linear-algebra

[stuck tendon]

For 3c, recall ker L = { (u1,u2) in R^2: L(u1,u2)=0}. You can use that L(u1,u2) = (u1,u1+u2,u2,u1-u2). Hence if (u1,u2) is in ker L, we have (u1,u1+u2,u2,u1-u2) = (0,0,0,0), whence u1=u2=0. So ker L = {0}

[spare widget]

it's the solution set of the homogeneous equation

i.e. all vectors that get mapped to 0 by your linear operator

solution as in set of vectors

let's have an example

n dot v = 0

then the solution set of the above is all vectors v perpendicular to n

say you have now

A1 dot v = 0
A2 dot v = 0
...
Am dot v = 0

then the solution set of the above is all vectors v perpendicular to A1, A2, ..., An

if you stack A1,...,Am in a matrix as rows you have

A v = 0

A = [A1; A2; ...; Am]

[stuck tendon]

From 1st comp, u1 =0, from third comp, u2 = 0. Check with others: u1+u2=0+0=0. u1-u2=0-0=0. So it works.

[spare widget]

the solution is the intersection of all hyperplanes

with normals A1, ..., Am

[stuck tendon]

No. You can read off the solution

To find the kernel, you need to solve L(x) = 0 for x in the domain.

For example, for b, you would say (u1 + u3, u2 + u4) = (0,0). This would give u1 = -u3, u2 = -u4.

So the kernel is {(-u3,-u4,u3,u4)} = {u3(-1,0,1,0) + u4(0,-1,0,1)}

[spare widget]

usually you don't have such a nice problem structure

so it's good to practice some row reduction

it's essentially what 1345631 is doing but he's doing it without the matrix

the benefit of row reduction is that you do not have to think

[stuck tendon]

You didn't post the actual task for exercise 4. The picture just shows a sentence of information

If the question is to calculate L(1,1,1) then use linearity. L(1,1,1) = L(1,0,0) + L(0,1,1)

Then compute it.

[spare widget]

rref with [A | I]

Then at the end you will have [I | A^{-1} ]

since you are technically computing A X = I

Then X = A^{-1}

Do you know how to solve a system of linear equations with an augmented matrix i.e.

Ax = b -> [A | b]

Well the only difference is that instead of vectors we put matrices in there

AX = I

[A | I] -> [I | X]

And the X that you get in the second part is the inverse A^{-1}

You form the augmented matrix [ A | I] and then rref

Once you do rref you end up with [ I | X]

X is A^{-1} provided the inverse exists

[wintry steppe]

y'all talkin to yourselves?

[stuck tendon]

No, it was to Student212. They suspiciously deleted all their messages

[wintry steppe]

i know

[clear kettle]

nope they deleted their messages because no one was answering others

[clear kettle]

What does the eigenvector help you do

[wintry steppe]

diagonalization is one

[clear kettle]

yeah ik

subtracting by lambda and finding the determinant

[plush dust]

Hi, is there a fast way to write a similar matrix to give matrix A?

[meager harness]

how to to 4

[spare widget]

Lagrange multipliers

[restive raft]

If this is a linear algebra course/problem they probably want you to find the eigenvalues and eigenvectors of the symmetric matrix associated with the quadratic form then the maximum/minimum values are the max/min eigenvalues and the eigenvectors(normalized since it's on a sphere) are where they occur.

[wintry steppe]

secret lagrange multipliers

[meager harness]

how to do 2c) and 2d)

[wintry steppe]

probably by diagonalizing

[subtle gust]

why is it that a homogeneous linear system of equations having more unknowns that equations always has infinitely many solutions?

i understand it's cuz the coefficient matrix is not invertible

but why didn't we draw the same conclusion for homogeneous linear systems having more equation than unknowns

[spare widget]

because the rank of a mxn matrix can at most be min(m,n)

[subtle gust]

oh

yeah got u

ty

[spare widget]

x^TB^TBx = (Bx)^TBx = y^Ty > 0

y = Bx

since B is invertible, if x!=0 then y=Bx != 0

that's necessary to provide positive definiteness instead of positive semi-definiteness

if B was not invertible

then B^TB is still symmetric and positive semi-definite

in fact B doesn't even need to be symmetric for B^TB to be symmetric and positive semi-definite

I don't know why the mention that B is symmetric actually, not sure what the point of that is

[tame pond]

I think the point is that real symmetric matrices are diagonalizable

so in the diagonal form B^2 only has non negative eigenvalues

then the invertibility gives positive definiteness

[spare widget]

I don't think I need symmetry for B^TB to be symmetric positive semi-definite

so the symmetry of B itself felt redundant

i.e. I don't know why it is in the problem statement

to confuse the students?

[restive raft]

why are you talking about B^TB the question is about B^2

[spare widget]

Because the question could have as well been for B^TB without B being symmetric and it would have been more general

[tame pond]

My impression of the problem was that they work on F = R only and they have a theorem about real symmetric matrices, namely that they are diagonalizable.

Which is admittedly a very negative view of the way they are being taught by being so presumptious

[spare widget]

In practice you'll run into B^TB all the time, even with B non-square. And B^TB is symmetric positive semi-definite even then. Having B symmetric is a wasted opportunity (and confusing why you would even need it in the derivation except for them using B^2 there, but B^2 = B^TB for B symmetric anyways).

[vivid field]

How would I set up number 3 ?

[dawn ocean]

multiply each matrix by a variable, add them together and set them equal to zero then you have a system of 3 equations that you solve if there is a solution other than all of your scalar multiples are 0 then they're not linearly independent

they're not linearly independent becuase you can do like 6 the first vector -5 the second +2 the third and that will equal zero

[vivid field]

Would this be how you work it out

[dawn ocean]

it's redundent to make a matrix with all those zeros and also you messed up with the first vector you have an extra one in htere otherwise your work is fine

[vivid field]

dang. yeah your right

how would the work be when trying to determine span ?

[dawn ocean]

it's a yes/or no question so you don't need to do any work

the reason is because D(2,2) is four dimensional

and you have 3 vectors

so the answer is no

you need at least 4 vectors

and also the other reason you don't need to do work

is because you already check the linear independence even if there were four vectors they wouldn't be independent and wouldn't span all of your space

[vivid field]

Thanks!

[vivid field]

I am confused on what this is asking me?

[wintry steppe]

what's not clear?

i feel like what it's asking you is pretty explicit

[stoic python]

TTerra

[vivid field]

What would I show to prove it ? just that ? that what I am not sure about

[wintry steppe]

you figure out what a subspace is and you prove that this is one

[restive raft]

Proof? What is this “proof” thing you speak of

[burnt rapids]

You would show that all linear combinations of 4x4 diagonal matrices are also 4x4 diagonal matrices

Wait. It tells you to show it's a subspace of M_4x4.. Well in that case you show that all linear combinations of 4x4 diagonal matrices are 4x4 matrices.

[stuck tendon]

$W$ being a subspace of $\mathbb{M}(4,4)$ means that $W$ is a non-empty subset of $\mathbb{M}(4,4)$ (or equivalently, one that contains the zero vector) that satisfies:

• closure under addition: for any $v,w$ in $W$, $v+w \in W$.
• closure under scalar multiplication: for any $\lambda \in \mathbb{F}$, where $\mathbb{F}$ is the field, we have that $\lambda v \in W$.

[stoic python]

1345631

[stuck tendon]

So your first statement was correct. You need to show any linear combination of diagonal matrices is a diagonal matrix

[burnt rapids]

And the second statement was incorrect?

[stuck tendon]

I'd say it's imprecise. It's technically not wrong since 4x4 diag matrices are 4x4 matrices. What's needed is that the linear comb of diag matrices are still diag

[worn hawk]

Linear algebra awesomeness

So you just started to learn linear algebra. You see simultaneous equations, you are happy. You tell your self this is a territory I know. You see matrices, another old friend. You see vectors, another childhood friend.

By the time you are comfortable, you start using the Gaus elimination. And then you slowly start moving into familiar yet uncharted territories. You visit determinants, you move into all kind of spaces, vector spaces, abstract spaces.

From two dimensions, to three dimensions to n dimensions, you start observing these spaces, what is normal, what is not. And does this really apply to higher dimensions.

You start transforming spaces, you start asking questions. How does a 3d vector look like in a 2d space.

Now when you look at vectors, you can see them as matrices.

You start splitting vectors, to learn the truth within.

You are nearly reaching the end of your journey. You are asking a lot of questions.

You reached the end. You ask where to go from here? What can I do with the power in my hands?

PS : How do I tell someone that linear algebra is awesome?

[spare widget]

Show them a tangible application of it that they are interested in

I doubt poetry will do it

[worn hawk]

Mostly computer science students

[spare widget]

Then pick CS applications that the students may be interested in

I personally do graphics and image processing, so for instance I would show such students applications from graphics and image processing

e.g. image reconstruction by solving a discretised multiharmonic equation

[worn hawk]

The only one which I have an idea is data science

[spare widget]

Or transformations in a graphics engine

Or finite element methods for light transport

[worn hawk]

Can you point to some resources on these?

[spare widget]

https://youtu.be/7UaF-Xvg5Eg

https://youtu.be/UwPkWXQ6ORY

https://youtu.be/TbWQ4lMnLNw

Light transport (based on the rendering equation) is linear, i.e. L = Le +TL in operator notation

one considers the L2 space and the solution per pixel is an inner product <We,L> where W is the sensor fubction and L the radiance function

Image reconstruction from sparse data using the harmonic equation (esssentially the solution of a linear system): https://www.shadertoy.com/view/3dGSDw

I don't think it works on mobile though

This is also linear: https://www.shadertoy.com/view/MlcfRM

You have to click to make a wave

Intersections with linear or affine primitives (e.g. triangles): https://www.realtimerendering.com/intersections.html

transformations in graphics engines: https://learnopengl.com/Getting-started/Coordinate-Systems

[worn hawk]

Wow. This is so interesting

Even I am interested

[spare widget]

Basically pick showy stuff that the students may like, and show them where and how linear algebra is used.

It's not very useful for core CS stuff like compilers or algorithms and data structures though.

Physics simulations solve linear system often even when the model is nonlinear (e.g. it gets linearized)

So simulations like this: https://youtu.be/IPayi38vQws

[spare widget]

interactive image processing using fft (it's linear): https://david.li/filtering/

Light transport in 2d also interactive (the rendering equation is linear and the light transport operator too): https://benedikt-bitterli.me/tantalum/

This is also probably good as an intro for CS students: https://youtu.be/2c8XQlQApx8

For students interested in neural networks (matrix calc): https://youtu.be/tIeHLnjs5U8

For physics students you can show them einstein's equations and explain the tensors there etc

[spare widget]

for example: https://www.youtube.com/watch?v=8ptMTLzV4-I&list=PLJHszsWbB6hrkmmq57lX8BV-o-YIOFsiG

this also: https://youtu.be/kGXr1SF3WmA?list=PLJHszsWbB6hpk5h8lSfBkVrpjsqvUGTCx&t=408

[slate hatch]

can a polynomial a+bx+cx² be written as a vector like this: (a,b,c)

?

[wintry steppe]

sure. rigorously speaking, the space of second degree polynomials (with coefficients in R, say) is isomorphic to R^3 by the map you've written down

[spare widget]

You can use $\frac{1}{k!}\left(\frac{d}{dx^k}(\cdot)\right)(0)$ to extract the k-th component (counting from 0). Just substitute dot with the polynomial.

[wintry steppe]

divide by k factorial

[spare widget]

Yeah forgot

[stoic python]

criver

[spare widget]

This is in one direction, in the opposite direction just make the linear combination of the monomial basis.

[clear kettle]

Hey guys can you explain how to do this

for number 1 I was thinking of a basis for ker L [1 0 ] but i'm not sure

[wintry steppe]

what does $P_{2}$ stand for ?

[stoic python]

AimaneSN

[slow scroll]

in the context of linear algebra, P_n is usually the vector space of polynomials of degree <= n

[wintry steppe]

I see

[wintry steppe]

For the kernel of L suppose you have to find which vectors span polynomials P such that L(p)=0 e.g. $t^{2} p'(t)=$0 for all t

[stoic python]

AimaneSN

[wintry steppe]

looks like Ker L = all constant polynomials

[clear kettle]

so we just set it to 0

idk

t^2 + 2t + 1?

[wintry steppe]

I think Ker L = span{1}, but not sure.

[wintry steppe]

wdym

[clear kettle]

idk an example of a polynomial

[wintry steppe]

but this is not a constant polynomial

[clear kettle]

for p'(t)

no but what would be p'(t)

[wintry steppe]

do you know what the notation p'(t) means?

[clear kettle]

the derivative of a function p(t)

[wintry steppe]

correct

now what does it mean for p(t) to be in the kernel of L

[clear kettle]

span of the vectors for p(t) = 0.

[wintry steppe]

i don't understand what you mean

what is L? you should answer in terms of that

[clear kettle]

i'm referring to this defintion

L is a linear function right

[wintry steppe]

what does it mean, in terms of L, for a polynomial p(t) to be in the kernel of L?

that's right

[clear kettle]

I think it is a set of vectors that are equal to 0

[wintry steppe]

no

please say it precisely

do you know what the kernel of a linear map is?

[clear kettle]

Not yet

[wintry steppe]

then how are you supposed to do the question asking you to find the kernel of a linear map?

you should review your notes and come back to this one

[clear kettle]

refer to the examples from the book and do the same

[wintry steppe]

you can't just copy the steps from an example in the book and expect everything to work out

you should actually learn what the things you're being asked to compute are

[clear kettle]

The kernel of L, ker L, is the subset of V consisting of all elements v of V
such that L(v) = Ow.

[wintry steppe]

so just L(v) = 0, right?

[clear kettle]

yeah

[wintry steppe]

so p(t) in the kernel of L means...

[clear kettle]

p(t) = 0

[wintry steppe]

no

[clear kettle]

I want to learn the definition but I don't understand it enough

[wintry steppe]

the definition is "v is in the kernel of L if L(v) = 0"

in your exercise, L(p(t)) = t^2 p'(t). what does it mean for p(t) to be in the kernel of L?

[clear kettle]

all elements in the equation should make the function 0?

[wintry steppe]

i don't understand what that means

i want you to say L(p(t)) = 0

do you agree?

[clear kettle]

yes

[wintry steppe]

alright

[spare widget]

what is the vector v in the above (when you compare to L(v) = 0), think about that

[wintry steppe]

now, L(p(t)) = t^2 p'(t), so we have t^2 p'(t) = 0

alright?

[wintry steppe]

I think i's important we should add for all t. but whatever

[wintry steppe]

it doesn't matter

[clear kettle]

alright

[wintry steppe]

cool

[clear kettle]

so L(p(t)) = 0

[wintry steppe]

so now you have to figure out what this means in terms of p. this is where you might have to do a little bit of working

[clear kettle]

ok

I want to imprint this definition in my mind

[wintry steppe]

you should, it's one of the most important in linear algebra

"things that a linear map sends to zero"

the problem tells you that p(t) is a quadratic, so you might as well write it as p(t) = at^2 + bt + c. what you should do is figure out what you can say about a, b, c if t^2 p'(t) = 0

alternatively, if you know (and are allowed to use) some calculus, you don't have to do this nonsense with the coefficients (what does it mean for a function to have derivative zero everywhere?)

[clear kettle]

You know it's a quadratic because of t^2 right?

[wintry steppe]

i know it's a quadratic because the problem says L is a map from P_2 to something

P_2 being the space of quadratic polynomials (or less, before some pedant corrects me)

[clear kettle]

I wish to use calculus but i don't know i have to ask the professor

how can i visualize a linear map

[wintry steppe]

in many ways

[spare widget]

like so for R^2: https://www.youtube.com/watch?v=kYB8IZa5AuE

[wintry steppe]

i don't know if it counts as visualization, but any linear map can be written as a matrix, after choosing bases

[worn hawk]

https://www.3blue1brown.com/lessons/inverse-matrices

[wintry steppe]

jesus christ

3b1b shills in chat otday

[clear kettle]

What would it look like as a matrix , how many columns and rows do i know to put.

[spare widget]

L : V -> W -> your matrix has dim(W) rows and dim(V) columns

you need to pick bases in V and W to produce a matrix for L though

[spare widget]

if you prefer I can "shill" Bazett instead: https://www.youtube.com/watch?v=On6wkamacRE

[clear kettle]

Do i pick random numbres

[spare widget]

no, you do not pick random numbers

[clear kettle]

how do i solve the equation for 0 and what about the derivative of the function.

[wintry steppe]

by doing this

you don't need to write anything as a matrix to do this one

(not to dismiss the usefulness of matrix representations)

[clear kettle]

it could be easily be 0 but I don't think it is

[wintry steppe]

well, 0 is an option, but of course it is (0 is in the kernel of every linear map)

if p(t) = at^2 + bt + c, figure out what p'(t) is, then figure out what t^2 p'(t) = 0 implies about a, b, c

[clear kettle]

how did you know you had to write the original equation without the derivative first

or would it make more sense that way because we need to know first how it looks like before

[wintry steppe]

i don't know what you mean

which equation?

[spare widget]

TTerra picked an arbitrary vector from P_2

p(t) = at^2 + bt + c

[clear kettle]

why did you need to write the function without it being derived for p(t) = at^2 + bt + c i mean

[spare widget]

is an arbitrary vector from P_2

[wintry steppe]

.

[spare widget]

if the exercise said P_n

they would have picked

$p(t) = \sum_{k=0}^n a_kt^k$

[stoic python]

criver

[spare widget]

just by the definition of what P_n is

[clear kettle]

so you chose the equation to find the kernel?

okay so the derivative is 2at + b ?

[wintry steppe]

no

[spare widget]

you pick an arbitrary vector from the space

and then check what constraints you need on it

[wintry steppe]

still not right

[spare widget]

for L(p) = 0

[clear kettle]

haven't done derivatives in a while

[wintry steppe]

there you go, it's 2at + b. now substitute that into t^2 p'(t) = 0.

what does that imply about a and b?

[clear kettle]

but now i remember lol

a + b = 0?

[wintry steppe]

no

what is t^2 p'(t)? you know p'(t) = 2at + b

[clear kettle]

yes

do we multiply by t^2?

[wintry steppe]

yes/

and what happens if the result of that equals zero?

[clear kettle]

you get a polynomial

or vector

[wintry steppe]

well, sure, but which one?

please write out t^2 p'(t) in terms of a and b

[clear kettle]

2at^3 + t^2b = 0

[wintry steppe]

YES

and what does this imply about a and b?

[clear kettle]

they are 0? scalar multiples?

[wintry steppe]

the first one.

so a and b are zero

[clear kettle]

oh. I have a question

[wintry steppe]

so you've proven that if p(t) is in the kernel of L, then it only has a constant term.

[clear kettle]

Why cant t be 0 as well

[spare widget]

you have to have the above hold for all t

not just t = 0, but t can be zero

[clear kettle]

so for the definition of the kernel when they say elements they mean the ones in the equation

[wintry steppe]

the ones satisfying L(p(t)) = 0, yes

[spare widget]

they mean all polynomials that satisfy this, and you just found out that such polynomials have a = b = 0

[wintry steppe]

can you show yourself that every constant polynomial is in the kernel of L? i think this should be straightforward enough to do yourself. then you've successfully shown that the kernel of L is precisely the set of all constant polynomials.

[clear kettle]

the p(t) in L is what the derivative was right?

[wintry steppe]

no, but it's something we're taking the derivative of (which the definition of L demands we do)

[clear kettle]

that's what i mean we're taking the derivative of

so yes?

[wintry steppe]

if that's what you mean, yes.

[clear kettle]

yes, ok.

[wintry steppe]

now this is how you should proceed

you've done the hard step

[clear kettle]

So we still have to prove

[wintry steppe]

once you've convinced yourself that the kernel of L really is the space of all constant polynomials, you need to find a basis of this

[clear kettle]

idk the dimensions to find it

i'm assuming it needs to be 2 columns at least

[wintry steppe]

.

[clear kettle]

i thought a basis was a set of vectors

lol

[spare widget]

your vectors are polynomials here

[clear kettle]

what does it leave if a and b are 0

[wintry steppe]

constnat polynoamisl

[clear kettle]

hmm

t^3 + t^2?

[spare widget]

you have p(t) = a t^2 + b t + c, you set a=b=0, what do you get

[clear kettle]

p(t) = c

[spare widget]

then what's in the kernel of L?

[clear kettle]

c

a and b are 0

[spare widget]

example q(t) = 3 is in the kernel, r(t) = -5 is in the kernel, s(t) = 0 is in the kernel

i.e. any constant polynomial is in the kernel

the kernel is the set of constant polynomials (all of them)

[clear kettle]

i thought it was a vector equals to 0

[spare widget]

check L(q) =? 0

or more generally L(c) =? 0

the kernel is associated with the linear map

the kernel itself is the set of vectors

which the linear map sends to 0

[clear kettle]

L(c) = 0?

[spare widget]

and now you need to find a basis for this space of constant polynomials that constitute the kernel

[clear kettle]

p(t) =c right?

[spare widget]

I don't get the question

[clear kettle]

.

[spare widget]

well that's clear

[clear kettle]

aren't you left with c

[spare widget]

I think we established that

and that this is the set of vectors in the kernel

now you need to find a basis for the kernel

for that you have to remember the definition of basis

[clear kettle]

so p(t) = c was the kernel because L(p(t)) = 0

ok

[wintry steppe]

if p(t)=c then p is just one element of the kernel, not all of the kernel

[clear kettle]

ok

[spare widget]

I think it would have been easier on you if you had tried a problem that involves vectors from R^n beforehand. Directly jumping to polynomials while notions like kernel are not solid is probably difficult.

[clear kettle]

maybe yeah.

what is the definition of the basis for the kernel

How do you determine the basis from what we know

[spare widget]

what is the definition of a basis?

[clear kettle]

set of vectors

[spare widget]

more precisely

the vectors need to be linearly independent and span the space

[clear kettle]

i want to imagine linearly independence on a graph

[spare widget]

what vector(s) span the space of constant functions

[clear kettle]

and also do you mean the space in this case P2 -> P3

[spare widget]

remember than span ${v_1,...,v_n} = {\sum_i \alpha_i v_i ,:, \alpha_i \in \mathbb{R}}$

[stoic python]

criver

[spare widget]

so what vector(s) span the kernel (the space of constant functions)

[clear kettle]

v1 v2?

[spare widget]

what are v1,v2?

don't just try random answers

if something's unclear ask

[clear kettle]

2at^3 and t^2b

yeah

[spare widget]

why did you pick 2at^3 and t^2b

I'll give you a hint

[clear kettle]

I don't know

It's because a and b would have to be multiplied by a vector

?

[spare widget]

if you have a 1D space, then it looks like this ${\alpha v_1 , : , \alpha \in \mathbb{R}}$

[stoic python]

criver

[clear kettle]

so cv1?

You determined it was one vector because there is only one variable c?

[spare widget]

in your case the kernel is ${p(t) = c ,:, c\in\mathbb{R}}$

[stoic python]

criver

[spare widget]

what can you choose as a basis vector

what polynomial multiplied by a scalar gives you a constant polynomial

what function multiplied by a scalar gives you a constant?

[clear kettle]

t^2?

[spare widget]

let's multiply t^2 with a scalar

alpha * t^2

is this the constant function?

[clear kettle]

no

[spare widget]

ok, so what function multiplied by a scalar gives you the constant function

[clear kettle]

would it be 0

[spare widget]

let's try

alpha * 0 = 0

[clear kettle]

or 1

[spare widget]

it gives you one constant function, namely the 0 one, but the issue is that it doesn't give you any other constant function

let's try 1

[clear kettle]

a constant is when it doesn't change right?

[spare widget]

alpha * 1 = c

what is alpha then

so one option is to pick 1

another option is to pick any constant except 0

let's say I pick d != 0

then alpha * d = c -> alpha = c / d

and I can reproduce any constant

I highly suggest scaling back to R^n

figuring out notions like linear combination, span, basis, kernel, etc. there

[clear kettle]

what is a constant lol

[spare widget]

a constant function is one that has the same value for all t

e.g. p(t) = 5

or p(t) = 3.14

or p(t) = 0

or p(t) = the same real number for all t

[clear kettle]

how do you know in this case

[spare widget]

how do I know what?

[clear kettle]

t

what t is to get those numbers

[spare widget]

a constant function has the same value for any t

e.g. p(0) = p(1) = p(t) = c

[clear kettle]

how do i find the constant lol

other than p(t) = c

[spare widget]

Let's say I pick as a basis the constant function v(t) = 2

[clear kettle]

.

The constant thing was just confusing

[spare widget]

Then let's say I have a vector p(t)=c in the kernel

Then c/2 * v(t) = p(t)

Since c and consequently p(t) was arbitrary, then v(t) spans the kernel

I could have picked any other constant basis function not equal to 0

e.g. w(t) = 1

Then c * w(t) = p(t)

Then w(t) is another possible basis

[clear kettle]

so what doesn't work is what i'm asking

and how would i know in this problem , what would constitute a basis

[spare widget]

Scale back to R^n and solve some problems

[clear kettle]

can you give an example like what.

[spare widget]

Do you have a textbook?

[clear kettle]

yes

[small vigil]

I am trying to show the following assertion:
Suppose that any vector of ${y_1,\dots,y_{m+1}}$ over the vector space $X$ can be denoted by the linear combination of vectors ${x_1,\dots,x_m}$ over $X$. Then show that ${y_1,\dots,y_{m+1}}$ are linear dependent.

Could you help me?

[stoic python]

keith_1

[clear kettle]

so basically I should have solved problems by hand

[spare widget]

Show that a1 y1 + ... + a_{m+1} y_{m+1} = 0 for (a1,...,a_{m+1}) != 0 by expanding the yi in terms of the xi

[spare widget]

What's the title of the textbook?

I don't know it, but I suppose it has exercise on R^n

Solve some for linear combination, independence, span, basis, kernel

[clear kettle]

ok

[worn hawk]

Kernel, just a term used for solving AX=0

Also check these at your own time for visualisation. https://youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

[spare widget]

@small vigil here's a hint, if yi = 0 -> {y1,...,y_{m+1}} lin. dep., so you are left with the case where none of the yi are 0

expand yi = \sum_j b_ij x_j

Then study \sum_i a_i yi = 0 and find a case where not all a_i == 0

[small vigil]

Yes, I tried following zero of the linear combination as you commented but I have no idea what to do next.

[spare widget]

expand the expression I mentioned

[small vigil]

the linear combination of yi can be changed into that of xi with \sum_j a_j b_ij. If the y_1 to y_{m+1} are lin. indep., all the terms must be zero. But this does not imply a_j == 0 immediately.

[spare widget]

expand the expression properly and group terms by xj

[small vigil]

\sum_i a_i y_i = \sum_i a_i (\sum_j b_ij x_j) = \sum_j (\sum_i a_i b_ij ) x_j?

[spare widget]

Yes

And that whole thing must be equal to 0

But x1,...,xm are lin indep

So what follows for the coefficients in front?

[small vigil]

From the definition of L.I., \sum_i a_i b_ij = 0 for all j?

[spare widget]

Yes

But that is equivalent to B^T a = 0

And B^T is a m x (m+1) matrix

What follows for the kernel of B^T

[small vigil]

There exists a nontrivial solution a?

[zinc timber]

#calculus

[spare widget]

Yes the kernel is non-trivial

So there's an a != 0 that satisfies it

So y1,...,y_m+1 are lin dep

[small vigil]

Thanks a lot for your kind advice. I got it!

[clear kettle]

can anyone give me the solution to 1?

[clear kettle]

.

[zinc timber]

what have you tried

you can't just ask for a solution

[wintry steppe]

i spent almost an hour helping you with (i) and you just want the solution?

[crystal parrot]

Plays the war drum

[spare widget]

It's clear the user didn't take away much from the explanations. My understanding is that they jumped to a problem they were not ready for without solving simpler problems beforehand. So even if we try to explain it, they get stuck on the simpler concepts used to explain the problem.

[wintry steppe]

that's true

not knowing what kernel was was probably a red flag

[spare widget]

That's why I recommended that they just solve some simpler problems from their textbook beforehand, as it is close to impossible to solve a problem asking for bases of the kernel and range when they don't know what a basis is.

[wintry steppe]

assuming people even know what the words mean in the questions they post is surprisingly too high an expectation

[fair wren]

since {x_1, ..., x_m} is a generator of X and with only m elements then its a basis for X?

[spare widget]

Now that you mention it, I noticed that there's nothing about x1,...,xm being linearly independent in the problem statement #linear-algebra message of @small vigil. This doesn't change the proof by much though.

[fair wren]

what I said was right tho?

[spare widget]

If one assumes that $x_1, \ldots, x_m$ are linearly dependent, then without loss of generality let $x_1, \ldots, x_k$ be linearly independent, and any of $x_{k+1},..., x_m$ can be expressed in terms of $x_1,\ldots,x_k$. Then the above system simplifies further, and one gets a matrix $B^T$ with size $k \times (m+1)$, so the null-space is at least of dimension $(m+1)-k$.

[fair wren]

Hm?

[stoic python]

criver

[clear kettle]

wtf have you even read the messages, i can ask for whatever I want , i've been attempting to do this problem for a while and you haven't even contributed.

[fair wren]

idk wym by null-space , u talking 'bout $X = {0}$?

[stoic python]

alexisreen

[spare widget]

@small vigil you should probably modify your solution if these are not assumed to be linearly independent - I guess I didn't read the problem statement properly

[spare widget]

I am talking about #linear-algebra message

[clear kettle]

The explanations won't make sense if it wasn't applied to the whole problem , part of the question was answered.

[fair wren]

I still don't understand what you mean by null-space?

[spare widget]

i.e. if one assumes $x_1,...,x_m$ linearly independent, with $y_i = \sum_j b_{ij} x_j$ then $\sum_i a_i y_i = \sum_j x_j (\sum_i a_i b_{ij}) = 0$ and from $x_1, \ldots, x_m$ lin. indep. it follows $\sum_i a_i b_{ij} = 0$ or equivalently $B^Ta = 0$. But $B^T\in\mathbb{R}^{m\times (m+1)}$ so the null-space is at least of dimension 1, and then there is $a\ne 0, , a\in Null(B^T)$ such that the sum is zero, thus $y_1, \ldots, y_{m+1}$ are linearly dependent.

[stoic python]

criver

[clear kettle]

I think i found the basis for ker L

[fair wren]

@spare widget since {x_1, ..., x_m} is a generator of X and with only m elements then its a basis for X? is this correct?

[clear kettle]

isn't it 2t^3 and t^2?

[spare widget]

that's not in the discussed problem statement, I don't even know where it came from

[clear kettle]

?

[fair wren]

But is it wrong?

they said that every vector in X can be denoted as a linear combination of the {x_1, ..., x_m}

[spare widget]

we haven't assumed that X is of dimension m, if you assume that dim(X) = m, and that x1,...,xm is a spanning set, then it follows that x1,...,xm are linearly independent, and thus it is a basis

[clear kettle]

Is my answer right?

[spare widget]

it's not

[clear kettle]

why not ?

[spare widget]

trying to guess it won't work

[clear kettle]

does it have to include c?

[spare widget]

no, they say that the vectors y1,...,y[m+1] can be written as linear combinations of x1,...,xm

Suppose that any vector of ${y_1,\dots,y_{m+1}}$ over the vector space $X$ can be denoted by the linear combination of vectors ${x_1,\ldots,x_m}$ over $X$

[fair wren]

but the y_i can be written by the combinations of the x_i

also we can deduce that the dimension of X is finite

[stoic python]

criver

[clear kettle]

So what is the basis can you give a hint

i ran out of guesses

[spare widget]

sure, but I can pick y1=y2=...=y_{m+1}, and x1=x2=...=xm

[fair wren]

how is this relevant?

[spare widget]

the above satisfy the problem statement and the xi do not form a basis

the problem statement doesn't say that every vector in X can be written as a linear combination of x1,...,xm

it only says that the vectors y1,...,y[m+1] can

[fair wren]

every vector in {y1,...,y_m+1}

yes but how does falsify that {x_1,...,x_m} forms a basis?

[spare widget]

yes every vector in y1,...,y[m+1] are just m+1 vectors

[spare widget]

why would an arbitrary set of vectors form a basis?

[clear kettle]

wait i think i got it

[fair wren]

{y1,...,y_m+1} is also a vector space right?

[spare widget]

no

[clear kettle]

(t^3-2)+(t^2-1)?

[fair wren]

Why?

[spare widget]

y1,...,y[m+1] are just a set of vectors

it's m+1 vectors

it's not a vector space

[clear kettle]

Is it right?

[spare widget]

are we talking past each other, or am I tripping?

[fair wren]

@spare widget

[clear kettle]

I want to know if my answer is right

[fair wren]

which condition does {y_1,...,y_m+1} breaks?

[spare widget]

potentially all conditions

[fair wren]

Hm?

[spare widget]

or at the very least addition and multiplication by a scalar

if I pick arbitrary m+1 points, they do not form a vector space unless they are all 0

then they form the trivial space {0}

[fair wren]

ok ok i see

[spare widget]

for them to form a vector subspace I would need span(y1,...,y[m+1])

[fair wren]

pls explain this to me

whats a span and a generator?

how would they differ?

[spare widget]

A generator of a vector space is also known as a spanning set.

[fair wren]

and what is a spanning set then?

[spare widget]

it's the same thing as far as I understand

[fair wren]

Hm?

a span(y1, ... , y_p) of a vector space X is a generator right?

[spare widget]

again, I think they are exactly the same thing

[fair wren]

Is this correct?

[spare widget]

again, I think they are exactly the same thing

[fair wren]

So its correct?

its a yes / no question lol

[spare widget]

I think means I am not certain, but I know that in some books they are used interchangeably

however generator seems to have many other meanings: https://en.wikipedia.org/wiki/Generator_(mathematics)

[fair wren]

then i assume its correct

@spare widget since X is a vector space of atleast m+1 element

isn't {x_1, ..., x_m} a generator of {y_1, ..., y[m+1]}?

[spare widget]

there's nothing saying that X is a vector space of at least m+1 elements

[fair wren]

Ye typo

[spare widget]

X can as well be {0}

y1,...,y[m+1] is not a vector space

[fair wren]

ah rip

[spare widget]

unless they are all 0

even this doesn't have to be true:

span(x1,...,xm) = span(y1,...,y[m+1])

since span(y1,...,y[m+1]) may be a proper subspace of span(x1,...,xm)

[clear kettle]

no one is able to explain why the answer i had was a basis or not

rip

[fair wren]

so $\exists \lambda_1, ..., \lambda_{m+1} \in \mathbb{K} : \sum_{i=1}^{m+1} \lambda_i y_i = 0 \iff \lambda_i = 0 ,$ this is what to prove

[spare widget]

what to prove?

what are you trying to prove?

[clear kettle]

To prove my answer

[fair wren]

they asked to prove {y_1,...,y[m+1]} are linearly independent

[spare widget]

no, they want a proof that y1,...,y[m+1] are linearly dependent

dependent

[fair wren]

ah dependent

[spare widget]

but we already went over the proof

[fair wren]

im trying to do it too

so $\exists \lambda_1, ..., \lambda_{m+1} \in \mathbb{K} : \sum_{i=1}^{m+1} \lambda_i y_i = 0 \iff \lambda_i \neq 0 ,$ this is what to prove

[stoic python]

alexisreen

[spare widget]

this is also wrong

[fair wren]

hm?

[spare widget]

vectors are linearly dependent if there exists (a1,...,an) such that

sum_i a_i v_i = 0 for (a1,...,an) != 0

[fair wren]

what?

so all the coefficients should be non-zero?

[spare widget]

no the opposite

it's enough that a single coefficient is non-zero

[clear kettle]

i have a question

[spare widget]

what you wrote is lambda_i != 0, which would imply for all i

[fair wren]

I said there is a lambda_i which is non-zero

[clear kettle]

how would you find the basis from just one polynomial

[fair wren]

bro u blind?

XD

[spare widget]

yes but you didn't specify the i

[fair wren]

i said there is a lambda_i from 1 to m+1

[clear kettle]

he isn't reading so yeah

[fair wren]

i is in {1, ...,m+1}

u can see that in the sum range

[wintry steppe]

there'd be less problems if just one person helped at a time

[clear kettle]

Can you help me TTerra

[spare widget]

this says exists lambda1, ... lambdam+1 in K ... <-> lambda_i !=0, but maybe I missed where you specify that there exists i in {1,...,m+1}

[wintry steppe]

no, i'm busy

[clear kettle]

😦

[fair wren]

kek

[fair wren]

ah gotcha

[clear kettle]

like you already know the answer i just want to know if what i did was right @wintry steppe

[spare widget]

the point is that the negation of x_i lin indep <-> sum_i a_i x_i = 0 only for (a1,...,a[m+1]) = (0,...,0) is:
x_i is lin dep <-> there exists (a1,...,a[m+1]) != (0,...,0) such that \sum_i a_i x_i = 0

[fair wren]

so $\forall y_p \in {y_1, ... , y_{m+1} } , \exists \alpha_1 , ..., \alpha_m \in \mathbb{K} : y_p = \sum_{k=1}^m \alpha_k x_k$

this is true?

[stoic python]

alexisreen

[spare widget]

well I would write it as $y_i = \sum_{k=1}^{m} \alpha_{ik} x_k$

[stoic python]

criver

[fair wren]

why?

[spare widget]

so I would be able to differentiate the coefficients corresponding to different yi

[wintry steppe]

oh my god lmao

at this point it's just you two fighting

[clear kettle]

the basis is [1 0]

[fair wren]

oh ok

[clear kettle]

isn't it!!!

??

[spare widget]

they wants to do a proof I already did, we are not fighting

it's not their homework, don't worry

[clear kettle]

Can you answer me then? :/

[wintry steppe]

i don't care if it's homework or not, it's just funny

[clear kettle]

the basis is [1 0] or [2 1]

for the kernel of L

[spare widget]

just use this

[clear kettle]

u guys blame not being to explain things by blaming it on others lool

[fair wren]

then $\sum_{i=1}^{m+1} \lambda_i y_i = \sum_{i=1}^{m+1} \lambda_i \sum_{k=1}^{m} \alpha_{i k } x_k = \sum_{i=1}^{m+1} \sum_{k=1}^{m} \lambda_i \alpha_{i k } x_k$

is this correct @spare widget ?

[spare widget]

yes

[fair wren]

niceee

[stoic python]

alexisreen

[fair wren]

still true @spare widget ?

[clear kettle]

people out here acting like they got phds

[spare widget]

yes from linearity

[fair wren]

ok cool

[spare widget]

You could have spent all those hours practicing problems, and by now you would know what a basis is, what a kernel is, and what a range is. I can suggest textbooks if you want.

[fair wren]

can u suggest me textbooks too?

[clear kettle]

I could practice problems but I have the hw due tonight lol

[spare widget]

unfortunate

[clear kettle]

I get to drop three so it doesn't matter anyway

[spare widget]

hoffman and kunze, friedberg, axler

[halcyon spindle]

Linear algebra done wrong.

[fair wren]

$\sum_{i=1}^{m+1} \sum_{k=1}^{m} \lambda_i \alpha_{i k } x_k = 0 \iff \sum_{i=1}^{m+1} \sum_{k=1}^{m} \lambda_i \alpha_{i k } = 0$?

[clear kettle]

kolman and hill?

[stoic python]

alexisreen

[clear kettle]

They suck ass tbh

[fair wren]

lmfao this name is sus

[fair wren]

@spare widget can u check?

[spare widget]

how did you get this?

[fair wren]

idk just intuition ?

[clear kettle]

Why are germans so interested in linear algebra

[spare widget]

then stop relying on intuition

[clear kettle]

intuition is sometimes stupid

[fair wren]

idk how to progress than

[clear kettle]

I heard someone once give the dumbest proof by relying on their intuition

[fair wren]

still a proof

[spare widget]

the idea is, that if x1,...,xm are assumed to be linearly independent, then c1x1 + ... + cm xm = 0 <-> c1=0, ..., cm=0

now find out what terms are the ci in the above

[clear kettle]

a proof that doesn't work

[fair wren]

why u assume they are linearly independent?

[fair wren]

idc

[spare widget]

because it's the easier case, later on you can prove the case where they are linearly dependent using a small modification of the above

[clear kettle]

don't respond then , your ego is hurt

[spare widget]

it reduces to it

[fair wren]

oh i see so two cases

[clear kettle]

You could have spent all this time practicing problems

[fair wren]

wym? im saying its still a proof

[spare widget]

you'd just have to single out linearly independent vectors from x1,...,xm, wlog x1,..,xk, and the rest should be in their span, then it reduces to the first case with some extra notation

[clear kettle]

It's not a proof , proving nothing isn't counted , is what i mean.

[fair wren]

gotcha

[spare widget]

ignore Student212, they are just having a bad day

[clear kettle]

yeah i'm having a hard time solving this problem you can't explain. My bad for asking a guy with a potato for their pfp.

[spare widget]

welp I can't explain it so, you'll have to find someone that can

[clear kettle]

I am in the process of that.

[fair wren]

i see btw if {x_1, ...,x_m} is linearly independent , then it forms a basis in some way?

[clear kettle]

honestly you should get a life.

[gray dust]

what a lousy convo

[spare widget]

only if x_1,...,x_m spans X does it form a basis for X, otherwise it forms a basis for some subspace

[wintry steppe]

hi rokabe!

[clear kettle]

hi rokabe!

[fair wren]

maybe {y_1,...,y_{m+1}} is the subspace that forms a basis for?

[spare widget]

potatoes grow in the earth, it's our life

[clear kettle]

you are a couch potato

[fair wren]

This is #linear-algebra bruh not potato channel kek

[gray dust]

@clear kettle drop the attitude

[fair wren]

_ i feel the ban hammer_

[gray dust]

@fair wren dont minimod

[clear kettle]

Who do you think you are bruh? I'm dropping my nut sack on this convo.

[gray dust]

the question involves prerequisite knowledge

[fair wren]

wym minimod

[gray dust]

knowledge that tterra & criver have spent way too long trying to convey

[spare widget]

not necessarily, span(y1,...,y[m+1]) could be a proper subspace of span(x1,...,x[m]) depending on how the y1,...,y[m+1] are chosen (equivalently how the aij coefficients are chosen)

[clear kettle]

ching chong ching ba da bing

[fair wren]

ok but it isn't relevant in this problem ig?

[spare widget]

it isn't, but you asked, so I answered

[gray dust]

and now you say they cant explain the problem

[fair wren]

yes i see

[gray dust]

this spoiled behavior, asking to be spoonfed like this, wont fly here

[oblique prairie]

very productive discussion in #linear-algebra today

[clear kettle]

spoonfed my asshole

[fair wren]

u can tell

woah , something is rysing but isn't the sun its the ban hammer

[wintry steppe]

rokabe not banning yet is so funny to watch

[clear kettle]

I love rokabe

[gray dust]

go read whatever book is assigned to ur class, stop pestering tterra/criver, and grow the hell up

[clear kettle]

I did all three, I'm growing the hell up but if my answer was explained then none of this would have happened

Like instead of complaining the problem is too hard to explain and saying to do other problems lol

[gray dust]

based on how much info you had going into the convo, fully explaining the problem wouldve taken half a school semester's worth of time

so no thats bullshit

[fair wren]

i think it was necessary to assume the dependence(resp independance) of {x_1, ..., x_m} since we have no info to use in the main problem right?

[clear kettle]

bullshit excuses. they questioned basic shit and didn't get to the point/

[gray dust]

yeah basic shit that you had a hard time answering

[clear kettle]

Since when does linear algebra end peacefully?

[gray dust]

i wouldnt insult you for not knowing the basics

[wintry steppe]

unfortunately for you, linear algebra pervades mathematics

[clear kettle]

no i didn't have a hard time answering , we just learned this lol

[gray dust]

but i will for repeatedly asking me to clarify things that were better gained from reading the book

[clear kettle]

Honestly who uses the book anymore

[fair wren]

first year engineering's student pain kekw

[clear kettle]

They summarize it in class notes

[gray dust]

all the same

those are resources provided for ur understanding

[spare widget]

in the main problem they can be linearly dependent, but you can reduce this case to the linearly independent case by picking out a set of say k linearly indepndent vectors, wlog x1,...,xk such that span(x1,...,xk) = span(x1,...,xm). Then you need some extra terms in the summation, but the case reduces to the linearly independent one. Then dim(Null(B^T)) >= m+1 - k.

[clear kettle]

It's so easy to criticize something you knew before

Like I would like to bully a chinese person for not understanding english

[wintry steppe]

your shitty attitude is being criticized, not your level of linear algebra knowledge

[clear kettle]

shitty attitude for asking questions and not getting a response? You had an attitude by making fun and not answering me.

[gray dust]

i wouldnt insult you for not knowing the basics
but i will for repeatedly asking me to clarify things that were better gained from reading the book

[clear kettle]

The book is so long

[gray dust]

that sucks but we're not a substitute for the book

[fair wren]

still i don't understand this span(x1,...,xk) = span(x1,...,xm)

[gray dust]

ok im done making my point

[fair wren]

ban hammer activate

[gray dust]

ur muted for 24h, go read the notes in the meantime

[spare widget]

it just means that x[k+1],...,x[m] can be expressed in terms of x1,...xk.

[gray dust]

come back if u want clarification on any concepts afterward

[spare widget]

If I have a set of linearly dependent vectors: x1,...,xm, I can start taking away vectors such that the dimension of the span is not reduced. This procedure will end when I cannot take away anymore vectors without reducing the span. So at the end I will end up with a subset of the initial linearly dependent vectors, and that subset will be linearly independent.

[gray dust]

also hi tera, late reply but w/e

[fair wren]

so ur saying that every vector in {x1,...,xm} can be denoted by a linear combination of span(x1,...xk) ?

[spare widget]

if span(x1,...,xk) = span(x1,...,xm) then you can write x[k+1],...,xm as linear combinations of x1,...,xk

[fair wren]

ah since x[k+1],..., xm are dependent?

[spare widget]

yes

[fair wren]

nice

[spare widget]

dependent on {x1,...,xk}

so the problem reduces to the linearly independent case, with some extra terms popping out

[fair wren]

exactly, so to get into the proof, i shall suppose that {x1, .., xm} is independent right?

[spare widget]

if anything the above can only increase the nullspace of B^T

[fair wren]

u see i don't understand this nullspace of B^T we can discuss it later

lets do the first case

[spare widget]

in an actual proof you would have to rigorously deal with the linearly dependent case, since it pops in extra terms

[fair wren]

yes we do first case linear independent

right?

[stoic python]

alexisreen

[spare widget]

yes, though for each notation can be better

[fair wren]

so this RHS will automatically get reduced to 0

for all lambda_i whether they are zero or non-zero

that means {y_1, ..., y_m+1} are linearly dependent

right ?

[spare widget]

you need to group the terms

[fair wren]

which ones?

[spare widget]

so they have the same form as

[fair wren]

I don't quite understand what u mean tho?

[spare widget]

you need something of the form c1 x1 + ... + cm xm = 0

[fair wren]

ok i see

so we have $\sum_{i=1}^{m+1} b_i y_i = \sum_{i=1}^{m+1} b_i \sum_{k=1}^m a_{ik} x_i = \sum_{i=1}^{m+1} b_i (a_{11} x_1 + ... + a_{ii} x_i + .. + a_{im} x_m)$

what do i group here @spare widget ?

I see

[spare widget]

$\sum_{i=1}^{m+1}b_i\sum_{k=1}^ma_{ik}x_k = \sum_{k=1}^m \left(\sum_{i=1}^{m+1}b_i a_{ik}\right) x_k = \sum_{k=1}^m \alpha_k x_k = 0$

[stoic python]

alexisreen

[fair wren]

hm??

how'd u get that?

[spare widget]

now use the linearity to get that the alpha_1=0,...,alpha_m = 0

[spare widget]

linearity

[stoic python]

criver

[spare widget]

noticed a typo of xi

[fair wren]

https://cdn.discordapp.com/attachments/540211747613704221/969736838883454977/324736074750623748.png

This correct right?

so this is of type c_1 x_1 + .. + c_m x_m

[spare widget]

first term should read ai1

[fair wren]

ye

[spare widget]

but beyond that it is correct

[fair wren]

tysm

[spare widget]

I am not sure why you wrote it though

[spare widget]

this is all that you need

[fair wren]

we have this

https://cdn.discordapp.com/attachments/540211747613704221/969735135069110322/324736074750623748.png

https://cdn.discordapp.com/attachments/540211747613704221/969736838883454977/324736074750623748.png

we got this

[spare widget]

use this instead

[fair wren]

ok sure but lemme see my idea

[spare widget]

I am going to go sleep, so I'll just leave you a sketch of what to do next

[stoic python]

alexisreen

alexisreen

alexisreen

[fair wren]

whatchu think?

[spare widget]

From this you get the equations $\sum_i b_i a_{1i} = 0$, ..., $\sum_i b_i a_{im} = 0$, which are equivalent to $A^Tb = 0$. Where $(A^T){ji} = a{ij}$. The matrix $A^T$ is of size $m \times (m+1)$ so its null-space is at least one-dimensional. Then there exists $b\ne 0$ such that $A^Tb = 0$, so $y_1,\ldots,y_{m+1}$ are linearly dependent.

[stoic python]

criver

[spare widget]

I gotta go sleep.

good night

[fair wren]

u2

[slate hatch]

If we're to check if vector A exist in a plane in R3, we take the cross product of the plane vectors. We then take the scalar product of that and A. If it is zero, A belongs to the plane.

I don't get this. Aren't there an infinite amount of planes in R3, that are perpendicular to the normal?

[fringe fjord]

The plane thus defined is in particular the one that contains (0,0,0).

[pseudo cairn]

could someone suggest approaches for this question or point me in the right direction with books etc?

[fair wren]

https://cdn.discordapp.com/attachments/268882317391429632/969768033050562620/324736074750623748.png

@sharp idol can you look at this please

[native rampart]

No?

[fair wren]

why

[native rampart]

F=span{(1,0,0)} G=span{(0,1,0)}
(1,1,0) is in F+G

[sharp idol]

Beat me to it

[native rampart]

Not in F \inter G

[sharp idol]

Or F U G for that matter.

[fair wren]

oh gotcha

well i gotta prove something

Let E be a vector space and F,G subspaces of E

prove F inter G is a subspace

so i did this

since F, G are subspaces they contain 0

so F inter G is non-empty

now i need to prove x-y is in F inter G

and a*x is in F inter G

this is what i gotta prove right @native rampart ?

[sharp idol]

Wouldnt F n G be the vector space that shares the same basis elements as F and G?

[fair wren]

How can u addup two spans

[fair wren]

you see i still don't understand the basis stuff very well tho

[sharp idol]

Okay. Give me a second. I gotta get these dishes done

[fair wren]

sure take ur time senpai

[fringe fjord]

If F and G are explictly given by bases, there are not necessarily any basis elements in the intersection.

[stoic python]

dackid

[fair wren]

ye i don't understand

[sharp idol]

Perhaps Troposphere can explain it more thoroughly.

[fair wren]

the span here is span{(1,0);(0,1)}?

[fringe fjord]

For example, if F = span{(1,1,0),(1,-1,0)} and G=span{(1,0,1),(1,0,-1)}, then their intersection is nontrivial -- it contains all vectors of the form (t,0,0) -- but you cannot see that simply by intersecting the two bases.

[sharp idol]

Fair point. My mistake

[fair wren]

What mistake?

[fringe fjord]

Sorry, I meant to reply to Dackid's post below yours instead.

[fair wren]

Its all good

[fringe fjord]

But you're right, you need to prove that a·x and x+y (or x-y if you want it that way around) are in the intersection.

[fair wren]

oh well to prove the second thing - x+y -

i need to the 2 cases

there is no way i can do it directly

so Let x,y in F n G

[fringe fjord]

Huh? It feels pretty direct to me.

[fair wren]

what

how

[sharp idol]

Strange. Dont (1,0,0) and (0,1,0) also span F?

[fringe fjord]

Suppose x is in the intersection and y is in the intersection.
Because x is in F and y is in F we have x+y in F.
Because x is in G and y is in G we have x+y in G.
Because x+y is in F and it is also in G, it is in the intersection of F and G.

[fringe fjord]

Yes, for each subspace there are many different possible sets that span it.

[fair wren]

This is 2 cases ig?

No?

[fringe fjord]

I only see one case there. There's no division into cases.

[fair wren]

because u divided the first case

is where u studied x and y in F

secondly in G

then u deduced

[fringe fjord]

No, every line in the proof is relevant to every choice of x and y.

[fair wren]

at the end

[fair wren]

exactly

thats two choices

i mean i found a new way to this instead of doing both ax and x+y we can just prove ax+by in F n G

[fringe fjord]

No. Whenever you have chosen an x and y, you need all the lines to reach the conclusion. Division into cases in when only some of the lines are relevant, based on properties of your choice.

[fair wren]

oh right

ok then its done by now

[sharp idol]

That's what I thought. So was my reasoning wrong, or does it just need to be a bit fine tuned to have merit?

[fair wren]

Let $x,y \in F \cap G$ Then $$x+y \in F$$

[stoic python]

alexisreen

[fair wren]

thus its also in G

so x+y in F n G

right?

[sharp idol]

You can skip to x-y in FnG, which covers two pieces right off the bat

Really the only thing I'd change is your wording, but you got it

[fair wren]

tysm

but how is this related to spans and basis stuffs @sharp idol ?

[sharp idol]

It wasnt actually that relevant to the problem. That was my bad

[fair wren]

would be lovely to understand them tho

[sharp idol]

I was just trying to develop an intuition for what was being asked, but my logic was a bit off.

A basis of F is a collection of linearly independent vectors that span F

[fringe fjord]

It's true that given F and G you can always find bases for them such that the intersections of the bases form a basis for the intersection. But if you already have bases for the subspaces, those bases won't necessarily be ones with that proper. And the only good way to tell is to figure out the intersection first.

[sharp idol]

Got it! So we'd basically be going in circles to deduce that, making it somewhat irrelevant to this

[fair wren]

that means any vector of F, can be written as a linear combination of the vectors from the basis?

[sharp idol]

For example, $\Q(\sqrt{2})={ a+b\sqrt{2}: a,b \in \Q}$. Here, a basis for $\Q(\sqrt{2})$ is ${1,\sqrt{2}}$.

[sharp idol]

Bingo!

[fair wren]

ugh when u did this example of Q(sqrt(2))

got me more confused

baka

[stoic python]

dackid

[sharp idol]

No worries. Q(rt(2)) is the set I described. And since rt(2) is not rational, there is no a in Q s.t. a*1=rt(2)

Yea, vector spaces are a lot more general than you may think

[fair wren]

oh wow

so its the vector (1, sqrt(2))?

[sharp idol]

No, if you want to think of it that way, the vectors that span Q(rt(2)) are (1,0) and (0,rt(2)).

[fair wren]

how did u get those?

ah ye got it

its a(1,0) + b(0, rt(2))

[sharp idol]

You got it :D

[fair wren]

senpai u explain stuff so ez

ok continue please

is a span the same thing as a generator?

[sharp idol]

Hmmm, they kind of have the same idea going for it

Because all of the basis elements that span our vector space generate every element in said vector space.

[fair wren]

whats the relation between a basis and a span tho?

sum a_i x_i = 0 iff forall a_i = 0?

[tribal willow]

can someone explain condition (b) of the lemma that is necessary to say that W = W_1 + ... + W_k is a direct sum of W_1, ..., W_k?

[fair wren]

bcz they are independent ig?

[sharp idol]

This here shows us that each basis element $x_i$ is linearly independent of the others.
Suppose it we not the case, the $\exists x_i$ s.t. $-\frac{1}{a_j}\sum_{i\neq j} a_i x_i =x_j$, which tells us $x_j$ is a linear combination of the other basis vectors.

[stoic python]

dackid

[fair wren]

Yes

[sharp idol]

A key condition is that the elements that span a subspace are linearly independent of one another

[fair wren]

so span(v_1, ..., v_n) of a subspace of dimension n , each v_k is independent of others?

[sharp idol]

Correct

[fair wren]

Damn

but no

so a span is a basis in this case?

[sharp idol]

The reason why span isn't quite generate is because you need to equip the vectors with scalar multiplication

[fair wren]

oof i don't understand now

a basis is a span with this property

right?

[sharp idol]

A basis of F is a "collection" of elements ${v_1,v_2\dots, v_n}$ s.t. each $v_i$ is linearly independent of the others (as your picture states above) and $F=\text{Span}{v_1,\dots ,v_n}$. The span is the collection of all linear combinations of $v_1,\dots, v_n$.

[stoic python]

dackid

[fair wren]

same vectors?

[stoic python]

dackid

[fringe fjord]

A "span" is always an entire subspace. A "basis" is just a few strategically selected elements of the (sub)space. The subspace can be the span of a basis, though.

[fair wren]

wtf

so for example

R^2

the span of it

is span((1,0); (0,1))?

[fringe fjord]

The span of (1,0) and (0,1) is R^2.

The span of R^2 is also R^2.

[fair wren]

so R² = span((1,0); (0,1))?

[fringe fjord]

You can take the span of any set of elements of a vector space.

[sharp idol]

Exciting stuff really

[fringe fjord]

Yes.

[fair wren]

WOAH

so sus

and whats the basis of R²?

[sharp idol]

There are many. The typica set is (1,0) and (0,1)

[fringe fjord]

There are many possible bases. One of them is {(1,0), (0,1)}.

Another is, for example, {(24,33),(sqrt 2, pi)}.

[fair wren]

why

[fleet sun]

a set of vectors ${v_1,\ldots,v_k}$ is a basis for $\mbox{Span}(v_1,\ldots,v_k)$ if and only if they are linearly independent

[stoic python]

ManifoldCuriosity

[fringe fjord]

Because every vector in R^2 can be written as a linear combination of (24,33) and (sqrt2,pi) in exactly one way.

If you pick two random vectors in R^2, chances are overwhelming that putting them together will form a basis for R^2.

[fair wren]

wait wait im confused

R² is the vector space right?

[fringe fjord]

Yes.

[fleet sun]

that's the relationship between basis and span

[fair wren]

span({v1,...,vk}) is also vector space?

[fleet sun]

it's always a subspace of the whole space yes

[fringe fjord]

It's a subspace, but a subspace is also a vector space in its own right.

[fair wren]

lmao

a vector space is a collection of vector right

{v1, ..., vk} isn't this also a vector space too?

[fringe fjord]

No.

[fair wren]

woah now im confused

[fleet sun]

a vector space is not just any set of vectors

[fair wren]

why?

[fair wren]

a set of vectors over a field

[fringe fjord]

A "vector space" must be closed under addition and scalar multiplication.

[fair wren]

hm i see

i see why it isn't a vector space

[fringe fjord]

The notation {v1,...,vk} suggests there are only k vectors in your set, which is not enough for it to be closed under multiplication with every scalar (unless in the boring case that all of the vi's are the zero vector).

[fair wren]

bcz if u add up two vector from that set we're not sure if the result is in there

right?

[fringe fjord]

That too.

[fair wren]

Also wait

i've seen a theorem saying

there is always a basis

for a vector space of a finite dimension

[fringe fjord]

Yes.

[fair wren]

isn't a span of infinite dimension then?

oh wait no

im so confused about the dimensions

$\dim_{\mathbb{R}} (\mathbb{R}^2) = 2$

[stoic python]

alexisreen

[fair wren]

right?

[fringe fjord]

"Dimension" is now many vectors there are in a basis, so it's pretty trivial that a vector space of finite dimension must have a basis.

[fleet sun]

yes

[fair wren]

but what about $\dim_{\mathbb{R}} (Span{v_1, ..., v_k})$

[stoic python]

alexisreen

[fringe fjord]

That's somewhere between 0 and k.

[fleet sun]

more info is needed for that

[fair wren]

what info?

[fleet sun]

how many of the vectors in that set are linearly independent

[fringe fjord]

It's k if {v1,...,vk} are linearly independent.

[fair wren]

thats a fact

xd

[fair wren]

so its basically the dimension of the basis?

[fringe fjord]

You have the right idea, but "dimension of the basis" is not a wording we normally use. You can say it's the cardinality of the basis, if you want fancy words.

[fleet sun]

yeah, or simply size