Proving Subspace Dimension Inequality: W <= V in Vector Space V

[Dosmascerveza]

Ok I am doing some additional problems because my professor assigns evens and i do odd and evens so i can at least check my work in backs of book.


Prove that if W is a subspace of vector space V, dim(W) is less than or equal to dim(V).

Proof: We know W is a subspace of V and therefore has a basis. The basis in W is in V and we also know V has a basis. If the basis for W has same dimension as a basis for V then W is V and we are done. Since V contains W that means the basis for W must be a linear combination of vectors in basis of V. So since every w in W can be generated by a basis in V then then dimV cannot be smaller than dimW. there for dimW<=dimV.

 

[Dosmascerveza]

Linear Algebra Proof, (please check my reasoning)

Here is my questionable attempt at a proof for my LA class
my professor assigns evens so we can't even check our homework so i do both odds and evens but even then I can't check my proofs so easily.

Prove that if W is a subspace of vector space V, dim(W) is less than or equal to dim(V).

Proof: We know W is a subspace of V and therefore has a basis. The basis in W is in V and we also know V has a basis. If the basis for W has same dimension as a basis for V then W is V and we are done. Since V contains W that means the basis for W must be a linear combination of vectors in basis of V. So since every w in W can be generated by a basis in V then then dim(V) cannot be smaller than dim(W). there for dim(W)<=dim(V).

 

[Dosmascerveza]

bump please help! don't be shy

 

[emyt]

Ok I am doing some additional problems because my professor assigns evens and i do odd and evens so i can at least check my work in backs of book.Prove that if W is a subspace of vector space V, dim(W) is less than or equal to dim(V).

Proof: We know W is a subspace of V and therefore has a basis. The basis in W is in V and we also know V has a basis. If the basis for W has same dimension as a basis for V then W is V and we are done. Since V contains W that means the basis for W must be a linear combination of vectors in basis of V. So since every w in W can be generated by a basis in V then then dimV cannot be smaller than dimW. there for dimW<=dimV.

more rigourously, try wording it in terms of the replacement theorem. I think you may have to prove that the subspace is finite-dimensional as well, if V is finite dimensional

 

[HallsofIvy]

Ok I am doing some additional problems because my professor assigns evens and i do odd and evens so i can at least check my work in backs of book.


Prove that if W is a subspace of vector space V, dim(W) is less than or equal to dim(V).

Proof: We know W is a subspace of V and therefore has a basis. The basis in W is in V and we also know V has a basis. If the basis for W has same dimension as a basis for V then W is V and we are done. Since V contains W that means the basis for W must be a linear combination of vectors in basis of V. So since every w in W can be generated by a basis in V then then dimV cannot be smaller than dimW. there for dimW<=dimV.

I think it would be a little clearer to use the fact that, for V of dimension n, a set of more than n vectors in V cannot be independent.

 

[Petek]

Here is my questionable attempt at a proof for my LA class
my professor assigns evens so we can't even check our homework so i do both odds and evens but even then I can't check my proofs so easily.

Prove that if W is a subspace of vector space V, dim(W) is less than or equal to dim(V).

It can be challenging to give advice on such problems because I don't know what theorems/corollaries you've already proved, or even if your vector spaces are all finite dimensional. I'll assume that all spaces are finite dimensional.

Proof: We know W is a subspace of V and therefore has a basis.

You should consider the case W = {0}, although it's trivial.

The basis in W is in V and we also know V has a basis. If the basis for W has same dimension as a basis for V then W is V and we are done.

Remark: Bases don't have dimensions, only spaces and subspaces do. You probably meant to say something like: If the basis for W has the same number of elements as a basis for V, then W = V and we are done.

Since V contains W that means the basis for W must be a linear combination of vectors in basis of V. So since every w in W can be generated by a basis in V then then dim(V) cannot be smaller than dim(W). there for dim(W)<=dim(V).

Again, your terminology is inexact. A basis can't be a linear combination of vectors, but the elements of the basis (which are vectors) can be.

I find your statements in the last quote to be confusing. Have you already proved that, if dim (V) = n, then no set of linearly independent vectors in V can contain more than n elements? If so, I suggest that you try to base a proof on that.

Please post again if my comments aren't clear or if you have any questions.

Petek

 

[Petek]

Please note that the original poster also posed this question https://www.physicsforums.com/showthread.php?t=349098 in the homework thread.

@Dosmascerveza: Please post a question only in one thread to avoid duplication of efforts by those who are trying to help you. Thanks.

Petek