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 be regarded formally as rows, then the matrix representing the change-of-base from $\alpha$ to $\beta$ is denoted by $$\frac \alpha \beta \ \ \ \ \text{or}\ \ \ \ \beta \backslash \alpha \in K^{n\times n},$$
(but **not** $\alpha / \beta$) where $K$ is the base field. 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.

This notation has some benefits: for example we may write $v = \alpha x$ to say that $v$ is the linear combination $x_1 \alpha_1 + x_2 \alpha_2 + \cdots + x_n \alpha_n$ of the base $\alpha$ by the coefficient (column) vector $x = [x_1 \ x_2 \ \cdots \ x_n]^T$. Then it is natural to use symbols like
$$ x = \frac v \alpha \ \ \ \ \text{or} \ \ \ \ \alpha \backslash v. $$
Now suppose $y = [y_1 \ \cdots \ y_n]^T$ is the coordinate column vector of $v$ w.r.t. the base $\beta$. Then now we can write
$$ y = \frac \alpha \beta x, \ \ \ \ \text{that is,} \ \ \ \ \frac v \beta = \frac \alpha \beta \frac v \alpha \ \ \ \ \text{or} \ \ \ \ \beta \backslash v = (\beta \backslash \alpha)(\alpha \backslash v). $$
Be careful: perform cancellations only along the line $\backslash$, not $/$.

The above notaions $\beta \backslash \alpha$ and $\beta \backslash v$ are compatible in the sense that
$$ \frac \alpha \beta = \frac {(\alpha_1,\ \cdots,\ \alpha_n)} \beta = \left[ \frac {\alpha_1} \beta \ \frac {\alpha_2} \beta \  \cdots \  \frac {\alpha_n} \beta \right]. $$

We may define several similar symbols to make things involving change-of-base easy. Let $T: V \to V$ be a linear map. Denote by $T\alpha$ the ordered tuple $(T\alpha_1 ,\ \cdots, \ T\alpha_n) \in V^n$ of vectors. Then the matrix representation of $T$ is $\alpha\backslash T\alpha$.

With some more work, one may easily verify the formula
$$ \frac {T\beta} \beta = \frac \alpha \beta \frac {T\alpha} \alpha \frac \beta \alpha, $$
which means that the matrix of $T$ w.r.t. $\beta$ is the conjugate of  the matrix of $T$ w.r.t $\alpha$ by $\beta\backslash\alpha$.

**Even more benefit:** If the space $V$ is just $\mathbb R^n$, then the tuples $\alpha$ and $\beta$ are merely the matrices, and the three notations below denotes completely the same thing:
$$ \frac \alpha \beta = \beta \backslash \alpha = \beta ^{-1} \alpha, $$
and this is the reason why I recommended you not to use the symbol $\alpha / \beta$.

**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
$$ \alpha: x = [x_1 \ \cdots \ x_n]^T \mapsto \alpha x = \sum x_i \alpha_i. $$

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.