Let $A$ be an invertible matrix over $\mathbb{R}^{n\times n}$, $x, y$ vectors over $\mathbb{R}^{n \times 1}$ such that
$$Ax = y$$
Assume $(y_1, ..., y_n)$ are the coordinates of $y$ over a basis $B$. In solving $x$ we have
$$x = A^{-1}y$$
Is the vector
$$A^{-1}y = \begin{pmatrix} c_1\\ \vdots \\ c_n\end{pmatrix}$$
also in the basis $B$? In other words, if
$$y = b_1y_1+...+b_ny_n$$
is it also the case that
$$A^{-1}y = b_1c_1 + ... + b_nc_n$$
I know this is very simple, and I can also tell such a question reveals a somewhat pronounced confusion over very elementary concepts. I'm very new to linear algebra so please go easy on me! Thanks.
Since $A$ is invertible, there is linear transformation $\rho$ such that $\rho x=y$ and $x=\rho^{-1}y$. Now, we have $$[\rho x]_B=[y]_B=[y_1\ y_2 \ \cdots\ y_n]^T$$
$$[\rho]_B[x]_B=[y]_B$$ $$[x]_B=[\rho]_B^{-1}[y]_B=[\rho^{-1}]_B[y]_B=[\rho^{-1}y]_B=[c_1\ c_2\ \cdots\ c_n]^T$$ Hence, $$x=c_1b_1+\cdots+c_nb_n.$$