Vector space, linear transformations & subspaces

[ilyas.h]

Homework Statement

Let V be a vector space over a field F and let L and M be two linear transformations from V to V.

Show that the subset W := {x in V : L(x) = M(x)} is a subspace of V .

The Attempt at a Solution

I presume it's a simple question, but it's one of those where you just don't know where to start. This is my attempt;

Subspace:
-0 vector exists
-closed under addition
-closed under scalar multiplicationLinear transformation:
-x,y in V: L(x+y) = L(x) + L(y)
-a in F, x in V: L(ax) = aL(x)
Proving 0 vector exists:

L & M are linear transformations so:

L(x+y) = L(x) + L(y)

but since L(x) = M(x):

L(x+y) = M(x) + M(y) = M(x+y)

∴ L(x+y) + (-M(x+y)) = 0_v

Also:

L(ax) = aL(x)

but since L(x) = M(x):

L(ax) = aM(x) = M(ax)

∴ L(ax) + (-M(ax)) = 0_v

Therefore, 0 vector exists in W.That's all I've got. Don't know how to prove closed under addition and closed under scalar multiplication.

 

[fourier jr]

to show that it is closed you need to show that if x, y ∈ W and a ∈ F then x + y ∈ W and ax ∈W also but you've already done this by showing L(x + y) = M(x + y) etc above.

 

[ilyas.h]

to show that it is closed you need to show that if x, y ∈ W and a ∈ F then x + y ∈ W and ax ∈W also but you've already done this by showing L(x + y) = M(x + y) etc above

so essentially I've done the problem but in an "all in one" fashion?

 

[fourier jr]

I don't think that shows that 0 is in W though, since you need to have L(0) = M(0) for that to be true. I would start L(0) = L (x - x) = ...

& I would also skip the lines that say ∴ L(x+y) + (-M(x+y)) = 0_v because the previous lines are enough

 

[vela]

so essentially I've done the problem but in an "all in one" fashion?

No. It seems like you don't really have a clear idea of what you need to show, which makes writing a proof more difficult because you don't know where to start and where you're trying to get to.

To show that W is closed under vector addition, for example, you need to show that if x and y are in W, then x+y is in W. "x and y are in W" is your starting point, so you should start with the statement "Let x and y be in W." Note that in your attempts, you never said anything about what x and y are. For all we know from what you wrote, they are random vectors in V, so your assertion that L(x) = M(x) isn't justified. On the other hand, if you explicitly say x is in W, then by the definition of W, you know that L(x) = M(x).

"x+y is in W" is where you're trying to get to. This means you need to show that L(x+y) = M(x+y). When you get to that line, you can then conclude that x+y is in W based on the definition of W.

To prove that 0 is in W, what do you need to show? To prove W is closed under scalar multiplication, what do you need to show? Get what those mean clear in your mind, and the proofs will probably follow pretty easily, and you won't be wondering if you succeeded in showing what you were supposed to.

 

[Ray Vickson]

I don't think that shows that 0 is in W though, since you need to have L(0) = M(0) for that to be true. I would start L(0) = L (x - x) = ...

& I would also skip the lines that say ∴ L(x+y) + (-M(x+y)) = 0_v because the previous lines are enough

But, T(0) = 0 for any linear transformation T, so L(0)=M(0) (both =0).