linear-algebra

[wintry steppe]

I found the vector orthogonal to the plane is (3, 0, -7)

And I know if I can find a vector orthogonal to that one that passes through (4, -2, 3) I'll have my answer. Can someone help?

[half ice]

Dependent iff there exists a non-trivial solution

Independent if not dependent

[lapis fern]

thanks

[sick dragon]

how do i solve this? i get no consistent solution
i assume i need to turn it into an n x n matrix, but how

[zinc tapir]

@sick dragon do u have to solve this a specific way? i know we are in lin alg channel but if you can find two vectors that span the parallelogram you can cross product

[sick dragon]

i know that

but how do i solve it w/ linear algebra

i end up w/ a 3x4 matrix

which you cannot take the determinant

ive also done a bunch of other weird things, none of which give the correct answer

not helpful here

[half ice]

So a parallelogram is defined with two vectors. I don't see why you have 4

Could be that two of them just cancel out the other two. There's an easy way to find which ones:

P + S = (-8, 0, -2)
Q + R = (-8, 0, -2)

So if they do form a parallelogram, P and Q define it

Finally, the magnitude of the cross product P×Q gives the 3D parallelogram area

@sick dragon

[sick dragon]

thats not linear algebra

[hallow cliff]

you can find the determinant of the two vectors

wait how did you get vectors in R^3 kaynex?

[eternal finch]

Original problem is in R^3 https://discordapp.com/channels/268882317391429632/540211747613704221/716100097599078401

[hallow cliff]

oh I see

[eternal finch]

I hope somebody comes up with a clean solution. If we're not to use cross products, the only other solution I can think of is projecting the "left" side of the parallelogram onto the "bottom" side, getting the difference between the "left" side and that projection, and then multiplying the length of that difference with the length of the "bottom" side.

[hallow cliff]

uhh, you can find three vectors and take their determinate

[half ice]

You'll get 0, as they're in a plane

[sick dragon]

yeah, thats one of the things i did

[eternal finch]

Hm, if you take two that span the parallelogram and then take one that is perpendicular to both with a norm of 1...

But then you may as well use the cross...

[sick dragon]

is this relevant

[eternal finch]

It's the same ideas as the previous pictures you pasted.

[half ice]

No, sadly. You're working in R3 and so the determinant works with parallelepipeds

You don't have one of those here

[hallow cliff]

what if you turn it into a parallelepiped by adding a normal vector of norm 1

as red herrring mentioned

and to calculate that normal vector, you can find the nullspace of the transpose of the basis matrix

aka avoiding taking a cross product

?

[eternal finch]

Fk...

That's an interesting one. Lemme digest that.

So, suppose we take three vectors in the span of the parallelogram.

Make those the columns of a matrix A.

We want a vector orthogonal to the span of the parallelogram.

The span of the parallelogram is the column space of A.

The orthogonal complement of the column space of A is the left null space of A.

The left null space of A is the null space of the transpose of A.

Ah.

Ok.

Interesting approach.

So, once you find a vector in null space of the transpose of A, then you just normalize it.

Take P, Q, and that normalized vector, make them columns of a matrix B, and take the determinant of B?

[hallow cliff]

yep

[eternal finch]

Wao.

[half ice]

Alternate cross product if that works

[eternal finch]

True.

Or rather, the cross product and whatever vector found in what I just walked through are both vectors normal to the parallelogram.

No reason to believe it'd obey the laws of cross products?

[quartz compass]

$|\vec u \times \vec v \cdot \frac{\vec u \times \vec v}{|\vec u \times \vec v|}|$

beautiful

[eternal finch]

Er, sorry, I dun geddit...

[stoic python]

Merosity:

[quartz compass]

I messed up

I wasn't paying enough attention to what you're doing I guess

[hallow cliff]

red, that's the generalized way to find orthagonal vectors in R^n

[eternal finch]

Oh. Can you put a name to that formula?

[hallow cliff]

wait the one that merosity posted?

srry that's not what I meatn

[eternal finch]

Oh, were you not talking about it?

Yeah, I understood that's the generalized way to find orthogonal vectors in R^n.

Didn't think of it, tho.

Ok, I see what Merosity is saying.

[quartz compass]

wait I don't think I was saying that

[eternal finch]

Or, well, the formula?

[quartz compass]

what I've written is nothing more than the magnitude of the cross product of two vectors

[eternal finch]

Right.

[quartz compass]

but I only skimmed what was being written and wrote it as if it was dotted (to make a triple scalar product) with the cross product normalized

to get a determinant of 3 vectors

same thing, different perspective

[eternal finch]

Yeah, that's what I realized just now.

Didn't see it was a triple scalar product at first.

Because no parentheses => me confused.

[quartz compass]

well, the types force you to do it a certain way

$a \times b \cdot c$

[stoic python]

Merosity:

[eternal finch]

Right. You can't cross a vector with a scalar.

[quartz compass]

if they're all vectors, b dot c is a scalar crossed with a vector

yeah

well anyways, I don't know what the original question was, so not sure what's going on haha

[eternal finch]

beautiful
As my vector calculus teacher would say, isn't this booootifool?

well anyways, I don't know what the original question was, so not sure what's going on haha
It was find area of parallelogram in three-dee.

[quartz compass]

booti lol

[eternal finch]

"Using linear algebra".

By which asker meant "without cross products".

Are cross products linear algebra?

[quartz compass]

lol idk

[eternal finch]

Yeah, idk either.

[quartz compass]

to me they're determinant operators waiting to be dotted with a vector

to fill the last row

[eternal finch]

I think I saw a similar interpretation in 3blue1brown.

Interesting interpretation.

Meanwhile, me stick out hand and point out thumb ha ha.

[quartz compass]

idk as far as I know I just see it that way by how it's written

you expand along the i, j, k of fthe determinant to define the cross product, but if you dot a vector with it, you end up summing over that

and you can rewind it back a step to fill in the blanks of a full 3x3 determinant

[eternal finch]

Right.

[quartz compass]

I guess I see it more as an alternating antisymmetric thing more than anything

like you can do the same with n-1 vectors in n dim space

make a "cross product" of a single vector

in 2D

[radiant jasper]

Are quotient space and complementary space basically the same thing at a higher level?
For example:
Suppose some vector space V is a direct sum of subspace U and W. And v_1 + U, .. , v_m + U is a basis of V/U.
Then v_1, .. , v_m is a basis of W.

[sonic osprey]

Things get weird when you have infinite dimensional vector spaces that make this not true

[radiant jasper]

That's right, thanks!

[humble oak]

hello very quick question, if i have the determinants of 3 matrices, A B C, is the Determinant(ABC) = Det(A) * Det(B) * Det(C)?

[limber sierra]

this is the multiplicative rule of determinants, yes

det(AB)det(C) = det(ABC)

and det(AB) = det(A)det(B)

therefore

[humble oak]

Determinant(ABC) = Det(A) * Det(B) * Det(C) 😄

thanks

[elfin ingot]

quickie

is the determinant a linear functional on F^(nxn)

[sonic osprey]

no

[limber sierra]

linear functionals are necessarily linear.

det is multilinear in the rows of a matrix, but its not linear

[dusky epoch]

it's linear iff n=1

[wintry steppe]

If i have a plane of equation 5x-2y-2z=1

How can I find a normal vector of that plane ?

[dusky epoch]

look at the coefficients on x, y and z

(5, -2, -2)

and presto

[wintry steppe]

Thanks!

[sharp merlin]

How do i do this

[quartz compass]

one way is to put a vector with x,y,z in it and work out the matrix multiplication Av = 3v

then just solve for the components, whatever variables are left take the place of being arbitrary scalar multiples of your basis vectors

[wintry steppe]

https://cdn.discordapp.com/attachments/658293172128186368/716409368698028092/IMG_20200531_032451.jpg

Could someone check dis out?

[quartz compass]

,rotate

[quartz compass]

wrong channel should be in #prealg-and-algebra, but the reason is you are dividing by 0 when x=4

after you end up with |x-4| on the right side, you would still have to translate this back to information about 2/|x-4| > 1

since that's the original question being asked about

but 2 > 0 you can't divide by 0 to get 2/0 > 0/0

[wintry steppe]

But if the question is directly 2> |x-4| wouldn't it then make sense to take x not equal to 4 aswell?

[quartz compass]

in that case yeah

[wintry steppe]

Also sorry on that channel mistake and thank you for helping me out

[quartz compass]

I guess part of it is we're avoiding saying |1/0| is infinity, because infinity isn't a number

[wintry steppe]

Yeahh I understand that. I just wanted to know what happens if we shift it. In other question where the modulus is not in the denominator I think I included the x for it to be zero I guess I did wrong then :(

Anyways thanks again :D

[quartz compass]

yeah you're welcome

the main things to watch out for are usually the two things: When is the denominator 0? When are you square rooting a negative?

since in those cases you gotta exclude stuff

[wintry steppe]

I haven't really come across the square root kinda questions yet but will keep in mind thanks :D

[quartz compass]

yup you're welcome, if you have more questions I guess go to that other channel 👍

[wintry steppe]

Alright :D

[fickle tree]

I just want to know if I solved this the right way:
I know that A,B,C,D are points in the 3D space and that
A(1,1,1)
C(2,0,1)

I also know that L is the middle point of AB which is L(3,-1,3/2)

And M that is the middle point of CD
which M(5,1,-3/2)

I did the difference of L-A then x2 to find B and same with M and CD

[wintry steppe]

@fickle tree any chance you could increase the contrast on that image

[fresh acorn]

??

[dusky epoch]

?? indeed

what is this even meant to say

[wintry steppe]

??

[fresh acorn]

how can i solve ? i don't know your method

[dusky epoch]

before you ask "how can i solve this problem", you need the problem itself to be clearly stated

which yours isn't

[verbal bay]

LOL

[ocean sequoia]

why are two basis vectors given? Is there a reason?

wouldnt just(-2,1,0) be a basis of null psi(?)

is that just giving more examples?

[subtle walrus]

the dimension of a space is unique

so the basis cannot both have cardinality 1 and cardinality 2

and the book is correct

[ocean sequoia]

oh i never doubted the book is correct lol so (-2,1,0) is not a basis nor is (-3,0,1)? but both of them together?

[subtle walrus]

yes

[ocean sequoia]

because thats the only way you can span the whole space?

[subtle walrus]

its not the only way

but just 1 vector cannot span the whole space

[ocean sequoia]

i mean those span the space

yes need to be more precise with what i mean

working on that

[subtle walrus]

yes, they span the space

(and are linearly independent)

[ocean sequoia]

thanks!

[cobalt tartan]

If I have that Ax = b is consistent for some x for some fixed b, with A mxn, does this mean that [A | b] is consistent for every b in R^m?

[wintry steppe]

Not if any rows of A are dependent on other rows I don’t think

[wintry steppe]

hello there, im struggling to understand the definition of the characteristic... does anyone have any examples of when m wouldnt be equal to 0? (this is our def. appologies for it being in french)

[elfin ingot]

characteristic of what

ring?

[wintry steppe]

yes

[subtle walrus]

m is positive

0 is not

so m is not 0

big brain time

[wintry steppe]

ah, so like Z/7Z

[elfin ingot]

smallest number of times such that 1+1+1... = 0

[wintry steppe]

then m=7

[subtle walrus]

yes

[wintry steppe]

ah

[elfin ingot]

the char of an integral domain must be prime or 0

all fields are integral domains

[wintry steppe]

i mustve misread positive with strictly positive

thanks!

[limber sierra]

sorry this is

a linguistic thing

"positive" in english is generally called "strictly positive" in french

but sometimes

french sources use the english convention

which can get a bit confusing

[wintry steppe]

tbh i never know which is which in either language... i always assume "positive" means greater or equal to 0

[brittle juniper]

French positif is to be understood as a non strict inequality, English positive is usually to be understood as a strict inequality

I've seen someone here with a screenshot of their textbook where they used strictly positive for >0

mysterious things happen

[wintry steppe]

yes, its a bit confusing haha

Fn with adition is {0,1,2,...,n-1}
whereas Fn with multiplication is {1,2,...,n-1} right

here which are they asking for? Fn with multiplication or addition?

[sonic osprey]

I think you're getting confused here

F_n has both addition and multiplication on the elements {0,1,2,...,n-1}

[wintry steppe]

so we never ommit the zero?

[sonic osprey]

A field has two operations on it

addition and multiplication

along with a lot of axioms

[wintry steppe]

sous-espace vectoriel is vector space

the def. says it needs to be a group, in which case all elements needs an inverse

[sonic osprey]

Yes your vector space in this case is K^2

[wintry steppe]

in which case, 0 cant be part of the space, since nothing times zero is one

[sonic osprey]

which is an abelian group under addition

[wintry steppe]

as i understand

[sonic osprey]

However, you see that it also requires an action K x V -> V

and K is a field here

[wintry steppe]

yes so here we are dealing with Fn with addition, so {0,1}

the sub vector spaces are, 0,0 & 0,0;0,1 & 0,0;1,0 & 0,0;1,1

which is with the addition

but so does Fn with multiplication also have the element zero?

as in, just the vector space

[sonic osprey]

@brittle juniper you should probably just speak french to him

[wise harness]

oui le sev de Fn ca na pas l'element 0

tandis que Fn muni de l'addition, c'a l'element 0 comme tu la dit

[brittle juniper]

yay I can go to bed

[wise harness]

ici on a en effet affaire a Fn muni de l'addition

ils auraient surement ecrit Fn* pour signifier que c'est l'ensemble muni de la mult.

[wintry steppe]

ok merci

[little frigate]

vector w is (0,1)

so if we add all the vectors dw we get a vertical line i understand that,

but vector v is (1,0) and c is any integer so cv can be (-5,0) ... etc (0,8),

so basically points on the x axis

so can someone explain me how we get infinite parallel lines

no

can someone rather explain me how the sum of all vectors dw touches the points cv

is it because vectors are free?

for those who have gilbert strang book introduction to linear algebra it's the page 7

nvm i got it

[cold topaz]

is the answer to this problem correct?
Because I got 2x-2y, 3x+3y, not what is written here.
https://www.slader.com/textbook/9781118879160-elementary-linear-algebra-applications-version-11th-edition/464/exercises/11/

[limber sierra]

where did the 2y come from

can you show your work

[cold topaz]

I took T2, and replaced each component accordingly in T1.

so 2(x-y), 3(x+y)

[limber sierra]

what is T_1(x, y)

[cold topaz]

2x,3y

[limber sierra]

cool

what is T_2(2x, 3y)

[cold topaz]

that's my problem lol

[limber sierra]

T_2(x, y) = (x-y, x+y)

sooo

if instead of x, we have 2x

and if instead of y, we have 3y

T_2(2x, 3y) = ???

[cold topaz]

then we'll get the answer that's on slader

[limber sierra]

indeed

remember that function composition is

right-to-left

so in $T_2 \circ T_1$

[stoic python]

Namington:

[limber sierra]

you apply T_1 first

then T_2

simialrly, if it was $T_1 \circ T_2$, you'd apply $T_2$ first, then $T_1$

[stoic python]

Namington:

[limber sierra]

this might seem weird but its more natural if you think

$(T_2 \circ T_1)(x) = T_2(T_1(x))$

[stoic python]

Namington:

[cold topaz]

simialrly, if it was $T_1 \circ T_2$, you'd apply $T_2$ first, then $T_1$
@limber sierra what would be our solution in that case?

[stoic python]

sm:

[limber sierra]

in that case it'd be

T_1(x-y, x+y)

which is (2(x-y), 3(x+y))

as you said

[cold topaz]

oh

[limber sierra]

i.e. (2x-2y, 3x+3y)

so yeah, your problem is that you did the order wrong

function composition is right-to-left

apply the right function first

then the left function

[cold topaz]

it's annoying

so baisically, we take what we have in T2, and replace them accordingly with what we have in T1

[wooden forum]

So I'm looking at the linear transformation axioms and the second one looks completely redundant to me.

if f(a + b) = f(a) + f(b) then cf(a) = f(ca) by the very definition of multiplication, am I missing something here?

[wintry steppe]

take f(x) = x^3 for example

f(2*1) = 8 =/= 2 = 2 * f(1)

[wooden forum]

Yes but it's also not true for the first axiom

f(1 + 2) =/= f(1) + f(2)

is there any case where f(a + b) = f(a) + f(b) and cf(a) =/= f(ca)?

[subtle walrus]

yes

complex conjugation is what comes to mind

[wooden forum]

Oh

gonna have to look that up

[subtle walrus]

do you know what a complex number is?

the thing is over R there is no continuous example of such a map

[wooden forum]

yes

[subtle walrus]

ok so, consider a+bi a complex number

then complex conjugation is a function c, such that c(a+bi) = a-bi

[wooden forum]

Ohh

yeah I remember that

[subtle walrus]

you can confirm that this is closed under addition

but not scalar multiplication

[wooden forum]

because the scaler could be complex!

thank you

[subtle walrus]

on Q your argument works btw

so by extension it works for every continuous function on R

[fossil quest]

How can i show that there are linear transformations?

I know that additivity and homogeneity must be satisfied but idk what I'm supposed to do here

[desert ibex]

This is too big brain for me

[fossil quest]

What does v -> (v,0) mean? Or w -> (0,w)

[hybrid flint]

I'm guessing for any $v\in V$

[stoic python]

Neil_P:

[hybrid flint]

typically the element $a$ is implies to be in the set $A$

[stoic python]

Neil_P:

[hybrid flint]

@fossil quest Show that the (1) zero vector is included in the transformation, (2) the additive property (3) the constant multiplicative property

[fossil quest]

I dont know like where to start if I look at it :/ I kind of know what to do but not how to do

[hybrid flint]

Lol will my tex compile?

[fossil quest]

I'm very unfamiliar with these notations

[stoic python]

Neil_P:

Neil_P:

[fossil quest]

Oh and it's already defined that V and W are F-Vectirspaces

[hybrid flint]

afaik, This is all that's necessary to show it's a linear transformation.

And I'm a little out of my depth here

[fossil quest]

Ok but what's u and v in this case?

[hybrid flint]

That would be any element within the domain of V and W

[fossil quest]

u in V and v in W ? Or does that not matter

[hybrid flint]

Are i and j the mappings in question?

What are the definition of V, W?

[fossil quest]

V and W are F Vector spaces

Idk about I and j

[stoic python]

Neil_P:

Neil_P:

Neil_P:

[fossil quest]

It at least sounds easy lol

[hybrid flint]

Yeah, the tricky part is knowing what to prove

[stoic python]

Neil_P:

[hybrid flint]

My linear is a little rusty, sorry

[fossil quest]

Ok

Im not sure what you mean but (u + v ,0 ) ?

[stoic python]

slimvesus:

[fossil quest]

Yes

I dont know if i fully understand it but i remember that its a direct sum when adding 2 vector spaces and their intersection isnt 0

Oh ops

Yeah I dont think I got a clue

Sorry :/

(u,0) + (v,0) ?

(u+v,0)

Yeah wait that was quick

Wow ok that was simpler than I thought

Yeah now there's homogeneity

[hybrid flint]

@wintry steppe Was your class proof based?

Mine was only computational

I only know the outline of the proof from other classes

[fossil quest]

ci(u) = c(i,0) = (ci,0) = i(c*u) ?

Sorry for interrupting haha

Wow thanks that wasnt too hard! I hope my prof could also explain it just this easily without using fancy words and sentences

Ok I'll look it up

[hybrid flint]

https://link.springer.com/book/10.1007/978-3-319-11080-6

[fossil quest]

Lol wow thanks haha I just looked it up and it says 24 $

[hybrid flint]

No need for a sketchy intermediary lol

[wintry steppe]

Where did they get the

1 0
0 1

from?

[dusky epoch]

that's the identity matrix

[wintry steppe]

oh, so they are just using like

1 000
0100
0010
0001

like that kinda?

[dusky epoch]

i mean

[storm python]

yes, an identity matrix is an nxn matrix with 1 down the top left diagonal, 0 elsewhere

[dusky epoch]

they wrote down exactly what they did

i.e.

solve the matrix equation AX = I

I here refers to the identity matrix

and the identity matrix of size n is the n by n matrix which has 1s along the diagonal and 0s everywhere else

that's the defn

[sick robin]

So if "positive linear transformations" are self adjoint, and a normal linear transformation is positive if and only if each eigenvalue is non negative, wouldn't that imply all positive eigenvalue'd normal linear transformations are self adjoint?

Or is that what this proof is trying to guide me towards

Thats what I was thinking

A iff B B iff C, thus A iff C

Right

I'd need to double check the definition, but I don't recal self-adjoint matrices requiring the inner product to be positive

but again this may be a side effect of this proof

[sick robin]

And that corresponds to what I just refreshed on in the book

Self-Adjoint matrices must have positive eigenvalues, but the inner product of any x with some A isn't necessarily positive

So eigenvalue positive normal transforms are then logically self adjoint

Ope

no its real eigenvalues

Silly mistake

my b

Yes

That makes sense

Now onto this proof

Well -identity would work right?

tbh even just the identity

[outer tulip]

possible stupid question. Let $\mathbf x, \mathbf y \in \mathbb R^n$ and $A$ an $n \times n$ matrix. Notes that I'm looking at boil down to $${\mathbf x}^T A^T A \mathbf y = {\mathbf x}^T \mathbf y$$ iff $A$ is orthogonal. If $A$ is orthogonal, it's obvious but the converse I'm drawing a blank on

[stoic python]

George!:

[outer tulip]

(this is in proving that $(A \mathbf x, A \mathbf y) = (\mathbf x, \mathbf y)$ iff $A$ is orthogonal so we can't use that property here)

[stoic python]

George!:

[limber sierra]

suppose for contrapositive that A is not orthogonal, ie A^T A = B where B is not the identity matrix

then the equality isnt true; suppose for example that x and y are full of 1s

[outer tulip]

oh so it is pretty straightforward, was just wondering since the notes just said "iff A orthogonal" in passing

thanks!

[normal quiver]

whats an example of a transformation with a domain R^3 and codomain R^2?

[brittle juniper]

$$\fun f{\bbR^3}{\bbR^2}{(x,y,z)}{(0,0)}$$

[stoic python]

Tuong:

[steel bane]

k

[normal quiver]

how do you tell if something is a matrix transformation?
I guess for example T(x1,x2) = (3x2,x1,3)

[verbal bay]

lets say u transform a rectangle into another rectangle with an affine transformation, is that transformation unique?

[limber sierra]

a transformation can be represented by a matrix if and only if it is linear

so you check linearity

[verbal bay]

check image, kernel and linearity of the transformation @normal quiver

T(lambda.A + P) = lambda.T(a) + T(P)

[limber sierra]

why are you looking at the image and kernel?

[verbal bay]

cause for my exam i was told to always check linearity, whether the domainai....

nvm

thats checking whether

T is an endomorphism on V

then checkin that the domain of T = v and the image of V is a subgroup T

my bad, i got those mixed up

well continueing on my question, id say yes cause lets say u use other transformations M1,M2,M3 to transform the rectangle, the combination is the same in the end

so that there is really only 1 unique transformation, cause all the other 1s are combinations of other transformations that lead to the same thing

but not sure if im wrong

Am i supposed to @ helpers here as well after 15 mins ? Not familiar

[normal quiver]

@verbal bay so just check linearity?

[verbal bay]

Ye

[wintry steppe]

If $ U $ is a subgroup of $ V $ and $ a, b \in V $ how is $ a + U = b + U \Leftrightarrow (a-b) \in U $ ?

[stoic python]

milos:

[sonic osprey]

do you agree that a + U = b + U

implies that (a - b) + U = 0 + U?

[wintry steppe]

I'd want that

But hmm

[sonic osprey]

How exactly have you defined a + U?

[verbal bay]

Did i ask my question in a weird way?

[wintry steppe]

a + U is just a left coset

$ a + U = { a + u | u \in U } $

[stoic python]

milos:

[sonic osprey]

okay then it's easier to just say that

since a + U = b + U

one of the elements on the left side is a + 0 = a since 0 \in U

so this means that a \in b + U

which means that a = b + u for some u \in U

which gives you that a - b = u

[wintry steppe]

cooool

thanks

[normal quiver]

if a arbitrary matrix X multiplies a non-zero matrix, and the result matrix is all 0s, the arbitrary matrix would have to be singular because the determinant for that matrix should be 0

[sonic osprey]

Uh

But the other matrix could have determinant 0?

[gilded plume]

How would I go about this question?

Using the example of a 3x3 matrix, show the equivalence of the following methods for finding inverse matrices:
a.) Row operations b.) Elementary matrices

Im not too sure exactly what it is asking

[limber sierra]

do you know the "row operation" method of finding inverse matrices?

do you know the "elementary matrices" method of finding the inverse?

prove that they're equivalent; that is, whenever you're doing one, you're "basically just doing the other"

specifically in the 3x3 case

(hint: note that multiplication by elementary matrices corresponds to row operations! and that they're invertible)

[latent marten]

can someone explain why the answer is d

[quasi vale]

Can you find the direction vector of the given line?

Turn it from parametric form to vector form and you'll notice the direction vector straight away, or you can still notice it from here.

Then you can do dot products of each vector with the direction vector(doesn't take long) and see which dot product results to 0.

@latent marten

[latent marten]

thx @quasi vale

[radiant jasper]

Are there some other inner product on R^2 other than Euclidean inner product?
Is it possible to make (1,0) and (0,1) not orthogonal? <- Guess not cause <x,y> = 1/4 * ( <(x+y),(x+y)> - <(x-y),(x-y)> )

[subtle walrus]

you can define one for every symmetric positive definite matrix as <x,y> := x^TAy

[radiant jasper]

I haven't learned what positive definite matrix is but thanks anyway

[verbal bay]

lets say u have an endomorphism on R[x]6 so polynomials of max degree 6

is it possible kernel(T) = image(T)?

nvm lemme rephrase: is there an endomorphism on polynomials of max degree 6 for which kernel = image

id say no but idk if my method is right

cuz dim(image(T)) + dim(kernel(T)) = n

and n would be 7

so dim(image) = 7/2 = dim(kernel)

[eager burrow]

Yup sounds right

[verbal bay]

which is impossible cuz its a fraction?

Ok

wait u didnt confirm the rnd

end&

[eager burrow]

i saw where it was headed, lol

so yeah, there is no endomorphism with kernel = image on an odd-dimensional space

[dusky epoch]

dim(image) = 7/2

[verbal bay]

dimension(image(t)) = 7/2

[dusky epoch]

how the fuck can the dimension of any linear-algebraic object be 7/2

7/2 is not an integer

[eager burrow]

that is exactly what we have been establishing

maybe read the question again

[verbal bay]

thanks im so happy means i got it right on the exam

cheers

[shrewd saddle]

Can someone help me with this

I know that ||A||_2 = sqrt(l_max(A^T A)) but how does that become sigma_1

[eager burrow]

What's your definition of singular values?

[nimble raft]

svd?

Hevipelle seems afk

Can I ask a easier linalg qs?

[shrewd saddle]

That's the whole assignment there

I don't know more

[eager burrow]

Do you have some lecture notes or a book where "singular values" was defined? That should be the first thing to look into when you want to tackle an exercise with terms you don't understand

[shrewd saddle]

There's just some stuff about singular value decomposition

[eager burrow]

that sounds like it should be it, yes

[shrewd saddle]

So how does sqrt(l_max(A^T A)) = sigma_1

[fresh acorn]

Solution B

is my solution right ?

is there anyone to help

[nimble raft]

Is [AB] = AB * BA?

Oh wait thats a comma?

[A,B] = A*B - B*A?

If thats true then:
[kA + B, c] = (kA + B)c - c(kA + B) (Expanded weird operators)
= kAc + Bc - ckA - cB (Expanded brackets)
= kAc - ckA + Bc - cB (Associativity)
= k(Ac - cA) + Bc - cB (idk if this is legal, but factorize k?)
= k[A,c] + [B, c]

I'm not entirely sure what we're dealing with here though

Are these scalars? square matrices? idk.

[fresh acorn]

@nimble raft ty

[subtle walrus]

[A, B] is the (ring-theoretic) commutator of the square matrices A and B

[nimble raft]

Ring theoretic commentator, never herd of it

Square matrices make sense, seemed like only one that'd work

Was gonna ask some stuff on elementry linalg but frgot

[olive halo]

Hello, does anyone know the proof for the following claim (or know of a good reference for this)?
I'm talking about the direction that if <Av,v> is always real, then A is Hermitian (that is, A*=A)- the other direction is easier

[olive halo]

Never mind, I managed to solve it using the polarization identity 🙃

[last siren]

How to prove that the following inequality is always true for all values x. \
$ x^4 - x^2 - 2x + 3 > 0 $

[stoic python]

Niko:

[split heart]

to prove A is similar to B I should do: A = P^-1 B P?

[wintry steppe]

@last siren maybe you can find its minimum value and show it's >0

not linear algebra i think though

@split heart that's the definition of similarity yes, just exhibit some invertible matrix P for which that's true

[split heart]

And how can i find that P matrix?

I have A and B defined

Not P

or P^-1

[wintry steppe]

i don't know, you have to be more specific

[last siren]

@wintry steppe You mean show that the f(f'(x) = 0) > 0?

[wintry steppe]

for example, the matrices representing linear maps in different bases are similar

[split heart]

So, If I only have A nxn and B nxn I can't find P?

[dusky epoch]

you will have to be more specific

what do you know about A and B besides their being n by n matrices?

[split heart]

Their values

[dusky epoch]

oh so you're actually given two matrices

[split heart]

Yes

[wintry steppe]

@last siren something like that, you should have some "optimization theorems" that you might be able to apply (disclaimer i haven't worked the details, but that's the first thing i would try)

[split heart]

I obtained them from a previous exercise

[dusky epoch]

a good first step would be to compute their characteristic polynomials

and then, if necessary, the dimensions of their eigenspaces

they should match up

it'd be really helpful if you showed us the whole exercise exactly as stated though

[split heart]

Sure

I named both matrix T but I have to prove they are similar

I can see they're basis change matrix right?

a) is from canonic in domain a codomain

and b) B in domain and codomain

[dusky epoch]

well

ok

so your two matrices are actually matrices of the same transformation in different bases

so that's equivalent to similarity

[split heart]

Yes

I have to use A = P^-1 B P to prove?

[last siren]

@split heart You can use https://mathpix.com for scanning hand-written equations.

[split heart]

I'll check it out, ty

[wintry steppe]

pretty sure you use this result

it basically follows immediately

[toxic wraith]

Thank you so much!

[gritty frigate]

what is that symbol name

[limber sierra]

capital pi [assuming you're referring to the thing on the left]

it represents repeated product

like how capital sigma represents repeated sum

in latex it's $\prod_{i=1}^n a_{i, \sigma_i}$

[stoic python]

Namington:

[limber sierra]

the circle thing in the subscript is a lowercase sigma

if that's what you mean

[gritty frigate]

Ohhh it is like sigma but for products

Thanks !

[limber sierra]

yep.

[gritty frigate]

Do you guys have any explanation of the definition of determinants

[gritty frigate]

I understood the definition...

Where can I find the proof

Why is that happening

[subtle walrus]

the proof of what?

[eternal finch]

I think he means motivation.

I understood the definition...
Also, what is your definition of determinant?

[limber sierra]

based on his previous question, the product-of-cycles one

[gritty frigate]

The definition based of Cofactors

[fast mantle]

Hey! I've a question on pauli matrices and a specific notation I don't understand. Can I post it here? It's not for a class or anything

[zinc tapir]

Any hints

[fast mantle]

how do I evaluate the right side?

[wintry steppe]

can someone teach me how to do this if the det of a is 2020 and figure out det of b?

[torn hornet]

what did you try

[wintry steppe]

are u talking to me?

[torn hornet]

yes

[wintry steppe]

oh so i tried to solve it by case by case if that makes sense

like i put like each of them in different sets assuming the answer is this for each set

but i dont think its right

[torn hornet]

umm i am not sure what that means

[wintry steppe]

yeah i dont know what i was doing either hence why I am asking for help

[torn hornet]

ok what ideas do you have

[wintry steppe]

well that was one idea that i had which didn't make sense

the other i dont know if im being honest with you

[torn hornet]

ok so you know the determninant of the 2x2 right

and you want to know that of the 3x3

do you know anyway to compute 3x3 determinants

that rely on 2x2 ones

[wintry steppe]

uh i think

so

well wouldn't it rely on it anyway cause its using the same values and 2x2 is already in the 3x3

[torn hornet]

i mean just for general determinants

do you know how a nxn determinant can be reduced to (n-1)x(n-1) determinants

[wintry steppe]

yea

[torn hornet]

right so use that idea and see if you can figure out the problem

[wintry steppe]

yeah im already on that step

idk what to do after

[torn hornet]

think about it a bit

[wintry steppe]

nope dont know

[pallid rampart]

try expansion by minors on a specific row

[wintry steppe]

okay i figured it out tyy

and one more thing if a matrix's has characteristic polymonial written

to find the eigenvalues of it you just find the root right?

for example like this

[gray dust]

roots

[wintry steppe]

so it is?

roots of it are its eigenvalues? and how do u know what the size of the matrix is

[sonic osprey]

the size of the matrix is the degree of the characteristic polyomial

[wintry steppe]

oh ok so its a 4x4?

[gray dust]

roots of it are its eigenvalues?
yes
oh ok so its a 4x4?
yes

[wintry steppe]

just to confirm, if i were to inverse that matrix the characteristic polynomial will be the same right?

[gray dust]

what makes you think that?

[wintry steppe]

i think i read that somewhere in the textbook that no matter if u rearrange the values it will the same? idk might be on a different topic ig

[cyan linden]

can someone tell me if this is true and how to prove it? so If that every eigenvalue of a matrix A has algebraic multiplicity 1, then A is diagonalizable.

[sonic osprey]

What do you think?

[cyan linden]

i think its true i just dont kno how to prove it

[sonic osprey]

well, what does diagonalizable mean in terms of geometric multiplicity?

[cyan linden]

i have no idea

[sonic osprey]

you know what geometric multiplicity means right

[cyan linden]

yea

[sonic osprey]

basically a matrix is diagonalizable if the sum of the geometric multiplicities is n

the size of your matrix

[zinc tapir]

can someone explain how they went from line 1 to 3

[cyan linden]

ohhh okok got it nvm so the statement does not stand then

[zinc tapir]

shouldn't it be (x1,x2,-2x1+7x2)

[sonic osprey]

uh wait

why don't you think it stands

[zinc tapir]

any help

already checked if the zero vector is in W_3

[eternal finch]

You can isolate a_3 in 2a_1 - 7a_2 + a_3 = 0.

[zinc tapir]

yes

but shouldnt that be -2a1+7a2=a3

[eternal finch]

Oh, I see.

Yeah, should be.

[zinc tapir]

then line 3 is incorrect perhaps

[eternal finch]

shouldn't it be (x1,x2,-2x1+7x2)
Sorry, didn't see this, lol

[zinc tapir]

how would i check if this is a subspace in R3

W6={(a1,a2,a3)∈R3:5a_1^2−3a_2^2+6a_3^2=0}

i can't seem to put them in coords

like the previous

[zinc tapir]

<@&286206848099549185> 😒

i checked a) already

ah nvm

im dummy

[wintry steppe]

how do they mean when they say

"the subspace spaned by vector u"

[cursive narwhal]

It's the set of all linear combinations of the vector u, where you take elements from a given field

@wintry steppe

[shy atlas]

@cursive narwhal ur mom so fat, no finite set of vectors can be a basis for her

prank

chill

memes

shhh

[shy atlas]

@cursive narwhal vats where u at

[cursive narwhal]

oh shit rekt

[shy atlas]

[cursive narwhal]

I took a combinatorics class to count the number of ways I could do your mom

prank dude prank

don't worry about it

[shy atlas]

[zinc tapir]

I'm confused. Shouldn't this be a basis since x1 x2 x3 are linearly indep ?

[dire thunder]

where exactly

are u confused

@zinc tapir

[zinc tapir]

im trying to show that this set is linearly dep

[dire thunder]

this one?

[zinc tapir]

yea

not sure if i wasn't supposed to reduce the matrix to find if the coeff are zero

i see most ppl just use algebra to solve for a1 a2 a3 but i thought i could make it easier with reducing

maybe i messed up somewhere or something

[dire thunder]

calc give such rref

which shows that it is dep

[zinc tapir]

i can see that yeah

maybe i did mess up

how do i go from wht i have t othat rref

[dire thunder]

wtf

i also get some strange rref

ah no

messed calculation

how do i go from wht i have t othat rref
just proceed algorithm

[zinc tapir]

uhh

still not seeing it

2R2+R1=R1 yields 1 0 6 for the first row

[dire thunder]

what

check your correctedness of RREF algo

[zinc tapir]

not sure what u mean

[dire thunder]

nvm

[zinc tapir]

im attempting to get all zeros in the bottom

[dire thunder]

i thought it was your firs operation

[zinc tapir]

i have a 1 stuck there

[dire thunder]

i have
{-1, 2, 1}, {0, -1. 3} {0. 2, 0} one step befroe REF

[zinc tapir]

so

i messed up is what ur saying

could u check my process maybe pls

its hard to follow

from what ur saying

[dire thunder]

you lost - near 2 on the bottom

@zinc tapir

[zinc tapir]

ty i see now

i didn't copy down -2x^2a_2

i made it positive i guess

going to fix this

[misty basin]

Hi im new here, is this the right place to ask questions if i have questions about Lp spaces, p-norms and distance metrics?

[dire thunder]

#point-set-topology here maybe better

[misty basin]

i see, thanks!

[sonic osprey]

@misty basin I'll just answer your question here because thsi is probably more fitting

"which one to use" is kind of a weird question to ask

like in reality, you'll be presented with situations where different metrics make sense

like in quantum mechanics, L^2 spaces are always used

[misty basin]

i see, that makes much more sense >.< thanks!

cus i came to know it from computer science so i was kind of expecting like values i can just plug in like with most of the other stuff

[nimble raft]

smelp

I was tryin do some basic linealg and i have frogotton how to do this

$$\begin{pmatrix}1&-2&1\ 3&a&4\end{pmatrix}$$

[stoic python]

Soap:

[nimble raft]

What do I set "a", to be so I have no solutions and what do I set it so I have solutions?

I thought that if I make the matrix "dependent", I would have no solution.

And so my thought was:

$$\begin{pmatrix}1\ 3:\end{pmatrix}+\begin{pmatrix}1\ :4\end{pmatrix}=\begin{pmatrix}2\ :7\end{pmatrix}$$

[stoic python]

Soap:

[nimble raft]

So might as well just set "a" to -7, tada its not independent now.

But now I realize, dependent matrices are still solvable, and I got it mixed up with invertability.

[dire thunder]

a=-6

btw

[nimble raft]

Is that the only one?

It's cuz we multipled 3*-2 right?

[dire thunder]

,w RREF {{1, -2}, {3, a}

[stoic python]

Display results online and refine query

Requested by Commander Vimes#7532

[nimble raft]

hmm

where did the "a" go lol

[dire thunder]

,w RREF {{1, -2}, {3, -6}

[stoic python]

Display results online and refine query

Requested by Commander Vimes#7532

[nimble raft]

damn what?

[dire thunder]

,w RREF {{1, -2}, {3, 6+a}, a > 0

[nimble raft]

damn everythangs reducable

[stoic python]

Display results online and refine query

Requested by Commander Vimes#7532

[nimble raft]

,w RREF \begin{pmatrix}1&-2&1\ 3&-6&4\end{pmatrix}

lol rip

Oh its the determinant?

[dire thunder]

,w det {{1, -2}, {3, a}}

[nimble raft]

1*a - 2*3?

[stoic python]

Display results online and refine query

Requested by Commander Vimes#7532

[nimble raft]

So why is the determinant telling me what "a" needs to not have a solution?

[dire thunder]

well obviously when a = -6, second row is 2 times first row

but 7 is not 2 times 4

and you get 0 0 1

[nimble raft]

Yeah and it breaks it, so -6 part seems understandable

But how come determinant tells u that?

I don't see the link between determinant and solvability

Or is it just a coincidence

I guess we could test it out

[dire thunder]

Ax=b has a solution for every b <-> A is invertible

[nimble raft]

Ohhh

So i dungoofed it when I used the "b" column row, to make the matrix dependant

Instead I should've just looked at matrix "A", and not the "b" part

[dusky epoch]

Ax=b has a solution for every b <-> A is invertible
only under the assumption that A is square

[dire thunder]

well we have square matrix here

like invertible is for square matrices

[nimble raft]

It looked rectangular to me tbh

But thats cuz i was also including "b", bit

[dire thunder]

last column is vector of constants

[nimble raft]

Yea rip me..

So does this apply for all square matrices?

Including the determinant trick?

[dire thunder]

yes

[nimble raft]

Well for this one:

$$\begin{pmatrix}1&2&3\ ::4&5&6\ ::7&8&a\end{pmatrix}$$

[stoic python]

Soap:

[nimble raft]

det(A) = -3a + 27

So would setting "a" to -27? then scaling everything by -3 make it unsolvable?

[dusky epoch]

why -27 tho

108 ≠ 0

[nimble raft]

I was just going off since last time it was "a" + 6, and "a" = -6,

Tho I don't think its correct

Assuming, if the constants are unique...

[dusky epoch]

bruh

do you really not know how to solve the equation -3a + 27 = 0 for a

[nimble raft]

oh

lol

I dungoofed again

[wintry steppe]

Hi find the minimum of the function

I don't know how but it's linear algebra

[brittle juniper]

Think about orthogonal projections

[wintry steppe]

I'd consider the vector space (X,X^2)?

And the scalar product is the integral above ?

[haughty lagoon]

well your vector space could be the space of continuous functions on [0,pi], since sin(x) is not in R_2[x]. Then, you can write the space of continuous C[0,pi] = R_2[x] + R_2[x] orthogonal

[dusky epoch]

you'd consider, perhaps, the space L^2[0,pi] with inner product (f,g) |-> int[0,pi] f(x)g(x) dx

[wintry steppe]

Yeah thanks guys but I'd have to orthonormize the X,x^2 with the gram-shmidt process

And that's exhausting

I think I'll go with a system

(Sinx-(ax^2+bx)|x^2)=0

And (sinx-(ax^2+bx)|x)=0

[wintry steppe]

hello there, we have this problem (sorry for it being in french), just i dont get how only three vectors would be sufficient to make for a base in C3

how i see it, it should be (1,0,0) (i,0,0) (0,1,0) (0,i,0) (0,0,1) (0,0,i)

[dire thunder]

e_1 = [1 0 0]

[dusky epoch]

{(1,0,0), (i,0,0)} est linéairement dépendant

ton espace est un espace vectoriel sur C, pas sur R

[wintry steppe]

ah donc c un K espace vectoriel ou K c'est C

et si K cetait R

eh ben jaurais raison c ca?

[dusky epoch]

oui, si K était R, ton ensemble de 6 vecteurs serait une base pour V

[wintry steppe]

ok nice

autre question, quand on parle de C[t] la, t cest un element de C? ou cest genre un vecteur quelconque auquel il y a trois degres?

[dusky epoch]

t c'est t

$\bC[t]_{\leq 3}$ c'est l'espace des polynômes complexes en $t$ à degré au plus 3

[stoic python]

Ann:

[wintry steppe]

ah daccord

[nimble raft]

wack!

ann can speak french!

most impressive thing ive seen someone do on the server

[shy atlas]

she can speak russian too

[nimble raft]

hmm, idk which is cooler russian or french

tho it was pretty cool of her to answer one of those "non-english" questions with the language it was written in.

i usually have to use google translate

[shy atlas]

i trust her more than google

[dusky epoch]

russian is my native language

i'm at best passable in french but not much more

[nimble raft]

i only know blyat

and bonjour

[shy atlas]

ann taught me, cest quoi ce bordel

[dusky epoch]

ce bordel*

[shy atlas]

yeah that fancy e

[dusky epoch]

no

[shy atlas]

oo ce

[dusky epoch]

"what is this mess", if you translate literally

[shy atlas]

ive been saying it all wrong then

[nimble raft]

wait so ann u can speak 3 languages!!

;O

[shy atlas]

unbelievable

[dusky epoch]

more like two and a half

[shy atlas]

how did i read it wrong

[nimble raft]

tho being able to answer a math question is that language is enuf

[dire thunder]

russian is my native language
Ты что-то сказал(а), товарищ?

Русские вперед

[dusky epoch]

о боже

[dire thunder]

царя храни?

[dusky epoch]

можно пожалуйста без излишнего патриотизма?

только мне еще этого не хватало

[dire thunder]

патриотические вперед

[dusky epoch]

ну нахер

[dire thunder]

ладно, в любом случае не место, в лучшем случае #chill

[dusky epoch]

как раз хотела об этом упомянуть

[gilded plume]

Using the Leontief Model, If i have a 3x3 Economy Model and only 2 Out of the 3 sectors are profitable would the Economy still be productive?

[wooden forum]

the lowercase a on the top right is a typo right? lol

[dusky epoch]

yes

[misty sparrow]

(I know, off topic) but this is literally me 😳

[half ice]

You are Ann?

[dusky epoch]

TIL i have an alt

[storm python]

[wintry steppe]

Ann is the alt tbh

[zinc tapir]

Is my argument sound

[elfin ingot]

whats F(S,F)

@zinc tapir

but yea ig ur right

[zinc tapir]

second line i meant g(s)=0 btw

[elfin ingot]

yea

[wintry steppe]

why is $\lambda$ also nonzero?

[stoic python]

Alphonse [Maxwell]:

[karmic oracle]

because A^2 being the zero matrix means that A^2x is the zero vector

[wintry steppe]

but how does that say anything about $\lambda$ ? I get that $\vec{x}$ must be nonzero (definition), but where does that imply that $lambda$ must be nonzero?

[stoic python]

Alphonse [Maxwell]:

[karmic oracle]

because if lambda isn't 0, and x isn't zero, then lambda^2 x isn't zero.

[wintry steppe]

ah yes

now I get it

thanks ! 😄

[lunar cape]

guys, how do I prove that U = {A ∈ Mn(R) : A^t(transpose matrix) = −A} is a subspace of Mn(R)?

[torn silo]

how do you show generally that something is a subspace of something else?

[wintry steppe]

https://www.youtube.com/watch?time_continue=11&v=Eawc_ZuQI_8&feature=emb_logo

something like this?

[lunar cape]

how do i prove step 2 and 3 of this video tho? when i have a matrix with n dimensions?

[subtle walrus]

what is step 2 and 3?

oh

closed with respect to sum and closed with respect to scalar multiplication

you just take two arbitrary matrices from your set, add them and show that the sum is still in this set

that's step 2, similarly for step 3

[lunar cape]

okay, i will try

thanks

[zinc tapir]

any guidance on this proof

[eternal finch]

How'd you start?

[zinc tapir]

i dont know how to start is the thing xd

[eternal finch]

How would you prove A is a subset of B in general?

[zinc tapir]

well all the elements of A have to be in B

[eternal finch]

Right. To show that is true, you'd pick out an arbitrary element of A and show that it is also a member of B.

So, here, you do the same thing.

[zinc tapir]

so you are saying

the linear combinations of S1 have to be in S2

[slow scroll]

vectors in span(S1) have to be in span(S2)

[eternal finch]

the linear combinations of S1 have to be in S2
The linear combinations of S_1 have to be in the span of S_2.

[zinc tapir]

how would i say that

x1,x2,...,xn is in span(S_2)

[slow scroll]

what are x1, x2, ..., xn supposed to be

[zinc tapir]

there are elements in S1

[slow scroll]

well, they are vectors, and you need to show that for any arbitrary v in span(S1) that v in span(S2). So let v in Span(S1). Precisely speaking, what does it mean for v in Span(S1)?

[zinc tapir]

for a vector to be in span(S1) it has to be linear combination of vectors in S1, so if we let V=span(S1), can we say that there exists some linear combination vectors in span(S1) and since S1 is a subset of S2 those linear combinations of vectors in span(S1) are elements of S2 and therefore also exist in the span(S2)

[slow scroll]

i mean, yeah that's pretty much the idea. idk how precise you have to be. You have v = c1v1 + c2v2 + ... + cnvn where vi \in S1 for each 1<=i<=n. So what would be the corresponding linear combination of vectors representing v from S2?

[zinc tapir]

wouldn't it just be the same combination

since S1 is a subset of S2

[slow scroll]

sort of. v = v = c1v1 + c2v2 + ... + cnvn is in Span(S2) if it is a linear combination of the vectors in S2. i.e. v = c1v1 + c2v2 + ... + cnvn + .... + crvr. for vectors in vi in S2. You need to assign values for scalars ci so that you still have the same vector v in the end. So how do you handle the vectors which are in S2, but not in S1? (hint: don't overthink).

[zinc tapir]

i'm not quite sure, i can vaguely see how those combinations are an element of S1, and since S1 is a subset of S2 those same elements have to be an element of S2

are we supposed to say that c1,c2,...,cn is an element of R , and that we are only sure that x1,x2,...,xn are elements of S2 but not so sure of the scalars in S2?

[slow scroll]

well, like you said, we want essentially the "same" linear combination. How can we choose the scalars in such a way where we "throw out" the extra vectors from S2 we don't need?

[zinc tapir]

since c1x2+c2x2+...+cnxn is in S1 for sure

[slow scroll]

Basically, i am asking you if we have
v = c1v1 + c2v2 + ... + cnvn with vectors vi from S1

how can we write v as
v = c1v1 + c2v2 + ... + cnvn + .... + crvr
where we use all of the vectors in S2 we are given, so that it is truly a linear combination of vectors in S2, and not subset of S2?

[zinc tapir]

scratches head

where did crvr come from

[slow scroll]

i am indexing the vectors in S1 from 1 to n, and the vectors in S2 from 1 to r

Answer: ||v = c1v1 + c2v2 + ... + cnvn = c1v1 + c2v2 + ... + cnvn + .... + crvr where any of the scalars coming after cn are 0 ||

[wintry steppe]

you could even do a contradiction

Like suppose theres a vector in Span(S1) that is not in Span(S2)

[slow scroll]

thats a fantastic question. I assumed they were finite sets.

[wintry steppe]

yeah the infinite case would be difficult to solve

especially because the span could be uncountable as well

[zinc tapir]

so v=c1v1 + c2v2 + ... + cnvn + .... + crvr is in S2?

is that what u meant

[wintry steppe]

i think kxrider's proof holds up still though

just create a function s that returns one when a spanning vector is in S1 and zero when its not in S1

[slow scroll]

yes Chaf, and you take everything you don't need to be 0. i.e. c(n+1) = c(n+2) = ... = cr = 0

[wintry steppe]

and multiply it by all the constants

that way you don't need to index or anything

[eternal finch]

wouldn't it just be the same combination

Y'know, this is what I was thinking. I think just boils down to what your definition of linear combination is. The one I had in mind was this.

A linear combination of vectors in a set S is an expression a_1 s_1 + ... + a_m s_m, where the a's are in your field and the s's are in S.

[zinc tapir]

well yeah

didnt i mention that earlier i think

but i said x was in reals

is the real line considered a field

[wintry steppe]

yeah

with the typical operations of addition and multiplication

[eternal finch]

I was more talking about the fact that the definition I gave means a linear combination need not involve all the vectors in the set.

It's just a small thing.

[zinc tapir]

what do you mean by that

so you dont need to include the vectors in S2 that are missing in S1

[eternal finch]

Well, kxrider was talking about if you had a linear combination of S1, what linear combination of S2 corresponds to it?

If your definition requires you to use all of the vectors in S2, then you'd need to include all of them.

And that means you'd set the coefficients to 0 for the stuff you don't use.

[zinc tapir]

ok so basically you are saying your definition allows you to not have to include them

[eternal finch]

Yeah, pretty much. I mean, they're pretty much the same thing, anyways.

Like, I guess one benefit of using a definition where you don't have to use all of the vectors is that the definition admits linear combinations of an infinite set, as you guys were talking about earlier.

[zinc tapir]

Let S1 be a subset of S2, s be an element of span(S1), a be an element of R which implies that span(S1) is a linear combination expressed as s=a_1s_1+a_2_s_2+...+a_ns_n. Since S1 is a subset of S2, s_1+s_2+...+s_n are also elements of S2. Therefore the span of those vectors is a vector in span(S2). Therefore s=a_1s_1+a_2_s_2+...+a_ns_n is an element of span(S2) which implies that s is an element of span(S2). So the span(S1) is a subset of span(S2).

just realiuzed i messed up pretty bad

but i fixed it i think

well

this is like my second proof ever lol

so yeah im kinda ass at it

[eternal finch]

Let S1 be a subset of S2,
Ok.

x be an element of span(S1),
Ok.

a be an element of R
Just because a is an element of R doesn't mean that a_1, a_2, ..., a_n are. I get what you're trying to say. I'll give an alternative phrasing below.

which implies that span(S1) is a linear combination expressed as s=a_1s_1+a_2s_2+...+a_ns_n.
Don't say that span(S1) is a linear combination. s is not span(S1). Also, what the x's are is never stated.

You should say that an arbitrary element s of span(S1) can be expressed as s=a_1s_1+a_2s_2+...+a_ns_n, where the a's are in R and the s's are in S1.

Since S1 is a subset of S2, s_1+s_2+...+s_n are also elements of S2.
Well, s_1+s_2+...+s_n isn't necessarily, but s_1, s_2, ..., s_n are. You also say elements when you only state one element, so I'm guessing you meant commas in the first place.

Therefore the span of those vectors is a vector in span(S2).
The span of those vectors isn't a vector in span(S2). Any linear combination of those vectors is.

Therefore s=a_1s_1+a_2s_2+...+a_ns_n is an element of span(S2)
Which you state here.

which implies that s is an element of span(S2).
Ok.

So the span(S1) is a subset of span(S2).
Ok.

[zinc tapir]

i changed the x's to s

i shoulda used v but w.e

[eternal finch]

Sorry, forgot to change all of them.

It doesn't matter what symbols you use.

The problem is that you are getting your objects and types mixed up.

Well, there's also asserting that span(S1) is a linear combination and the span of vectors in S1 being a vector in span(S2).

Like, take this snippet.

for a vector to be in span(S1) it has to be linear combination of vectors in S1, so if we let V=span(S1), can we say that there exists some linear combination vectors in span(S1) and since S1 is a subset of S2 those linear combinations of vectors in span(S1) are elements of S2 and therefore also exist in the span(S2).

Clearly, you understand the ideas.

It's just a little mess up.

If you remove a bit, it'd be ok.

for a vector to be in span(S1) it has to be linear combination of vectors in S1, so if we let V=span(S1), can we say that there exists some linear combination vectors in span(S1) and since S1 is a subset of S2 those linear combinations of vectors in span(S1) are elements of S2 and therefore also exist in the span(S2).

You mix up being in S2 and being in span(S2).

[zinc tapir]

Let S1 be a subset of S2, an arbitrary element s of span(S1) can be expressed as s=a_1s_1+a_2s_2+...+a_ns_n, where the a's are in R and the s's are in S1. Since S1 is a subset of S2, s_1,s_2,...,s_n are also elements of S2. Therefore any linear combination of those vectors is a vector in span(S2). Therefore s=a_1s_1+a_2_s_2+...+a_ns_n is an element of span(S2). So the span(S1) is a subset of span(S2).

fixed it with ur suggestions

[eternal finch]

Yep, no errors now.

Or at least, I don't see any.

which implies that s is an element of span(S2)
This is kind of redundant since you say "s=a_1s_1+a_2s_2+...+a_ns_n is an element of span(S2)" already.