/sci/ - Science & Math » Thread #14169670

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Anonymous Mon 31 Jan 07:49:59 2022 No.14169670 View Reply Original Report

Quoted By: >>14169692 >>14169694 >>14169714 >>14169725 >>14169729 >>14170786 >>14170796 >>14170797 >>14171234 >>14171631 >>14171724

is a vector a complex number?

Anonymous

Anonymous Mon 31 Jan 2022 08:04:11 No.14169692 Report

Quoted By: >>14169741

>>14169670
complex numbers are syntactic sugar for vectors

Anonymous

Anonymous Mon 31 Jan 2022 08:06:38 No.14169694 Report

Quoted By:

>>14169670
If the question you're asking is if the complex numbers form a vector space over the complex numbers, then yes. But by the same token real numbers are also vectors, so are you asking if there is a bijection between the vector space formed over complex numbers and the vector space formed over the cartesian product of the real numbers with themselves, then also yes.

Anonymous

Anonymous Mon 31 Jan 2022 08:21:09 No.14169713 Report

Quoted By:

I absolutely cringe at you physicists' notion of a vector.

Anonymous

Anonymous Mon 31 Jan 2022 08:21:30 No.14169714 Report

Quoted By:

>>14169670
complex numbers are scalars.
complex numbers can be represented as a 2D real vectorspace.

they are both scalars and vectors, depending on how you think about them

Anonymous

Anonymous Mon 31 Jan 2022 08:30:10 No.14169725 Report

Quoted By:

>>14169670
Yes.

Anonymous

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Anonymous Mon 31 Jan 2022 08:34:11 No.14169729 Report

Quoted By:

>>14169670
no, the second value squared will not move it to the first position. There is no real equivalent. The Cauchy-reimann equations don't need to hold for differentiation.

Anonymous

Anonymous Mon 31 Jan 2022 08:44:17 No.14169738 Report

Quoted By: >>14169742 >>14170954

bro that's just a unit vector; the i hat, j hat, k hats and all the basis vectors. every unit vector can be expressed as a linear combination of these in that they are all linearly independent (i.e. each vector cannot be expressed as the linear combination of the other vectors)

Anonymous

Anonymous Mon 31 Jan 2022 08:48:34 No.14169741 Report

Quoted By: >>14170741

>>14169692
that's like saying polynomials are syntactic sugar for vectors

Anonymous

Anonymous Mon 31 Jan 2022 08:49:10 No.14169742 Report

Quoted By:

>>14169738
*are basis vectors

Anonymous

Anonymous Mon 31 Jan 2022 16:23:31 No.14170741 Report

Quoted By: >>14170746

>>14169741
And it would be true in both cases

Anonymous

Anonymous Mon 31 Jan 2022 16:25:39 No.14170746 Report

Quoted By: >>14170769 >>14170772 >>14170781 >>14170848

>>14170741
so it's okay if I start using "vectors", "complex numbers" and "polynomials" interchangeably, right?

Anonymous

Anonymous Mon 31 Jan 2022 16:30:53 No.14170769 Report

Quoted By:

>>14170746
yeah
retard

Anonymous

Anonymous Mon 31 Jan 2022 16:31:21 No.14170772 Report

Quoted By: >>14170897

>>14170746
Exactly. Just like both bicycles and trucks are vehicles, you can use those two words interchangeably.

Anonymous

Anonymous Mon 31 Jan 2022 16:33:58 No.14170781 Report

Quoted By:

>>14170746
They are data structures

Anonymous

Anonymous Mon 31 Jan 2022 16:34:42 No.14170786 Report

Quoted By:

>>14169670
Only a vector in the complex plane

Anonymous

Anonymous Mon 31 Jan 2022 16:37:21 No.14170796 Report

Quoted By:

>>14169670
Vector 1,i has magnitude 0. vector 1,1 has magnitude 1.

Anonymous

Anonymous Mon 31 Jan 2022 16:37:23 No.14170797 Report

Quoted By:

>>14169670
Vectors are vectors. Every field is a vector space over any of its subfields. In particular, C is a vector space over R. It's also a vector space over Q, but not a 2-dimensional one.

Anonymous

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Anonymous Mon 31 Jan 2022 16:49:52 No.14170848 Report

Quoted By: >>14170897

>>14170746
>so you're telling me that if..
>elephants are animals
>and snakes are animals
>THEN I can use "elephants", "snakes" and "animals" interchangeably!?
You can't make it up

Anonymous

Anonymous Mon 31 Jan 2022 17:03:00 No.14170897 Report

Quoted By: >>14170917

>>14170772
are bicykles and trucks syntactic sugar for vehicles?
>>14170848
are elephants or snakes syntactic sugar for animals?

Anonymous

Anonymous Mon 31 Jan 2022 17:05:54 No.14170917 Report

Quoted By: >>14170924

>>14170897
Whenever you say bicycles, you also imply vehicle. Whenever you say elephant, you also imply animal. Whenever you say polynomial, you also imply vector.

Anonymous

Anonymous Mon 31 Jan 2022 17:07:32 No.14170924 Report

Quoted By: >>14170936

>>14170917
are bicykles and trucks syntactic sugar for vehicles?
are elephants or snakes syntactic sugar for animals?
are polynomials syntactic sugar for vectors?
do you know what "syntactic sugar" means?

Anonymous

Anonymous Mon 31 Jan 2022 17:10:18 No.14170936 Report

Quoted By: >>14170949

>>14170924
You're simply being autistic. Everyone in this thread understood what anon was saying. He's saying that complex numbers are vectors and that that's not understood by loads of people who somehow believe that these are all completely distinct ideas.
It's not my problem that you're getting stuck up on "syntactic sugar" as if someone had raped your mother in front of you.

Anonymous

Anonymous Mon 31 Jan 2022 17:13:16 No.14170949 Report

Quoted By: >>14170951

>>14170936
>He's saying that...
He's saying that complex numbers are syntactic sugar for vectors. That's a complete nonsense. Seethe, cope, dilate.

Anonymous

Anonymous Mon 31 Jan 2022 17:13:43 No.14170951 Report

Quoted By: >>14170958

>>14170949
You're autistic. Answer: Are complex numbers vectors?

Anonymous

Anonymous Mon 31 Jan 2022 17:14:08 No.14170954 Report

Quoted By: >>14171241

>>14169738
>numbers can be expressed as other numbers
>so deep
This is your brain on linear algebra.

Anonymous

Anonymous Mon 31 Jan 2022 17:15:23 No.14170958 Report

Quoted By: >>14170965

>>14170951
Sure.

Anonymous

Anonymous Mon 31 Jan 2022 17:17:08 No.14170965 Report

Quoted By: >>14170978 >>14170980

>>14170958
Exactly. Complex numbers, which are vectors, can be represented in the form "a+bi". Syntactic sugar distracting from the nature of complex numbers.

Anonymous

Anonymous Mon 31 Jan 2022 17:22:07 No.14170978 Report

Quoted By: >>14170985

>>14170965
And yet it's a complete nonsense to say that complex numbers are syntactic sugar for vectors. Weird, isn't it?

Anonymous

Anonymous Mon 31 Jan 2022 17:22:49 No.14170980 Report

Quoted By: >>14170985

>>14170965
>the nature of complex numbers
Like what

Anonymous

Anonymous Mon 31 Jan 2022 17:25:05 No.14170985 Report

Quoted By: >>14170994

>>14170978
It's only nonsense if you take it literally, like an autist.
>>14170980
Linear transformations.

Anonymous

Anonymous Mon 31 Jan 2022 17:27:03 No.14170994 Report

Quoted By: >>14170999 >>14171316

>>14170985
>Linear transformations.
Those are just an array of numbers

Anonymous

Anonymous Mon 31 Jan 2022 17:27:49 No.14170999 Report

Quoted By: >>14171174

>>14170994
They are an array of numbers, yes, but not just an array of numbers.

Anonymous

Anonymous Mon 31 Jan 2022 18:21:02 No.14171174 Report

Quoted By: >>14171256

>>14170999
what even is an array?

Anonymous

Anonymous Mon 31 Jan 2022 18:37:31 No.14171234 Report

Quoted By:

>>14169670
A vector is a point with additional information attached, which forms a dynamic tensor.

This can be represented as a complex number, but also doesn't have to be.

Were it broken down into it's most fundamental pieces, yes, it is, but in the same way that your mom's vagina can be broken down into hydrogen.

Chad brain answer.

Anonymous

Anonymous Mon 31 Jan 2022 18:38:44 No.14171241 Report

Quoted By:

>>14170954
That's all of set theory you dumb fuck

Anonymous

Anonymous Mon 31 Jan 2022 18:42:02 No.14171256 Report

Quoted By: >>14171282

>>14171174
An ordered list

Anonymous

Anonymous Mon 31 Jan 2022 18:49:42 No.14171282 Report

Quoted By: >>14171567

>>14171256
no

Anonymous

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Anonymous Mon 31 Jan 2022 18:57:43 No.14171316 Report

Quoted By: >>14171326 >>14171338 >>14172403

>>14170994
no they aren't.

Let $L : U \rightarrow V$ be a linear transformation that obeys the linearity property $L(\vec{u} a + \vec{u}' a') = L(\vec{u}) a + L(\vec{u}') a'$

consider
$\vec{v} = L(\vec{u})$
given bases $\vec{u}_i \in B_u \subset U$ and $\vec{v}_j \in B_v \subset V$ expand the vectors in the equation
$\sum_j \vec{v}_j v^j = L\left(\sum_i \vec{u}_i u^i\right)$
apply linearity to the right hand side and basis expand the result in $B_v$
$L\left(\sum_i \vec{u}_i u^i\right) = \sum_i L(\vec{u}_i) u^i = \sum_i \left[\sum_j \vec{v}_j L^j_i \right] u^i = \sum_j \vec{v}_j \left[\sum_i L^j_i u^i\right]$
comparing the previous equations and assering uniqueness of basis expansion, you get
$v^j = \sum_i L^j_i u^i$

so they can be represented as arrays of numbers $w.r.t\; a\; choice\; of\; bases$

Anonymous

Anonymous Mon 31 Jan 2022 18:59:52 No.14171326 Report

Quoted By: >>14171344 >>14171581

>>14171316
>so they can be represented as arrays of numbers
That's what I said, they are just an array of numbers

Anonymous

Anonymous Mon 31 Jan 2022 19:03:03 No.14171338 Report

Quoted By: >>14171345

>>14171316
Who the hell gives components and bases the same name

Anonymous

Anonymous Mon 31 Jan 2022 19:05:15 No.14171344 Report

Quoted By:

>>14171326
that glosses over the important of bases when representing things in terms of arrays of scalars and doesn't tell you how to apply the array or how to transform it under a change of basis.

all that is clear by saying a linear transformation obeys linearity, and then deriving the multisimensional array representation directly. this approach works with tensors too (which are just multilinear maps).

Anonymous

Anonymous Mon 31 Jan 2022 19:06:16 No.14171345 Report

Quoted By:

>>14171338
it helps keep subscripts and variable names from getting out of hand when you have to expand several vectors in different bases.

Anonymous

Anonymous Mon 31 Jan 2022 19:57:33 No.14171567 Report

Quoted By: >>14171615

>>14171282
Yes

Anonymous

Anonymous Mon 31 Jan 2022 19:59:30 No.14171581 Report

Quoted By: >>14171642

>>14171326
>just an array of numbers
>just
No

Anonymous

Anonymous Mon 31 Jan 2022 20:07:43 No.14171615 Report

Quoted By: >>14171641

>>14171567
no

Anonymous

Anonymous Mon 31 Jan 2022 20:10:59 No.14171631 Report

Quoted By: >>14171642

>>14169670
what is a vector?

Anonymous

Anonymous Mon 31 Jan 2022 20:12:41 No.14171641 Report

Quoted By: >>14171863

>>14171615
Yes

Anonymous

Anonymous Mon 31 Jan 2022 20:12:54 No.14171642 Report

Quoted By:

>>14171631
Not an array of numbers.
Ask >>14171581

Anonymous

Anonymous Mon 31 Jan 2022 20:34:38 No.14171724 Report

Quoted By: >>14171779 >>14172362

>>14169670
A vector is a property which must have both mangitutde and length silly stupid person. I know this because unlike you I am an engineer :>

Anonymous

Anonymous Mon 31 Jan 2022 20:47:39 No.14171779 Report

Quoted By: >>14172378

>>14171724
>both mangitutde and length
this is wrong on at least 3 different levels

Anonymous

Anonymous Mon 31 Jan 2022 21:10:08 No.14171863 Report

Quoted By: >>14172338

>>14171641
no

Anonymous

Anonymous Mon 31 Jan 2022 22:43:22 No.14172338 Report

Quoted By: >>14172353

>>14171863
Yes

Anonymous

Anonymous Mon 31 Jan 2022 22:47:03 No.14172353 Report

Quoted By: >>14172392

>>14172338
no

Anonymous

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Anonymous Mon 31 Jan 2022 22:49:18 No.14172362 Report

Quoted By:

>>14171724
>teehee im pretending to be retarded cuz im dumb engineer xDDDD

Anonymous

Anonymous Mon 31 Jan 2022 22:51:16 No.14172378 Report

Quoted By: >>14172473

>>14171779
Why because of math autism? Calling elements of abelian groups with a field action "vectors" was a mistake, they should be "generalized vectors" or better yet a new word entirely

Anonymous

Anonymous Mon 31 Jan 2022 22:53:58 No.14172392 Report

Quoted By:

>>14172353
Yes

Anonymous

Anonymous Mon 31 Jan 2022 22:55:49 No.14172403 Report

Quoted By: >>14172674 >>14173119

>>14171316
What about linear transformations that don't obey that property though?

Anonymous

Anonymous Mon 31 Jan 2022 23:10:50 No.14172473 Report

Quoted By:

>>14172378
>Why?
1) mispelled magnitude
2) magnitude and length are the same thing
3) vectorspaces don't come equipped with a norm (the thing needed to define magnitude), you have to provide that separately

>What about linear transformations that don't obey that property though?
let me rephrase that: let $L : U \rightarrow V$ be a function that obeys the linearity property [yada yada]

the linearity property is what makes a function a linear transformation

Anonymous

Anonymous Mon 31 Jan 2022 23:57:15 No.14172674 Report

Quoted By:

>>14172403
Then it's not a LINEAR transformation. Kek

Anonymous

Anonymous Tue 01 Feb 2022 02:25:03 No.14173119 Report

Quoted By: >>14173223

>>14172403
>what about green paint that isn't green?

Anonymous

Anonymous Tue 01 Feb 2022 02:53:33 No.14173223 Report

Quoted By:

>>14173119
Bros..

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