S,T: V onto W are linear maps

1. Mar 8, 2008

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?

2. Mar 8, 2008

HallsofIvy

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)?

3. Mar 8, 2008

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?

4. Mar 8, 2008

HallsofIvy

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.

5. Mar 8, 2008

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!!!

6. Mar 9, 2008

mrandersdk

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.