Linearity and non-linearity in addition and multiplication

[Thytanium]

Homework Statement

I don't know how to prove the non-linearity of multiplication.

Relevant Equations

No relevant equations.

Hello friends. Excuse my ignorance. Why is addition linear and not multiplication?

 

[berkeman]

This Wikipedia article may help:

https://en.wikipedia.org/wiki/Linearity

Please have a read through that article, and let us know if you still have questions about it.

 

[fresh_42]

Homework Statement:: I don't know how to prove the non-linearity of multiplication.
Relevant Equations:: No relevant equations.

Hello friends. Excuse my ignorance. Why is addition linear and not multiplication?

Multiplication $R\times R\longrightarrow R$ is bilinear:
$$
(\alpha\cdot r+\beta\cdot s,t) \longmapsto \alpha \cdot r\cdot t+\beta \cdot s \cdot t\; , \;
(r,\alpha\cdot s+\beta\cdot t) \longmapsto r\cdot\alpha \cdot s+r\cdot\beta \cdot t
$$
So, what do you mean?

 

[Thytanium]

Multiplication $R\times R\longrightarrow R$ is bilinear:
$$
(\alpha\cdot r+\beta\cdot s,t) \longmapsto \alpha \cdot r\cdot t+\beta \cdot s \cdot t\; , \;
(r,\alpha\cdot s+\beta\cdot t) \longmapsto r\cdot\alpha \cdot s+r\cdot\beta \cdot t
$$
So, what do you mean?

Hello friend fresh_42. Thanks for answering me. I don't understand the result of this operation: $(\alpha\cdot r+\beta\cdot s,t)$ Please give me a link where I can study this.

 

[fresh_42]

Hello friend fresh_42. Thanks for answering me. I don't understand the result of this operation: $(\alpha\cdot r+\beta\cdot s,t)$ Please give me a link where I can study this.

Multiplication is a binary operation. It takes two inputs and generates one output: $m\, : \,(p,q) \longmapsto p\cdot q.$ A function $f$ is linear if $f(\alpha x+\beta y)=\alpha f(x) +\beta f(y).$ Now, let's look at the first input variable of multiplication: $m(\alpha x+\beta y, q)=(\alpha x+\beta y)\cdot q = \alpha x\cdot q +\beta y \cdot q =\alpha m(x,q)+\beta m(y,q).$ This means $m(\, . \,,q)$ is linear in the first argument. The same is true for the second argument, so $m(p,\, . \,)$ is linear, too. This means that $m(\, . \,,\, . \,)$ is linear in both arguments, i.e. it is bilinear.

 

[Thytanium]

Multiplication is a binary operation. It takes two inputs and generates one output: $m\, : \,(p,q) \longmapsto p\cdot q.$ A function $f$ is linear if $f(\alpha x+\beta y)=\alpha f(x) +\beta f(y).$ Now, let's look at the first input variable of multiplication: $m(\alpha x+\beta y, q)=(\alpha x+\beta y)\cdot q = \alpha x\cdot q +\beta y \cdot q =\alpha m(x,q)+\beta m(y,q).$ This means $m(\, . \,,q)$ is linear in the first argument. The same is true for the second argument, so $m(p,\, . \,)$ is linear, too. This means that $m(\, . \,,\, . \,)$ is linear in both arguments, i.e. it is bilinear.

Wonderful explanation fresh_42. Grateful to you friend. Good day.

 

[Thytanium]

 

[Orodruin]

Addition, on the other hand, is not linear in either argument because with $(a,b) \mapsto a+b$ then
$$
(a_1+a_2,b) \mapsto a_1+a_2 +b
$$
whereas it should map to $a_1+a_2+2b$ if addition was linear in the first argument.

(Addition of zero is linear in the other argument, but that’s just the identity map…)

 

[Thytanium]

Addition, on the other hand, is not linear in either argument because with $(a,b) \mapsto a+b$ then
$$
(a_1+a_2,b) \mapsto a_1+a_2 +b
$$
whereas it should map to $a_1+a_2+2b$ if addition was linear in the first argument.

(Addition of zero is linear in the other argument, but that’s just the identity map…)

Thank you Orodruin for this interesting information. I did not know that the sum was not linear in his arguments. That is very interesting. But I don't understand why if the sum were linear in the first argument it would produce $a_1 + a_2 + 2b$ and excuse my ignorance. Thanks again my friend. I will study more this information.

 

[Orodruin]

Thank you Orodruin for this interesting information. I did not know that the sum was not linear in his arguments. That is very interesting. But I don't understand why if the sum were linear in the first argument it would produce $a_1 + a_2 + 2b$ and excuse my ignorance. Thanks again my friend. I will study more this information.

If $f(a,b) = a+b$ would be linear in the first argument, then
$$
(a_1 + a_2) + b = f(a_1+a_2,b) = f(a_1,b)+f(a_2,b)
= (a_1+b) + (a_2+b) = a_1+a_2 + 2b
$$
This is only true if $b=0$ and not generally true for any $b$.

 

[fresh_42]

The concept length is linear:
The length of two trees is the sum of the lengths of each of them.
The length of half a tree is the half length of the original tree.

The concept of area is quadratic and non-linear.

The concept of volume is cubic and non-quadratic and non-linear.

However, length is not addition, area not multiplication, and volume not exponentiation. We have to distinguish between the operations and the interpretation of their results.

 

[Thytanium]

Addition, on the other hand, is not linear in either argument because with $(a,b) \mapsto a+b$ then
$$
(a_1+a_2,b) \mapsto a_1+a_2 +b
$$
whereas it should map to $a_1+a_2+2b$ if addition was linear in the first argument.

(Addition of zero is linear in the other argument, but that’s just the identity map…)

Thank you Orodruin for your PF Insights and especially for The Birth of a Textbook. Little by little I will read them. They are very interesting. Thank you friend.

 

[Thytanium]

If $f(a,b) = a+b$ would be linear in the first argument, then
$$
(a_1 + a_2) + b = f(a_1+a_2,b) = f(a_1,b)+f(a_2,b)
= (a_1+b) + (a_2+b) = a_1+a_2 + 2b
$$
This is only true if $b=0$ and not generally true for any $b$.

Thanks you my friend Orodruin. Happy Day.

 

[Thytanium]

The concept length is linear:
The length of two trees is the sum of the lengths of each of them.
The length of half a tree is the half length of the original tree.

The concept of area is quadratic and non-linear.

The concept of volume is cubic and non-quadratic and non-linear.

However, length is not addition, area not multiplication, and volume not exponentiation. We have to distinguish between the operations and the interpretation of their results.

Wonderful clarification fresh_42. Very grateful to you friend.

 

[WWGD]

@Thytanium : Can you test for the bilinearity of Inner-product, Cross -Product?

 

[Thytanium]

@Thytanium : Can you test for the bilinearity of Inner-product, Cross -Product?

Yes I can. It's simple. If I have two vectors A and B and do the inner product A.B, to prove bilinearity first with vector A, I do (kA).B where k is a scalar. So that's equal to k(A.B) = kA.B and then with vector B it's the same procedure so the inner product of two vectors is doubly linear. And for the cross-product AxB, where "x" is the cross-product symbol, I do the same.