Linear Transformation and changing the basis $T:\mathbb{R}^2 \to \mathbb{R}^2$ defined by:
$$ T(x,y) = (x+y,y) $$
Now we pick one vector $u_1 = (2,3)$, then- $$ T(u_1) = (2+3,3) = (5,3)\tag{1} $$
So far so good.
Now lets consider a Basis $B = \{u_1, u_2\}$ in $\mathbb{R}^2$. I want to do above calulaction in Basis B and then will convert back to standard basis.
$$[u_1]_B = (1,0)$$
$$T([u_1]_B) = (1+0,0) = (1,0)\tag{2}$$
Here $T([u_1]_B)$ is in Basis $B$, since transformation doesn't change basis and our input $[u_1]_B$ is in Basis B therefore output $T([u_1]_B)$ is also in basis $B$. 
Lets convert output back to standard basis -
$$\big[T([u_1]_B)\big]_B = (1,0)$$
$$T([u_1]_B) = (2,3)\tag{3}$$
Sorry for abuse of notation at eq. $(2)$. In equation $(2)$ LHS is $\big[T([u_1]_B)\big]_B$ .
My questions are- 


*

*Can we pass a vector w.r.t any arbitrary basis $B$ inside transformation definition? $\big($Is it even legal to do something like $(2)$ or we always need to pass a standard basis vector in $T(.)$ $\big)$.

*What does it mean to solve something like equation $(2)$ ?. 

*How Can we achieve $ T(u_1)$ given $\big[T([u_1]_B)\big]_B$? (If we change the basis and transform a vector, then how does it relate to transformed vector without changing the basis).
These questions might sound layman but I was not able to figure out answer. Any quick help would also be highly appreciated.
 A: You are misunderstanding the concepts. If $x$ is any vector in $\mathbb{R}^2$, and $\beta = \{v_1,v_2\}$ is a basis for $\mathbb{R}^2$, then $x$ can be written uniquely as $x = a_1v_1 + a_2v_2$ for some choice of constants $a_1$ and $a_2$, right? In this case, we define the coordinate vector of $x$ with respect to $\beta$ to be
$$[x]_\beta := \begin{pmatrix} a_1\\a_2 \end{pmatrix}.$$
Also, if $\gamma$ is another basis for $\mathbb{R}^2$, we define the matrix representation of $T$ with respect to $\beta$ and $\gamma$ (in that order!) to be the matrix $[T]_\beta^\gamma$ whose first and second column are $[T(v_1)]_\gamma$ and $[T(v_2)]_\gamma$, respectively.
Now, we have the relationship $[T(x)]_\gamma = [T]_\beta^\gamma [x]_\beta$. For example, let $x = (2,3)$, $\beta = \{(1,1),(1,0)\}$, $\gamma = \{(1,0),(0,1)\}$ and $T$ as defined above. Then, $T(x) = (5,3)$ and it is easy to see that
$$[T(x)]_\gamma = \begin{pmatrix} 5\\3 \end{pmatrix}.$$
On the other hand, since $T(1,1) = (2,1)$ and $T(1,0) = (1,0)$, the matrix of $T$ is given  by
$$[T]_\beta^\gamma = \begin{pmatrix} 2&1\\1&0 \end{pmatrix},$$
and, of course, since $x = 3(1,1) + (-1)(1,0)$,
$$[T]_\beta^\gamma [x]_\beta = \begin{pmatrix} 2&1\\1&0 \end{pmatrix} \begin{pmatrix} 3\\-1 \end{pmatrix} = \begin{pmatrix} 5\\3 \end{pmatrix}.$$
