# Thread: prove T(V) is a subspace

1. ## prove T(V) is a subspace

Let $T : R^n --> R^m$be a linear transformation, if V is a subspace of $R^n$ and
T(V) = {T(v): $v \in R^n$},
prove that T(V) is a subspace of $R^m$.
Sry, I know this is an easy question, but i am kindof confused with the definition of subspace

3. so is this right

If $v, w \in V$, then T(v) = v and T(w) = w so we get
T(v + w) = T(v) + T(w) = v + w
which means that $v+w \in V$ and condition (1) holds.
For any scalar c,
T(cv) = cT(v) = cv
so that $cv \in V$and condition (2) holds so V is a subspace

is $R^m$same as $R^n$?

4. Originally Posted by wopashui
so is this right

If $v, w \in V$, then T(v) = v and T(w) = w so we get
T(v + w) = T(v) + T(w) = v + w
which means that $v+w \in V$ and condition (1) holds.
For any scalar c,
T(cv) = cT(v) = cv
so that $cv \in V$and condition (2) holds so V is a subspace

is $R^m$same as $R^n$?
Not in general. If $n=m$, then $\mathbb{R}^n=\mathbb{R}^m$ ; they are not the same otherwise.

To show that $T(V)$ is a subspace of $\mathbb{R}^m$, you have to prove three conditions:

(1) $0_{\mathbb{R}^m} \in T(V),$
(2) If $u,v \in T(V)$, then $u+v \in T(V),$
(3) If $\lambda \in \mathbb{R}$ and $u \in T(V)$, then $\lambda u \in T(V)$.

For (1), $0_{\mathbb{R}^m}=T\left(0_{\mathbb{R}^n}\right) \in T(V)$.

For (2), if $u,v \in T(V)$, then pick $u_0,v_0 \in \mathbb{R}^n$ such that $T(u_0)=u$ and $T(v_0)=v$. Then

$u+v=T(u_0)+T(v_0)=T(u_0+v_0) \in T(V).$

And for (3), you just use preservation of scalar multiplication.

5. Originally Posted by Black
Not in general. If $n=m$, then $\mathbb{R}^n=\mathbb{R}^m$ ; they are not the same otherwise.

To show that $T(V)$ is a subspace of $\mathbb{R}^m$, you have to prove three conditions:

(1) $0_{\mathbb{R}^m} \in T(V),$
(2) If $u,v \in T(V)$, then $u+v \in T(V),$
(3) If $\lambda \in \mathbb{R}$ and $u \in T(V)$, then $\lambda u \in T(V)$.

For (1), $0_{\mathbb{R}^m}=T\left(0_{\mathbb{R}^n}\right) \in T(V)$.

For (2), if $u,v \in T(V)$, then pick $u_0,v_0 \in \mathbb{R}^n$ such that $T(u_0)=u$ and $T(v_0)=v$. Then

$u+v=T(u_0)+T(v_0)=T(u_0+v_0) \in T(V).$

And for (3), you just use preservation of scalar multiplication.
so for (3) if $u\in T(v)$, $c \in R$, then we have
cu = $cT(u') \in T(v)$

right?

6. Originally Posted by wopashui
so for (3) if $u\in T(v)$, $c \in R$, then we have
cu = $cT(u')$ = T(cu') $\in T(v)$

right?