Linear Algebra Coordinate Systems Isomorphism This is an excerpt from the book.
Let $B$ be the standard basis of the space $P_3$ of polynomials; that is let $B=\{1,t,t^2,t^3\}$.  A typical element $p$ of $P_3$ has the form
$p(t) = a_0 + a_1t+ a_2t^2 + a_3t^3.$
 Since $p$ is already displayed as a linear combination of the standard basis vectors, we conclude that $[P]B = [a_0,a_1,a_2,a_3]$.
This is the part I don't get. My understanding is if I were to change the basis of something, then I would have a matrix, for instance $B$ represented as $\{b_1,b_2\}$. And using the matrix, I would change $x$ to $[x]B$ by performing $B^{-1} x = [x]B$. None of these steps were taken for the above. Could anybody explain how the polynomial equation suddenly turned into a vector with respect to $B$?
 A: Your understanding applies only to a specific case of changes of basis.  In general, this is the idea: if we have bases $\mathcal A = \{v_1,\dots,v_n\}$ and $\mathcal B = \{w_1,\dots,w_n\}$, then our change of basis is the transformation $T$ that satisfies
$$
T([x]_{\mathcal A}) = [x]_{\mathcal B} 
$$
where 
$$
[x]_{\mathcal A} = \pmatrix{a_1\\ \vdots \\ a_n} \text{ means that }  
x = a_1v_1 + \cdots + a_n v_n\\
[x]_{\mathcal B} = \pmatrix{b_1\\ \vdots \\ b_n} \text{ means that }  
x = b_1w_1 + \cdots + b_n w_n
$$
Your approach works whenever the elements of $\mathcal B$ are specifically the column-vectors $\{[w_1]_{\mathcal A}, \dots, [w_n]_{\mathcal A}\}$.  In this case, the matrix of the transformation $T$ is given by $B^{-1}$, where $B$ is the matrix whose columns are those column-vectors of $\mathcal B$.
In the particular case where $\mathcal A$ is the standard basis (the vectors $(1,0,\dots,0),(0,1,0,\dots,0)$ and so on), $[w_1]_{\mathcal A}$ is literally the vector $w_1$ as we normally think of it.
In this case, however, the question they're asking is more fundamental.   Remember that if $\mathcal B = \{1,t,t^2,t^3\}$, then 
$$
[p]_{\mathcal B} = \pmatrix{a_0 \\ \vdots \\ a_3} \text{ means that }
p = a_0(1) + a_1(t) + a_2(t^2) + a_3(t^3)
$$
and so all we had to do was understand the information given to us.
A: The polynomial is not turned into a vector. The vector $[P]B=[a_0,a_1,a_2,a_3]$ shows the coordinate of $f(x)=a_0+a_1x+a_2x^2+a_3x^3$ when you choose $B=[1,x,x^2,x^3]$ as your basis. As I understand, there is no intention to change the basis.
