Linearity and non-linearity in addition and multiplication

AI Thread Summary
Addition is considered linear because it satisfies the linearity condition in its arguments, while multiplication is classified as bilinear, meaning it is linear in each of its two arguments separately. The discussion clarifies that if addition were linear in the first argument, it would yield incorrect results, such as producing an extra term when summing two inputs. Multiplication, on the other hand, maintains linearity in both arguments, allowing it to be expressed as a bilinear function. The participants also explore related concepts like the linearity of length, area, and volume, emphasizing the need to distinguish between operations and their interpretations. Overall, the conversation deepens the understanding of linearity in mathematical operations.
Thytanium
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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?
 
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Thytanium said:
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?
 
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fresh_42 said:
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.
 
Thytanium said:
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.
 
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fresh_42 said:
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.
 
:biggrin::biggrin:
 
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…)
 
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Orodruin said:
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.
 
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Thytanium said:
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##.
 
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  • #11
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.
 
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  • #12
Orodruin said:
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.
 
  • #13
Orodruin said:
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.
 
  • #14
fresh_42 said:
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.
 
  • #15
@Thytanium : Can you test for the bilinearity of Inner-product, Cross -Product?
 
  • #16
WWGD said:
@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.
 
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