You may use the division symbol. This division by a base has an important meaning: the duality. This notation has a lot of benefits, including reducing the amount of computations and explicitly showing the logic of linear algebra.
If $\alpha = (\alpha_1 ,\ \alpha_2 ,\ \cdots ,\ \alpha_n)$ and $\beta = ( \beta_1 ,\ \beta_2,\ \cdots ,\ \beta_n)$ are bases, which will always be regarded formally as row matricesrows, then the matrix representing the change-of-base from $\alpha$ to $\beta$ is denoted by $$\frac \alpha \beta \ \ \ \ \text{or}\ \ \ \ \beta \backslash \alpha.$$$$\frac \alpha \beta \ \ \ \ \text{or}\ \ \ \ \beta \backslash \alpha \in K^{n\times n},$$
where $K$ is the base field. (But not $\alpha / \beta$.) As I mentioned below, This notation is still valid even if $\alpha$ is not a basis, but just an ordered tuple of vectors. In this case $\beta\backslash \alpha$ may not be a change-of-base matrix, though.
Again more benefit: We may denote the coordinate map $V \to \mathbb R^n$ w.r.t the base $\alpha$ by just
$$ \frac 1 \alpha : v \mapsto \frac v \alpha, $$
and its inverse by just
$$ \alpha: x = [x_1 \ \cdots \ x_n] \mapsto \alpha x = \sum x_i \alpha_i. $$$$ \alpha: x = [x_1 \ \cdots \ x_n]^T \mapsto \alpha x = \sum x_i \alpha_i. $$
Note that the notations in this answer enjoys the associative law.
Here is another benefit. For a base $\alpha$, we may promise to denote the dual base $(\alpha_i^*)$ of $V^*$ by $\alpha \backslash 1$:
$$ \frac 1 \alpha := [ \alpha_1^* ,\ \alpha_2^* ,\ \cdots ,\ \alpha_n^* ]^T, $$
where the dual base is regarded as a column. This makes no conflict with `Again more benefit'. And our first notation
$$ \frac \beta \alpha $$
now may be regarded as a multiplication of column and a row, producing a square matrix:
$$ \frac \beta \alpha = \frac 1 \alpha \beta = \begin{bmatrix} \alpha_1^* \\ \alpha_2^* \\ \vdots \\ \alpha_n^* \end{bmatrix} [ \beta_1 \ \ \beta_2 \ \ \cdots \ \ \beta_n ], $$
which is exactly equal to the change-of-base matrix. This is the genuine meaning of our division by a basis.
So let's take a look at an example. How this notation may be used? Let me prove a famous theorem in linear algebra:
Theorem. Let $V$ be a fin dim vector space over a field $K$. Let $V^*$ be the dual space of $V$, and $V^{**}$ the double dual. Then there is a linear isomorphism of $V$ onto $V^{**}$ that is independent of the choice of base of $V$.
Proof. Let $n$ be the dimension of $V$ over $K$. Let $\alpha = (\alpha_1 ,\ \cdots,\ \alpha_n) \in V^n$ and $\beta \in V^n$ be bases of $V$, which are regarded as a row form. Then they induces the dual bases $\alpha^* = \alpha \backslash 1 = [\alpha_1^* ,\ \cdots,\ \alpha_n^*]^T$ and $\beta^* = \beta \backslash 1$ of $V^*$, which are regarded as columns. We use a lemma:
Lemma. The change-of-base matrix
$$ \frac {\alpha^*} {\beta^*} $$
from $\alpha^*$ to $\beta^*$ equals $\alpha \backslash \beta$:
$$ \frac 1 \alpha = \frac \beta \alpha \frac 1 \beta. $$
Remark: It is well-known that the resulting matrix must be the transposed inverse, but the reason why we got just the inverse is that we are regarding the dual bases as columns, not rows. So the lemma does not defy your background knowledge; if you regard the dual bases as rows, then you'll get the familiar transposed inverse.
So let's continue to working with the theorem. If we apply the lemma twice, then we know that the changeofbase matrix from $\alpha^{**}$ to $\beta^{**}$ equals $\beta \backslash \alpha$.
Now define a linear isomorphism $T:V \to V^{**}$ by the linear extension of the assignment $\alpha_i \mapsto \alpha_i^{**}$. We want to show that $T$ is independent of choice of the base; that is, $T\beta_i = \beta_i^{**}$ for all $i$. This is equivalent to showing that the matrix representation $\beta^{**} \backslash T\beta$ of $T$ equals the identity matrix. This is easily done by our notations:
$$ \frac {T\beta} {\beta^{**}} = \frac{\alpha^{**}}{\beta^{**}} \frac {T\alpha}{\alpha^{**}} \frac {\beta} {\alpha} = \frac \alpha \beta \cdot I \cdot \frac \beta \alpha = I. $$
This completes the proof. //
So our division notation allows us not to do enormous scalar computations. This is the other great benefit, I think.
