help-10

[grizzled shore]

what else does it need to pass

[daring steppe]

a(bv)=(ab)v

[grizzled shore]

sure

does it pass this?

[daring steppe]

i dont know how to asses that given it looks like

like i see aV = <ax, ay, az>

[orchid sand]

i'd encourage you to look again at the definition of a vector space

[grizzled shore]

a(bv) = a(bx, by, bz) = (abx, aby, abz) = (ab)(x, y, z) = (ab)v

[daring steppe]

what would b be here. we only have a and v?

[grizzled shore]

.

we need to show this is true

it is true.

[daring steppe]

yes but im confused how b is possible because we have 2 components a and v

i dont see another scalar multiple

[grizzled shore]

wdym

[daring steppe]

i see 1 scalar multiple and a vector

[grizzled shore]

where

[daring steppe]

r is the scalar multiple

[grizzled shore]

ok hold up

[daring steppe]

and <x,y,z> is a vector

if that is the case where is b?

[grizzled shore]

Define \begin{align*}\textcolor{green}{\odot}:\mathbb R\times \mathbb R^3 &\to \mathbb R^3\
\left(r, \begin{pmatrix}x\y\z\end{pmatrix}\right) &\mapsto ,r\textcolor{green}{\odot}\begin{pmatrix}x\y\z\end{pmatrix} \coloneq \begin{pmatrix}r\cdot x\r\cdot y\r\cdot z\end{pmatrix}\end{align*}

[daring steppe]

what is the comma meaning here?

[warm shale]

frosst

[grizzled shore]

the comma meaning we took a pair of things

1 from R and 1 from R^3

[daring steppe]

also what is RxR^3 refer to

[grizzled shore]

you take 1 copy of R and 1 copy of R^3 as inputs

[daring steppe]

ah

[grizzled shore]

r comes from R

[daring steppe]

yeah

[grizzled shore]

(x, y, z) comes form R^3

[grizzled shore]

the white dots on the right is the multiplication on R

[daring steppe]

what does the = with 2 dots mean?

ahh

[grizzled shore]

it means define

[daring steppe]

i still dont see an issue with that

[grizzled shore]

because im saying we can put this green circle dot operation between a real number, and a 3-tuple

and what does that mean?

it means i multiply the real number into each entry

it satisfies this

[daring steppe]

yeah

[grizzled shore]

it satisfies this

[daring steppe]

i get that

ok

[grizzled shore]

what else does it need to satisfy

[daring steppe]

what is b here hto

or is it just saying av=va

[grizzled shore]

a,b are elements of the scalars

no

they both come from R

[daring steppe]

but the scalar only has 1 element/

?*

[grizzled shore]

we need to ensure that this new green circle dot operation

needs to satisfy this

$a\textcolor{green}{\odot}(b\textcolor{green}{\odot} \mathbf v}) = (a\cdot b)\textcolor{green}{\odot} \mathbf v$ for all $a, b\in \mathbb R$ and $\mathbf v \in \mathbb R^3$

[warm shale]

frosst
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[daring steppe]

so, if r = 4 we get v= <4x,4y,4z>

[grizzled shore]

no we need 2 scalars

[daring steppe]

this can be written like 2(2v)

[grizzled shore]

no

we need this new operation to "play nice" with the multiplication on R

[grizzled shore]

you can think of it this way

[grizzled shore]

but this is really what we're tyring to say

[grizzled shore]

it's more like 2(2v) = (2 * 2)(v)

[daring steppe]

ok

[grizzled shore]

which sure, ends up being 4v

[daring steppe]

how do we get there from r

[grizzled shore]

we want to show that this is true

this is a statement

depending on how we define the green operation it can be true or false

[daring steppe]

and the dot means RxR^3 as in a is R and we are multiplying it into something R^3

im unsure of what that is doing

[grizzled shore]

which dot

there's 2 dots

one is white one is green

[daring steppe]

green

i dont mean the normal multiplication

[grizzled shore]

yes

$b\textcolor{green}{\odot}\mathbf v$ is an element of $\mathbb R^3$

[warm shale]

frosst

[daring steppe]

is it a cartesian product

[grizzled shore]

that's by definition, notice the codomain of this operation is R^3

[daring steppe]

i unfortunately dont know what a codomain is

[grizzled shore]

f:A -> B

f is the name of the function

A is the domain of f

B is the codomain of f

[daring steppe]

and do codomains have to be in the same R^n

[grizzled shore]

codomains are something about functions

it may have absolutely nothing to do with R^n

[daring steppe]

like a range of the output of the function

[grizzled shore]

close

[daring steppe]

or a set of all the outputs

[grizzled shore]

the range is a subset of the codomain

[daring steppe]

ah

[grizzled shore]

consider $f:\mathbb R\to \mathbb R$ that is defined by $f(x) = x^2$

[warm shale]

frosst

[grizzled shore]

the range of f is [0, inf)

but the codomain is all of R

[daring steppe]

gotcha

why cant f be less than 0?

[grizzled shore]

back to this, we want to show that for any a, b, and v we pick

[daring steppe]

i.e r = -1

[grizzled shore]

this equality holds

[grizzled shore]

x^2

[daring steppe]

i dont see any x^2?

[grizzled shore]

this is a completely different example not related to your question

[daring steppe]

ooh

[grizzled shore]

look

[daring steppe]

ahh i didnt see htis

[grizzled shore]

it's jsut an example about range vs codomain

[daring steppe]

yeah

ok i can see now

[o,inf)

[grizzled shore]

ok let's go back to this

we want to show for any choice of a, b in R, and v in R^3

this equality holds

[daring steppe]

but x ϵ R

[grizzled shore]

where do you see x

[daring steppe]

i'm extremely sorry for being so darn clueless, but i dont know how to prove this

not sure how to show it is true

[grizzled shore]

we use the definition given by

let $\mathbf v = \begin{pmatrix}x\y\z\end{pmatrix}$

[warm shale]

frosst

[grizzled shore]

then \begin{align*}
a\textcolor{green}{\odot} (b\textcolor{green}{\odot} \mathbf v) &=a\textcolor{green}{\odot} \begin{pmatrix}b\cdot x\b\cdot y\b\cdot z\end{pmatrix} \
&= \begin{pmatrix}a\cdot (b\cdot x)\a\cdot (b\cdot y)\a\cdot (b\cdot z)\end{pmatrix} \
&=\begin{pmatrix}(a\cdot b)\cdot x\(a\cdot b)\cdot y\(a\cdot b)\cdot z\end{pmatrix} \
&= (a\cdot b)\textcolor{green}{\odot} \mathbf v\end{align}

[daring steppe]

isnt that just the same thing tho?

[grizzled shore]

wtf is happenign tom y latex

[daring steppe]

acting funky tonight

[grizzled shore]

oh i shouldn't wrap align in math mode that's why

[daring steppe]

do we need the green dot between a and (bv)

[warm shale]

frosst

[grizzled shore]

of course

bv is in R^3

and a is in R

the only operation we have between 2 such objects is the green dot

[daring steppe]

yeah but isnt it v that is in it not b

[grizzled shore]

that's true

but (b green dot v) as an object is in R^3

[daring steppe]

like i could write it as a*greendot V times b

[grizzled shore]

you cannot

you dont know what V times b means

why is there a * before the greendot

[daring steppe]

yeah typo

A Greendot VB

[grizzled shore]

can't

i dont know what object VB is

[daring steppe]

but RxR^3 = R^3

if B is R and V is R^3

[grizzled shore]

that would be b greendot v

[daring steppe]

then BV is R3?

or

[grizzled shore]

bv is not b greendot v is not vb

these are not the same thing

[daring steppe]

ah

so BgreendotVgreendotA

[grizzled shore]

also can't

greendot takes R in the first argument then R^3 in the second argument

[daring steppe]

so greendot isnt associative?

[grizzled shore]

well i never said it was

(it turns out to be true)

but i never said that

you'd have to prove that it is associative

[daring steppe]

wiat

it is associative?

[grizzled shore]

it is

well

sort of

a vector space only cares about scalar multiplication from the left usually

[daring steppe]

a(b greendot v) = b(a greendot V)

[grizzled shore]

it does not have a concept of multiplication from the right

[grizzled shore]

what's the operation before the (

[daring steppe]

ah

multiply?

[grizzled shore]

can't

[daring steppe]

ok

[grizzled shore]

a is in R

b greendot v is in R^3

* is defined on R, between 2 elements of R

[daring steppe]

but isnt that what a scalar multiple is?

or would the greendot be implied

[grizzled shore]

no, a scalar multiplication is between R and R^3

yes

scalar multiplcation is the green dot

it is different to the multiplication on R

that's why i've been using a different symbol for it (namely the green circle dot compared to the white dot)

[daring steppe]

ok

[grizzled shore]

notice on the * line we used that the normal multiplcation is associative, without this we cannot actually show the statement holds

but it's fine, we know the normal multiplication is associative because R is a field with those operations

[daring steppe]

i dont yet know what a field is

[grizzled shore]

if you dont know what that means just take it to mean we do in fact know it's associative (it being the white dot)

[daring steppe]

ok

they define fields at the end of this chapter

so ill get there soon

[grizzled shore]

it's hilarious they do it in that order

i dont like it

but that's alright

a lot of the vector space operations rely on the interaction with the field

so not knowing what a field is sort of...makes it difficult to make precise what the relationships are

okay anyway, what else does this greendot need to satisfy?

[daring steppe]

so a field is more basal than the vector space?

[grizzled shore]

we have a multiplicative identity, and we also have that a(bv) = (ab)v

well you need a field to make a vector space so i'd say so

[daring steppe]

distributive?

[grizzled shore]

this is one of the distribution axioms

what's the other one?

[daring steppe]

if g and h are vectors and a is a scalar then a(g+h)=ag+ah

but we dont really have that ig

[grizzled shore]

aha

that's the issue right

a(g+h) is always a(0) = 0

wait

huh?

no this one is fine

ag + ah is always 0

it's 2 vectors being added together

0 = 0 for all scalars a, and vectors g, h

[daring steppe]

would ag and ah then be additive inverses?

or a =0

and its trivial

[grizzled shore]

well it does seem like every g is the additive inverse of every h

[daring steppe]

yeah

[grizzled shore]

so a(bv) = (ab)v and a(g+h) = ag + ah) are the 2 distributive laws for the green dot

there's more though

there's lastly (a + b)v = av + bv

[daring steppe]

additive identity: there exists some 0 ϵ V such that v+0 = v .

[grizzled shore]

actually i think people usually call a(bv) = (ab)v associativity

[daring steppe]

this seems true

[grizzled shore]

no we eneed this one

[daring steppe]

wasnt this a(g+h)

[grizzled shore]

that's different

[daring steppe]

that we showed worked with green dot

[grizzled shore]

a(g+h) is the scalar multiplication (green dot) distributing over the vector addition +

[daring steppe]

so a and b are scalars?

[grizzled shore]

(a + b)v is the scalar multiplication (green dot) distributing over the field addition

[grizzled shore]

yes

the last one

[daring steppe]

field addition is non-vector addition?

[grizzled shore]

field addition is the addition of real numbers

the field addition takes 2 things in R and adds them together

[daring steppe]

ok

[grizzled shore]

the vector addition takes 2 things in R^3 and adds them according to the rule you've set out

[daring steppe]

wait would a dual space then connect that field to some complex thingy

nvm

thats for later

[grizzled shore]

in your case it's this first one after the (a)

[daring steppe]

on my own time

[grizzled shore]

dual spaces will take some more time to develop

but yes it has to do with the field as well

[daring steppe]

ok yeah

i see that

[grizzled shore]

LA is usually the first time people see that there are different 0's and different + and * symbols

[daring steppe]

<x1+x2,y1+y2,z1+z2>=<0,0,0>

ah

[grizzled shore]

but it's usually hidden behind an awful use of the same symbol to represent different operations

[daring steppe]

ah

hense field and vector space

[grizzled shore]

hence the distinction between vector addition and the field addition

[daring steppe]

ok

[grizzled shore]

so show that this doesn't hold

[daring steppe]

(a+b) greendot v /= (A greendot v) + (b greendot v) ?

[grizzled shore]

yes

[daring steppe]

i thought green dot was distributive

is it just not distributive to a field

[grizzled shore]

we dont know that it's distributive over field addition

not a field, the field

[daring steppe]

ok

and why dont we know

[grizzled shore]

well we haven't proved that it is

or isn't

it turns out it isn't

and you need to prove that it isn't

[daring steppe]

do I need to know more about fields to do that

[grizzled shore]

you don't

you know what the addition on R looks like

you add 2 numbers

3 + 5 = 8

[daring steppe]

ah

i think i have it

i hope

[grizzled shore]

try it and see

[daring steppe]

(3+5) greendot V /= 8greendot V

im trying to figure out how to articulate why

i guess that doenst really prove anything

hm

its another statement

[grizzled shore]

that's not what we're showing btw

we want to show that
(3 + 5) greendot v ≠ 3 greendot v + 5 greendot v

ok but

[daring steppe]

yeah

[grizzled shore]

$\textcolor{green}{+}:\mathbb R^3\times \mathbb R^3 \to \mathbb R^3$

[warm shale]

frosst

[grizzled shore]

this should say (3 + 5) greendot v ≠ 3 greendot v green+ 5 greendot v

those aren't the same +'s

[daring steppe]

i guess i meant to say (3+5) greendot v = (8) greendot v != 3greendot v green+ 5 greendot v

[grizzled shore]

yes

that is what you mean to say

also, green + on the right

[daring steppe]

ah

so green + refers to adding R to R^3

[grizzled shore]

nope

we can't do that

adding 2 vector stogether

[daring steppe]

yeah ok

[grizzled shore]

green+ is vector addition

[daring steppe]

ah

gotcha

yeah you definitely cant <1x,2y> + <2x+3y+4z>

but 3v and 5v are vectors because 3 and 5 just scalar multiples

[grizzled shore]

<1x> doesn't make sense

[daring steppe]

yeah

3<x,y,z> + 4<x,y,z> = 7<x,y,z>

<3x,3y,3z> + <4x,4y,4z> = <7x,7y,7z>

dont these follow that rule

[daring steppe]

like (3+4)greendot V

where V = <x,y,z>

and 3+4 is field addition

[grizzled shore]

we don't know this

we're trying to show that 3<x, y, z> + 4<x, y, z> ≠ (3 + 4)<x, y, z>

[daring steppe]

yeah

but i cant find a place where 3<x,y,z> + 4<x,y,z> = 7<x,y,z> = (3 + 4)<x, y, z> isnt true

[grizzled shore]

you know that when you add 2 vectors together

it's always 0

that's what the vector addition does

[daring steppe]

?

[grizzled shore]

so you just need to show that <0, 0, 0> ≠ 7<x, y, z> for some <x, y, z>

well it says it right here

when i add 2 vectors i get 0

that's what this + does

[daring steppe]

<x,y,z>+<x,y,z> = <2x,2y,2z>

oh

[grizzled shore]

nope

we dont know this

[daring steppe]

wait we are talking about the problem or in general

in the problem i agree

in general that makes no sense to me

[grizzled shore]

"in general" implies a particular conventional version of +

[daring steppe]

i mean

[grizzled shore]

there is no "natural" addition of vectors

[daring steppe]

if we did <x,y,z>+<x,y,z> = <2x,2y,2z> outside of this problem

[grizzled shore]

we just use "entrywise addition" to be the "conventional" addition

[grizzled shore]

this does not define +

you need to tell me how to add 2 arbitrary vectors

not 2 of the same vector

[daring steppe]

yeah

like <1,4> + <3,6> = <4,10>

[grizzled shore]

that's not arbitrary

that's 2 specific vectors

(although it's wrong to call them vectors until you know they are vectors and not just 2-tuples)

[daring steppe]

tuple?

[grizzled shore]

ordered pair

n-tuple is an ordered list of n things

[daring steppe]

ah

[grizzled shore]

a vector space is formally a 3-tuple

(V, +, • ) is a vector space

[daring steppe]

ok

[grizzled shore]

it includes the + (vector addition) and • (scalar multiplication) functions

and technically, we more specifically say a K-vector space is... where K is the field

[daring steppe]

but a vector has an order

[grizzled shore]

but you'll understand that more when you get to the fields section

[daring steppe]

would that matter in a tuple

[grizzled shore]

a tuple is just an ordered list of things

[daring steppe]

oh

[grizzled shore]

they may not even be numbers

[daring steppe]

gotcha

[grizzled shore]

(V, +, • ) is a 3-tuple

it has a set V, and then 2 functions + and •

that's the 3 things it has

in that order

[daring steppe]

so if x,y, and z are components of a vector that belong to the set R then <x,y,z>+<3x,5y,2z> = <4x,6y,3z>

[grizzled shore]

the fact you used the word "vector" already indicates you've assumed a vector addition

[grizzled shore]

when the + and • are special, ie. they satisfy the vector space axioms

[grizzled shore]

that would be these ones

then the 3-tuple (V, +, • ) can be a called a vector space

you'll see later on that (F, +, • ) is a field if F is a (non-empty) set and + and • satify some rules, those would be the field axioms

[daring steppe]

adjusted: if x,y, and z belong to R and A is a 3-tuple <x,y,z> and B is also a 3-tuple <3x,5y,2z> then <x,y,z>+<3x,5y,2z> = <4x,6y,3z> only if it agrees with 8 axioms of vector space.

is this better

[grizzled shore]

no

it is not better

because we could've picked a different choice of vector addition

on top of that, we dont need a vector space structure for this either

perhaps i've muddled you in the formalism

Consider a function $\textcolor{yellow}{+}$ from $\mathbb R^3\times \mathbb R^3$ to $\mathbb R^3$, defined by $\langle a,b,c\rangle \textcolor{yellow}{+} \langle x, y, z\rangle \coloneq \langle a+x, b+y, c+z\rangle$ where $+$ is the addition on $\mathbb R$

[warm shale]

frosst

[daring steppe]

ah

[grizzled shore]

it turns out this yellow + is suitable for being a vector addition on R^3

that doesn't mean it'll definitely work

[daring steppe]

so back to the (a+b) greendot v ! = a greendot v + b greendot v then im confused because does 0greendot v != 0greendot v +0greendot v

[grizzled shore]

but as it stands, it doesn't violate any of the vector space axioms that require only vector addition

[daring steppe]

ok

[grizzled shore]

this particular choice of a and b is true

is equal

but we need it to be equal for all choices of a and b

and when you look at the answer they gave

[daring steppe]

isnt a + b just another way of representing some other scalar c

[grizzled shore]

they gave a counterexample where it doesn't work

[daring steppe]

yeah

i dont see what a and b are there

i just see the addition 2 vectors = to a zero vecotr

[grizzled shore]

where

[daring steppe]

where each of the vectors g and h are additive inverses

[grizzled shore]

we need

(a+b) greendot v = a greendot v + b greendot v

this to hold

we NEED it to hold if these operations are to lay the foundation of a vector space

we are showing by COUNTEREXAMPLE that it does not always hold

[grizzled shore]

hence R^3 is NOT a vector space under these operations

do you understand that + and • need to be special functions for you to have a vector space

they can't just be whatever you like

we are showing that for this choice of + and •, they are not special enough

they do not satisfy the requirements of being the operations of a vector space

and in particular, this choice of + and • does not satisfy the following requirement

how do we know it does not satisfy the following requirement?

we give a counter example

[daring steppe]

so the addition here isnt vector addition?

but there are 2 vectors being added

[grizzled shore]

it is not

[grizzled shore]

no, there are 2 3-tuples of real numbers being added together

we cannot technically call them vectors

vectors are elements of a vector space

this means you need a proper notion of vector addition and scalar multiplication FIRST

[daring steppe]

ok

[grizzled shore]

before you can start calling things vectors

[daring steppe]

so inorder for it to be a vector it must be true for all R^3?

[grizzled shore]

in order for it to be a vector it needs to come from a vector space

vectors can come in all sorts of shapes

functions can be vectors

polynomials can be vectors

matrices can be vectors

just to name a few

[daring steppe]

does x1 and y1 etc alone belong to R^3 or R

and the 3-tuple as a whole belongs to R^3

[grizzled shore]

R

[daring steppe]

ok

[grizzled shore]

$\mathbb R^3$ is understood as $\mathbb R\times \mathbb R \times \mathbb R$

[warm shale]

frosst

[grizzled shore]

it is a 3-tuple of real numbers

[daring steppe]

i still dont see how dthis doesnt satesfy the distributivity of scalar multiplication with respect to field addition

ah

like

what are a b and v

in this vector

[daring steppe]

yeah

i have no clue where 1 is coming from

[grizzled shore]

the left side says 2 * <1, 0, 0> which is <2, 0, 0>

from the definition of *

the right side is the addition of 2 vectors which is <0, 0, 0>

from the definition of +

well, <2, 0, 0> is not <0, 0, 0>

[grizzled shore]

but if it satisfied this

[daring steppe]

but do they have to be identical?

or does it need to work for everything in the set

[grizzled shore]

we would see that (1 + 1) * <1, 0, 0> = 1 * <1, 0, 0> + 1 * <1, 0, 0> = addition of 2 vectors = <0, 0, 0>

[daring steppe]

in that case any number in the product would be impossible

to be a vector space

also where does the 1+1 comefrom

[grizzled shore]

they picked it

[daring steppe]

in the problem it is just 2

[grizzled shore]

they picked a = 1, b = 1

and v = <1, 0, 0>

[daring steppe]

just 2 3-tuples being added

[grizzled shore]

and showed it does not work

[daring steppe]

i dont see any addition?

field

[grizzled shore]

yes

you could chose (1 + 2)

or (3 + 5)

or (10 - 3)

none of them will work

[daring steppe]

like where is the a and b here?

[grizzled shore]

you just need to show 1

[daring steppe]

it doesnt make any statement about field addition here?

[grizzled shore]

that's the definition of how to add 2 vectors together

[grizzled shore]

no it does not

[daring steppe]

so why is ther (1+1)

which is field definition

in their answer

[grizzled shore]

this

we need this to be true

for the operations to be a vector space

[daring steppe]

ok

but that has 1 vector

and 2 scalars

[grizzled shore]

yes

[daring steppe]

our problem has 0 scalars

and 3 vectors

[grizzled shore]

where

[daring steppe]

where are the scalar values?

a and b

i see v1 +v2

[grizzled shore]

this + sign

[daring steppe]

2 vectors

[grizzled shore]

is the + between av and bv

[daring steppe]

but the 2 v's there arent the same

so is a = 1

[grizzled shore]

av = <x_1, y_1, z_1> and bv = <x_2, y_2, z_2>

[daring steppe]

and b = 1

[grizzled shore]

in here yes

[daring steppe]

I just dont see how we can use that equation if we have 2 different vectors

given that there is only 1 in

[daring steppe]

this 1

[grizzled shore]

but av + bv is 2 vectors

well

we shouldn't call them vectors

[daring steppe]

wouldnt it be av_1 +bv_2

[grizzled shore]

but, let's call it that for the sake of simplicity

no

[daring steppe]

v isthe same in both

[grizzled shore]

this tells us how to add 2 vectors together

av is a vector

bv is a vector

we can add them together

[daring steppe]

yeah

but x1 != x2

[grizzled shore]

(the use of the word "vector" here is very wrong)

so what

[daring steppe]

so

v != v

so

v_1

[grizzled shore]

any 2 vectors are added together like this

[daring steppe]

v_2

[grizzled shore]

whether they are the same or not

it does not matter

nowhere did anyone mention that there would be different conditions for adding same vectors

we dont mention it because it is not important to us

[daring steppe]

1v + 1v != (1+1)v.

[grizzled shore]

we pick a = b = 1

and v = <1, 0, 0>

[daring steppe]

why can we just say v = <1,0,0>

[grizzled shore]

and show that this does not hold

[grizzled shore]

because this needs to hold

[daring steppe]

yeah

[grizzled shore]

for EVERY a, b in R

and for EVERY v in R^3

if we can so much as find 1 singular example where it doesn't hold

the entire thing falls apart

[daring steppe]

if we had <x3,y3,z3> would it hold

[grizzled shore]

even just 1 counterexample is enough

we dont know

depends on the choice of x y z

the problem here is if you picked v = <0, 0, 0>

it looks like it holds

but it only holds for <0, 0, 0>

not for EVERY choice of <x_3, y_3, z_3>

[daring steppe]

so no sum works ever?

[grizzled shore]

now you could otherwise also make it "look" true by picking a = -b

then you have (a - a)v = 0v = 0 and av + (-a)v = 0

so it "looks" like it holds

[grizzled shore]

this now holds for all v

[grizzled shore]

this now holds for all a, b

but there's a combination of a, b and v you can pick to break the axiom

[daring steppe]

this is a vector space

[grizzled shore]

with the set of all 2x2 matrices over real numbers yes

[daring steppe]

im just confused

that if scalar multiplication distributes *over scalar addition

is one of the rules

[grizzled shore]

$(M_{2\times 2}(\mathbb R), +, \cdot)$ is a $\mathbb R$-vector space

[warm shale]

frosst

[grizzled shore]

where does it say that

[grizzled shore]

ok what about it

(the word distribute is very important)

oh you edited it

[daring steppe]

yes

i dont see

how if we have. a problem

not with v but with v1 and v2

we can just represent those 2 separate vectors as 1 v

i also am still confused because i dont see any scalar addition between the scalar values

i see 2, i guess 3-tuples being added

and summing to a 3rd 3 tuple

where is the distribution

[grizzled shore]

how if we have. a problem
not with v but with v1 and v2
we can just represent those 2 separate vectors as 1v
can you give an example

[grizzled shore]

here

[daring steppe]

i could say 2+3 = 7 thats wrong

like i see the <1,0,0>

and (1+1)

[grizzled shore]

r and s are from the field

they are scalars

[daring steppe]

yeah

but the arent getting added

[grizzled shore]

r + s is scalar addition

[daring steppe]

yes

[grizzled shore]

it says right there (r+s) • v

[daring steppe]

but r and s arent getting multiplied anywhere

[grizzled shore]

that's r + s

[daring steppe]

in the law

[grizzled shore]

no they aren't

[daring steppe]

not in the equation

[grizzled shore]

who said they need to multiply together

[daring steppe]

1<x1,y1,z1>+1<x2,y2,z2> = (1+1)<x2,y2,z2>

[grizzled shore]

but we dont know that this is true

also where does it multiply scalars

in that

[daring steppe]

1 is a scalar

it is being multiplied into <x1,y1,z1>

to get <1x1,1y1,1z1>

[grizzled shore]

ok

what about that

[daring steppe]

that to me is the same, in the equation

and distribution <x,y,z> over (1+1)

how is that not

[grizzled shore]

what youve just said isn't a question

it's not even a proper sentence

[daring steppe]

like

1 * V = V

[grizzled shore]

sure

[daring steppe]

(1+1)V = V+V

[grizzled shore]

woah

who said that's true

[daring steppe]

i dont see how thats not true

[grizzled shore]

we never said it was true

[daring steppe]

ik

[grizzled shore]

you can't assume it to be true

in particular they gave the example in the answer

how is it not true?

well, when V = <1, 0, 0>, it is not true

[daring steppe]

how

[grizzled shore]

on the left side

[daring steppe]

1* v1 = v1

+v1 = 2v1

[grizzled shore]

(1+1)V = 2V = <2, 0, 0>

[daring steppe]

V1 +v1 =. 2v1

[grizzled shore]

on the right side

[daring steppe]

yeah

[grizzled shore]

1V + 1V = <0, 0, 0>

left and right sides are not equal

[daring steppe]

but <1,0,0> +<1,0,0> = <2,0,0>

[grizzled shore]

so this statement does not hold for all V

[daring steppe]

so i dont see your point

[grizzled shore]

no it is not

[daring steppe]

the original statement is just impossible

but the actual equation

[grizzled shore]

that's your conventional addition

we are not talking about your conventional addition

[daring steppe]

its like saying 2+3=7

thats just not true

[grizzled shore]

we are talking about this addition

[daring steppe]

yes this is like 2+3=7

[grizzled shore]

there's nothing wrong with that

2 + 3 = 5 is perfectly valid

[daring steppe]

yes

lmoa

[grizzled shore]

okay 2 + 3 = 7 is also perfectly valid

[daring steppe]

i mean 7 wrote 5 2 times

to reference what i wrote earliier

[grizzled shore]

• is a function that sends the ordered pair (2, 3) to 7

[daring steppe]

how

[grizzled shore]

great

what's wrong with that

[daring steppe]

well

if + is addition

[grizzled shore]

woah

[daring steppe]

as i used in 2+3 = 5

then 2+3 !=7

[grizzled shore]

we're just "calling it" addition

it doesn't have to be the same addition that you're thinking about

as i said earlier

there are multiple different +'s

if i have 1 apple

and i get another apple

i dont have 1 apple 1 apple

i have 2 apples

that's because when we want to add <number> <noun> we dont add the nouns together

we just add the numbers

but this is also only true if the nouns are the same

if i have 1 apple, and i get an orange

i dont have 2 apples

[daring steppe]

i mean that equation doesnt make sense tho

[grizzled shore]

these are sort of "rules" of "english addition"

[daring steppe]

i think you knew what i meant when i said 2+3 != 7

[grizzled shore]

i know what you mean

and im telling you it's fine

[daring steppe]

i wasnt suggesting a new form of addition

i was saying using the field addition we have been talking about

[grizzled shore]

you're saying that this idea of addition does not line up with your preconcieved notion of addition

you are correct

it does not

[daring steppe]

under those rules itdoesn twork

just as the 3-tuples above

= <0,0,0>

[grizzled shore]

but this is not necessarily needed

[daring steppe]

but that, to me doenst suggest that the addition of 2 vectors

is not in a vector sapce

space

like

then how can you have 2 vectors in R^3 get added

[grizzled shore]

remember

[daring steppe]

if <0,0,0> exists

[grizzled shore]

a vector space includes the operations + and •

if you pick terrible choices for + and • it may not be true that you have a vector space

[daring steppe]

yes

[grizzled shore]

when you pick entrywise addition, entrywise multiplication, then yes, R^3 with those choices of + and • is a vector space

[daring steppe]

i see how this instance doesnt work

but i dont see

how it relates to the (a+b)

[grizzled shore]

the point is that R^3, with the choices given in (a) is not a vector space

[daring steppe]

yes (a+b)V ! = 0 vector but (a+b)v = av + bv

[grizzled shore]

it's not that the addition is wrong

it's that the addition DOES NOT work with the scalar multiplication given

and when i say doesn't work

[daring steppe]

if thats the case

[grizzled shore]

i mean it doesn't satisfy one of these

[daring steppe]

how do u add any 2 vectors

[grizzled shore]

depends what you mean by "add"

[daring steppe]

1*v=v

right?

so if i want to add vectors

<0,0,0>

still exists out there

and if they cant sum to that

they aren't vectors?

[grizzled shore]

what?

that's not a question

[daring steppe]

ok

[grizzled shore]

let me give you an example of a vector space that may make you realise it's very general

[daring steppe]

ok

forget it = <0,0,0>

can v1+v2 ever be a vector space

if not im extremely confused

like

R^3

should be a vector space

is what im trying to say

like vector a + vector b = <a1+b1,a2+b2,a3+b3>

must be a vector space

right?

[grizzled shore]

Let $A = {\text{apple}, \text{banana}}$, define 2 operations $\oplus$ and $\odot$.
\begin{align*}
\oplus:A\times A &\to A \
(a, b) &\mapsto a \oplus b
\end{align*}
$$\text{apple} \oplus \text{apple} = \text{apple}\qquad \text{apple} \oplus \text{banana} = \text{banana}$$
$$\text{banana} \oplus \text{apple} = \text{banana} \qquad \text{banana} \oplus \text{banana} = \text{apple}$$

\begin{align*}
\odot:A\times A &\to A \
(a, b) &\mapsto a \odot b
\end{align*}
$$\text{apple} \odot \text{apple} = \text{apple}\qquad \text{apple} \odot \text{banana} = \text{apple}$$
$$\text{banana} \odot \text{apple} = \text{apple} \qquad \text{banana} \odot \text{banana} = \text{banana}$$

[warm shale]

frosst

[grizzled shore]

$(A, \oplus, \odot)$ is a vector space.

[warm shale]

frosst

[daring steppe]

yeh

but

[daring steppe]

is this true

this is at the route what i think is confusing me

because

would <0,0,0> not be part of the set of products of vector a +vector b

I think It would just never come up

[grizzled shore]

that's not enough information

[grizzled shore]

i have no idea what this means

what is the "set of products of the sum of 2 vectors"

[daring steppe]

co domain?

[grizzled shore]

you've only given me the +

[daring steppe]

codomain of vector a +vector b

[grizzled shore]

we need the • as well

[daring steppe]

would <0,0,0> be in that

• being vector addition

[grizzled shore]

codomain is something of a function

[daring steppe]

oh

[grizzled shore]

vector a + vector b is not a function

• is a function

[daring steppe]

ok

if i made a list of all the products of the addition of vector a and vector b

would i ever encounter <0,0,0>

[grizzled shore]

"all the products"

you mean all the outputs?

product is a special word

[daring steppe]

oh

outputs

[grizzled shore]

yes

[daring steppe]

ok

so

are vectors a and b actually vectors

if so

how is that different

[grizzled shore]

that's not a sentence

[daring steppe]

from v1+v2= <0,0,0>

[daring steppe]

im asking if i have vectors a and b. and a + b = <0,0,0>

to be a vector they have to be in vector space

how is that different

from v1 +v2 = <0,0,0>

and why are those not vectors

[grizzled shore]

because this is not the same as the 8 axioms of a vector space

[daring steppe]

but this is?

[grizzled shore]

also

you need quantifiers

[grizzled shore]

this is also not enough

[grizzled shore]

this is the "smartass" answer to what's a vector

"a vector is something in a vector space."

[daring steppe]

I'm trying to say

[grizzled shore]

the followup question would be, "what does it mean to be in a vector space?"

[daring steppe]

that if a+b = <0,0,0> valid

[grizzled shore]

for all a and b?

for some a and b?

there exists a and b?

you need quantifiers

[daring steppe]

i thought here you were saying that it was a valid output for <0,0,0>

[grizzled shore]

<0, 0, 0> is not a function

it cant have an output

it is the output to the + function

for some a and b

[daring steppe]

*List of sums of the 2 vectors"

[grizzled shore]

and in the axioms we will say that

for all a in V, there exists b in V such that a + b = 0

this is the inverse axiom

[daring steppe]

I'll try to rephrase is there no v1+v2 combo that can make <0,0,0>

[grizzled shore]

there will always be a v2 for every choice of v1 that when you add them, gives <0, 0, 0>

[daring steppe]

yeah

[grizzled shore]

this is guaranteed by this axiom

[daring steppe]

so why does v1+v2 =<0,0,0>

not work in the problem

[daring steppe]

if this is the case

[grizzled shore]

this is under the assumption that a, b are vectors, that they come from a vector space

[daring steppe]

ok

[grizzled shore]

in the problem they are not vectors

[daring steppe]

so

[grizzled shore]

namely because they don't come from a vector space

[daring steppe]

there are no vectors where that is true

[grizzled shore]

that's not what it says

[daring steppe]

so no 2 vectors can = <0,0,0>?

[grizzled shore]

that's not the oppposite of this statement

[grizzled shore]

also, just because you aren't a vector space doesn't mean you can't satisfy this

a non-vector space operation can satisfy this axiom

a vector space operation satisfies all of the axioms

not just 1

not just 2

all of them

[daring steppe]

yeah

[grizzled shore]

at the same time

[daring steppe]

<1,1,1>+<-1,-1,-1> = <0,0,0>

are these vectors?

[grizzled shore]

i dont know

that's an equation

[daring steppe]

how would you write them as cvectors then

[wild swallow]

where are the elements coming from

are we using the + defined in your original question?

[grizzled shore]

an equation is a true or false statement

[daring steppe]

this is why im confused

i can find vectors that satesfy that

how can i then say those vectors are not in vector space

[wild swallow]

they can't be vectors if your space isn't even a vector space

[daring steppe]

ok

frosst has been trying to explain to me the distributive rule that says: (a+b)v=av+bv

and that in the case of our question (a+b)v != av+bv

and i dont understand why that is the case

[grizzled shore]

the rule says it's true for all a, b and v

if we can find 1 combination of a, b and v where the equality fails

then it's not distributive

[wild swallow]

yeah, it isn't true because there exist choices of a, b, and v that violate the equality

[daring steppe]

right

so no vector addition can = a zero vector?

[grizzled shore]

that's not the oppposite statement

[daring steppe]

so what is?

if its not a vector space

[wild swallow]

in your case, you have

2<1, 0, 0> = <2, 0, 0>
but simultaneously
<1, 0, 0> + <1, 0, 0> = <0, 0, 0>

[daring steppe]

then i cant have a vector?

[wild swallow]

a vector is an element of a vector space

[daring steppe]

yeah i agree this is a problem

[wild swallow]

if you don't have a vector space in the first place, there are no vectors to speak of

[daring steppe]

but i dont see for example

[grizzled shore]

the negation of $$\forall a,b\in R,, \forall v\in \mathbb R^3, (a+b)v = av + bv$$
is $$\exists a, b\in R,, \exists v\in \mathbb R^3, (a+b)v \neq av + bv$$

[warm shale]

frosst

[daring steppe]

<x1,y1,z1>+<x2,y2,z2> = <x1+x2,y1+y2,z1+z2>

how is this not a vector space ^

[grizzled shore]

that is not enough information to know if it's a vector space or not

[grizzled shore]

you see this

[daring steppe]

what information do i need to give

[wild swallow]

this isn't the + that you've got in your question though

[grizzled shore]

this vector space as 2 elements

and i've written alot to describe the vector space

[daring steppe]

how is it different than the +\

they both look like vector addition

[grizzled shore]

you need to give similar levels of information on what to do for me to be able to tell you if it's a vector space or not

[daring steppe]

just the 1 in the problem doenst work

[wild swallow]

your + defines
<x1,y1,z1>+<x2,y2,z2> = <0,0,0>

[daring steppe]

yeah

that doens't work

[grizzled shore]

1 counter example is all you need

[daring steppe]

i agree

but

[wild swallow]

well the one in your problem is not vector addition

[daring steppe]

yeah

and how can i determine that

because

[wild swallow]

it does not look like vector addition

[wild swallow]

we determined it here

[daring steppe]

this doesnt look wrong

[wild swallow]

how does the not equals sign not look wrong?

[daring steppe]

it is only wrong when <2,0,0> = <0,0,0>

[wild swallow]

?

[daring steppe]

the not equal sign does look wrong to me

thats wat im saying

[wild swallow]

in a vector space, you must have equality

[daring steppe]

yeah

thats the answer

[wild swallow]

here, you do not have equality

[daring steppe]

not the problem

[wild swallow]

so you failed to make a vector space

the very fact that it's not equal means you don't have a vector space

[daring steppe]

yeah

i dont see how that doesn't equal

[wild swallow]

?

you claim that <2,0,0> = <0,0,0>?

[daring steppe]

besides <2,0,0> != <0,0,0>

[wild swallow]

well that's precisely the problem

[daring steppe]

yeah

thats what im trying to say

but i also dont get

how i can create a vector that = <0,0,0>

[wild swallow]

in a vector space, 2<1,0,0> = <1,0,0> + <1,0,0>

[daring steppe]

yeah

i agree

[wild swallow]

in your space, this fails to be true

[daring steppe]

is this impossible?

[wild swallow]

so it's not a vector space

i don't know what you're talking about

[daring steppe]

i meant vector addition*

are there any 2 vectors that add to <0,0,0>

[wild swallow]

of course there are

[daring steppe]

ok

so

then they have to exist in a vector space

[wild swallow]

but not <1,0,0> and <1,0,0>

[daring steppe]

yeah

exactly

yes

[wild swallow]

<1,0,0> cannot add to itself to become <0,0,0>

[daring steppe]

but the problem never specifies <1,0,0>

[wild swallow]

so?

you are meant to produce the counterexample

[wild swallow]

you are the one to come up with <1,0,0> or any other example

[daring steppe]

yeah

[wild swallow]

so what if the question didn't say anything about <1,0,0>

[daring steppe]

but if no vector in that form exists

then how can any 2 vectors exist in that form

[wild swallow]

?

you're saying <1,0,0> doesn't exist in R^3?

[daring steppe]

if v1 + v2 = <0,0,0> is not a vector space

how can you have 2 vectors

that add to <0,0,0>

[wild swallow]

this is an equation

not a vector space

[daring steppe]

ok

[median dome]

What you are given in the question is meant to be a rule, not an equation

Meaning it has to be satisfied by all possible vectors

[daring steppe]

then v1 and v2 are tuples and not vectors?

[wild swallow]

what?

what are v1 and v2

[daring steppe]

frosst had told me, which i probably have lost in translation at this point, that a tuple is and ordered list

[wild swallow]

sure

[daring steppe]

vectors

[wild swallow]

a tuple is an ordered list of numbers

[daring steppe]

i was saying can any 2 vectors v1 and v2, sum to <0,0,0>

[wild swallow]

in a vector space, they can

[daring steppe]

because we said v1+v2 =<0,0,0> is not a vector space

[wild swallow]

again, this is an equation

[daring steppe]

so how do i find a vector space where they can

[wild swallow]

a vector space is a set V equipped with vector addition and scalar multiplication

it's not just some equation

[daring steppe]

ok

[wild swallow]

R^3 with its usual addition and multiplication is a vector space

<1,0,0> + <-1,0,0> = <0,0,0> (under the usual addition in R^3)

but it also happens that
2<1,0,0> = <2,0,0> under the usual multiplication in R^3

[daring steppe]

yeah

[wild swallow]

in your question, you are equipping R^3 with a different addition operation

[daring steppe]

and how do i tell this?

[wild swallow]

read the question

[wild swallow]

it tells you

[daring steppe]

"the intended operations are the natural ones."

[wild swallow]

no?

[daring steppe]

wrong cp

[wild swallow]

it doesn't say that in your question

[daring steppe]

oops

Prove or disprove that R3 is a vector space under these operations.

yes

[wild swallow]

yes

[daring steppe]

but

the + symbol

[wild swallow]

"under these operations"

[daring steppe]

is the same

as the + symbol

[wild swallow]

so what?

[daring steppe]

so how do i differentiate between vector addition and scalar

[wild swallow]

they are redefining what it means

you just have to accept that fact

[daring steppe]

how tho? they never mention that it is scalar additon>

[wild swallow]

well they don't operate on the same elements

[daring steppe]

right, but <x1,y1,z1> and <x2,y2,z2>

[wild swallow]

presumably, you are working over R as the base field

[daring steppe]

look like vectors

[wild swallow]

so?

[daring steppe]

so how do i know if its vector addition

or scalar addition

[wild swallow]

you check the axioms

[daring steppe]

if the look is the same

right

[wild swallow]

?

[daring steppe]

but the axioms dont answer the problem

[wild swallow]

scalar addition is between scalars not vectors