Proof about isomorphism (Prove that T is an isomorphism if and only if $T(\beta)$ is a basis for W) 
Let V and W be n-dimensional vector spaces, and let $T:V \rightarrow W$ be a linear mapping. Suppose that $\beta$ is a basis for V. Prove that T is an isomorphism if and only if $T(\beta)$ is a basis for W.
My first question is since basis is a linearly independent set that spans V, why can it be written as $T(\beta)$, where it is in the input position?

Here is some work I have done:
$\Rightarrow$: We want to show that $T(\beta)$ is a linearly independent set and $T(\beta)$ spans $W$.
Let $x \in V$, then $x= \sum_{i=1}^n a_iv_i$, and let $w \in W$.
Since T is injective,N(T)={0},$\sum_{i=1}^n a_iv_i$ $\in N(T)$, so $\sum_{i=1}^n a_iv_i$=0. 
Since {$v_i$} is a basis for V, we have got unique scalars $a_i=0$.
$T(\sum_{i=1}^n a_iv_i)$=0, by linearity, $\sum_{i=1}^n a_i T(v_i)$=0. 
Since T is surjective, we have  $\sum_{i=1}^n a_i T(v_i)$=w.
Since w is written as a linear combination of $T(v_i)$, span $(T(\beta))$=W. Hence $T(\beta)$ is a basis for W.
$\Leftarrow$: For this direction, is it trivial?
 A: Write $\beta=\{b_1,...,b_n\}$.
Suppose $T(x)=0$. Since $T(\beta)$ is a basis for $W$, there exists $c_1,...,c_n$ such that $c_1 T(b_1)+...+c_n T(b_n)=T(x)=0$. Clearly, $c_1=...=c_n=0$. Hence, $x=c_1b_1+...+c_nb_n=0$. This shows $T$ is injective.
By a similar method, let $y\in W$. Then there exists $d_1,...,d_n$ such that $d_1T(b_1)+...+d_nT(b_n)=y.$ Hence $x=d_1 b_1+...+d_nb_n$ solves $T(x)=y$. Therefore, $T$ is surjective.
A: You wrote:

My first question is since basis is a linearly independent set that spans V, why can it be written as $T(\beta)$, where it is in the input position?

I'm going to try to re-state your question so that I can understand it better. I hope I haven't misinterpreted your question. To me, it sounds like you're asking the following:

Since $T:V\to W$ and $\beta$ is a subset of $V$, (not an element of $V$), why can we plug $\beta$ into $T$? In other words, what is $T(\beta)$?

In general, if $X$ and $Y$ are sets, and $f:X\to Y$ is a function, then for any $A\subseteq X$, we can define $f(A)$ as follows:
$$f(A)=\left\{y\in Y\,\vert\,y=f(a)\text{ for some }a\in A\right\}.$$
$f(A)$ is called "the image of $A$ under $f$".
Since $\beta=\{v_1,\ldots,v_n\}$ for some $v_1,\ldots,v_n\in V$, we have that
$$T(\beta)=\{T(v_1),\ldots,T(v_n)\}.$$
As for your proof: your proof of the forward direction may have been adequate, but because of the wording I wasn't able to tell.
An alternate approach to the problem would to break it up as follows: let $V$ and $W$ be $n$-dimensional; let $T$ be linear; and let $\beta$ be a basis for $V$. Show the following:
(1) $T$ is injective iff $T(\beta)$ is linearly independent.
(2) $T$ is surjective iff $T(\beta)$ spans $W$.
If you would like me to post solutions to either of these parts, then let me know.
Addendum: Below I'll post proofs for problems (1) and (2) above.
(1) Our goal is to show that $T(\beta)=\left\{T(v_1),\ldots,T(v_n)\right\}$ is linearly independent iff $T$ is injective.
Suppose $T(\beta)=\left\{T(v_1),\ldots,T(v_n)\right\}$ is linearly independent. Let $x\in V$, and let $T(x)=0$. Since $x\in V$ and $\beta$ is a basis for $V$ we have that
$$x=\sum_{i=1}^n a_iv_i$$
for some scalars $a_1,\ldots,a_n$. So we have that
$$\sum_{i=1}^n a_i T(v_i)=T\left(\sum_{i=1}^n a_iv_i\right)=T(x)=0.$$
It follows that $a_1=\ldots=a_n=0$. Hence $x=0$. And $T$ is injective. $\quad\Box$
Now suppose that $T$ is injective. and let $\sum_{i=1}^n a_i T(v_i)=0$. It follows that
$$T\left(\sum_{i=1}^n a_iv_i\right)=\sum_{i=1}^n a_i T(v_i)=0.$$
Since $T$ is injective, this implies that $\sum_{i=1}^n a_iv_i=0$. And since $\beta=\{v_1,\ldots,v_n\}$ is a basis, this implies that $a_1=\ldots=a_n=0$, which shows that $T(\beta)=\left\{T(v_1),\ldots,T(v_n)\right\}$ is linearly independent. $\quad\Box$
(2) Our goal is to show that $T(\beta)=\left\{T(v_1),\ldots,T(v_n)\right\}$ spans $W$ iff $T$ is surjective.
Suppose $T(\beta)=\left\{T(v_1),\ldots,T(v_n)\right\}$ spans $W$. Hence for every $w\in W$, there exist scalars $a_1,\ldots a_n$, such that
$$w=\sum_{i=1}^n a_i T(v_i)=T\left(\sum_{i=1}^n a_i v_i\right),$$
which shows that for each $w\in W$ there is a $v\in V$ such that $w=T(v)$. Hence $T$ is surjective. $\Box$
Suppose $T$ is surjective, and let $w\in W$. Our goal is to show that $w$ is in the span of $T(\beta)=\left\{T(v_1),\ldots,T(v_n)\right\}$. Since $T$ is surjective we have that $w=T(v)$ for some $v\in V$. And since $\beta=\{v_1,\ldots,v_n\}$ is a basis for $V$, we have that
$$v=\sum_{i=1}^n a_i v_i$$
for some scalars $a_1,\ldots,a_n$. Hence we have that
$$w=T(v)=T\left(\sum_{i=1}^n a_i v_i\right)=\sum_{i=1}^n a_i T(v_i),$$
which shows that $w$ is in the span of $T(\beta)=\left\{T(v_1),\ldots,T(v_n)\right\}$. $\quad\Box$
A: Let's clarify the notation. If the basis is $\beta=\{v_1,v_2,\dots,v_n\}$, then
$$
T(\beta)=\{T(v_1),T(v_2),\dots,T(v_n)\}
$$
The rank-nullity theorem says that
$$
\dim V=\dim\ker(T)+\dim\operatorname{Im}(T)
$$
without any further assumption than $T$ being linear. In particular, $\dim\operatorname{Im}(T)\le\dim V$.
Suppose you know that $T(\beta)$ is a basis of $W$, where $\beta$ is a basis of $V$.


*

*If the set $T(\beta)$ is linearly independent in $W$, you know that $\dim\operatorname{Im}(T)\ge\dim V$ and therefore $\dim\operatorname{Im}(T)=\dim V$.

*If the set $T(\beta)$ is a spanning set of $W$, you know that $\operatorname{Im}(T)=W$.

*If the set $T(\beta)$ is a basis, you get from the two points above that $\dim W=\dim V$. Moreover $W=\operatorname{Im}(T)$ ($T$ is surjective) and $\dim\ker(T)=0$ ($T$ is injective). Therefore $T$ is an isomorphism.
Let's attack the converse. First I state two general facts about linear maps and the third point is the conclusion from the first two.


*

*An injective linear map maps linearly independent sets to linearly independent sets.

*Any linear map maps a spanning set of the domain onto a spanning set of the image.

*If $T$ is an isomorphism, $T(\beta)$ is linearly independent and spans $\operatorname{Im}(T)=W$. Hence $T(\beta)$ is a basis of $W$.
