Linear Algebra: Vector Spaces, Subspaces, etc.

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Homework Help Overview

The discussion revolves around identifying subsets of R³ that qualify as vector subspaces. Participants are examining specific forms of vectors and the properties that must be satisfied for these subsets to be considered subspaces, including closure under addition and scalar multiplication.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to verify whether given subsets of R³ contain the zero vector and are closed under addition and scalar multiplication. There is a focus on understanding the definitions and properties of vector spaces and subspaces.

Discussion Status

Some participants are clarifying the problem statement and discussing the necessary conditions for subsets to be subspaces. There is an ongoing exploration of the implications of the properties of vector spaces, with some participants questioning their understanding of the axioms and how to apply them.

Contextual Notes

There is a noted confusion regarding the algebraic versus geometric interpretations of vector spaces, which may affect participants' ability to grasp the concepts fully. Additionally, some posts indicate a lack of clarity in the original problem statement, which may hinder the discussion.

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Homework Statement


Which of the following subsets of R3? The set of all vectors of the form

a) (a, b, c), where a=c=0
b) (a, b, c), where a=-c
c) (a, b, c), where b=2a+1

Homework Equations


A real vector space is a set of elements V together with two operations + and * satisfying the following properties:

I. If u and v are elements of V, then u+v is in V (i.e., V is closed under the operation +).
(a) u+v=v+u, for u and v in V.
(b) u+(v+w)=(u+v)+w, for u, v, and w in V.
(c) There is an element 0 in V such that u+0=0+u, for all u in V.
(d) For each u in V, there is an element -u in V such that u+(-u)=0.

II. If u is any element of V and c is any real number, then c*u is in V (i.e., V is closed under the operation *).
(e) c*(u+v)=c*u+c*v, for all real numbers c and all u and v in V.
(f) (c+d)*u=c*u+d*u, for all real numbers c and d, and all u in V.
(g) c*(d*u)=(cd)*u, for all real numbers c and d and all u in V.
(h) 1*u=u, for all u in V.

Let V be a vector space and W a nonempty subset of V. If W is a vector space with respect to the operations in V, then W is called a subspace of V.

The Attempt at a Solution



I am having a VERY hard time grasping such abstract concepts. I think I understand the definitions of a vector space and subspace, but I don't really understand how to 'reason' with them, and put them into practice. I have seen lectures on opencourseware (MIT) with Dr. Strang and his use of geometric illustrations is much more easier follow, but my prof (and my textbook) uses only algebraic equations and I think that's where my problem is. :(

This also stands in the way of my learning of other concepts like Linear Independence, Basis, Dimension and Homogeneous Equations...
 
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You are missing some words in your problem statement.
Which of the following subsets of R3? The set of all vectors of the form

a) (a, b, c), where a=c=0
b) (a, b, c), where a=-c
c) (a, b, c), where b=2a+1
Should the first sentence be "Which of the following subsets of R3 are subspaces?"
 
Mark44 said:
You are missing some words in your problem statement.

Should the first sentence be "Which of the following subsets of R3 are subspaces?"

thats correct. sorry about the confusion.
 
OK, presumably you already know that R^3 is a vector space, so you don't have to verify that all ten axioms are satisfied.

For each of your problems verify that:
  1. The subset contains the zero vector
  2. The subset is closed under addition. IOW, if v1 and v2 are in the subset, then v1 + v2 is in the subset, tool
  3. The subset is closed under scalar multiplication. IOW, if v1 is in the subset, and c is a real number, then c*v1 is in the subset.
 
Mark44 said:
OK, presumably you already know that R^3 is a vector space, so you don't have to verify that all ten axioms are satisfied.

For each of your problems verify that:
  1. The subset contains the zero vector
  2. The subset is closed under addition. IOW, if v1 and v2 are in the subset, then v1 + v2 is in the subset, tool
  3. The subset is closed under scalar multiplication. IOW, if v1 is in the subset, and c is a real number, then c*v1 is in the subset.

ok, correct I get this.

So for a):

The subspace is (0,b,0) and the zero vector is obtained by setting the scalar c=0 and multiplying through.

But, since v1+v2=(a1+a2, b1+b2, c1+c2)
v1+v2=(0, b1+b2, 0) is not true for R3, then it is not in R3

In the same way does this subset not satisfy the scalar multiplication axiom?
 
00PS said:
ok, correct I get this.

So for a):

The subspace is (0,b,0) and the zero vector is obtained by setting the scalar c=0 and multiplying through.

But, since v1+v2=(a1+a2, b1+b2, c1+c2)
v1+v2=(0, b1+b2, 0) is not true for R3, then it is not in R3
it defintely is in R3,

00PS said:
In the same way does this subset not satisfy the scalar multiplication axiom?

i think you're approaching the subspace axioms the wrong way
for your case (0,b,0), assuming b \in \Re

then
zero
pick b = 0, clearly this is in this is in b \in \Re, and so the zero element is in (0,b,0): b \in \Re

closed under addition
u = v1 + v2 = (0, b1 + b2, 0)
is b = b1 + b2 in R? if so (0,b,0) is closed under addition. ie the result of addition of 2 elements of the set is a member of the set


and similar for multiplication... then you've shown its a subspace as Mark said
 
Last edited:
lanedance said:
it defintely is in R3,



i think you're approaching the subspace axioms the wrong way
for your case (0,b,0), assuming b \in \Re

then
zero
pick b = 0, this is in the space defined by (0,b,0): b \in \Re

closed under addition
u = v1 + v2 = (0, b1 + b2, 0)
is b = b1 + b2 in R? if so is within (0,b,0): b \in \Re, can you find a b = b1 + b2

and similar for multiplication... then you've shown its a subspace as Mark said

This seems very similar to what I just wrote...I can't see much of a difference?

So if b1+b2=0 then the subspace is in R3, but that would mean the vector is (0,0,0) which makes the subspace trivial?
 
It's not at all similar. You tried to argue that the set of vectors {(0, b, 0)} was NOT closed under addition, and was therefore not a subspace of R^3.

What you need to do is to show that this set IS closed under vector addition and IS closed under scalar mulitiplication. What you have done that, you will have shown that the set IS a subspace of R^3.
 
Let U and V be subspaces of R^n. Show that if W:={w E R^n: w=u+v for some u E U and v E V} then W is a subspaceof R^n.
 
  • #10
How would I do this?
 
  • #11
Hi sqenshal

you would go about it the same as above, try and show it is closed under addition & scalar multiplication based on the properties of the original subspaces, so

\forall w_1, w_2 \in W \Rightarrow a.w_1 + b.w_2 \in W, for a,b \in \Re
 

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