ah, I see. That makes quite a bit more sense. I guess that's what allows us to show that the axioms are true, defining O in the given set. matt grime
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I realize I was a little lax, there. I should have defined a new addition, @, and let + be the old addition. Thus

(O+x)@(O+y)=O+x+y.

We're just defining addition on the cosets of O in the obvious way.

I know this is kind of a dumb question, but what's a coset? This is my first math course beyond the introductory things they have us take... I'm still trying to get a grasp of the more rigorous language and terminology.