How to show $T$ is linear transformation and bijective if $T(x)=[x]_{\alpha}$ and $\alpha =\{v_1,...,v_n\}$ is basis for $V$? Here $V$ is a vector space and $\alpha =\{v_1,...,v_n\}$ is a basis for $V$. Let $T:V\rightarrow \Bbb{R}^n$ defined by $T(x)=[x]_{\alpha}$ for every $x\in V$. How to show $T$ is linear transformation and bijective?
To show it's a linear transformation, we have to show that $T(ax+y)=aT(x)+T(y)$, but how are we gonna show this in this question as $[x]_a$ is present?
To show bijectivity, we show $\dim V= \dim \Bbb{R}^n$, and then show either injective or surjective. But in this case I think $\dim \Bbb{R}^n=n$ but how about $x$? We don't know the basis of $x$.
 A: To show that, e.g., $T(ax) = aTx$ for $x\in V$ and $a\in\mathbb R$, let $x\in V$, $a\in\mathbb R$. Then $x = c_1v_1 + \ldots + c_nv_n$ with some $c_1,\ldots,c_n\in\mathbb R$. You have $[x]_\alpha = (c_1,\ldots,c_n)^T$. Since $ax = ac_1v_1 + \ldots + ac_nv_n$, you also have that $[ax]_\alpha = (ac_1,\ldots,ac_n)^T$. Hence $T(ax) = [ax]_\alpha = (ac_1,\ldots,ac_n)^T = a(c_1,\ldots,c_n)^T = a[x]_\alpha = aT(x)$. Now, do similarly to prove that $T(x+y) = T(x) + T(y)$.
As to the second task, you are right that you only have to show injectivity. For this, let $x\in V$ such that $T(x) = 0$. Now, if $x = c_1v_1 + \ldots + c_nv_n$, what can you infer about $[x]_\alpha$?
A: T is injective since $T(x_1)=T(x_2) \implies [x_1]_\alpha = [x_2]_\alpha$ which suggests that if $x_1=\sum_{i=0}^na_iv_i$ and $x_2=\sum_{i=1}^nb_iv_i$ then by the given condition $a_i=b_i$ for each $i$ in one through $n$ which implies that $x_1 = x_2$. Therefore $T$ is injective.
Also observe that $V = L(\alpha)$ i.e. $V$ is the set of all linear combinations of elements of the set $\alpha$. Now consider any $x:=(c_1, c_2, ..., c_n)$ in $\mathbb{R^n}$. Since V is the set of all linear combinations of elements of the set $\alpha$ the linear combination $\sum_{i=1}^nc_iv_i$ is contained in $V$ and its image under the linear transformation $T$ is $x$. Therefore every vector in $\mathbb{R^n}$ has a pre-image in V under the linear transformation $T$ which is why $T$ is surjective and hence bijective.     
