Let $V,W$ be vector spaces, $T:V\to W$, a linear transformation. If $U$ is a subspace of $V$, then $T(U)=\{ T(x)\mid x\in U\}$ is a subspace of $W$.

The question is as follows.

Let $V$ and $W$ be (real) vector spaces and let $T:V\to W$ be a linear transformation. Given a $U$ such that $U$ is a subspace of $V$, let $T(U)$ denote the set of all images of the form $T(x)$ where $x \in U$. Show that $T(U)$ is a subspace of $W$.

Here's my answer. Can anyone tell me if this is right?

Since $U$ is a subspace of $V$, if $u,v \in U$, $cu+dv \in U$ where $c,d \in \mathbb{R}$.

Now if $T(U)$ is a subspace of $W$, then that means that for $T(u),T(v) \in T(U)$ where $u,v \in U$, $T(cu+dv) \in T(U)$. This is what we must show.

Let us assume $u,v \in U$. Since $W$ is a vector space, if $T(u),T(v) \in W$, $cT(u)+dT(v) \in W$ for $c,d \in \mathbb{R}$. This is one of the vector space axioms. Since $T$ is linear, $T(cu+dv)=cT(u)+dT(v)$ which fulfills the criteria for a vector subspace, since the zero vector is created by $0u+0v$. So $T(U)$ is a subspace of $W$.

1 Answer

This is somewhat acceptable as a proof but superfluous and unilluminating. You need to show that if $u,v \in T(U)$, then $cu + dv = T(U)$. What can you say about $u\in T(U)$? About $v\in T(U)$? Can you create some $x\in U$ such $T(x) = cu+dv$?

Rigorous solution (spoilers):

Let $c,d \in \mathbb R$, $u,v \in T(U)$.
As $u\in T(U)$, $\exists a \in U$ such $T(a) = u$.
As $v\in T(V)$, $\exists b\in U$ such $T(b) = v$.
As $U$ is a vector space, $ca+db \in U$, so $T(ca+db) \in T(U)$.
By linearity of $T$, we have $T(ca+db) = T(ca)+T(db)=cT(a)+dT(b)=cu +dv$. So we can set $x=ca+db \in U$ and see $cu+dv=T(x) \in T(U)$ as desired.