I have the following question.
Let $V$ and $W$ be vector spaces, let T: $V \to W$ be a linear transformation, and let $B = \{v_1, v_2, ..., v_n\}$ be a basis of $V$. Prove that if $T$ is injective, then $V$ is isomorphic with $Range(T)$.
I have assumed that $T$ is injective so we know $T(v_1) = T(v_2)$, which implies that $v_1 = v_2$, and also that the $\ker(T) = \{0\}$.
I am not sure how to proceed from here.
On
If $f : V \to V$ is injective and $\{v_1, \dots{}, v_n\}$ is a base for $V$, then a base of $\operatorname{im} f$ is $\{f\left(v_1\right), \dots{}, f\left(v_n\right)\}$. (In general, injective linear functions map linear independent vectors to linear indipendent vectors.) Thus this subspace must coincide with the whole $V$, being $V$ finite-dimensional.
On
It's fairly easy to prove that if $v_1.\dots,v_n$ is a basis of $V$, then $Tv_1,\dots,Tv_n$ spans range$(T)$. It is also easy to prove that if $T : V\to W$ is injective and $v_1,\dots,v_n$ is a linearly independent list in $V$, then $Tv_1,\dots,Tv_n$ is linearly independent in $W$.
Using these two results, we can conclude that $Tv_1,\dots,Tv_n$ is a basis of range$(T)$ and thus the dimension of range$(T)$ is the same as $V$. Now use the result that two finite dimensional vector spaces are isomorphic if and only if they have the same dimension to conclude that $V\cong \text{range}(T)$.
Or you can make an argument that $T$ is a bijection from $V\to \text{range}(T)$ to conclude that $T$ is an isomorphism.
By definition of range, we know $T$ is surjective from $V \to $ Range($T$). So $T$ is bijective (not from $V \to W$!). All we need to check for isomorphism is see if Range($T$) is a vector space.
Since Range($T$) $\subseteq W$, checking for closure is enough. Let $w, w'$ be arbitrary in Range($T$) and $a$ be a scalar from the field of $W$. Then,
$$ \begin{align*} aw + w' &= aT(v) + T(v') \tag{By injectivity} \\ &= T(av) + T(v') \\ &= T(av + v') \\ & \in \text{Range}(T) \end{align*} $$ This completes the argument.