Is M a Subspace of V if S and T are Linear Maps from V onto W?

  • Context: Graduate 
  • Thread starter Thread starter adottree
  • Start date Start date
  • Tags Tags
    Linear
Click For Summary

Discussion Overview

The discussion revolves around whether the set M, defined as the collection of vectors x in V such that Sx is in the range of T, is a subspace of V. Participants explore the necessary conditions for M to be a subspace, including closure under addition and scalar multiplication, within the context of linear maps S and T from V onto W.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant outlines the requirements to show that M is a subspace of V, including the need to demonstrate that the zero vector is in M, closure under addition, and closure under scalar multiplication.
  • Another participant suggests that showing closure under scalar multiplication is sufficient to conclude that the zero vector is in M, as the zero vector can be expressed as a scalar multiple of any vector.
  • A participant clarifies that M consists of vectors x in V for which there exists a vector y in V such that Sx = Ty, indicating a relationship between the mappings S and T.
  • Several participants discuss the implications of linearity of S, noting that S(x1 + x2) can be expressed in terms of S(x1) and S(x2), and how this relates to the elements of M.
  • One participant points out that referring to M(x1) or M(x) is incorrect, as M is not a linear map but a subspace.
  • Another participant suggests a more complete expression of the relationship between S(x1 + x2) and the range of T, emphasizing the importance of showing that S(x1 + x2) is in the range of T through the combination of y1 and y2.

Areas of Agreement / Disagreement

Participants generally agree on the need to demonstrate closure properties to establish that M is a subspace. However, there are differing views on the necessity of showing that the zero vector is in M, and some participants clarify terminology and expressions used in the discussion.

Contextual Notes

Some assumptions about the properties of linear maps and the definitions of subspaces are implicit in the discussion. The relationship between the mappings S and T and their ranges is a central focus, but the discussion does not resolve all aspects of these relationships.

adottree
Messages
4
Reaction score
0
S,T: V onto W are both linear maps. Show that M:={x out of V s.t. Sx out of Range(T)} is a subspace of V

I know that to show M is a subspace of V I must show:

i. 0 out of M
ii. For every u, v out of M, u+v out M
iii. For every u out of M, a out of F, au out of M.

I just don't know how to start it, can someone help?
 
Physics news on Phys.org
You are using "out of" where I would use "in" but I understand what you need.

First, as I have pointed out before. You do not need to show that 0 is in M. Since 0v= 0 (the 0 vector), show that M is closed under scalar multiplication immediately gives you that.

M is (in my language!) the set of all vectors, x, in V such that Sx is also in the range of T: there exist some y in V such that Sx= Ty.

Okay, suppose x1[/sup] and x2 are in M- that is, there is y2 in V such that Sx1= Ty1 and y2 such that Sx2= Ty2. What can you say about S(x1+ x2).

Now suppose x is in M- that is, there is y in V such that Sx= Ty- and a is in F. What can you say about S(ax)?
 
S(x1 + x2) = S(x1) + S(x2) (this is because S is a linear map)
= T(y1) + T(y2) = M(x1) + M(x2)

and

S(ax) = aS(x) = aT(y) = aM(x)


Thanks for your help, is this kind of right?
 
adottree said:
S(x1 + x2) = S(x1) + S(x2) (this is because S is a linear map)
= T(y1) + T(y2) = M(x1) + M(x2)

and

S(ax) = aS(x) = aT(y) = aM(x)


Thanks for your help, is this kind of right?

You should not be saying "M(x1)", "M(x2)", or "M(x)" since they are meaningless. M is not a linear map, it is a subspace of V.
 
S(x1 + x2) = S(x1) + S(x2) (this is because S is a linear map)
= T(y1) + T(y2) therefore (x1 +x2) in M

and

S(ax) = aS(x) = aT(y) therefore (ax) in M


I think this is better (I hope)! Thanks you've been a great help!
 
just to make it complete you should write:

S(x1 + x2) = S(x1) + S(x2) (this is because S is a linear map)
= T(y1) + T(y2) = T(y1+y2)

so you can se that S(x1+x2) is in the range of T, namly hit by y1+y2 under T.
 

Similar threads

Undergrad Explanation of a Line of a proof in Axler Linear Algebra Done Right 3r
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Question from a proof in Axler 2nd Ed, 'Linear Algebra Done Right'
  • · Replies 3 ·
Replies
3
Views
3K
MHB Is $\phi(U)$ a Subspace of $\mathbb{R}^m$?
  • · Replies 10 ·
Replies
10
Views
2K
Undergrad Trying to get a better understanding of the quotient V/U in linear algebra
  • · Replies 10 ·
Replies
10
Views
3K
MHB Find Linear Mapping: Help Solving $V \setminus (W_1 \cup \cdots \cup W_m)$
  • · Replies 22 ·
Replies
22
Views
4K
Undergrad Confused about small detail in rank-nullity theorem
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Showing that a set of differentiable functions is a subspace of R
  • · Replies 19 ·
Replies
19
Views
6K
MHB Rank of composition of linear maps
  • · Replies 7 ·
Replies
7
Views
2K
Null space of dual map of T = annihilator of range T
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Meaning of "passage to the quotient"?
  • · Replies 3 ·
Replies
3
Views
1K