Subspace test involving linear transformations

[p3forlife]

Homework Statement

Determine whether the subset W of the vector space V is a subspace of V.
Let V = L(Q⁴) (the set of linear transformations from rational numbers with 4 coordinates to rational numbers with 4 coordinates).
Let W = { T in V = L(Q⁴) | { (1,0,1,0) , (0,1,0,-1) } is contained in N(T) }
where N(T) is the nullspace of T

Homework Equations

Subspace test involves checking if the zero vector is in W, whether W is closed under addition, and whether W is closed under scalar multiplication.

The Attempt at a Solution

1. Check if zero vector is in W
From definition, N(T): {x in V | T(1,0,1,0) = 0 , T(0,1,0,-1) = 0 }
So the zero vector is in W

2. Check if closed under addition
From linearity of T, T(x+y) = T(x) + T(y)
where x = (a, b, c, d) in W
y = (p, q, r, s) in W

3. Check if closed under scalar multiplication
From linearity of T, T(cx) = cT(x)
where x = (a, b, c, d) in W

So W is a subspace of V.

Is it enough to use the linearity of T to check for closure under addition and multiplication? Is there something else I should include in my solution? I'm unsure if I'm supposed to do the subspace test on the actual vectors x, y, etc. or if I'm supposed to test the transformations, T(x), T(y), etc. For example, should closure under addition be asking whether x + y is in W or if T(x) + T(y) is in W?

Thanks in advance!

 

[Mark44]

Your set W consists of all the transformations from Q^4 to Q^4 that send (1,0,1,0) and (0,1,0,-1) to (0, 0, 0, 0). This fact has ramifications on what the matrices for this set of transformations looks like.

I don't see that your work uses the two given vectors.
For the first part, you need to show that T(0, 0, 0, 0) = (0, 0, 0, 0)

 

[p3forlife]

Okay, so checking if T(0,0,0,0) = 0 is asking if (0,0,0,0) is in N(T). Since (0,0,0,0) is not in N(T), then the zero vector is not in W. Is this right?

 

[HallsofIvy]

Homework Statement

Determine whether the subset W of the vector space V is a subspace of V.
Let V = L(Q⁴) (the set of linear transformations from rational numbers with 4 coordinates to rational numbers with 4 coordinates).
Let W = { T in V = L(Q⁴) | { (1,0,1,0) , (0,1,0,-1) } is contained in N(T) }
where N(T) is the nullspace of T

Homework Equations

Subspace test involves checking if the zero vector is in W, whether W is closed under addition, and whether W is closed under scalar multiplication.

The Attempt at a Solution

1. Check if zero vector is in W
From definition, N(T): {x in V | T(1,0,1,0) = 0 , T(0,1,0,-1) = 0 }
So the zero vector is in W

2. Check if closed under addition
From linearity of T, T(x+y) = T(x) + T(y)
where x = (a, b, c, d) in W
y = (p, q, r, s) in W

W is a space of linear transformations. "Closed under addition" is not a question of 'x+ y' but of T+ S where T and S are in W. If T and S are in W is T+S in W?

3. Check if closed under scalar multiplication
From linearity of T, T(cx) = cT(x)
where x = (a, b, c, d) in W

Again, the question is not about cx but about cT. If T is in W is cT in W?

So W is a subspace of V.

Is it enough to use the linearity of T to check for closure under addition and multiplication? Is there something else I should include in my solution? I'm unsure if I'm supposed to do the subspace test on the actual vectors x, y, etc. or if I'm supposed to test the transformations, T(x), T(y), etc. For example, should closure under addition be asking whether x + y is in W or if T(x) + T(y) is in W?

Thanks in advance!

 

[HallsofIvy]

Okay, so checking if T(0,0,0,0) = 0 is asking if (0,0,0,0) is in N(T). Since (0,0,0,0) is not in N(T), then the zero vector is not in W. Is this right?

Once again, the question is not about T(0,0,0,0)= 0. It is about the 0 transformation: T(x,y,z,w)= (0,0,0,0) for all x,y,z,w. If T= 0, the linear transformation that maps every (x,y,z,w) to (0,0,0,0), is it in W? That specifically, is asking whether T(1, 0, 1, 0)= (0,0,0,0) and whether T(0,1,0,-1)= (0,0,0,0) and the answer to that is trivial.

 

[p3forlife]

Once again, the question is not about T(0,0,0,0)= 0. It is about the 0 transformation: T(x,y,z,w)= (0,0,0,0) for all x,y,z,w. If T= 0, the linear transformation that maps every (x,y,z,w) to (0,0,0,0), is it in W? That specifically, is asking whether T(1, 0, 1, 0)= (0,0,0,0) and whether T(0,1,0,-1)= (0,0,0,0) and the answer to that is trivial.

Thanks for your reply.
So it's obvious that the zero vector is in W.

Check if T + S is in W (closure under addition):
T(1,0,1,0) + S(0,1,0,-1) = 0 + 0 = 0
Since T + S is in N(T), W is closed under addition.

Check if cT is in W (closure under scalar multiplication):
cT(1,0,1,0) = c * 0 = 0
cT(0,1,0,-1) = c * 0 = 0
Sin cT is in N(T), W is closed under scalar multiplication.

Is this right or am I missing something?