Suppose we have a vector space $V$ over a scalar field $\mathbb{F}$ and two different bases $\mathcal{B}=\lbrace\mathbf{v}_{i}\rbrace_{i=1,\ldots , n}$ and $\mathcal{E}=\lbrace\mathbf{w}_{i}\rbrace_{i=1,\ldots , n}$. A given vector $\mathbf{x} \in V$ can be described in terms of either of these bases, by the (unique) linear combinations $$\mathbf{x}= \sum_{i=1}^{n}a_{i}\mathbf{v}_{i} \quad \text{or}\quad \mathbf{x}= \sum_{i=1}^{n}b_{i}\mathbf{w}_{i}$$ where $a_{i} \neq b_{i}$.
Now, suppose that we have a linear operator $\mathcal{S}$ which maps the basis $\mathcal{E}$ to the basis $\mathcal{B}$, defined in the following manner, $$\mathcal{S}\left(\mathbf{w}_{j}\right)= \mathbf{v}_{j}$$ As $\mathbf{v}_{j} \in V$ it can itself be described in terms of a (unique) linear combination of the basis vectors $\mathbf{w}_{i}$ $\left(i=1,\ldots , n\right)$ and as such, we have that $$ \mathbf{v}_{j}=\sum_{i=1}^{n}\left[\mathcal{S}\left(\mathbf{w}_{j}\right)\right]_{i}\mathbf{w}_{i}=\sum_{i=1}^{n}S_{ij}\mathbf{w}_{i}$$ where we have defined $\left[\mathcal{S}\left(\mathbf{w}_{j}\right)\right]_{i} \equiv S_{ij}$, which is the $i^{th}$ component of the $j^{th}$ basis vector $\mathbf{v}_{j} \in \mathcal{B}$ with respect to the basis $\mathcal{E}$.
Is this a correct description of how to change between two different bases for the same vector space? (The equation $\mathbf{v}_{j}=\sum_{i=1}^{n}S_{ij}\mathbf{w}_{i}$ is given in a text book that I've been reading as a formula for switching between two different basis sets for a given vector space and I'm trying to justify its form).