How much like reciprocals are reciprocal vectors? Is there a matrix division that allows $\mathbf{A} = 1 / \mathbf{a}$ in three or two dimensions?

Question

A quote within the question reciprocal vectors reads:

$$ {\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot{\bf b}\times {\bf c}}, $$

plus cyclic permutations, are said to be reciprocal vectors.

To the OP's comment

...but I'm still not sure what they mean by reciprocal vectors

found under this answer, and interesting reply by^† @KimJongUn says:

It's just a name. Looking at ${\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot{\bf b}\times {\bf c}},$, if you "cancel" $\bf b\times \bf c$, then you are left with $\bf A=\frac{1}{\bf a}$, the "reciprocal" of . Of course, you can't legitimately carry out this cancellation but it does help to explain the name.

Question: How much like reciprocals are reciprocal vectors? Is there a matrix division that allows $\mathbf{A} = 1 / \mathbf{a}$ in three or two dimensions?

I think we can write it like that at least for 1D, but I'm not sure if there are issues with the sign.

---

^†okay maybe it's not them

Answer 1

The operation $x \mapsto \frac{1}{x}$ has the property that when you perform it twice (on real numbers) you get back $x$, and when performed on the number $1$, leaves it invariant (which also happens to be true for $-1$). "Negation" is another such operation (which leaves $0$ invariant, instead of $1$).

The source of the operation in the "reciprocal vectors" question has developed an operation on triples of independent vectors in 3-space with the property that when performed twice, brings you back to the same triple-of-independent-vectors. In that sense, it's a little bit analogous to the $x \mapsto \frac{1}{x}$ operation on the real line. And if you regard the standard basis as a particularly nice triple of vectors (because when you stack them up into a $3 \times 3$ matrix, you get the identity), this operation has the property that it leaves invariant this 'special' triple. (I believe that it leaves invariant any positively-oriented orthonormal basis, but I could be wrong --- I didn't look all that closely.)

Note that this is an operation on triples of vectors. If you have a single vector $a$, but lack $b$ and $c$, you can't even define $A$. So it's certainly not very strongly analogous to reciprocals of single numbers in the real number line.

One last point: there are situations in which you have lists of non-zero numbers, and it makes sense to invert each one of them. This essentially never comes up in linear algebra, so it doesn't get a name there, but it comes up in some applications where we have lists of numbers, so it appears in some programming languages. In matlab, for instance, if a is a list of (nonzero) numbers, then 1 ./ a is the list whose $i$th element is the reciprocal of the $i$th element of a. When might this be useful? Suppose that the $i$th element indicates the number of cars (uniformly randomly) crossing a "measuring point" on road $i$ in each hour. Then the $i$th element of the reciprocal indicates the expected waiting time for a car on road $i$, in hours.

Answer 2

The reciprocal vectors are defined if the three given vectors are linearly independent, because otherwise $\mathbf{a}\cdot\mathbf{b}\times\mathbf{c}=0$ (and also for all permutations). If the three vectors are linearly independent, then $\mathbf{b}\times\mathbf{c}$ is orthogonal to both $\mathbf{b}$ and $\mathbf{c}$, so it cannot be orthogonal also to $\mathbf{a}$.

If we take the three coordinate vectors $\mathbf{i},\mathbf{j},\mathbf{k}$, the first of the reciprocal vectors is $$ \dfrac{\mathbf{j}\times\mathbf{k}}{\mathbf{i}\cdot\mathbf{j}\times\mathbf{k}}=\mathbf{i} $$ and similarly you get $\mathbf{j}$ and $\mathbf{k}$ for the other two.

More generally, if you take the scalar products of the first reciprocal vector with the vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ you get $$ \frac{\mathbf{b}\times\mathbf{c}}{\mathbf{a}\cdot\mathbf{b}\times\mathbf{c}}\cdot\mathbf{a}=1, \quad \frac{\mathbf{b}\times\mathbf{c}}{\mathbf{a}\cdot\mathbf{b}\times\mathbf{c}}\cdot\mathbf{b}=0, \quad \frac{\mathbf{b}\times\mathbf{c}}{\mathbf{a}\cdot\mathbf{b}\times\mathbf{c}}\cdot\mathbf{c}=0 $$ and similarly for the cyclic permutations. So the set of reciprocal vectors is nothing else than the dual basis with respect to the identification of the dual space with $\mathbb{R}^3$ itself by means of the scalar product.