# Thread: prove T(V) is a subspace

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

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

3. so is this right

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

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

4. Originally Posted by wopashui
so is this right

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

right?

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

right?