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**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.