linear-algebra

[novel hamlet]

but the problem is proving it

well yes but actually no

since w=t0 or v=t0 wich makes w=v

[lavish jewel]

if you treat vectors point vectors, they have their tail at the origin

any two vectors then define 3 points: the origin, and the 2 points they are pointing toward

those 3 points lie on a plane, so the same interpretation of an angle on that plane holds

[novel hamlet]

okay

well i think it is just enough if i select v=0 and calculate <w,0> and //w//*//0//

and suprisingly 0=0

and the njust same for w=0 and then say w=t0 and v=t0 whjere t is any real number

[lavish jewel]

the trivial case doesn't help much

a vector in any direction can be made 0 by multiplying by 0

you actually only care about nonzero ones here

[wintry steppe]

how do i do this question

im very confused

[lavish jewel]

what's the definition of a rectangle

[wintry steppe]

the anwser is (6, 18, -4)

[wintry steppe]

two pairs of equal sides

[lavish jewel]

yep

[wintry steppe]

so what do i do to solve this

[lavish jewel]

you can make vectors from one of the points to the other two

[wintry steppe]

so I do vector AB = CD

[lavish jewel]

in length, sure, but possibly in the opposite direction

[wintry steppe]

so AC = BD

ive just been stuck on this homeowrk problem for like 40 minutes now

and i dont get the textbook anwser

[lavish jewel]

i'll try for like 5 mins and see

[wintry steppe]

okok

wait

nvm

i got it

@lavish jewel

[lavish jewel]

ok

probably had the wrong directions of vectors?

[wintry steppe]

no to solve it we have to split up the rectangle into two triangles

and make the vectors side lengths

and the hyptonuse is the diagonal vectors

and solve for the point that way

[lavish jewel]

aight

you don't really need to do that tho

B + BA + AD = BD can be done fully with vectors

and at that point you should already have c1 and c3 from the side equalities, too

it saves you a square root 😛

[novel hamlet]

I have (V<g,h>) and (W<a,b>) that are inner product spaces that have dim V <= dim W and U⊂ V (u is subspace for v). I need to show that f:U->W can be expanded to isometry F:v->W (if f:U->W then there exist isometry F:V->W such that F|U = f)

cant make that last one correct for some reason

[cobalt lodge]

anyone mind helping me with a problem, ive been stuck on this for a whileeee

[nocturne jewel]

the answer to the question you posted is 7

[cobalt lodge]

stuck on 1C

[wintry steppe]

convince yourself that the columns of A would span all of R^4 if and only of those of R would span all of R^4 (e.g. "row operations do not change column rank")

can the columns of R span all of R^4?

@cobalt lodge

[cobalt lodge]

yes??? @wintry steppe

[wintry steppe]

why is that?

(that's not correct)

is (0, 0, 0, 1) in the span of the columns of R?

[cobalt lodge]

no?

[wintry steppe]

right, so the columns of R don't span all of R^4

[cobalt lodge]

therefore the columns of A doont span all of R^4 either

[wintry steppe]

ya

[cobalt lodge]

gothcya gothcya

thank u!

[wintry steppe]

i said "convince yourself" for that one since i think there are a few ways to approach it

as long as it makes sense why

[cobalt lodge]

ive seen some youtube videos and there are a bunch that have different approaches

[pure tangle]

can someone please help me fuund where I'm making a mistake with this question

it wants me to find the original matrix based off the solution and a column

to solve for column 2, I used the point (?) part of the solution and did [-1,2,0] + 2 C2 = [-1,2,0]

I got [1,1,2]

for c3 I looked at the null space and used 2c1 + c2 + c3 = 0

and got [5,-1,6]

and finally i used the t parameter equal to 0 and got [7, -2, 8]

but when I row reduce those columns with the given solution d I dont get the same equation

[soft sun]

start by stating what rank is your matrix

[pure tangle]

2

[soft sun]

by using the theorem for relation of ranks and number of free variables

[pure tangle]

what do you mean? isnt rank = number of variables - number of free variables

[wintry steppe]

2/(2i) = -i but the conjugate of 2i is -2i

the expression you expect is $\bar{z}z=|z|^2$, not $|z|$

[stoic python]

(T*(Terra), -dτ)

[restive moon]

ty

[steel moon]

hi

[pure tangle]

<@&286206848099549185>

[steel moon]

what does it tell you when each row has a pivot column

that the system is consisntet?

[wintry steppe]

|u| = 4, |v| = 5, |u-v| = 6, what is the magnitude and direction of u+v?

[quartz compass]

write them out in terms of dot products

[wintry steppe]

and then what

i was only able to write out one: u*v=(u1v1+u2v2+...+u(n)v(n))

[quartz compass]

ah, don't write out the components

write like |u|^2 = u dot u and |u-v|^2 = (u-v) dot (u-v)

[west shoal]

help

how do i do this problem

not sure what to do

<@&286206848099549185>

i'd really like some helo

if i start with e= (1 0 ) and e2 = (0 1

then when i do the x1 + x2 / -x2

should e1 become 1 0 or 2 0, and will e2 become (-1 0)?

[nocturne jewel]

What's T[e_1]?

[wintry steppe]

bruh we needed your help in ou xD

[nocturne jewel]

cool

[spice storm]

Hello

Let f:R^3->R^2 and g:R^2->R^3 be given by:

f(e1)=e1+e2, f(e2)=2e1-e2, f(e3)=e2;

g(e1)=e1-e2+e3, g(e2)=e1+3e2-2e3.

Compute the matrix of the composition fog. Then use it to calculate (fog)(3e1+2e2).

[west shoal]

not given

that's all im given

[nocturne jewel]

Yes im asking you to do some work and tell me what it is

[spice storm]

Is my work corret?

[west shoal]

well if we have e1 = (1 0) and e2 = (0 1)

would that mean that T(e1) = e1 + e2, T(e2) = -e2?

[nocturne jewel]

No

[west shoal]

im so confused

[nocturne jewel]

what's e_1

[west shoal]

1 0

[nocturne jewel]

[1,0]^T right

[west shoal]

mhm

[nocturne jewel]

so what's x_1 of e_1?

[west shoal]

1

[nocturne jewel]

and x_2?

[west shoal]

0

[nocturne jewel]

so what's T[e_1]?

[west shoal]

1 0

[nocturne jewel]

yes

[west shoal]

and then the -x2 right

so it's still 1 0 regardless

[nocturne jewel]

so, not a trick question, how do you express e_1 as a linear combination of e_1 and e_2

(dont over think it lol)

[west shoal]

i got it

thanks

[nocturne jewel]

[spice storm]

I think for my A matrix is wrong

I got it!! Matrix A was wrong

🙂

[tame mural]

There should be an available RREF property which allows you to easily say this

[wintry steppe]

If i have a matrix in r3 and have only one independent column then my column space would be a line right ?

And if i have vector (0,0,0) would i be wrong to say the column space is a point in r3

[knotty raven]

Say I have a nilpotent transformation L:v->v where L^1=L, L o L^n-1 = L^n and L^n=0. I believe that each composition of the function produces a subspace of the last iteration's image. I am wondering how I can prove whether there will never be an iteration such that L^k+1 is not a subspace of L^k.

I think it should follow something like, since L is nilpotent, if an iteration was not a subspace of the previous then it would not equal 0 for some n.

but that feels wrong.

@wintry steppe in R3 if you have one independent vector then its span is a line yes

and 0 is always dependent

since column space is a span of vectors, its a vector subspace, and that means it must go through the origin, so if you think of the 0 space as a point, the point is on the origin.

[naive whale]

,w nilpotent

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jolimath#8528

[compact relic]

I've been working through some Linear Algebra problems and one that Is asking

Columns of AB are linearly dependent, then are columns of B also linearly dependent?

I worked backward that if B is linearly dependent, then Bx=0, thus adding A to both side of ABX= 0A , would get you ABx = 0, that'd mean AB are also linearly dependent right?

However, I'm not sure if you can work it reversely like that

or if there's another way for me to prove since AB are linearly dependent, B also linearly dependent

[dusky epoch]

there isn't because it's clearly false

take B = the identity and A = 0

@compact relic

[limber sierra]

not directly like this, no

as an example of this, consider "if x^2 > 0, then x > 0"

this is obviously false

for example, (-1)^2 = 1 > 0

but -1 < 0

but if we work "backwards":
x > 0, therefore squaring both sides, x^2 > 0

the point is, you cant just reverse the direction of implication statements

(you can prove a statement by proving the contrapositive, however)

(but in this case that wont help you since, as ann mentioned, your statement is just false)

[compact relic]

I see, how would I go about proving that since columns of AB are linearly dependent, thus colums of B also linearly dependent?

or ^that is simply false and because?

[limber sierra]

it's false, so you cant prove it

and the way you prove such a statement false is by providing a counterexample

as Ann did

$\begin{pmatrix}0&0\0&0\end{pmatrix}\begin{pmatrix}1&0\0&1\end{pmatrix} = \begin{pmatrix}0&0\0&0\end{pmatrix}$ which has linearly dependent columns, but $\begin{pmatrix}1&0\0&1\end{pmatrix}$ has linearly \textit{in}dependent columns

[stoic python]

Namington

[compact relic]

oh okay, that made a lot of senses. Thank you for the helps

[tame mural]

Does anyone have any book recommendations for a survey or introduction into vector spaces over finite fields?

[nocturne jewel]

Show the product of 2 linear operators is invertible iff both operators are invertible.
Im pretty sure the reverse direction is just show the product is injective by kernel, but not sure how to do the forward

[gray dust]

@nocturne jewel assume A,B:V->V bijective. show AB is injective & surjective by defns

[nocturne jewel]

Yeah but showing AB is injective is sufficient?

V is finite dim* but idk why V being finite/infinite dimensional matters/affects anything

[foggy field]

Hi everyone nice to meet you?

I would know if someone already studied with this book?

Translating the book name: "linear algebra with applications "

Author :Jeffrey holt

[gray dust]

i see it as a general function thing, not linalg-specific. if f:X->Y and g:Y->Z are bijective then gof is bijective. showing gof is injective & surjective is easy

[nocturne jewel]

Yeah but AB being injective is equivalent to AB being invertible

So I just need to show AB is injective which is obvious since A and B are injective

But idk how to start with AB being invertible means A,B are

[gray dust]

so do it

[nocturne jewel]

I did do it, Im confused on the other direction lol

[gray dust]

that's not the forward

[nocturne jewel]

Ik

I said Ik how to do the reverse

Show the product of 2 linear operators is invertible iff both operators are invertible.
Im pretty sure the reverse direction is just show the product is injective by kernel, but not sure how to do the forward

(AB)(v) = 0 iff v = 0 -> AB is invertible (obviously flushed out more) is the reverse

[wintry steppe]

because this is false in infinite dimensions. for example, on a space of sequences, left shift \circ right shift = id is invertible, but are the shifts invertible?

it is true that if A and B are invertible, then so is AB. as rokabe says, this is just a generic function thing.

since that's true even for infinite-dimensional spaces, it should serve as a hint that proving the other direction requires finite-dimensionality

that is, showing that if AB is invertible then A and B are should use finite-dimensionality.

[misty storm]

do u guys know of a tool for visualization in r³?

[wintry steppe]

geogebra 3d

[misty storm]

thx

fellers, what the heck is this?

2 sheets hyperboloid?

[blissful vault]

how do i prove that the dimension of the intersection of two subspaces have to be smaller or equal to the smallest of the 2 subspaces?

intuitively i think about dim(U)+dim(W) = dim(U+W) - dim(U intersection W)

[limber sierra]

suppose for contradiction theres a basis for U intersect V that's larger than the dimension of (WLOG) U

then this set must be linearly independent in U, since its a basis for U intersect V and constraining our purview just to U won't change its linear independence

so we have a linearly independent set that's larger than U's dimension

problem!

[blissful vault]

let me read this slowly

ok got it thx!

[pure tangle]

Are there any video series or websites you all recommend for learning why different applications in linear algebra fundamentally work? I feel like between lectures, studying, and homework I'm just memorizing processes and rules without being able to build intuition around what's trully going on

[slow scroll]

You could read a book

I read sergei treil’s “linear algebra done wrong” personally, and I thought it did a good job of doing what you’re asking for.

It might take more time than just memorizing the processes and rules though.

@pure tangle

[ruby loom]

I just want to confirm if I'm thinking about this correctly, I solved this integral for the potential values of a0, a1, a2, a3, which gave mee

which would seem to indicate that my basis is (a0, 0, 9a2, 0)

[frosty vapor]

?

[ruby loom]

Okay, so revising that earlier statement, I could attempt to construct a basis of all such polynomials as (a0, 0, 0, 0), (0, 0, 0, 0), (0, 0, 9a_2, 0), (0, 0, 0, 0), but these are not linearly independent?

[frosty vapor]

uh

if thats ur basis then ur polynomials are all of the even degreed ones

[ruby loom]

I am a bit confused, so forgive me if my questions don't make sense

[frosty vapor]

which is not going to satisfy the constraint

with the integral

[ruby loom]

hm? Don't my coefficients for the odd degree terms become 0?

Oh, or should it be that they can be anything whatsoever because it's the vector values that become 0?

I'm really just trying to get a handle on what's going on here, and how to relate the results of my computation to the construction of a basis

[slow scroll]

which would seem to indicate that my basis is (a0, 0, 9a2, 0)
a basis is a set of linearly independent vectors which span a space. You wrote down one vector in R4.

[ruby loom]

okay, so what is the significance of my scalar coefficients in constructing my basis?

I'm aware, for example, that a basis for R3 would be (1, 0, 0), (0, 1, 0), (0, 0, 1),

[slow scroll]

okay, so you have the equation a_0 + 9a_2 = 0. These are the constraints you have to deal with. The standard basis for P_3(R) is 1, x, x^2, x^3. You should sort of think about how to modify this so that a_0 = -9a_2 for any polynomial

[ruby loom]

why, for example, is a basis for P_3(R) not (1, 0, 0, 0, (0, x, 0, 0), (0, 0, x^2, 0), (0, 0, 0, x^3)? Just from following a similar logic to how I would construct one for R4

Is it because (1, 0, 0, 0) and so forth are not elements of P(R)?

[slow scroll]

You are correct that you can identify 1 -> (1,0,0,0) , x -> (0,1,0,0), .... under an isomorphism P_3(R) to R4, but if we are talking about a basis for P_3(R), and not R4, its probably a little easier (communication wise) to actually work with polynomials.

For example, by (1,0,0,0), its not necessarily clear if you are identifying this vector with 1,x,x^2, or x^3, or some other polynomial.

[frosty vapor]

ordered basis or sumn idk

inded

[ruby loom]

I see, so...

[slow scroll]

i mean, this identification is something you do a lot for calculations and stuff, but to say a "basis for P_3(R) is (a0, 0, 9a2, 0)" is a bit worse than just an abuse of notation/language.

anyhow, you want the space of polynomials such that the coefficient of 1 is 9 times the coefficient of x^2

[ruby loom]

Oh, alright, so it's just 9a_3, x, a_3*x^2, x^3, if I'm understanding correctly?

I was under the impression my scalar coefficients were to be separated from my vectors

Thanks btw, your points have really helped! I'm revisiting how this construction works

[slow scroll]

@ruby loom i think you might need to check your calculation. I am getting a different equation from a0 + 9a2 = 0. Also, 9a_3, x, a_3*x^2, x^3 is not the correct basis

[ruby loom]

oh woops, yeah I see, it should be 3a_0

that first term

[slow scroll]

i get a0 + 3a2 = 0. Now, 3, x, x^2, x^3 still doesn't work

the polynomial 1 + x^2 is in the space spanned by those vectors while 1 + 3 != 0

[ruby loom]

3a2? Sorry, but that integral is even so it becomes 2*a2 * the integral from 0 to 3 of x^2 which is 9

or am I going crazy?

And then the 2 can be factored out

oh, disregard that!

I see what you did. Sorry about that

[slow scroll]

np. anyway, as a hint, the basis vector that controls whether a0 + 3a2 = 0 is going to be a sum of monomials.

a polynomial which satisfies a0 + 3a2 = 0 itself

[ruby loom]

But this will still necessarily be a polynomial in the space of P(3)?

[slow scroll]

yea it will

[ruby loom]

so clarifying question, does this mean that my coefficients for the x1 and x3 terms can be anything?

[slow scroll]

x and x^3 have no constraints, so we can include them in our basis w/o any issue

[ruby loom]

alright

thanks for your help!

[slow scroll]

@ruby loom did you find the basis?

and np

[ruby loom]

Oh, well I thought it would be -3a_2, x, (a_2)x^2, x^3, though your monomial comment makes me think perhaps I'm misunderstanding how to find a polynomial that satisfies the coefficients

[slow scroll]

nope. -3a_2, x, (a_2)x^2, x^3 spans the same space as 1,x,x^2,x^3

we know that x^3 and x belong in the basis. There is one more vector that we need to add. its certainly not x^2 (or scalar times x^2) since the integral of x^2 from -3 to 3 is not 0, and similarly for f(x) = 1.

(this is the sanity check that the basis elements for your space should actually belong to your space)

[ruby loom]

ah! I see so (0, x, 0, x^3)

[slow scroll]

hmm? what does that mean? x + x^3 is not a basis element, if that's what that means

and x, x^3 are not the only basis vectors, if that's what you meant

[ruby loom]

okay, so I'm tempted to just make it (-a0 - 3a2, x, x^3) so as to cancel out the (potential) non zero values of a0 and 3a2

[slow scroll]

you have the right idea, but the space spanned by -a0 - 3a2, x, x^3 is the same as the space spanned by 1,x,x^3

name a polynomial other than linear combinations of x and x^3 that should belong to this space

[ruby loom]

how is that so? Because a0 and a2 are arbitrary so they can serve the same purpose as 1?

[slow scroll]

-a0 - 3a2 is just a number. We can divide (assuming nonzero) by -a0 - 3a2 to get 1, so 1 is in the space, and we can replace -a0 - 3a2 with 1 in that basis.

That's also why I am not writing redundant scalars next to monomial terms in a basis. for example, the space spanned by a, bx, cx^2, dx^3 is the same as the space spanned by 1, x, x^2, x^3 (assuming a,b,c,d are nonzero)

anyway, do this

name a polynomial other than linear combinations of x and x^3 that should belong to this space

[ruby loom]

alright. x^2 - 3?

[slow scroll]

yep.

[ruby loom]

Alright!

[slow scroll]

x^2 - 3 is not in the span of {x, x^3}. Thus, the set {x, x^3, x^2 - 3} is linearly independent. Does this span U?

[ruby loom]

yes

therefore, it's a basis for U

[slow scroll]

ye

[ruby loom]

thanks for your patience!! I haven't done LA in awhile

[slow scroll]

npnp

[spice dock]

I'm having two vectors: u = (6, 4, 2) and v = (1, -3, 3). I'm supposed to find a third vector w = (x, y, z) so that they generate a cuboid, with volume -180. So far I've gotten to this equation : 18x - 16y - 22z = -180 from calculating determinant. I asked this same question here a few days ago but I can't figure it out. I know they all are 90 degrees, but what to put in for x,y,z..?

[slow scroll]

im not sure why you have the = -180, but
in 18x - 16y - 22z, x should be (1,0,0), y should be (0,1,0), and z should be (0,0,1).

and when you sum up 18x - 16y - 22z, you will get the "w" you need.

oh i see why -180

you mean || 18x - 16y - 22z|| = 180

wait no

[spice dock]

Yeah, both kinda, I'm just saying what the task says.. :). But it's rare to have negative volume.

What I figured was if I took the lenght of u and v, with squareroot and sum of them squared, you know, the u * v * w should be -180 right?

[slow scroll]

im guessing it wants ||u|| ||v|| ||w|| = 180, but idk where the negative comes from

[spice dock]

But it just rings ''faulty'' somhow

Ye, I'm just writing what the task says, i would print it, but it is in norwegian so i dont know if it would be any good.

[slow scroll]

here's what i know: since u and v are perpendicular to each other, ||u x v|| = ||u|| ||v||
u,v,w represent the sides of a cuboid, and u x v gives you the side perpendicular to both u and v. You need to choose the vector of appropriate length in the direction of u x v so that the magnitudes all multiply to 180

can't say where the negative comes from though

[spice dock]

Ok, what I did a earlier was put in some numbers for x, y and z(just trial and error), so that the equation 18x - 16y - 22z = -180 was correct, i believe it was x = 3, y = 5, and z = 7. But then the product of x * y * z =/(didn't equal) -180. But maybe thats because i didn't do the absolute value as you mention.

Like when I take the lenght of u = 7.48 and v = 4.35.

Then w should be lenght = 5.51 to be 180.

[slow scroll]

you computed a determinant like $\begin{vmatrix} x&y&z \ 6&4&2 \ 1&-2&3 \end{vmatrix}$, right?

[stoic python]

kxrider

[spice dock]

-3 instead of -2 but yes, that is correct.

[slow scroll]

oops yea. anyway, in this, x,y,z are vectors. x = (1,0,0), y = (0,1,0), z = (0,0,1). i.e. 18x - 16y - 22z = (18,-16,-22)

this is the cross product of (6,4,2) and (1,-2,3)

you don't get to choose what x,y,z are here

[spice dock]

Ok, I see, so how to calculate the vector w then? I'm sorry if I should see it myself. Very new to this.

[slow scroll]

its np. first, do you understand why u and v are perpendicular (orthogonal)?

[spice dock]

Yes, because the dot product is 0.

[slow scroll]

alright, good. So the visual is this

u, v, and u x v form the sides of a cuboid

[spice dock]

Yes, as I have seen in geogebra 3d as well 🙂

[slow scroll]

u x v has the right "direction," but we need to extract the correct length, so that the volume is 180. that would be our w

[spice dock]

Correct.

I've been working with - 180

should maybe be working with +

[slow scroll]

||u|| ||v|| ||w|| = 180 so ||w|| = 180/(||u|| ||v|| ).

yea idk about the whole negative thing. i don't think there is a standard interpretation of "negative volume" here

[spice dock]

Yes, that's true, and I've been doing that, though with the negative, and with the negative I came up with ||w|| being 5.51 in lenght.

[slow scroll]

okay, that's actually approximately correct

[spice dock]

From w, but I cant figure out to get the lenght into a vector. it would be something like 5.1 = squareroot x^2 + y^2 +z^2

Maybe writing something wrong here, a lot of numbers in my head, I've been trying to solve this for more than 10 hours over the weekend.

[slow scroll]

we want something with length 5.51 in the direction of u x v. The standard way to do this is to "make u x v length 1" by computing (u x v)/(||u|| ||v||). This vector has length 1, and points in the direction of u x v. Now, how would we make it 5.51 in length?

[lavish jewel]

isn't the magnitude of the vector u x v also the area of the parallelogram they describe?

[slow scroll]

yea, it is

[spice dock]

yepp

[lavish jewel]

so you just need $| u \times v |$*L = 180

[stoic python]

Edd

[spice dock]

Yes, and the L is 5.51.

[lavish jewel]

oh, you found it

ok, good

[spice dock]

But I need the L = 5.51 to be put into x,y and z :). And there's where we are now I believe 🙂

[lavish jewel]

it's as they said above

normalize u x v and multiply it by L

[slow scroll]

The standard way to do this is to "make u x v length 1" by computing (u x v)/(||u|| ||v||). This vector has length 1, and points in the direction of u x v. Now, how would we make it 5.51 in length?
referring to this

[lavish jewel]

precisely

you could also just normalize that by $\Vert u \times v \Vert$

[stoic python]

Edd

[lavish jewel]

you make it length 5.51 by multiplying by 5.51 😛

[spice dock]

Ye, I could do the ''lenght to a point'' equation.

As I call it lol

[lavish jewel]

idk what that is

[spice dock]

It's the (u x v)/(||u|| ||v||).

I believe

Distance from a plane to a point.

And set distance to 5.51

Or am I way off.

[slow scroll]

lets say i have a vector v. Do you know what i mean when i say something like 3v points in the same direction as v?

[spice dock]

Yepp, it's 3 times as long.

[lavish jewel]

your distance formula is wrong btw

[spice dock]

Could I just write 5.51*(x,y,z)

[lavish jewel]

$\frac{u \times v}{\Vert u \Vert \Vert v \Vert}$ has length $\sin \theta$

[stoic python]

Edd

[slow scroll]

im going to bed, take it away, edd

[lavish jewel]

D:

ok, so what's the remaining question

[slow scroll]

The cross product of u and v was (18, -16, -22) to save you the scrolling

Ok sleep now

[lavish jewel]

aight

[spice dock]

Ok, let's go. I shall translate the task from norwegian to english as best I can.

Find the vector w e R^3 so that the vectors u, v and w generate a cuboid( = straight prism) with oriented volume of - 180. Where: U = (6,4,2) and V = (1,-3,3)

[lavish jewel]

mhm

well

you already know that $\Vert u \times v \Vert$ is the area of the base, so $\Vert u \times v \Vert \cdot L = 180$ lets you find the length of the third vector

[spice dock]

Yes

[stoic python]

Edd

[lavish jewel]

the third vector goes in the direction of $u \times v$. that direction is given by the unit vector $\frac{u \times v}{\Vert u \times v \Vert}$

[stoic python]

Edd

[lavish jewel]

so $w = \frac{u \times v}{\Vert u \times v \Vert} \cdot L$

[stoic python]

Edd

[lavish jewel]

i don't remember how to calculate signed volumes from this, so it might be -L instead

that's about it 😛

[spice dock]

Still taking this in xD

So, say, if I want the value of x in the w vector, how do i do that with that equation? I'm in deep here xD

[lavish jewel]

bruh

u x v will give you 3 numbers

say w = (x,y,z)

[spice dock]

Ahh, ok, so i do the equation 3 times to get the values, got it.

[lavish jewel]

no

this equation already gives you the whole vector

u x v is a vector

L is a scalar

$\Vert u \times v \Vert$ is a scalar

[stoic python]

Edd

[lavish jewel]

so w is already a full vector, all 3 of its components are in there

[lavish jewel]

this is all

[spice dock]

Ok, so I divide the 3 vectors by the scalar and mulitply them by L.

Please say thats the correct way 😄

[lavish jewel]

what 3 vectors

wtf

[spice dock]

sorry

x,y,z

[lavish jewel]

i don't understand what you don't understand lol

do you know how it works when you multiply a vector by a scalar?

[spice dock]

Yes, you just put the scalar in front and mulitply it by all the numbers in the vector.

[lavish jewel]

ok

that's what you do, then

[spice dock]

So, what I got is w = (3, -2.56, -3.52)

I believe that is correct. Thanks so much @slow scroll and @lavish jewel, really appreciated!

[uneven cypress]

Hello I needed some help in Algebra and was wanting to know if I can just post the question or do I need to ask first.

[steady fiber]

[uneven cypress]

Stephanie receives a salary of​ $650 per month plus a commission of​ 5.5% on the first​ $3,000 of​ sales, and​ 7% of all sales over​ $3,000. Find​ Stephanie's gross pay for the month if her sales were ​$11,190

[steady fiber]

that's not linear algebra

#prealg-and-algebra

is what you're looking for

[uneven cypress]

Thanks

[hearty meteor]

can someone help me understand b more?

is it asking to prove that the dimension of T is equal to its image?

[wintry steppe]

"dimension of T" makes no sense. it's asking you to prove that the rank of T equals (thing described)

the rank is the dimension of the image of T

ugh don't post the same question in multiple channels

[hearty meteor]

Yeah I'm sorry about that, I didn't intend to spam, I just thought it fit better in lin alg. Thanks though

[magic light]

I know if T(v1)... T(vn) are linearly independent then v1...vn are also linearly independent because suppose we have a1v1+...+anvn=0

We know the T(a1v1...+anvn) = T(0)=0 and by linear properties
a1T(v1)...+anT(vn)=0 thus a1..an=0

Is this also true the other way around?

v1...vn are independent then T(v1)... T(vn) are independent?

[slow scroll]

nope

[magic light]

I thought so

But why?

I mean I guess the zeroing transformation shows this is wrong right?

Because even though a1...an=0 the solution isn’t trivial since T(v) =0 is possible meaning its scalar could be non-zero while still satisfying the equation

[slow scroll]

its true that the 0 map is a counter example. More generally, you can always define a linear transformation by where it maps a basis. So you can just define T to be a map which takes a basis to some linearly dependent collection of vectors in the codomain.

[magic light]

What if KerT={0}?

[slow scroll]

then it does take linearly independent sets to linearly independent sets

[magic light]

I thought so - but is it because of the 0?

I mean

v1...vn lin indep thus non-zero, therefore T(vi)!=0 and T(vi) != T(vj) for i !=j

[slow scroll]

think about reversing this for a proof

I know if T(v1)... T(vn) are linearly independent then v1...vn are also linearly independent because suppose we have a1v1+...+anvn=0

We know the T(a1v1...+anvn) = T(0)=0 and by linear properties
a1T(v1)...+anT(vn)=0 thus a1..an=0

[magic light]

Right so we get a1T(a1v1)... and a1=an=0 is the only solution because of the quote here?

I’m having a hard time justifying it being the only solution

Just because T(v1) != T(v2) doesn’t mean they are independent, is what I’m trying to say

[slow scroll]

v1...vn lin indep thus non-zero, therefore T(vi)!=0 and T(vi) != T(vj) for i !=j
this is true, but I don't think it helps much

Suppose that $v_1, \dots, v_n$ are linearly independent and $\ker T = 0$. Suppose that $a_1T(v_1) + \cdots + a_nT(v_n) = 0$. Then $$T\left( \sum_{i=1}^n a_iv_i \right) = 0 $$

[stoic python]

kxrider

[slow scroll]

can u finish the argument?

[magic light]

Hmm

Ahhhh

The vector inside the T must be 0

Because only T(0)=0

So sum(ai vi) = 0 only has the trivial solution

I think that means aiT(vi) only has one solution as well

[slow scroll]

yes, good. sum a_i v_i = 0 and linear independence of v_1, ..., v_n shows that a_1 = ... = a_n = 0. Thus, the T(v_i) are linearly independent.

[magic light]

Nice.

It didn’t work before because sum(ai vi) could be “anything” without KerT={0}

[slow scroll]

yup, pretty much. In particular, when ker(T) != 0, you can always pick a nonzero vector v from the kernel, and use that to form nontrivial linear combinations of T(v_i) which are 0

[magic light]

Nice

Thanks, I can go to sleep now lol

This was keeping me awake

[slow scroll]

np. good questions. i think understanding these things goes a very long way in linear algebra

[magic light]

Thanks again, have a good day

[zealous junco]

if we're talking about A having an eigenvector

Av = lambda v

must this mean A has an associated norm that is induced by a norm on V, where v in V?

[slow scroll]

V is a normed vector space?

[zealous junco]

nah

[slow scroll]

what is the norm you are suggesting then?

[zealous junco]

Actually maybe

so here's the problem

my question is regarding c)

is there anything implying that R^n here is a NVS?

[slow scroll]

are you concerned about convergence of the series expansion for e^A or something?

[zealous junco]

yea

i mean I want to use

submultiplicativity of the norm

but not all matrix norms are submultiplicative

[slow scroll]

hmm

[zealous junco]

So i'm just confused what is the norm assumed to be used here yea

[slow scroll]

the most natural norm here is the operator norm, which is submultiplicative. I've never really thought about this tho tbh

you should be able to do something like $$|e^A| = \left|\sum_{k=1}^\infty \frac{1}{k!} A^k \right| \leq \sum_{k=1}^\infty \frac{1}{k!} |A^k| \leq \sum_{k=1}^\infty \frac{1}{k!} |A|^k < \infty$$

[stoic python]

kxrider

[zealous junco]

yea I did that

thanks

So i'll just assumed it's an operator noem

[slow scroll]

np. and the choice of norm technically doesn't affect convergence for matrices

[zealous junco]

you mean the choice of norm for the v right

wait nvm

oh cuz all norms are equivalent

[slow scroll]

ye

[zealous junco]

Also I wonder why I can say this?

if p_N(A) is the partial sum for the infinite series e^A

like I know this is true fo real numbers but I kind of don't know when this is the case in general

haven't formally learned any things abt limits in metric space/generalization of limit

(If I showed p_N(A) = Pp_N(\Lambda)P^-1 already)

[slow scroll]

is this for analysis? For a diagonalizable matrix, I don't think you have to go through all of this

[graceful vortex]

matrix multiplication and inversion is continuous, since it is given in coordinates by polynomials

[zealous junco]

it's linear algebra in a numerical analysis class lol

[zealous junco]

I see, thanks!

yea I just wanted like a brief overview of when this can be done, we don't need to prove to that much detail ( I think)

[tawdry bramble]

I was trying to find out how two subspaces were equal

[slow scroll]

you were asked to show something like <v1, v2, v3> = <w1, w2, w3>?

[tawdry bramble]

Yeah

[slow scroll]

the idea is to show that <v1, v2, v3> is a subset of <w1, w2, w3> and <w1, w2, w3> is a subset of <v1, v2, v3>.

the <v1, v2, v3> just means span{v1,v2,v3}, the space of all linear combinations of v1,v2,v3 btw, that is correct

[tawdry bramble]

Oh, so those are just the vectors you use to make all the other vectors in the set, by adding and scalar multiplication?

[slow scroll]

right, every vector in span{v1,v2,v3} has the form c1v1 + c2v2 + c3v3 for some scalars c1, c2, c3

[tawdry bramble]

So to show that either subspace is a sunset of the other, I have to show that each element of one set is a linear combination of elements from the other set

And vice versa

[slow scroll]

yep. In practice, there might be a shortcut with matrix tricks, but that's the idea in theory.

[tawdry bramble]

Sweet. You say matrix tricks and that made me think of the theorems the professor was talking about directly after subspace talking about linear dependence and independence in regards to a number of vectors being larger or smaller than the number of dimensions your working or something like that where inequalities of amounts were involved, and I wondered if any of that could be uses to show equality of subspaces

[slow scroll]

sometimes. Lets say for example we are in R3, and I ask you to show that <v1,v2,v3> = <w1,w2,w3>. If v1,v2,v3 are linearly independent, then they are a basis for R3, so <v1,v2,v3> = R3. Then, to show that <v1,v2,v3> = <w1,w2,w3>, all you have to do is show that w1,w2,w3 are linearly independent as well.

I was more referring to tricks like making v1,v2,v3,w1,w2,w3 the columns of certain matrices and computing rref.

[tawdry bramble]

In either case, it sounds very interesting.

Thanks for the help. You including many others have been a great help. I'm glad a discord server like this exists.

[slow scroll]

npnp

[waxen flume]

https://i.imgur.com/k9626C3.png

i got always, always, sometimes, always

[slow scroll]

wrong on c and d

[marble lance]

Why is c wrong?

[slow scroll]

for c, take T to be the 0 matrix. 0 is in the range of the 0 matrix, but that doesn't imply in any way that the zero matrix is onto

[dusky epoch]

d is wrong

... wait

[marble lance]

The said sometimes for c

[slow scroll]

oh, oops

[waxen flume]

0.0

im right for a, b, and c?

[slow scroll]

yea

just take a closer look at d

[waxen flume]

one to one means there is a pivot in every column

not sure where to go after that lol

[slow scroll]

My understanding of the question is that A can be basically any matrix. Not every matrix has a pivot in every column of its rref, does it?

[waxen flume]

no, it does not

sometimes

[slow scroll]

i agree with sometimes.

[waxen flume]

thanks

[slow scroll]

np

[waxen flume]

https://i.imgur.com/a90L3D3.png

okay so

i think this might be a reflection over x-axis

b/c of the -e_2

im familiar a little with [cos, -sin; sin cos]

but im not sure how that plays a role in this question

[slow scroll]

a reflection over the x axis would look like T(e1) = e1, T(e2) = -e2

write down what T does to a vector (x,y) in the standard basis.

[waxen flume]

1 sec

I made a few random points
(1,4) => (-4,1)
(2,3) => (-3,2)

basically the x and y get flipped

but the x becomes negative

how do i describe this geometrically?

[slow scroll]

what does this transformation do to (1,0) and (0,1)? See if that helps at all

[waxen flume]

OH

rotates pi/2

@slow scroll

[slow scroll]

ye, which direction?

[waxen flume]

counterclockwise

[slow scroll]

why?

[waxen flume]

actually i dont know lol

i just assumed

[slow scroll]

well i disagree

[waxen flume]

why?

[fickle citrus]

Rate my hint

[slow scroll]

take (1,0) -> (0, -1). maybe draw it out

[fickle citrus]

In case it isn't clear, my $g$ isn't the same as your $T$

[stoic python]

ShatteredSunlight

[fickle citrus]

To be super clear, my $g$ would be the answer to this

[stoic python]

ShatteredSunlight

[fickle citrus]

If it's too clear then maybe I shouldn't give this hint

[waxen flume]

its counterclockwise

lol

[fickle citrus]

If you're saying your T is counterclockwise, then I, like rider, disagree

Maybe my hint was confusing after all

[slow scroll]

i think it maybe kinda was . the map shattered posted is a counterclockwise rotation, but its a different map than what you have vici

[waxen flume]

L

so it was clockwise

why though

[fickle citrus]

That's....how the map was defined?

[waxen flume]

mine?

or his?

[slow scroll]

wow it looks like (0,1) is at (0,2), don't mind my artistry

[waxen flume]

my (1,0)

becomes (0,1)

its the other direction

your arrow is wrong lol

[slow scroll]

the transformation was T(e1) = -e2, T(e2) = e1, right?

[waxen flume]

yes

[slow scroll]

yeah, exactly. i.e. T(1,0) = (0, -1), T(0,1) = (1,0).

(1,0) gets rotated 90 degrees clockwise around the origin and (0,1) gets rotated 90 degrees clockwise around the origin

[waxen flume]

i thought (e1, e2)

(1,0) -> (0, 1)

[slow scroll]

e1 = (1,0), e2 = (0,1), the standard basis vectors

for example, when you know what T does to e1, and e2, you can write a matrix for T : $\begin{bmatrix} 0&1\ -1&0 \end{bmatrix} = [T(e_1) \quad T(e_2)]$

[stoic python]

kxrider

[slow scroll]

and you can write things like $(a,b) = a\mathbf{e_1} + b\mathbf{e_2}$.

[stoic python]

kxrider

[slow scroll]

but um, anyway, do you see why T(e1) = -e2, T(e2) = e1 is the same as saying T(1,0) = (0, -1), T(0,1) = (1,0)?

[stoic python]

mirzathecutiepie

mirzathecutiepie

[marble lance]

Your basis vectors from the original space might not be in the subspace.

[dusky epoch]

$v = a_1v_1,..., a_nv_n$?

[stoic python]

Ann

[dusky epoch]

suppose for the sake of contradiction that W fails to have a finite spanning list

...wait

i dont know where i was going with this

damn you axler

for every linearly independent list of vectors w_1, w_2, ..., w_n in W, there exists a vector w_{n+1} in W that is not in their span

is axler non-sadist enough to allow us to use induction

to say that this implies we can construct a sequence of lists of ever-growing length, each one linearly independent

and then look at the first one that has more vectors in it than the dimension of V

[pure tangle]

can someone please help me fuund where I'm making a mistake with this question

it wants me to find the original matrix based off the solution and a column

to solve for column 2, I used the point (?) part of the solution and did [-1,2,0] + 2 C2 = [-1,2,0]

I got [1,1,2]

for c3 I looked at the null space and used 2c1 + c2 + c3 = 0

and got [5,-1,6]

and finally i used the t parameter equal to 0 and got [7, -2, 8]

but when I row reduce those columns with the given solution d I dont get the same equation

[pure tangle]

<@&286206848099549185>

[desert rapids]

It seems that you went wrong for working out c_3

A better way would be to again utilize the solution set. You can simply set the t equal to zero and s equal to 1, and get the particular solution $x = \begin{bmatrix} 3 \ 3 \ 1 \ 0 \end{bmatrix}$.

[stoic python]

LifeSource

[desert rapids]

Then, (you got $c_2$ right), we have that $3 \begin{bmatrix} -3 \ 0 \ -4 \end{bmatrix} + 3 \begin{bmatrix} 1 \ 1 \ 2 \end{bmatrix} + c_3 = \begin{bmatrix} -1 \ 2 \ 0 \end{bmatrix}$.

[stoic python]

LifeSource

[desert rapids]

Then it is easy to solve for $c_3$

[stoic python]

LifeSource

[desert rapids]

And you can use a similar strategy for c_4

Got it?

@pure tangle

[acoustic path]

Hey i have a small question. We know that the range of a linear map is surjective if it equals the co-domain, doesnt that mean the range also holds 0 cuz the co-domain must contain the 0 vector to be vector space?

What im trying to say is that the range should map vectors to 0 too. But then its the null space not the range

[desert rapids]

For it to be a vector space then it has to contain the identity element

Correct

Also note that a linear map by definition preserves the origin

[acoustic path]

Yeah i thought about it some more i get it now

[desert rapids]

Good!

@pure tangle Did you get it?

[pure tangle]

@desert rapids thanks so much for the response. Sorry let me take a second to look over what you said

@desert rapids thank you so much so your solution makes perfect sense, do you know what was wrong with my solution for c3? I looked at the null for s = 1 and t = 0 and solved for (2,1,1,0) = 0and thats where I differed from you

[tawdry bramble]

What is the column space of any 3x3 non-singular matrix?

[pure tangle]

@desert rapids when I use your method for c3 I get the same answer I did with my solution

[strange crystal]

im not crazy for realizing that solving a constant coefficient homogenous linear differential equation is just solving for the eigenvalues right?

dont remember that too greatly from when i took linalg but it seems really familiar

[lone tide]

whats the best website to learn math

other than Khan Academy

i mean i said website

fatwa kya hai?

mai pakistani houn

ah

[magic light]

If I have a matrix A
do the matrices
A^TA and AA^T have the same determinant?

it should be that way because B = (A^TA)^T = A^TA so Det(B-xI) = Det(B^T-xI)

[steel plover]

In general, det(AB) = det(A)det(B) = det(B)det(A) = det(BA)

[magic light]

sorry I meant the same eigenvalues

because I have a matrix where it doesn't work and I'm wondering if I am misunderstanding it

[steel plover]

Can you share your example?

[magic light]

https://i.imgur.com/0rI9gpD.png

https://i.imgur.com/wuly2C6.png

that's A^TA(idk why the professor writes it like that)

https://i.imgur.com/EOXwNhW.png

that's AA^T

their eigenvalues aren't equal hmm

https://i.imgur.com/w4F397j.png

that's the answer

to A^TA

[steel plover]

Ah yes, there's a mistake in your argument:
(A^TA)^T = A^TA

And in this case, it is not the same as A A^T

Since in particular, those two matrices have different sizes

[magic light]

shouldn't AA^T and A^TA have the same eigenvalues though?

#help-6 message

[spiral star]

eigenvalues are the same, yea

[magic light]

ah I see

just without 0s

[steel plover]

Yeah, I'm just talking nonsense.
Sorry I'm just too tired tonight to actually do maths

[weak glade]

Hi! Can I prove that a set îs a direct summand without a direct complement? I have this problem

Firstly, i think i need to prove that A is a subspace of R^2 but then? My text books if A and B are two subspaces of vector space V and if A+B(direct sum) = V and A intersected with B={0} then A and B are called direct summands

Can I assume that if a set is a subspace then it will definitely exist another subspace such that their direct sum Will be the vector space and intersection the null space?

[sick zodiac]

Is anyone smart at Linear Algebra online that could help me out understanding some prompts?

[steel plover]

It is true in finite dimension. In infinite dimension you need the axiom of choice to prove it.

Here I believe that they just want you to give the direct complement of A

[weak glade]

I work only in finite dimentions, so i don't need to prove it?

I was a bit confused because it says 'if so'

[steel plover]

Well, it is a fact, but if it hasn't been proved in class yet, you probably would need to prove it in order to be allowed to use it.

Also, they say "if so" in order not to hint at what is the answer.

[weak glade]

Ok. Thank you! It hasn't been proved in class and it won't be. How could i prove it? I have no idea

[steel plover]

Maybe you know of a theorem that says that if $(a_1,...,a_k)$ is a free family of $\mathbb{R}^n$, then there exists vectors $a_{k+1},...,a_{n}$ such that $(a_1,...,a_n)$ is a basis of $\mathbb{R}^n$ ?

[stoic python]

Dagnyr

[nocturne jewel]

I asked this the other day and still confused on how to go about it:

Let S and T be linear operators on the finite-dimensional vector space V. Show that ST is invertible iff both S and T are

I can do the reverse direction pretty easily, since you just show the ker(ST) = {0} by considering ST(v)=0, but cant see where to start on the forward direction

[wintry steppe]

as i said

you need to use finite-dimensionality in some way, since the forward direction is false in infinite-dimensional spaces

what are some tools that finite-dimensionality gives you? rank-nullity, choosing a basis, etc.

[nocturne jewel]

ik the image of ST is V and kernel is {0}

But I dont see how dim(V) = n (for example) would help to show that S and T are

Wait, probably wrong but:

If dim(V) = n, then dim(Im(ST)) = n
but dim(Im(ST)) <= min{dim(Im(S)),dim(Im(T))}
but dim(Im(S)) and dim(Im(T)) cannot be bigger than n

@wintry steppe would that work / at all right? lol

[wintry steppe]

nice

[nocturne jewel]

YAY

[wintry steppe]

so they're both onto. why does that imply they're also one-to-one?

[nocturne jewel]

what's onto

[wintry steppe]

surjective

this will show that they're both invertible

[nocturne jewel]

isnt showing surjective or injective sufficient for finite dimensional?

[wintry steppe]

for operators between spaces of the same finite dimension, yeah

i just wanted to make sure you knew that lol

[nocturne jewel]

yeah I was digging through my notes cause I always forget which page characterization of invertible operators is written on

and my argument already holds that dim(Im(S))=dim(Im(T)) since if they didnt the min function wouldnt return n

[idle bloom]

anyone know how to do this? i have no idea

[graceful vortex]

Draw two random half-spaces. Is the union convex ? In what position do they have to be for the union to be convex ?

[wintry steppe]

could i have help with this

im so confused by how would you do this

<@&286206848099549185>

[lavish jewel]

what's confusing you

[wintry steppe]

i dont get how I am suppoused to know if its a vector, scalar or meaningless

you dont solve for anything?

@lavish jewel

[digital bough]

Well begin with the first one. Will a cross product always yield a vector?

[wintry steppe]

yes

[digital bough]

And will addition of vectors always yield a vector (kernel being a vector aswell)?

[wintry steppe]

yes

oh

[lavish jewel]

there you go, then

[wintry steppe]

so this is a vector

[lavish jewel]

yep

[wintry steppe]

right

is this a scalar then

[lavish jewel]

yyes

[wintry steppe]

what about this then

is it asking for two cross products

[lavish jewel]

well

do the one in parenthesis first

[wintry steppe]

thats a vector

[lavish jewel]

all right

and how about the cross product of 2 vectors

[wintry steppe]

its also a vector

so its just asking for the cross product of vector a with the cross product of vector BC

so its a vector

overall

right

[lavish jewel]

yep

[wintry steppe]

one last question cause this is my last homework question for the day

for this

i got the cross product to be (20, -24, 19)

right

i dont get what perpendicular means

in this case

[lavish jewel]

if you draw them in 3D space, they form 90° angles to each other

[wintry steppe]

when you cross 2 vectors you get a perpendicular vector to those 2 vectors you crossed

(20, -24, 19) x (x, y, z) = 0

right

[lavish jewel]

what's that

[wintry steppe]

isnt that what i have to do?

[lavish jewel]

that's a dot product

that's still wrong

[wintry steppe]

oh

so what should i do

they just want <20, -24, 19>

[lavish jewel]

,w cross product of {-5, -1, 4} and {-1,-4,-4}

[digital bough]

You can use the dot product using a system of equations but they want you to use cross product

[stoic python]

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Requested by Edd#5182

[lavish jewel]

you were already done

[wintry steppe]

this vector is orthogonal to (-5,-1,4) and (-1,-4,-4)

<20, -24, 19>

u already did it

wait what?

how

im confused

it says use cross product and you did the cross product

[lavish jewel]

that is the definition of the cross product

[wintry steppe]

and it is orthogonal to the 2 vectors

oh

[lavish jewel]

a = b x c yields a vector a that is orthogonal to both b and c

[wintry steppe]

so the cross product means the perpendicular vector of two vectors

[lavish jewel]

you can test if you want to convince yourself. NOW you can use a dot b = 0 and a dot c = 0

[wintry steppe]

so if you <20, -24, 19> dot <-5,-1,4> it should = 0

,w <20, -24, 19> dot <-5,-1,4>

[stoic python]

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Requested by Moonlight ⌬ ∰#9250

[wintry steppe]

oh

so the anwser was just 20, -24, 19

[lavish jewel]

yep

[wintry steppe]

whenever you cross 2 vectors the vector you get is always perpendicular to the original 2 vectors

oh im big dumb i feel stupid now for asking this question lmao

,w <12,9> dot <13,14>

[stoic python]

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Requested by Moonlight ⌬ ∰#9250

[digital bough]

And the magnitude of that vector that isn’t in the plane represent the area of a parallelogram

[wintry steppe]

ah okok that makes alot of senese thank you for the help guys @digital bough @lavish jewel

[lavish jewel]

all good

[meager tinsel]

Alright so suppose I have a a_1x_1 + a_2x_2 + .. +a_nx_n = 0, a_i is integer
so this gives a n-1 dimensional int lattice
how do I find its basis?
Like I can find something, i.e.

1. (-a_2/gcd(a_1,a_2), a_1/gcd(a_1,a_2), 0, ...0)
2. (same for all other pairs with 1
   but I m nit sure if this is actually the lattice and not just sublattice

<@&286206848099549185>
sorry for bothering since its already in the channel(

[desert rapids]

Hm. Your calculations must be wrong because the method is clearly valid.

I didn’t go through the calculations myself

[pure tangle]

@desert rapids my answer was right. I was misinterpreting the final answer because it's a plane formula that's equivilant, but different, from the original equation

For part c can I just do something like a = alpha1 + alpha2 ....
b = alpha2 - alpaha3 ... or does the question imply a different type of equation?

so my attempt at equation was the red

[dense gorge]

can anyone hep me with a question?

[meager tinsel]

Wow I actually cannot find anything about how to solve integer homogeneous linear equation

Is that like...less trivial than what I thought it is?

[limber sierra]

nah but it falls more under the purview of number theory than linear algebra

@meager tinsel

youre asking how to solve a degree 1 diophantine equation

such systems are mostly solved by the chinese remainder theorem

see https://en.wikipedia.org/wiki/Diophantine_equation#System_of_linear_Diophantine_equations

[meager tinsel]

Yes I'm asking exactly that and I know CRT

Lemme check this part

I think I looked through it but maybe I didn't see

@limber sierra okay yes it describes what I want but I have some trouble thinking this late
Can you sort of push me in the direction of
What this turns into when m=1

As in my case

[sonic osprey]

https://en.wikipedia.org/wiki/Bézout's_identity

[meager tinsel]

I know this one

But how do I apply this method described in the article exactly

[sonic osprey]

what method are you talking about

[meager tinsel]

The one linked in the article

Not yours

But previous one

[sonic osprey]

Uh

Okay the method in that article is pretty complicated because it relies on smith normal form which is a kind of generalization of euclidean row reduction

[meager tinsel]

Yes

It's hard to imagine without a paper

Which I don't have atm

But if m=1

Like in my case

I expect it to be feasible?

[sonic osprey]

since the integers are a PID like described. You can follow the process in the smith normal form article to do that

I mean, if m = 1 in your case, its much easier to just apply bezout's identity

[meager tinsel]

Okay then can you explain how do I apply it

To get a basis

[sonic osprey]

read the examples in the bezout article and euclidean algorithm articles

[pure tangle]

sorry i know this is annoying but can someone please help with my previous question after this person? That's the last problem I have to finish before turning something in that's due soon

[meager tinsel]

@sonic osprey
I can see how to solve it for d /= 0

But I knew that already

I.e. when it's not homogeneous

But I cannot find anything about how to solve the homogeneous part

[sonic osprey]

Uh what?

If I'm understanding you correctly

you can solve the case when d = 0 in exactly the same way?

You have the trivial solution of all zeroes

[meager tinsel]

I think it will yield 0 0 0 ... solution

[sonic osprey]

and from there, you can find all the solutions by using

[meager tinsel]

Yes for 2 variables

But what is the formula for n variables

(Which is the question I was asking the whole time)

Because the formula for n variables is a basis of lattice in question basically

[sonic osprey]

It's the same

[meager tinsel]

Wdym the same

[sonic osprey]

You can just use the fact that gcd(a,b,c) = gcd(a, gcd(b,c)) for example

to do bezout's/euclidean algorithm for n variables

[meager tinsel]

...
Okay maybe I'm extremely stupid but can you help me out and write that formula for say 3 variables

[sonic osprey]

what formula?

[meager tinsel]

This

I dunno how else can I possibly explain what I want

[sonic osprey]

So the idea is that you can rewrite ax + by + cz = d as

[meager tinsel]

Ye I understand that part
You can get to the ax+by+cz=0

And solve it

[sonic osprey]

ax + g((b/g)y + (c/g)z) = d if we let g = gcd(b,c). So you can introduce a dummy variable say k and let k = (b/g)y + (c/g)z so your equation becomes ax + gk = d

Now it's just two variables and you know how to solve that

[meager tinsel]

Okay and you are saying that will work with d=0 just fine

[sonic osprey]

the d = 0 case isn't special

[meager tinsel]

Well I don't want 000 solution

[sonic osprey]

Well you can use the picture I sent to find all the solutions by using the (0,0,0) solution

[meager tinsel]

I don't see how exactly to extract 2 dimensions from this

[meager tinsel]

This

Because the solutions space is supposedly 2 dimensional

[sonic osprey]

uh yeah

like in my picture, the solution space is 1 dimensional since you have a single arbitrary integer k

[meager tinsel]

Lemme try it
So I will have basically
ax + g((b/g)y + (c/g)z) = 0

And for that I ll have to do
l = gcd (a, g)

So I'll have
x = -g/l*coef
And

(b/g)y + (c/g)z) = a/l*coef

Which I then solve.....?

I guess

I'm unsure

I did an edit

[sonic osprey]

Yeah thats the right idea

[meager tinsel]

But how do I solve this

(b/g)y + (c/g)z) = a/l*coef

For every coef

Or whatever

Is there a way explicitly write 2 vectors?

Like I could do with ax+by=0

[sonic osprey]

No, you can only explicitly write the vectors in that case because you already start with the (0,0) solution. Otherwise, like the picture states, you need to start with some solution for (y,z) to be able to write down the vectors

[meager tinsel]

Okay that makes sense

[sonic osprey]

but if you find a solution, then you can write down the vectors

[meager tinsel]

I guess the idea here is that I need to solve this
(b/g)y + (c/g)z) = a/l
And than multiply solutions by coef

[sonic osprey]

in this case, gcd(b/g, c/g) = 1 so you can find a single solution to (b/g)y + (c/g)z = 1 and then have a solution for all (b/g)y + (c/g)z = k

[meager tinsel]

Oh hm

Yea you right they are already coprime

That's good
And so say I will get like
Cy0 + pn,C z0 -qn
Solution
For this
(After multiplying by whatever is needed)
And then I'll have this * some other coefficient
And something for x
It does feel like I'll get some product of parameters somewhere instead of liner sum

But maybe not..?

I ll think about it

[zealous junco]

I dont understand 2 parts of this proof

first is why the 1st line becomes second line

[pure tangle]

<@&286206848099549185>

[zealous junco]

oh yea this makes sense thx

other part is

how did they get the last line from 2nd last line

oh wait i see it

some linearity stuff

[stoic python]

slimvesus

[zealous junco]

yep, thanks!

[pseudo thicket]

can someone explain this part?

[wintry steppe]

it's just checking that the space of polynomials of degree at most n form a vector space

with the "obvious" operations

[pseudo thicket]

what does it mean by p(t) = a0 != 0,degree of p?

[wintry steppe]

if p is a nonzero constant then you say it has degree 0

[pseudo thicket]

wait so it's just saying purely p(t) is just = a0

not p(t) = a0+a1t+a2t^t... ?

[wintry steppe]

IF it so happens that a polynomial is a nonzero constant

the thing on the right is defining degree

[pseudo thicket]

what is pt(t) = 2 + 3t + 4t^2 for example?

the degree is 2?

[wintry steppe]

what is the degree of that?

yes

[pseudo thicket]

then simply p(t) = 4 the degree is 0

[wintry steppe]

yes

[pseudo thicket]

is that what the slide is conveying?

[wintry steppe]

that's what the definition of degree is, yes

[pseudo thicket]

thank you so much

[wintry steppe]

[pseudo thicket]

Is the difference between Vector space and subspace

for example Vector Space it is 'larger' than the subspace.. well because subspace is a sub of the larger vector space?

[wintry steppe]

"subspace" implies there's a larger vector space in which it lives

[pseudo thicket]

is a vector space bigger than subspace?

R2 is not a subspace of R3, right?

[wintry steppe]

right

since R^2 isn't even a subset of R^3

any subset of a vector space that's also a vector space with the same operations and zero vector is called a subspace.

[pseudo thicket]

Okay got it

So vector space has the 3 properties:

0 vector
scalar multiplication
and two vectors u and v added lives in the space

and a subset of a vector space that also has these 3 properties is called a subspace

is that right?

[wintry steppe]

a vector space also places some restrictions on those operations of scalar multiplication and vector addition, but largely yes

(e.g. associativity, commutativity, distributivity)

[pseudo thicket]

got it thank you guys

Then R^3 is a subspace and a vector space itself, in this case?

[digital bough]

Holy shit I only just realised this now. Why in gods name did I think (1,0) is the same as (1,0,0)

[pseudo thicket]

R3 is a vector space

if we have 2 vectors in R3, the span of these 2 linearly independent vectors is a plane

the plane is a 2D, it is a subspace

is this right so far?

but doesn't R2 vectors create a plane too?

so why is R2 not a subspace of R3?

Is it because we cannot set the "level" of the plane created by R2 in the Z axis?

i am trying to get a geomteric understanding

since technically R3 has a "copy" of R2 with the z being 0

[pseudo thicket]

does this even make sense lol

2d plane is just 2d plane, but we cannot specify which level the plane is at (z axis)

lets say this is a R2

this is r3

the yellow plane isn't "flat" like the blue one.. we can choose the "angle" of the plane

hence why R2 isn't a subspace of r3, but span of 2 LI of R3 is

Thank you

this is so confusing

[digital bough]

Don’t worry I just learned that now and I’ve been doing LA since 20 january lol. The mistake I guess was that I kept thinking of sub spaces of r3 and r^2 geometrically although I never thought of r2 being a subspace to r3 but for some reason I thought it was a subset. All exercises regarding subspaces all just stated “this is a subset of V, prove it is a subspace” etc, i never actually encountered an ex asking if V is subset of V2.

[weak glade]

How can I find the Dimension and a base of this supspace? For T i can find, but i don't know how to work with S

[lavish jewel]

can you recognize the definition that is given there?

i'll rewrite it for you to see if that helps

$S = {v \in \mathbb{R}^3 , \vert , \langle v, n \rangle = 0 }$

[stoic python]

Edd

[weak glade]

And how can I write the vectors? I havent worked with this form yet

[lavish jewel]

they're just generic vectors in R3

this one has infinite solutions

first think about the dimension

do you recognize the definition you were given?

$\langle v, n \rangle = 0 = x_1 - x_2 + 2x_3$

[stoic python]

Edd

[weak glade]

do you recognize the definition you were given?
@lavish jewel i dont understand what is n

[lavish jewel]

well, ignore that and look only at the definition you were given. what is that the equation for?

[weak glade]

So the Dimension is 3 and a base can be the canonic base?

[lavish jewel]

no

the equation they gave you is the equation of some geometric shape

[weak glade]

Hmmm

[lavish jewel]

do you know the definition of a plane?

[weak glade]

Ooo yes

The vector space is a plane s-o Dimension is 2?

[lavish jewel]

yes

[weak glade]

And how can I find a base?

I need two liniar independent vectors but i dont know how to find them

[lavish jewel]

there are infinitely many, but my thinking is that you can pick any one vector that satisfies the equation

literally any

and then take the cross product of the normal vector and this vector you just made up

if v = (x1,x2,x3), the normal vector should be evident in what they gave you, which is why i gave you the definition of the plane as a dot product

[weak glade]

The normal vector is (1,-1,2)

[lavish jewel]

mhm

now make up any vector that is orthogonal to that, which would mean that vector is on the plane

-1, 1, 0 could work, for instance

[weak glade]

Thanks

[urban egret]

Hi I have a question trying to conceptually understand transformations. Why does the transformation matrix made from a basis transform onto the standard basis? and not from the standard basis to the basis?

[mossy lodge]

Check briefly that $\alpha(v) = \lambda v$ is equivalent to $( \alpha - \lambda \cdot I )(v) = 0$.

[stoic python]

PandaMan-AMB

[mossy lodge]

There we go, I dont understand this notation, if the mapping is defined by \alpha, then how can you make a mapping for (\alpha - \lambda * I)

This is new to me

[dusky epoch]

if you have two maps f and g then their sum f+g is the map which sends x to f(x)+g(x)

pointwise addition, as you may call it

[mossy lodge]

Ah ok, but sadly I dont have two maps, The map is F x V -> V

[dusky epoch]

??

what are you talking about

[mossy lodge]

I only have one map

[dusky epoch]

you have alpha and -lambda*I

those two maps are the ones being added

[mossy lodge]

I dont have a map for lambda

[dusky epoch]

I refers to the identity map

accordingly, λI is the map which scales every input by λ

[mossy lodge]

Oh ok, I get it now