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My question is "is ##\frac { 1 } { \vec { A } }## is a vector or not and if yes then what is it's components?"

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- Thread starter Hawkingo
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In summary, the conversation discusses the concept of taking the inverse of a vector and whether it is possible or not. The idea of finding an inverse for a vector is not commonly used in physics or engineering, and in order to define a multiplicative inverse, the multiplicative structure of the vector needs to be examined. This includes considering objects such as vector spaces, rings, algebras, division rings, and fields. However, for most of these structures, the concept of dividing vectors does not exist. f

- #1

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My question is "is ##\frac { 1 } { \vec { A } }## is a vector or not and if yes then what is it's components?"

- #2

Science Advisor

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

My question is "is ##\frac { 1 } { \vec { A } }## is a vector or not and if yes then what is it's components?"

- #3

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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

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We could view the cross product as a multiplication between vectors, however

- #5

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To define a mutliplicative inverse, you need to investigate the multiplicative structure. Objects which have vectors as elements are:

My question is "is ##\frac { 1 } { \vec { A } }## is a vector or not and if yes then what is it's components?"

- 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)##

- #6

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Can you explain this in a simpler way because I have no idea about the ring or quaternions.To define a mutliplicative inverse, you need to investigate the multiplicative structure. Objects which have vectors as elements are:

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.

- 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)##

- #7

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My question is "is ##\frac { 1 } { \vec { A } }## is a vector or not and if yes then what is it's components?"

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

- #8

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I tried to get the answer by mathpix app(powered by wolfarmalpha engine) and got this answer.Suppose you have ##\frac { 1 } { \vec { A } }##, what do you expect ##\frac { \vec{A} } { \vec { A } }## to equal?

- #9

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Can you explain this in a simpler way ...

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

I thought the quaternions were the example you have chosen:... because I have no idea about the ring or quaternions.

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

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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.I tried to get the answer by mathpix app(powered by wolfarmalpha engine) and got this answer. View attachment 234497

- #11

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I tried to get the answer by mathpix app(powered by wolfarmalpha engine) and got this answer. View attachment 234497

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

- #12

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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} ##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.

- #13

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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.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} ##

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