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Matrix multiplication.

Anonymous Sun 07 Mar 19:12:54 2021 No.12802999 View Reply Original Report

Quoted By: >>12803016 >>12803163 >>12803168 >>12803211 >>12803288 >>12803786 >>12803815 >>12803845 >>12806032 >>12806061 >>12806120 >>12807712 >>12808010

Why do we multiply matrices this way? Why not:

a1*b1 a2*b1
a1*b2 a2*b2

Is there a rigorous mathematical reason it's not done this way? Or is it just lame semantics/tradition?

Anonymous

Anonymous Sun 07 Mar 2021 19:14:31 No.12803003 Report

Quoted By: >>12803154

take one linear system and plug it into another linear system. the pattern emerges

Anonymous

Anonymous Sun 07 Mar 2021 19:17:50 No.12803016 Report

Quoted By: >>12803154

>>12802999
it's because of linear mappings. if you don't know what that is, think of rotation, scaling, shearing, reflection through an axis, orthogonal projection. linear maps can be composed: you reflect through an axis and then rotate by an angle etc. matrices are a way of encoding linear maps. multiplying a column vector by matrix corresponds to applying the corresponding transformation. multiplying two matrices together corresponds to composition of the corresponding transformations. that's why.

Anonymous

Anonymous Sun 07 Mar 2021 19:51:40 No.12803154 Report

Quoted By: >>12803158

>>12803003
>>12803016
Still doesn't explain why

a1*b1 a1*b2
a2*b1 a2*b2

is correct while

a1*b1 a2*b1
a1*b2 a2*b2

is not. I could cook up any space or system where we multiply column-wise first.

Anonymous

Anonymous Sun 07 Mar 2021 19:53:39 No.12803158 Report

Quoted By:

>>12803154
convention
You could just as well define matrix multiplication in a different way

Anonymous

Anonymous Sun 07 Mar 2021 19:55:15 No.12803163 Report

Quoted By:

>>12802999
Take a linear algebra class.

Anonymous

Anonymous Sun 07 Mar 2021 19:56:00 No.12803168 Report

Quoted By: >>12806051 >>12806053

>>12802999
This thread is a good example of how mathfags don't know how to communicate. If you take an nx1 vector and multiply it to a 1xn vector, you get a unique operation that returns a scalar. This operation must remain the same when generalizing to nxm times an mxn matrix.

Anonymous

Anonymous Sun 07 Mar 2021 19:56:42 No.12803171 Report

Quoted By:

Let AB=C
and Bx=y

now C*x must behave like A*y

Anonymous

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Anonymous Sun 07 Mar 2021 20:07:41 No.12803211 Report

Quoted By: >>12807577

>>12802999

Anonymous

Anonymous Sun 07 Mar 2021 20:24:11 No.12803288 Report

Quoted By:

>>12802999
the Jew fears the straightforward a1*b1 a2*b1 [...] multiplication

Anonymous

Anonymous Sun 07 Mar 2021 22:15:51 No.12803786 Report

Quoted By:

>>12802999
Because cool properties emerge from that definition and from other definitions it doesnt. Every linear aplication corresponds in a biyective way with a matrix and thats cool

Anonymous

Anonymous Sun 07 Mar 2021 22:21:44 No.12803815 Report

Quoted By:

>>12802999
because doing it that way would be fucking boring

Anonymous

Anonymous Sun 07 Mar 2021 22:28:14 No.12803845 Report

Quoted By:

>>12802999
>Is there a rigorous mathematical reason it's not done this way
Yes, it's obviously if you look at it as linear transformations

Anonymous

Anonymous Mon 08 Mar 2021 11:02:54 No.12806026 Report

Quoted By:

Bump

Anonymous

Anonymous Mon 08 Mar 2021 11:04:42 No.12806032 Report

Quoted By:

>>12802999
It's Imposed to be this way so that it is consistent with linear map composition.

Watch this video for intuition
https://youtu.be/XkY2DOUCWMU

Anonymous

Anonymous Mon 08 Mar 2021 11:08:55 No.12806051 Report

Quoted By:

>>12803168
dumb engineering monke

Anonymous

Anonymous Mon 08 Mar 2021 11:10:04 No.12806053 Report

Quoted By:

>>12803168
>If you take an nx1 vector and multiply it to a 1xn vector, you get a unique operation that returns a scalar. This operation must remain the same when generalizing to nxm times an mxn matrix.
ahahahha. yeah bro, this is the reason why matrix multiplication is defined this way.

Anonymous

Anonymous Mon 08 Mar 2021 11:12:16 No.12806061 Report

Quoted By:

>>12802999
Because the matrix represents a linear map. Do a mapping and then another mapping, that's a third equivalent mapping and it matches matrix multiplication

Anonymous

Anonymous Mon 08 Mar 2021 11:35:43 No.12806120 Report

Quoted By: >>12806129

>>12802999
Because technically, row vectors are linear functionals (dual vectors) that take column vectors to an element in R.

Anonymous

Anonymous Mon 08 Mar 2021 11:40:19 No.12806129 Report

Quoted By: >>12807672

>>12806120
>Because technically, row vectors are linear functionals (dual vectors) that take column vectors to an element in R.
You could define it the other way around though.

Anonymous

Anonymous Mon 08 Mar 2021 18:52:36 No.12807577 Report

Quoted By: >>12807629

>>12803211
Which book is this?

Anonymous

Anonymous Mon 08 Mar 2021 19:03:03 No.12807629 Report

Quoted By: >>12807667

>>12807577
Not the OP, but that's Linear Algebra Done Right by Axler, page 74.
Best math textbook I've ever read.

Anonymous

Anonymous Mon 08 Mar 2021 19:10:34 No.12807667 Report

Quoted By:

>>12807629
Thx anon!!

Anonymous

Anonymous Mon 08 Mar 2021 19:11:44 No.12807672 Report

Quoted By: >>12807761 >>12807804

>>12806129
Yes, but there are other more general cases when a space doesn't have a natural isomorphism to its dual space.

Anonymous

Anonymous Mon 08 Mar 2021 19:22:12 No.12807712 Report

Quoted By: >>12808047

>>12802999
you want the ring of matrices nxm to be isomorphic to the ring of linear maps from F^n to F^m

Anonymous

Anonymous Mon 08 Mar 2021 19:34:27 No.12807761 Report

Quoted By: >>12807804

>>12807672
How do those cases relate to row and column vectors then?

Anonymous

Anonymous Mon 08 Mar 2021 19:42:58 No.12807788 Report

Quoted By: >>12807812

In a more physical intuition, it's bc rotational invariance. I do an experiment on this desk. If I rotate my desk, I better get the same results. Any operation in physics better obey this property. Essentially, that is what a vector means. A vector is anything that changes if you rotate it. That's why we vector addition works in the way it does. Matrix multiplication is just the dot product. It is bc rotational invariance that we multiply it that way, at least in physics.

Anonymous

Anonymous Mon 08 Mar 2021 19:48:13 No.12807804 Report

Quoted By: >>12807826

>>12807761
>>12807672
no isomorphism between V and V* is needed to express the idea. the thing is that if you choose a basis for V, then in V* you need to work in the corresponding dual basis. evaluation of a form on a vector becomes matrix multiplication of a row of coordinates with a column of coordinates.

Anonymous

Anonymous Mon 08 Mar 2021 19:49:58 No.12807812 Report

Quoted By:

>>12807788
>A vector is anything that changes if you rotate it.
>Matrix multiplication is just the dot product.
shhhhh, some mathematician might hear you

Anonymous

Anonymous Mon 08 Mar 2021 19:55:33 No.12807826 Report

Quoted By:

>>12807804
>the thing is that if you choose a basis for V, then in V* you need to work in the corresponding dual basis.
Yes, but what stops me from using row vectors as basis and adapting the multiplication rule accordingly?

Anonymous

Anonymous Mon 08 Mar 2021 20:40:14 No.12808010 Report

Quoted By: >>12808038

>>12802999
Go ahead and try it, you just get a different algebra which satisfies none of the theorems we would like.

Anonymous

Anonymous Mon 08 Mar 2021 20:47:52 No.12808038 Report

Quoted By: >>12808047

>>12808010
But only because they're build using this convention.
Or do you have a counterexample?

Anonymous

Anonymous Mon 08 Mar 2021 20:49:19 No.12808047 Report

Quoted By:

>>12808038
the "theorems we would like" is exactly this one: >>12807712

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