Question

Let V,W

be vector spaces, T:V→W
, a linear transformation. If U
is a subspace of V
, then T(U)={T(x)∣x∈U}
is a subspace of W
.

Answer 1

Yes, T(U)={T(x)∣x∈U} is a subspace of W.

Step-by-step explanation:

When U is a subspace of V, T(U)={T(x)∣x∈U} is a subspace of W. To prove this, we need to show that T(U) satisfies the three conditions for being a subspace. First, T(U) contains the zero vector because T(0)=0. Second, for any u1,u2∈U and scalar c, we have T(cu1+u2)=cT(u1)+T(u2), which shows closure under addition and scalar multiplication.

Finally, since T is a linear transformation, it preserves vector addition and scalar multiplication, ensuring that T(U) is closed under these operations. Therefore, T(U) is indeed a subspace of W.

In summary, the image of a subspace under a linear transformation remains a subspace in the codomain. This is due to the properties of linear transformations, which ensure that the image set satisfies the necessary conditions for being a subspace. Thus, T(U)={T(x)∣x∈U} is indeed a subspace of W.

The proof involves demonstrating that T(U) satisfies the three conditions for being a subspace: containing the zero vector, closure under addition and scalar multiplication. By leveraging the properties of linear transformations, we can establish that T(U) is indeed a subspace of W.

Answer 2

To determine if T(U) is a subspace of W, one needs to verify that T(U) is closed under vector addition and scalar multiplication, which are intrinsic properties of a linear transformation.

Step-by-step explanation:

The question refers to the properties of linear transformations between vector spaces and subspaces. To show that T(U), the image of a subspace U under the linear transformation T, is a subspace of W, we must verify that T(U) satisfies the subspace criteria. These criteria include being closed under vector addition and scalar multiplication, containing the zero vector from W, as well as associativity and commutativity of vector addition in vector spaces. It is a fundamental property of linear transformations that if T is linear and U is a subspace of V, then T(U) is indeed a subspace of W. This can be shown by taking any vectors x, y in U, and a scalar c; since T is linear, T(cx + y) = cT(x) + T(y), which belongs to T(U), hence satisfying closure under addition and scalar multiplication.