# B Is 1/vector also a vector?

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1. Nov 23, 2018

### Hawkingo

Let $\vec { A }$ = $a \dot { i } + b \hat { j } + c \hat { k }$
My question is "is $\frac { 1 } { \vec { A } }$ is a vector or not and if yes then what is it's components?"

2. Nov 23, 2018

### A.T.

Isn't this a math question, rather than physics? You can define the operator to mean whatever you want.

3. Nov 23, 2018

### hilbert2

The $\frac{1}{\hat{A}}$ doesn't mean anything if only addition and scalar multiplication of vectors is defined. You also need a multiplication between vectors, making it an algebra instead of a simple vector space. For instance, real numbers are 1-dimensional vectors, but there is also a multiplication operation which makes it possible to find an inverse $\frac{1}{a}$ for any nonzero real number $a$.

For $N\times N$ matrices there is both an addition and multiplication operation, but there exist more than one matrix that is not invertible, making the situation more difficult than with real numbers where only $0$ was uninvertible.

4. Nov 23, 2018

### Delta2

I think one reason that mathematicians haven't yet defined the inverse of a vector, say, $\vec{v}\in \mathbb{R}^3$ , is that it isn't of any use in physics or engineering, or other applied sciences. If you are faced with a problem in physics or engineering and you thinking about taking the inverse of a vector then this usually implies that you are thinking in a wrong way about problem, and you need to change your way of thinking in order to arrive at a successful solution.

We could view the cross product as a multiplication between vectors, however it needs to be modified in order to be able to find a unique identity element $\vec{e}$, and with it to define the inverse of a vector such that $\vec{v} \times \vec{v^{-1}}=\vec{v^{-1}}\times \vec{v}=\vec{e}$. But cross product is fine and has wide applications the way it is, I doubt that its modification would be about as wide and useful.

5. Nov 23, 2018

### Staff: Mentor

To define a mutliplicative inverse, you need to investigate the multiplicative structure. Objects which have vectors as elements are:
• vector space: no multiplication at all
• ring: in general no neutral element $1$ and if so, only a few units
• algebra: same as ring, often not even commutative, e.g. Lie algebras
• division ring: e.g. the quaternions, where we have two inverses - left and right, so the notation $1/\vec{A}$ is still undefined
• field: e.g. $\mathbb{C}$ as $\mathbb{R}-$algebra, where we have $1/\vec{A} = (a - b \cdot i)/(a^2+b^2)$
That means, that up to very few exceptions for which we do have the concept of dividing vectors, the question does not occur in this generality.

6. Nov 23, 2018

### Hawkingo

Can you explain this in a simpler way because I have no idea about the ring or quaternions.

7. Nov 23, 2018

### PeroK

Suppose you have $\frac { 1 } { \vec { A } }$, what do you expect $\frac { \vec{A} } { \vec { A } }$ to equal?

8. Nov 23, 2018

### Hawkingo

I tried to get the answer by mathpix app(powered by wolfarmalpha engine) and got this answer.

9. Nov 23, 2018

### Staff: Mentor

I thought the quaternions were the example you have chosen:
https://en.wikipedia.org/wiki/Quaternion
They are a non commutative extension of complex numbers.

In case you were only referring to a three dimensional, real vector space, then the answer is: either there is no multiplication at all, and ergo no division, or if we choose the cross product as multiplication, we will lose associativity, have no neutral element $1$ and are not commutative.

10. Nov 23, 2018

### Staff: Mentor

But what does $1$ here mean? The real number $1$ or a vector $\vec{1}$? As a real number, this wouldn't be a division, since we left the vector space, and as a vector, we will have trouble to define it.

11. Nov 23, 2018

### PeroK

So, what about $\frac{1}{\vec{A}} = \frac{\vec{A}}{|A|^2}$? And use the scalar product for your multiplication?

12. Nov 23, 2018

### Hawkingo

I understood that we can't regard cross product as multiplication but I think we can operate a scalar(here 1 in numerator) with a vector (in denominator) algebrically, just like $5*\vec { A } =5\vec { A}$

13. Nov 23, 2018

### Staff: Mentor

You can define $\dfrac{\vec{A}}{\vec{A}}=1\in \mathbb{R}$ for $\vec{A}\in V-\{\,\vec{0}\,\}$ but this isn't a division, because it leads outside the set of vectors. In addition we consequently get $\vec{A}\cdot \vec{B} \in \mathbb{R}$ which will lead to the concept of inner products, but is still no division.

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