# How to define the inverse of a vector?

Most physical situations in mechanics can be modeled using a combination of derivatives - specifically, derivatives of position: velocity and acceleration. But physical situations can also be modeled other ways. Consider the scalar equation for velocity in one dimension:

$$v = \frac{dx}{dt}$$

it can be modeled just as well by a different quantity, called "slowness" which is described as:

$$s = \frac{dt}{dx} = \frac{1}{v}$$

Which is used commonly in day-to-day life. For example, runners usually measure distance in minutes per mile, or minutes per km. Walkers go slow enough such that it doesn't make sense to measure speed in miles/hour or a similar unit, but instead to say they take ~20mins per mile. However, this is rarely, if ever, used in physics.

In order to model slowness physically, there are a few basic things to know. Firstly, when we want to add velocities, they add quite nicely. If I stand on top of a flat bed truck and run at 5m/s while the truck is going 10m/s (disregarding special relativity) my total speed is 5+10=15m/s. For a slowness, we have to take advantage of knowing this fact to figure out how to "add" slownesses together.

We know $$v_1+v_2=v_t$$, so if $$s_1=\frac{1}{v_1}$$ and $$s_2=\frac{1}{v_2}$$ and $$s_t=\frac{1}{v_t}$$ then $$\frac{1}{s_1}+\frac{1}{s_2}=\frac{1}{s_t}$$ and therefore:

$$\frac{1}{\frac{1}{s_1}+\frac{1}{s_2}}=s_t$$

defining an operation associated with this (called "oplus" - discussed in detail here):

$$x \oplus y := \frac{1}{\frac{1}{x}+\frac{1}{y}}$$

and its inverse ("ominus")

$$x \ominus y := \frac{1}{\frac{1}{x}-\frac{1}{y}} = x \oplus (-y)$$

we have

$$s_1 \oplus s_2 = s_t$$,

so while velocities add, slownesses oplus.

It seems slowness is not easily measured in vector form, even though it represents the same physical quantity as velocity and therefore has both magnitude and direction.

The vector version of slowness should (I think) fulfill three requirements:

1. Preserve direction (point same direction as velocity vector)
2. Invert magnitude (magnitude of slowness should be 1/speed)
3. Coordinate independence (slowness in the x-direction doesn't effect y-direction, etc)

There is only one vector which satisfies the first two is the vector $$\frac{\vec{v}}{|\vec{v}|^2}$$ which unfortunately doesn't also satisfy the third requirement, because the magnitude of $$\vec{v}$$ is affected by all coordinates of $$\vec{v}$$. For example, if $$\vec{v}=\langle1,2\rangle$$ then $$\frac{\vec{v}}{|\vec{v}|^2}=\langle\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\rangle$$ but if $$\vec{v}=\langle1,3\rangle$$ then $$\frac{\vec{v}}{|\vec{v}|^2}=\langle\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\rangle$$ which means just changing the y coordinate also changed the x coordinate of the slowness.

In order to preserve coordinate independence, as well as stay consistent with the one-dimensional definition of slowness, one can define a "vector" for slowness as taking the reciprocal of each component. So if $$\vec{v}=\langle x,y,z\rangle$$ then $$\vec{s}=\langle\frac{1}{x},\frac{1}{y},\frac{1}{z}\rangle$$.

The initial problem is that it appears to not preserve direction or invert the magnitude of the velocity vector. However it requires changing a fundamental vector property.

It comes down to how we measure distance in a coordinate system, and the operations we use on vectors. We all know that vectors add together, which makes sense since velocity and position do the same, and those things add when they are scalars.

One problem with defining slowness as a vector may be that slowness does not satisfy vector properties, even in one dimension! Slownesses do not add, they oplus (as shown above).

So instead of defining slowness, which is the reciprocal of slowness, as a vector, why not define it as something else? Something which is like a vector, but more readily taking advantage of its properties. For example, it could be different in how we measure its magnitude, as well as other properties:

Given $$\vec{v}=\langle x,y,z \rangle$$

$$\frac{1}{\vec{v}} := [x^{-1},y^{-1},z^{-1}]$$ <-- not a vector, instead could be called an "inverse vector" or "invector", denoted by square brackets

$$|\frac{1}{\vec{v}}| := \sqrt{(\frac{1}{x})^2\oplus(\frac{1}{y})^2\oplus(\frac{1}{z})^2}$$

This appears to behave quite nicely, since

$$|\frac{1}{\vec{v}}|=\sqrt{(\frac{1}{x})^2\oplus(\frac{1}{y})^2\oplus(\frac{1}{z})^2} = \sqrt{\frac{1}{x^2+y^2+z^2}} = \frac{1}{\sqrt{x^2+y^2+z^2}} = \frac{1}{|\vec{v}|}$$

Which satisfies requirement #2.

Through this definition, requirement #1 can also be satisfied, provided we change how we measure distance. Most people are familiar with a graph on a logarithmic scale. This helps visualize data which grows exponentially by changing where the numbers are located geometrically on an axis.

In our new idea of distance, we will work with reciprocal space (I am aware that term is used to describe a crystal lattice, but I am not using the same thing here). BOTH the x and y axes (we'll start working in 2 dimensions) will be re-scaled such that x=1/x and y=1/y. The origin will be replaced with a single "point at infinity" similar to projective geometry.

Graphing reciprocal vectors in a space which is measured this way satisfies requirement #1 - preserving direction. So all three requirements are satisfied with this new definition, provided we say that this isn't a vector, and it lives in a different space.

The vector $$\langle 2,3 \rangle$$ (shown right) when graphed in Cartesian space looks the same as the vector $$\langle \frac{1}{2},\frac{1}{3} \rangle$$ (shown left) graphed in reciprocal space.

An amazing thing about this space is that the geometric "tip to tail" addition of vectors applies to this new space as well, except it corresponds to oplussing of invectors instead of addition of vectors. And, assuming this represents a slowness, it is entirely consistent with vector addition of velocity!

We define (invectors will be denoted with a *)

given $$\vec{a}* = [a_1,a_2]$$ and $$\vec{b}* = [b_1,b_2]$$

$$\vec{a}* \oplus \vec{b}* := [a_1 \oplus b_1, a_2 \oplus b_2]$$

For instance:

$$\vec{v_1} = \langle x_1,y_1 \rangle$$ and $$\vec{v_2} = \langle x_2,y_2 \rangle$$

$$\vec{v_1}+\vec{v_2} = \vec{v_t} = \langle x_1+x_2,y_1+y_2 \rangle$$

as a slowness invector, that would be

$$\vec{s_1}* = [\frac{1}{x_1},\frac{1}{y_1}]$$ and $$\vec{s_2}* = [\frac{1}{x_2},\frac{1}{y_2}]$$

$$\vec{s_1} \oplus \vec{s_2} = \vec{s_t} = [\frac{1}{x_1+x_2},\frac{1}{y_1+y_2}]$$

which is consistent because $$\frac{1}{[\frac{1}{x_1+x_2},\frac{1}{y_1+y_2}]} = \langle x_1+x_2,y_1+y_2 \rangle = \frac{1}{\vec{v_t}}$$

Using the fact that vector addition is the same as invector oplussing, it is possible to prove that since lengths in reciprocal space are 1 divided by their geometric lengths, there is a new Pythagorean theorem for reciprocal space.

$$c^2=a^2 \oplus b^2$$

because we know that $$\frac{1}{length(x)}=x$$ in reciprocal space (where length(x) is the length you measure normally, if you took out a ruler, for example and x is the "actual length") and we know that $$length(c)^2 = length(a)^2+length(b)^2$$

so $$\frac{1}{c^2} =\frac{1}{a^2}+\frac{1}{b^2}$$

and

$$c^2 =\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}} = a^2 \oplus b^2$$

which explains the equation above for magnitude of an invector.

We can also see that if we use the inverse operation of $$\oplus$$, $$\ominus$$ (o-minus) we can define a linear distance function along the axes in one dimension. We can call this function "closeness" because it is how close one object is to another. A small closeness is a large distance and a large closeness is a small distance (because they are reciprocals).

$$closeness(x,y) = c(x,y) := |x \ominus y|$$

and for two dimensions

$$c((x_1,y_1),(x_2,y_2)) := \sqrt{(x_1 \ominus x_2)^2 \oplus (y_1 \ominus y_2)^2}$$

The three dimensional formula is similar.

We can see that under this closeness (distance) formula in 1 dimension, the distance between the reciprocal of any integer and the reciprocal of the next is 1. The distance between 1 and 1/2 is $$|1\ominus\frac{1}{2}|=1$$. The distance between 1/2 and 1/3 is 1, the distance between 1/3 and 1/4 is 1, and so on. The distance function here is translation-invariant - if we move the axes the lengths of lines do not change.

I have skirted along without mentioning the fact that the identity under the oplus operation is $$\infty$$. I've found that $$\infty$$ is an integral part of this system. It effectively works just like zero in the Cartesian system. $$a\oplus\infty=a\ominus\infty=a$$ for all a and in general, $$\infty = -\infty$$.

As far as I can tell, this makes sense physically - a body with $$0$$ velocity has $$\infty$$ slowness. A body which has $$0$$ distance between it and something else has $$\infty$$ closeness to it. The invector of acceleration can be found to have physical meaning as well, and to work perfectly well in reciprocal space.

I have not included everything which I have found about this system, including the dot product of two invectors $$(\vec{a}*\cdot\vec{b}*=[a_1\cdot b_1,a_2\cdot b_2])$$ and how they relate to derivatives describing motion in a reciprocal way.

As an amateur with only high school level math training, I'd simply like to ask, does this make any sense? Is anybody aware of a vector way to describe slowness or other inverse vector quantities which is different (or the same) from my own work? I'd like to understand how this relates to mathematics in general, and if my ideas and work are valid. A vector is generally defined as a mathematical object with both magnitude and direction, but it seems to me that even though that suits this type of idea, a vector is unable to describe the type of object I'm dealing with here. Is there another way to do this that is already accepted by the math community? Is my work new or does this exist somewhere? Is defining the inverse of a vector even possible? If vectors are independent of coordinate system, why does changing to reciprocal space change anything? Basically, I would just like to know more details about how to define the inverse of a vector, in a mathematical sense and a physical sense.

• There's a lot to read here, but here's a quick piece of information you might find helpful. Typically, taking the derivative of a scalar quantity with respect to a vector quantity gives you a vector quantity which is called the gradient. I suspect that a traditional treatment of your "slowness" would make it some kind of gradient vector. Commented Aug 5, 2016 at 19:23
• Use ">" (without an dollar signs) followed by an important paragraph to help highlight important parts or lines that users should focus on so that they may better understand what is being asked. Commented Aug 5, 2016 at 19:44
• I'm going to write an "answer" to this describing dual vector spaces. I haven't read through the entire question, so my answer is going to be incomplete. Commented Aug 5, 2016 at 20:15
• Dividing by the square of the speed satisfies all three conditions. The magnitude is inverted, the direction is the same, and coordinate independence. Actually this is the only definition that even meets conditions 1 and 2. Commented Aug 5, 2016 at 21:23
• @MattSamuel Unfortunately, it's actually not coordinate-independent; at least not physically as it relates to the velocity. I show above that if you change the y-velocity then it has an effect on the x-slowness if you define it that way Commented Aug 5, 2016 at 21:26

First off: yes, everything you've said makes sense, and this is a very interesting question.

Let $Vel$ be the vector space of velocities. Examples of vectors in $Vel$ are "3 miles north per hour" and "4 miles west per hour".

What is the reciprocal of a velocity? It's easy to write such things down: "1/3 hours per mile north", "1/4 hours per mile west", and so forth. But what are they?

The mathematical concept of a dual vector space doesn't exactly answer your question, I think, but I feel like it comes close. Let's talk about what a dual vector space is and how it applies here.

If $V$ a vector space (any vector space), then the "dual" of $V$, denoted $V^*$, consists of all linear functions $V \to \mathbb{R}$. (A function $f : V \to \mathbb{R}$ is linear if for all vectors $x$ and $y$ and scalars $s$, $f(x + y) = f(x) + f(y)$ and $f(sx) = s\ f(x)$.)

(Of course, if the scalar field of $V$ is some other field $F$ instead of $\mathbb{R}$, then replace $\mathbb{R}$ with $F$ in the above definition.)

The definition of addition and multiplication for $V^*$ is probably what you would expect:

• Given elements $f$ and $g$ of $V^*$, their sum $f + g$ is defined as that function $h$ such that for all vectors $x$ in $V$, $h(x) = f(x) + g(x)$.
• Given an element $f$ of $V^*$, and a scalar $a$ in $\mathbb{R}$, the product $af$ is defined as that function $g$ such that for all vectors $x$ in $V$, $g(x) = a\ f(x)$.

Notice that applying a function in $V^*$ to an argument in $V$ behaves a lot like multiplication. The definition of addition in $V^*$ looks a lot like the distributive property, and the definition of scalar multiplication looks a lot like commutativity of multiplication.

Now, on to our example. What does the dual space $Vel^*$ look like? We could call its elements "dual-velocities", but what is a "dual-velocity"?

The boring answer is that a "dual-velocity" is a linear function $Vel \to \mathbb{R}$. But that's not very informative. So let me show you can example of a "dual-velocity".

I'm defining the "dual-velocity" $f$ as that function such that

$$f(\text{1 mile north per hour}) = 2,$$ $$f(\text{3 miles north per hour}) = 6,$$ $$f(\text{1 mile east per hour}) = 1,$$ $$f(\text{4 miles west per hour}) = -4,$$ $$f(\text{1 mile north and 2 miles west per hour}) = 0,$$

and so forth.

There's no single obvious way of interpreting this dual-velocity value, but we can certainly come up with some interpretations. It could represent the effect that wind has on a wind turbine: a faster wind in a given direction makes the turbine spin faster, but depending on the direction of the wind, the turbine might spin faster or slower, or not at all, or in the wrong direction. Or it could represent the amount that a wind helps or hinders an airplane flying in a particular direction.

### Plotting dual-velocities

How can you plot a dual-velocity?

You can plot a dual-velocity on the same coordinate plane that you plot velocities on. But don't plot them the same way! Usually, you would plot a velocity as an arrow starting at the origin and ending at a point. But it doesn't make much sense to plot a dual-velocity the same way on the same coordinate plane.

A velocity is represented by a point on our coordinate plane. There's a set of velocities which our dual-velocity maps to 0. It happens that if you look at all these velocities on the coordinate plane, they form a straight line passing through the origin. (Unless the dual-velocity is the zero vector—see below.) So draw that line and write a 0 next to it. Likewise, there's another set of velocities which are mapped to 1, and this set is also a straight line. Draw that line and write a 1 next to it. Do the same for all of the integers. You'll end up with a bunch of equally spaced parallel lines, labeled with the integers.

The zero dual-velocity is the exception to the above. The zero dual-velocity maps all velocities to the number 0, so you'll have to think of another way to plot it.

### A basis for $Vel^*$

Like all finite-dimensional vector spaces, the vector space $Vel$ (just velocities, not dual-velocities) has a finite basis. For every velocity $x$, there are scalars $a$ and $b$ such that

$$x = a (\text{1 mile north per hour}) + b (\text{1 mile east per hour}).$$

Of course, you can write this more concisely as $\langle a, b \rangle$.

The vector space $Vel^*$ has a finite basis, too. Let $N$ be the dual-velocity $N(\langle a, b \rangle) = a$, and let $E$ be the dual-velocity $E(\langle a, b \rangle) = b$. Then any dual-velocity can be written in the form $cN + dE$, or, alternatively, as $[c, d]$.

### Out of steam

This answer is going on a lot longer than I wanted it to, so I'm going to leave off with a couple of things to think about.

• What units can dual-velocities be measured in? As I mentioned above, applying a dual vector to a vector from the original space acts a lot like multiplication. This means that you can "multiply" a dual-velocity with a velocity to end up with a scalar. Since velocity has units of miles per hour, it probably makes sense to imagine dual-velocities as having units of hours per mile.
• Given a nonzero velocity, there are lots of different dual-velocities that you can "multiply" with that velocity to get 1. However, out of all of these dual-velocities, one of them will have the smallest magnitude (and that magnitude will be the reciprocal of the magnitude of the velocity). You could define the "reciprocal" of a velocity as referring to this dual-velocity. Find a formula for this "reciprocal" and see if it matches what you've come up with already.
• What a beautiful answer. I wish textbooks would give such deep insights, as opposed to mere computational rote. Commented Jul 25, 2017 at 21:02