linear-algebra

[thorn robin]

you could view that as living in R^2 or R^3, it doesn't make much difference

the three vectors lie in a common plane regardless

[gloomy arrow]

If I'm trying to find if a vector is in a span of vectors and I get a free variable in the matrix does that mean it does span?

[gray dust]

if a matrix A's columns are the vectors whose span you're checking, and b is the vector you're checking that lies in those vectors' span, you're trying to find at least one x that satisfies Ax=b. this system being consistent guarantees at least one x exists, so b lies in the span. having at least 1 free var means there are many possible x, ie infinitely many linear combos of A's cols that produce b

[devout pine]

how do i describe the range of the matrices?

[empty copper]

An (m x n)-matrix multiplied with an (n x 1) vector yields a (m x 1) vector

[zealous lagoon]

Apparently the answer is the system is always consistent for all h values, why is that so?

[devout pine]

Its actually (2h+4)x2=0

Remember that 2h+4 should be a coefficient

what happens with the matrix in 1c? i was row reducing it and i ended up with an inconsistent matrix

do i say codomain and domain are r3 but the range is undefined? or

[smoky lagoon]

to show that 3 functions are linearly dependent, is it enough to show that two of them are linearly independent, so if you multiply whatever the third is by zero you will still satisfy the condition that c1V+c2U...+cN(arbitrary vector)=0?

[slow scroll]

Yes. A set of vectors is linearly dependent if one of the vectors can be written as a linear combination of the others

Even if some of the vectors have 0 contribution in this linear combination, it still counts as a linear combination @smoky lagoon

{(1,0,0), (2,0,0), (0,1,0)}
Is a linearly dependent set for example

this is just because (2, 0, 0) = 2(1, 0, 0)

[smoky lagoon]

thank you

i thought they all had to be dependent on each other, but that is much easier

[slow scroll]

npnp

[wintry steppe]

If there exists an invertible linear transformation P: V -> V such that PTP^-1 is diagonalizable (with T: V -> V), does that mean T is also diagonalizable?

(I'm trying to figure out if the (similarity) condition for a matrix to be diagonalizable can be generalized to all linear transformations on a finite dimensional vector space)

[slow scroll]

is the proof for matrices any different from the proof for generic linear transformations tho? @wintry steppe

[wintry steppe]

is the proof for matrices any different from the proof for generic linear transformations tho? @wintry steppe
I assume it follows from the matrix representation of T?

[slow scroll]

keep in mind: the conditions that go into showing that a matrix is diagonalizable have nothing to do with the choice of basis.

[wintry steppe]

I see

[thin bloom]

Diagonalizable == Could be made into diagonal matrix with base change
IIRC.

[wintry steppe]

Hmmmm

I'll see what I can do from here

Thanks for your inputs @slow scroll @thin bloom

[slow scroll]

that also just means that the geometric multiplicity of eigenvalues equal the algebraic multiplicity of eigenvalues. the characteristic polynomial is invariant under change and basis and same with the dimension of eigenspaces.

[sonic osprey]

I feel like you're overcomplicating it here kx

[slow scroll]

im just tryna say the proof is the same whether ur talking about matrices or not.

[sonic osprey]

I mean yeah that's true

[bright otter]

Hey can anyone help me answer 5?

[thin bloom]

Use Gauss-Jordan Elimination

Proceed and you're fine

[bright otter]

,w row reduce {{1,2,1,6},{3,7,6,37}}

[stoic python]

Display results online and refine query

Requested by Brother E#2684

[bright otter]

So now how do I express this in terms of (x,y,z)?

Is it
x-5z=-32
y-3z=19
z=z

x=5z-32
y=3z+19
z=z

Would this be final answer?

[thin bloom]

They said express in terms of a

[bright otter]

oh right so
x = 5a-32
y = 3a+19
z = a

Is that correct @thin bloom ?

[thin bloom]

I think so but put it in to check it out

[idle echo]

Is it a property of matrices that, given AB = C => A = CB^(-1)?

[wintry steppe]

B has to be invertible

[quartz compass]

or you could also require C is invertible

[slow scroll]

doesn't C invertible only make B left invertible and A right invertible tho
EDIT: thats only when A and B aren't square.

[vital raft]

could somebody help me with linear equations

?

[restive shuttle]

How would one approach this?

The online answers show Eigenvalue solutions but I have not learned it yet

I've been staring at this for hours

I also haven't learned characteristic equations

[hidden frigate]

Teacher gave us a "for fun" problem to stretch our brain, curious on how to solve.

[slow scroll]

not linear algebra ^

[hidden frigate]

Pre-? Class is listed as Algebra

[slow scroll]

#precalculus or #prealg-and-algebra yea

[hidden frigate]

K, thansk

[slow scroll]

maybe try solving $$\begin{pmatrix} 1 & z \ 4 & 3 \end{pmatrix} \begin{pmatrix}a \ b \end{pmatrix} = \begin{pmatrix}a \ b \end{pmatrix}$$
for $z$

[stoic python]

kxrider:

[slow scroll]

@restive shuttle

[restive shuttle]

what is the reasoning behind that?

since it's R2 transformation?

i got zb = 0

so z must be 0?

[slow scroll]

By "fixing a line" we mean that T(v) = v for every point v on the line
so solve for the z that makes that happen and be sure to exclude those values.

[restive shuttle]

oh

so all values except z = 0

thanks!

[slow scroll]

np

[restive shuttle]

linear algebra is hard for me

so much stuff going on, all these concepts, very conceptual

really overwhelming

[slow scroll]

@restive shuttle hold on im dumb :/

[restive shuttle]

what is it

[slow scroll]

T still fixes the line 2a + b = 0.

so like, T(-1, 2) = (-1, 2) for example

oh yea nvm thats only if z=0 :p

[restive shuttle]

how might one approach this one

[slow scroll]

draw a pic

oh wait area preserved

you can choose to map one of the points to the origin, and then you just have to make sure you end up with a right triangle w/ side lengths sqrt2

[heavy glacier]

what is this T^*? does not seem to be conjugate transpose

[prisma prawn]

@restive shuttle preserved area means that the determinant of your linear mapping needs to be 1

map (0,0,1) to (0,0) and the rest such that the determinant is 1

[slow scroll]

There is no determinant for a linear map from R3 to R2.

Indeed, the geometric interpretation of determinant fails here. In R3 It measures the volume of the parallelepiped spanned by the columns, while the determinant in R2 is the area spanned by the columns.

[viscid kernel]

Is there a way to visualise the kernel and the image of a linear transformation ??

[coral ferry]

does anyone have any recommendations for rigorous proof based linear algebra textbooks

im looking for smth with some tough problems to do

[sonic osprey]

Hoffman Kunze

[coral ferry]

ill check that out

[viscid kernel]

Uhm anyone Im dieing here

[pallid swallow]

@viscid kernel yeah, interpret kernel as a vertical line and image as a horizontal line, both crossing at the origin.

[winged shoal]

Is f(x)=1/2x+1 linear?

If x is a function of f

[viscid kernel]

@winged shoal
nope it isnt, cuz look when the denominator is 0 the function is going towards infinity that way your function gets curved , so when x = -1/2 there you have your vertical assymptot the function gets closer and closer to so that means you are using limits so lim x ---> 0. Linear functions go to infinity when the input of that goes to infinity, they dont get blocked by any kind of barrier in this case the vertical assymtot if

example: f(x) = x+5 so the only way the function goes infinity if x goes to infinity

But I dont think the question belongs here go to "algebra" 🙂

[icy osprey]

I put 1 soln, correct?

[vast torrent]

@winged shoal if you mean (1/2)x +1 then f is a straight line

[winged shoal]

Not sure if I wrote it the right way.

Let me write and take a picture in case there is a difference

You still say no?

[agile gyro]

Depends what you mean by linear

If you mean y=mx+b then ya

If we have

$V=\mathbb R^3$, $\beta={(1,0),(0,1)}$

[stoic python]

wasd:

[agile gyro]

Then $(x,y)=x(1,0)+y(0,1)$ so $[(x,y)]_\beta=(x,y)$

[stoic python]

wasd:

[agile gyro]

But you could rewrite $\beta$ as $\beta={(0,1),(1,0)}$ so $[(x,y)]_\beta=(y,x)$?

[stoic python]

wasd:

[agile gyro]

Because $(x,y)=y(0,1)+x(1,0)$

[stoic python]

wasd:

[slow scroll]

I don’t see why not lol. Usually bases are indexed by their coordinate number so it’s not ambiguous like this

[agile gyro]

So I can write that?

But it’s supposed to be bijective

[slow scroll]

So I can write that?
Please don’t actually do that xd

It is bijective. It’s just a permutation of the same collection of vectors (I.e. it’s inverse is is just switching the vectors back to how they were)

[agile gyro]

So the order of the basis matters?

[slow scroll]

In matrix algebra, switching rows to solve a system is an elementary operation for example.

It matters for the way it looks. It’s like choosing your y axis to be horizontal and x to be vertical. You can do it, but it’s just unconventional and weird lol

[agile gyro]

But then it’s not bijective

[slow scroll]

Sure it is let f be the map that maps x to y and y to x
f(x,y) = (y, x).

f(f(x,y)) = (x,y). So f is it’s own inverse

[agile gyro]

Then the map $\phi_\beta:\mathbb R^2\to \mathbb R^2$ given by $\phi_\beta(x)=[x]_\beta$ is not bijective

[stoic python]

wasd:

[agile gyro]

Because phi(4,2)=phi(2,4) but (4,2) \neq (2,4)

[slow scroll]

Do u know what bijective means here?

[agile gyro]

It means it’s injective and surjective

[slow scroll]

Phi(4,2) is not phi(2, 4)...

Phi(4,2)=(2,4) and phi(2,4)=(4,2)

[agile gyro]

We just showed $[(x,y)]_\beta=(x,y)=(y,x)$

[stoic python]

wasd:

[slow scroll]

Wut....

[thin bloom]

Er what is this

Can't find the problem involved with this

[slow scroll]

I leave this to u cuz I gtg

[thin bloom]

O no

[agile gyro]

Lol

[thin bloom]

Is this whose question

[agile gyro]

Mine

[thin bloom]

And the question being?

[agile gyro]

Just scroll up a little

[thin bloom]

Ah, it seems like it's about base set

[agile gyro]

Yeah

[thin bloom]

No, in this context base doesn't form a set, because the order is important

As you see, if you change the order sth sinister happens

[agile gyro]

I thought sets weren’t ordered?

[thin bloom]

Base doesn't form a set, that's what I meant

It's more of ordered list

[agile gyro]

Oh

Okay i see

Thanks @thin bloom

[thin bloom]

Np

[idle echo]

Is there a name for the approach to solve this so that I can read about it?

[agile gyro]

Maybe try inverting C

If it’s invertible

[thin bloom]

That would be hard

Prob you could look into Gaussian elimination process

[idle echo]

Aren't two 3x3 matrices multipliable?

I'm getting that E = A(C^(-1)) is undefined.

but that C^(-1) isn't

[thin bloom]

Don't focus in C^(-1)

Look into Gaussian elimination firat

First*

[idle echo]

I thought that was just a method for solving for REF?

[thin bloom]

It is relevant here

It also involves elementary matrices to reduce it into that

Utilize The process

[idle echo]

Do you want me to solve both matrices for REF? @thin bloom

[thin bloom]

No

Try it on C

Oh wait no, try it on A

Sorry

[idle echo]

ok

[thin bloom]

And show me matrices on each step

[idle echo]

Thanks!

[brittle fog]

Hello, could I get some help on a problem? It should be simple but we arent allowed to use the theorems that make it easy

Prove A invertible iff Ax=b has unique solution. I did => but I am stuck on <=

Or should I move this to one of the question channels

[idle echo]

@thin bloom

I got stuck

actually no

[thin bloom]

Make one row (0 1 2)

[idle echo]

For the last matrix on the bottom, R_3 + -4R_2 -> R_3

[warm flicker]

if you have more rows than columns in an augmented matrix, can you still have a consistent solution if each row is linearly dependent?

because if you think of them as equations, yeah you can have an infinite number of equations that intersect at a point.

[smoky lagoon]

how to do C?

or at least what should i investigate?

I know that sin and cos are linearly indepdendent, so i need to show one of them with the third vector

[viscid kernel]

Just turn the cos(x) into sin(x-pi/2)

[smoky lagoon]

o shit ur right

im very very bad with trig stuff

what does that tell me

[viscid kernel]

Dude its important, just learn that whenever you have time, it always trick people who dont know it 🙂

[smoky lagoon]

i know

it still hurts me

some day ill get around to memorizing the unit cirle

[viscid kernel]

Ahahahhaha dont memorise it

[smoky lagoon]

anyways, can i show that sin(t+pi/10)/(sin(t-pi/2)) is some constant

and if so how

[quartz compass]

you can't, that won't do it

try to represent it as a linear combination of sin and cos

[smoky lagoon]

isn't that the whole goal?

just to show that their ratio is a constnat

is a non-zero constant

[quartz compass]

there exists an a,b such that $a\sin(t)+b\cos(t) = \sin(x+\pi/10)$

[stoic python]

Merosity:

[quartz compass]

what you're saying is not a constant "show that sin(t+pi/10)/(sin(t-pi/2)) is some constant"

so you can't show something that's false

[smoky lagoon]

there exists an a,b such that $a\sin(t)+b\cos(t) = 0$

[stoic python]

leo1:

[smoky lagoon]

where a and b are non-zero constants

isn't that the definition of linear dependence?

[quartz compass]

no there isn't

[smoky lagoon]

wait no that wouldnt work

[quartz compass]

why are you saying that

[smoky lagoon]

i mesed up my syntax

[quartz compass]

the only way that is 0 is when a,b are 0 because sin and cos are linearly independent

[smoky lagoon]

to show that two vectors are linearly dependent dont you just have to show that cv-->+bu-->=0 where u and v are vectors and c and b are non-zero constants

i know that cos(t) and sin(t) are indepdendent so it has something to do with that third vector and i just have to show that it's linearly dependent to one of the first two

[dry spear]

https://cdn.discordapp.com/attachments/409596215697866773/677000833992228879/unknown.png

if I take any point on l(t) and any point on H, subtract them to get a new vector, then project that onto n, it should give me the same result everytime right?

but it doesn't

[boreal crescent]

how do i do b

i have it written out but it doesnt make sense

when i apply the transformation on both sides of V = c1y1 + .. + cmym

its not necessary that every element in V maps to something in W right

part a i got, part c is just concluding from the previous two parts

[subtle walrus]

its not necessary that every element in V maps to something in W right
what

[boreal crescent]

surjectivity implies that every element in W has atleast one element in V

[subtle walrus]

yes

[boreal crescent]

but that doesnt mean that every element in V has to map to W

right?

[subtle walrus]

that's what a function is

[boreal crescent]

ohhh

[subtle walrus]

a function has to assign an element in W to every element in V

[boreal crescent]

but i mean

they havent said its a function

just a transformation

[subtle walrus]

that's synonymous

[boreal crescent]

is it though? like we learnt that there can be transformations that might not map every element in V to W. Like if V and W were both R2, then the function sqrt(x) will not map all negative x values

even though all negative x values are in V

[subtle walrus]

you would have to look up your definition of transformation then

but that would be very uncommon

[boreal crescent]

this is a generalized proof though right. the example i gave shows what im trynna say

[subtle walrus]

i mean you don't even need it

for b) you should start with an element in W

then it has a pre-image in V because T is surjective

and go from there

[boreal crescent]

ohhhh ok

[subtle walrus]

if you can write every element in W as a linear combination of T(y_1), ..., T(y_m), you are done

admittedly you need that at least T(y_1), ..., T(y_m) lie in W

but yeah, ask your teacher about what a transformation is

[toxic timber]

Not entirely sure if this is the right channel (still in high school), but I needed help understanding a proof involving lemniscates and linkages

https://www.parabola.unsw.edu.au/files/articles/2010-2019/volume-52-2016/issue-1/vol52_no1_2.pdf

I understand everything until the top of page 4. I don't understand where that quadratic equation came from

nor do I understand what the shape being an antiparallelogram has to do with anything

[zealous lagoon]

These are the matrix I reduced, anyways, for a I was asked to find the trivial solution and B I was asked to find the non trivial . I’m so confused on what’s the difference

So for a I just said, h=1/2 and b I said h can be any value

Can someone clarify ahaha I’m a lost

[wheat prairie]

can someone explain to me how these two steps happened (displayed by the arrows)

[autumn oar]

(AB)^T = (B^T)(A^T)

For the first one

[pine patrol]

https://math.ryerson.ca/~danziger/professor/MTH141/Handouts/Slides/elementary.pdf

Look at theorem 2

They're just properties of the transpose of a matrix

[autumn oar]

And (A + B)^T = A^T + B^T also for the first one

The second one is probably in the doc node linked but i cant figure out which properties it is

[quartz compass]

since $y^TX\theta$ is a 1x1 matrix it's equal to its own transpose

[stoic python]

Merosity:

[autumn oar]

Wait how do u know its 1x1

[quartz compass]

magic

[autumn oar]

Ty, very cool

[wheat prairie]

thank you guys 😄

[autumn oar]

Anyways can anyone explain the proof for jacobi iteration converging on diagonally dominant matrices it makes no sense

[wheat prairie]

and @quartz compass is correct it's a scalar because of the dimensions of theta, X, and y dictated by linear regression.

[autumn oar]

Ignore my question actually it's way more involved than i thought

It requires the Banach fixed point theorem which i know nothing about and i dont know most of the concepts used in the banach fixed point theorem

[wheat prairie]

https://tenor.com/view/iknow-some-of-these-words-mhmm-clueless-words-iknow-gif-5609882

[hollow ridge]

Wth..

Idk what I am missing

Number 5

I’m trying to find a determinant of A

Here is my work

The answer is 3

But I got the negative sign

I eliminated the first row and fourth column

The cofactor of C14 is negative

<@&286206848099549185>

I did another way and got 3

But the first time i got is -3

[autumn yacht]

@hollow ridge

you fking idiot

[hollow ridge]

Sup

[autumn yacht]

line 2

[hollow ridge]

What

Where is line 2

[autumn yacht]

-1... + -3

your work

you took the +/- wrong

= -1 ... + -3 ... + 2 ... + 1 ...

the one I bolded, should have the opposite sign

[hollow ridge]

Fuckkkk

God damn it

It’s always happen to me

Every time I get stuck with it, I ask someone to check and guess what... the fking signs

[autumn yacht]

lel

good luck man

[hollow ridge]

Thanks man...

Idk how to stop forgetting the signs

And I got some points off on my past quizzes too

The fking signs

[south pollen]

is it always possible to get a reduced row echelon form

[half ice]

Yes

[south pollen]

fml

[half ice]

Not always possible to reduce to identity though

[south pollen]

do you start at the top then go down or bottom to top

wait holdup

I’m stupid I went backwards in a step

[half ice]

I go left to right

[south pollen]

this would be reduced right?

[half ice]

Ya

[south pollen]

then I could conclude the third and fifth variables are free?

because they have no pivot in their column

thanks

[south pollen]

not sure how I went wrong

o

eehhhh

[viscid vale]

will the first column in a matrix always be in the basis of its column space? since its guarunteed to be a pivot column?

[sonic osprey]

the first column could be all zeroes

is it still guaranteed to be a pivot column?

[viscid vale]

hm i see

[long lion]

fuys

[radiant prawn]

Could anybody give me a hand with part c? I've got the injectivity proof, but am struggling to prove it's not surjective and doesn't contradict part A. I've got a general idea that if I let p(x) = 1, then R will never map to that value (I think this is a valid counter example) and my first thoughts for not contradicting A

are that P isn't finite dimensional, however I was thinking if I chose a finite dimensional P, where its all functions less than or equal to degree two, that my counterexample of surjectivity still holds. Obviously I'm confused, and any help would be greatly appreciated!

[empty copper]

R never maps to a constant polynomial

Hence it's not surjective

Done

[subtle walrus]

the zero polynomial gets mapped to itself

[empty copper]

R never maps to a non-zero constant polynomial

[subtle walrus]

also for your concern

if you consider the space P_n of all polynomials with at most degree n

it's finite-dimensional vector space

but R is not a map P_n -> P_n

[empty copper]

tfw P is inifinite-dimensional

[radiant prawn]

That makes a lot of sense guys, Thank you.

[subtle walrus]

also there is a typo in the exercise

"Show that S is injective..."

[radiant prawn]

Is it not injective?

[subtle walrus]

i mean S

i assume they mean the map R

[radiant prawn]

Lol, yup that's true

[spiral sonnet]

what's the trivial solution for an augmented matrix where b is the 0 vector?

[slow scroll]

^ i.e. What is always a solution to Ax = 0? @spiral sonnet

[severe magnet]

suppose we have a vector space which contains all functions

would $U:=\lbrace f|f \text{ is improperly integrable} \rbrace$ be a subspace of that vectorspace?

[stoic python]

Nabil | GMT+1:

[severe magnet]

I thought about using the fact: $f$ is improperly integrable $\iff$ $\sum_{k=1}^\infty f(k)$ converges

[stoic python]

Nabil | GMT+1:

[severe magnet]

multiplying a scalar to the series/adding another convergent series to the initial series wouldn't result in divergence

so imo it is a subspace

am I wrong on that one?

ah wait that would only go for positively defined functions

nvm

[brittle juniper]

this characterisation of the convergence of improper integrals looks super super suspicious

[severe magnet]

yes since it is for functions $f:]1,\infty[ \to ]0,\infty[$

[stoic python]

Nabil | GMT+1:

[severe magnet]

which I kind of didn't take into account

[brittle juniper]

even with this additional info

[severe magnet]

and it has to be monotonically decreasing

[brittle juniper]

now that's a bit better

[severe magnet]

since otherwise defining a series over the function would be kinda nonsense yea

this goes for specific functions I guess

[brittle juniper]

you don't have to introduce any series for this

improper riemann integrals already behave nicely with linearity

[severe magnet]

I totally forgot about that

didn't think that was the way for improper integrals aswell but it makes sense

[cloud cairn]

if given something like 2 vectors in terms of x: <x+1, 4-x, x> , and <3-x^2, 2+x^2, x^3>
how would you go about finding values of x, such that the 2 vectors are parralel. One thought was evaluating the cross product and setting it equal to 0.
(2 vectors parallel have a cross product of 0)

[devout pine]

can anybody lend a hand with 4d? i keep on getting different vectors

like when i calculate T[3,4] im getting [26,1]

so i multiply S with [26,1] and then i get [28,-78]

but then when i calculate ST([3,4]) i get [-23,51]

forget it, the only reason why i cant get part d correct was because my part c was wrong (I did the composition TS instead of ST)

[wintry steppe]

not sure if this is the right chat, but could someone help me with this?

https://gyazo.com/b87dde5b668e193d19a66a2d5a1017d3

[half ice]

@wintry steppe
Indeed your question isn't related to linear algebra.

The idea is to compare the rates in walls/hour. That gives:
1/x + 1/(x+3) = 1/3

[native river]

Hi, if the ckmposition of two functions is T¤S injective and we are told to find if S is injective. How do you ho about doing this?

$\textbf{What i have in mind is to show that since T¤S is injective then S is injective by definition}$

[stoic python]

joseph2531:

[native river]

Does anyone know if this is correct or wrong? <@&286206848099549185>

[wintry steppe]

Nope it's not true by definition

You should use the definition of injectivity, though

[native river]

Thank you

Has anyone of Lagrange polynomial?

[radiant prawn]

Can anybody give me a hand with part d? I've proven all the others, but this one has me stumped. I've been given the hint to incorporate part b and c, and will include a tenative proof for the backwards direction, but am still quite unsure.

[agile gyro]

Or you just prove dim(U+W) = dim(U) + dim(W) - dim(U \cap W)

[radiant prawn]

That would make sense, although would that actually be an easier proof? I'll give it a go. Thank you.

[agile gyro]

It’s not that bad

Hint: ||start with a basis of U cap W then extend it to a basis of U and W||

[gloomy arrow]

Do you learn about jordan form in a first year linear class

[sonic osprey]

sometimes, sometimes not

[wintry steppe]

Hello, can someone help explain this to me. I kinda understand it, but the last part is a little confusing

[wintry steppe]

<@&286206848099549185>

[real wedge]

what in the world is dual space

I understand its the set of linear functionals from V to F but... theres a lot of things i dont understand. Like the basis for it

And whats the motivation behind it anyway

[vast torrent]

Given a basis e1,e2,...,en there is a dual basis consisting of linear functionals such that

$\ell_i(e_k)=\delta_{ik}$

[stoic python]

gfauxpas:

[vast torrent]

Kronecker delta

[pallid swallow]

I still don't get the intuition why trace is basis-independent (in terms of linear functionals)

[merry jewel]

FAM, help

how do i start this proof?

and what does Bii and Aii have dimension mi x mi?

[fast pivot]

try Gaussian elimination

[merry jewel]

for A?

but these are hypothetical values

[fast pivot]

sure

[merry jewel]

okay, i don't even know where to start since aint A11 an entry?

So when doing gaussian elimination i'm not sure how the B's will come into play

[fast pivot]

maybe start with B and try to find its inverse in terms of B_ij

[merry jewel]

hmm, i saw someone do a similar proof

he made use of the fact of A's eigen values

[boreal crescent]

need some help with part c

a,b is easy

if kerT = 0

that proves that v1...vn,w1,...,wm is a linearly independent set of vectors

ok i think i got it

so you can write any vector v in V as c1v1 + ... + cnvn

if you could write v in V as a linear combination of w1,...,wn, then that would imply that vi,...,vn,w1,...,wn is LD

which is a contradiction

idk the reverse though

nvm got it

works with rank nullity theorem

DIm(IM(T)) = m+n

dim(R(m+n)) = m+n

so dimkerT = 0

so ker t = 0

[wintry steppe]

Hey guys, I have a small problem... I have given a matrix M that - multiplied by x vector - should be equal to zero (Mx = 0).

https://cdn.discordapp.com/attachments/665152019589365760/678179583564185630/Screenshot_20200215-110217__01.jpg

The solution is this: https://cdn.discordapp.com/attachments/665152019589365760/678179583329042432/Screenshot_20200215-110217__02.jpg

My solution has inverted signs

What did I do wrong?

Or is the solution still correct? since the vectors are just inverted 🤔

[gray dust]

you're ok. your vectors are the book's vectors scaled by -1. they span the same space

[wintry steppe]

Aight, thanks 🙂

[mild flicker]

Hey guys, I have a question regarding my approach to a problem.
You are tasked with finding the distance between a the point B(1,02) and the line with direction vector [-1,1,0] and which goes through the point A(3,1,1).
This is an example from the book "Linear Algebra: A modern introduction".

While I do get the same answer to this question as my book, my method is really different.

And my question is if my method is actually valid to use.

I made a plane perpendicular to the line. By definition, the line will be a normal vector on the plane.
The normal vector to plane ax+by+cz = d is [a,b,c].
Therefore, the plane -1x+1y+0z = d has the normal vector [-1,1,0]
The plane is -x+y =d
At point B(1,0,2), we have -1+0=d. d = -1. This gives the plane -x+y = -1

K: -x+y = -1
L: [3,1,1] + [-1,1,0]t
-(3-t) + (1+t) = -1
-3+t + 1+t = -1
-2+2t = -1
2t = 1
t = 1/2

P: [3,1,1] + [-1,1,0]*1/2
P: [3,1,1] + [-1/2,1/2,0]
P: [5/2, 3/2, 1]

The distance between P(5/2, 3/2, 1) and B(1,0,2) is √(11/2)

My book mainly uses projection for these types of questions, but I personally really dislike the idea of using projections

[mild flicker]

Can somebody explain in very simple terms how the formula for projection is derived?

[sonic osprey]

You're basically using projections anyways

[wintry steppe]

So I have a semi random vector which can be some fixed u mapped with f=Ku+e, where K is some nxm matrix and e is a normal distribution in n-dimensions with fixed variance across all the various dimensions. Why then if I take the SVD of K and then have U^Tf that the norm of this increases with the variance of the noise?

[mild flicker]

Wut? How is this a projection?

Because all I did was taking a plane perpendicular to the line and just move that plane over to where it intersected B

[sonic osprey]

And that's what projection does yes

[wintry steppe]

Oh, I'm an idiot - random shit is more likely to be in gaps far away from the column space of the left singular vectors given that f is defined by K.

[mild flicker]

Here's my current informal understanding of what a projection is. It might be totally wrong.
“You pretend that U is a full line, take the length of vector V at the point where V is perpendicular to U, and multiply that length by the original vector U”.

[sonic osprey]

Yeah but what does projection do geometrically

[mild flicker]

I just see it as scaling vector U

[sonic osprey]

https://en.m.wikipedia.org/wiki/Projection_(linear_algebra)

[mild flicker]

Wait... so you can simply just see a projection as a linear transformation, which only applies to a specific vector?

[sonic osprey]

I have no idea what you mean

[stoic python]

Anesthetic:
Compile Error! Click the reaction for details. (You may edit your message)

[sonic osprey]

you need to take the absolute value of y_i but yes

(in case you work over complex)

[brittle juniper]

on $\bbR^n$ the functions
$$y\mapsto\sqrt{\sum_{i=1}^n|y_i|^2}\quad\text{and}\quad y\mapsto\sum_{i=1}^n|y_i|$$
are both norms

[stoic python]

Tuong:

[mild flicker]

@sonic osprey Thank you so much for helping me out. I personally would have never thought to use a linear transformation to prove projection (at least for R2). Thank you so much 😄

[wintry steppe]

yikes

im in 8th grade and need some help with my graph math

education where i live sucks

[wintry steppe]

Hi

I have some questions about the simplex method

What do you do if there is a zero ratio or a negative ratio

Do any of you guys know about span?

<@&286206848099549185>

[ivory shore]

uhm wat

no i dont

i help kids

cause i am a kid

[cursive narwhal]

Do you have a specific question?

@wintry steppe

[wintry steppe]

Yes I do

@cursive narwhal

[cursive narwhal]

sure, what have you tried?

[wintry steppe]

This is what I did for the first problem

@cursive narwhal

Does it look right ?

[cursive narwhal]

certainly looks correct

fuck i'm too tired for this. Someone will have to help you check your work. I'm sorry 😦

[sage mauve]

yeah its correct

[wintry steppe]

Thanks

Would it be better if I PM you?

@sage mauve

This is what I did for trying to find the equation of a hyperplane

@wintry steppe brute force it with parametrics

if you aren't allowed to use the cross product

How do I do that. And what does that mean ?

its been a year, but if (1,1,0) and (0,3,1) span a plane, then you can form the equation with 3 distinct points that lie on the plane, which would be a linear combination of each of them

these three points would satisfy the same equation

you would equate them and solve for what x, y and z is

you could use parametrics, RREF, or sheer luck

I did use RREF

dunno, but if I was told not to use the cross product, that is what I would do

Do you know if what I did is right

[wintry steppe]

So you don't need to use a cross product

@wintry steppe The way I would approach something like this is to recall that an (n-1) dimensional hyperplane can be written as all of the points that are orthogonal to a certain normal vector that one has to find

So we if we find the normal vector $\mathbf n$, we can write out the equation $\mathbf n \cdot \mathbf x = 0$, and if you expand that out, that's exactly the equation

[stoic python]

Sadly Lachrymose:

[wintry steppe]

Of course, finding the normal vector is the hard part. The cross product is a cheap trick for three-dimensional vectors, but you can do this in general

You can use the fact that the normal vector should be orthogonal to all vectors within the plane, so you have $\left[\begin{matrix}1 & 1 & 0\end{matrix}\right] \mathbf n = 0$ and $\left[\begin{matrix}0 & 3 & 1\end{matrix}\right] \mathbf n = 0$. You can combine these into a single matrix equation: $\left[\begin{matrix}1 & 1 & 0\ 0 & 3 & 1\end{matrix}\right] \mathbf n = \mathbf 0$. Now, you should see that the normal vector is simply an element of the null space of this matrix, and you know how to find that using the reduced-row-echelon form.

[stoic python]

Sadly Lachrymose:

[narrow haven]

;l

[wispy perch]

hi, how do i go about proving this?

[fast pivot]

how is matrix multiplication defined? start with that

[noble wing]

What’s the name for a matrix that equals 0 after A^(n), for n above a positive integer?

[sonic osprey]

nilpotent

[noble wing]

thx

[wintry steppe]

Hi, I have no idea what this question is asking

Any help would be much appreciated

[vast torrent]

Are you more familiar with i and j or e1 and e2for the standard basis

Or the same comfort

Hehexd

[wintry steppe]

e1 e2

[vast torrent]

$(x,y) = x \hat{e_1} + y \hat{e_2}$

[stoic python]

gfauxpas:

[vast torrent]

That's what it means to be written in terms of the standard basis

If {b1,b2} were another basis

$(x,y)=\gamma_1 b_1 + \gamma_2 b_2$

[stoic python]

gfauxpas:

[vast torrent]

And so we would say that

$(x,y)=(\gamma_1,\gamma_2)_{{b_1,b_2}}$

[stoic python]

gfauxpas:

[vast torrent]

That's what the notation means

[scarlet ermine]

i'm not sure exactly how to do this but this might be a linear algebra problem so i'm posting it here

[vast torrent]

Chat is busy

[wintry steppe]

@vast torrent i think i manage to figure it out

so because its the basis vector

essentially just plug the vectors into -2 and 5

[vast torrent]

You got it

Just like my equation but working backwards

Okay elf girl

You're up

[scarlet ermine]

essentially i have a primary vector and i have a second list of vectors
i want to find the "similarities" between the primary vector and the list of vectors and get a "score" for each matchup

not super clear on the correct terms, but thats how im visualizing the problem

[vast torrent]

Example with small vectors?

[scarlet ermine]

imagine the core vector is something like [1; 2; 3] and you are given a list of vectors like [2; 3; 4] [4;5;6] etc

and you want to rank the vectors on the list that's most similar to the given initial vector

[vast torrent]

Similar in what sense

[scarlet ermine]

lemme explain the initial problem

there is a list of candidates who are asked to assign a score of importance to a list of issues, a potential voter does the same
i want to order the candidates whose issue priorities most closely match the voter priorities

right now the way i have the solution visualized is that the scores assigned are all vectors and i'm trying to mathematically calculate the similarities between said vectors

but i could be wrong

[vast torrent]

Does it make sense to add these vectors

Because if not then they're not vectors in the sense of linear algebra

Can you add them and multiply them by -1?

[scarlet ermine]

i'm not sure to be honest

i'm still trying to visualize the problem clearly

[noble wing]

What’s a good shorthand notation for switching rows in a matrix? I’ve used (scalar)W & W += or -= for type 1 & type 3 operations

[vast torrent]

I think this is more statistics or discrete math

$r_i \leftrightarrow r_j$

[stoic python]

gfauxpas:

[scarlet ermine]

like if this were to be in three d, im imagining a bunch of arrows, and we're trying to measure the distance between the tips

the closer the distance the more similar the vectors are

[vast torrent]

Oh that sounds like linear algebra

But

[scarlet ermine]

because the sum of the scores is always the same

[vast torrent]

Yeah im not the one to ask about what youre trying to do soz

$r_i \to r_i + \lambda r_j$

[stoic python]

gfauxpas:

[vast torrent]

$r_i \to \lambda r_i$

[stoic python]

gfauxpas:

[vast torrent]

Is what i use

[noble wing]

Thx

[vast torrent]

For the three ero

@scarlet ermine sorry idk, wait 15 minutes then ping helpers

[wintry steppe]

Is it just A and D?

like it can't be B because its not a set

[scarlet ermine]

@vast torrent i figured it out i think

i just take the dot product of the matchups and sort based on score

[wintry steppe]

wdym

[noble wing]

What do you call a matrix that’s diagonal from right to left?

[quartz compass]

antidiagonal matrix

[noble wing]

🙏, kinda wish it had a spiffy upper case abbreviation. For laziness sake

[quartz compass]

say what haha

[noble wing]

Like U for unitary matrix or H for Hermitian

I know it’s not as interesting of a matrix but I am profoundly lazy

[vast torrent]

What are they useful for

Antidiagonal matrices

[fast pivot]

they are like a time reversal operation for vectors

[vast torrent]

Like that movie with Christopher Lloyd and Michael J Fox?

[river jasper]

Need help with a linear algebra question

part e

[half ice]

Lots of FOIL

[river jasper]

I ended up with a 2x1 vectors multiplied by a 2x1 vector

[half ice]

Expand as you normally would

[river jasper]

how can I multiply that?

[half ice]

Oh fk, you're right

Yeah that doesn't work

[river jasper]

can you transpose one of the matrix and then multiply?

[half ice]

You can't get an answer to a broken question

[solar osprey]

Anyone can help

With a linear in/dependence problem

If v1, v2, ...., vn belong to R^n

And they are linearly independent

How can we show that none of the v’s is the zero vector

I thought about something but I dont know if it makes sense

[dusky epoch]

show that a set that does contain the zero vector is guaranteed LD

[solar osprey]

Suppose one was the zero vector

then for any lambda

Different than 0

L1.V1 + L2.V2+.... +Ln.Vn = 0

if one of the Vs was zero

Then it doesnt matter if L is 0 or not

[dusky epoch]

overcomplicated.

very overcomplicated.

[solar osprey]

Which contradicts the defintion

Of linear independence

Why?

I thought it was simple tbh

[dusky epoch]

you could just say

suppose one of the vectors was zero

then make a linear combination of your vectors where the coefficient on the zero vector is 1 and the coefficients on all the rest are zeroes

does it sum to the zero vector? yes

is the linear combination trivial? no

therefore, linearly dependent.

that's it.

[solar osprey]

Thats exactly

What I said

Lol

[dusky epoch]

yeah but you said it obtusely

barfing up a lot more symbols than needed

[solar osprey]

Yea Im french educated

Hard time communicating

Thanks tho

[dusky epoch]

on te fait tout écrire avec un maximum de "rigueur" ?

[solar osprey]

Ouias

[dusky epoch]

ce qui pour certains veut dire "faire une grande soupe de symboles qui servent presque à rien"

[solar osprey]

Personnellement, , j’adore la rigueur

[round badge]

Loving rigor is something new to me

[dusky epoch]

la rigueur est bonne en modération
tout argument peut être rendu aussi rigoureux qu'il devient impossible à suivre

[solar osprey]

When something is rigorous it really makes u dive into the thing

Thats why I enjoy books a lot

And I dont follow my professor’s notes

Id rather study books

[round badge]

But it's like Ann said, it's gotta be in moderation

[dusky epoch]

it is very much possible to have an unhealthy amount of rigor

[round badge]

Otherwise you follow unnecessarily complicated paths and so on

[dusky epoch]

if you deconstruct every argument until you're working directly with the axioms of ZFC

you're doing something very very wrong

[solar osprey]

I dont do that

I do that for practice sometimes

[dusky epoch]

you kind of veered into that territory with your proof, is what i'm saying.

[solar osprey]

So I know that I understood my stuff

I feel so safe

In this server

[solar osprey]

Um im gonna send my solution now

what do u think

[wintry steppe]

can someone help me for 35.5

[gray dust]

what's the defn of basis?

[wintry steppe]

a basis of a subspace is a linearly independent set of vectors

I thought the answer would be yes because its linearly indepdent, but the answerkey says its 'no'

[gray dust]

a set S is a basis for a vector space V if span(S)=V and S is lin indep

recall the defn of span, rethink whether $\Span\brc{u,v}$ is lin indep

[stoic python]

RokettoJanpu:

[wintry steppe]

wait i'm confused

isn't span u,v independent tho

[gray dust]

no, $\brc{u,v}$ is LI

[stoic python]

RokettoJanpu:

[wintry steppe]

LI?

[gray dust]

lin indep

[wintry steppe]

it is independent right..

[gray dust]

but $\Span\brc{u,v}$ is a different beast

[stoic python]

RokettoJanpu:

[wintry steppe]

what do you mean by beast

[gray dust]

i'm saying span{u,v} is not the same as {u,v}

[wintry steppe]

okay i think i get it. Basically a basis is a SET of VECTORS, but the span contains a lot/infinitely many vectors which is dependent. Therefore its NOT A BASIS

[gray dust]

a set S is a basis for a vector space V if span(S)=V and S is lin indep
drill this into your head first

now span{u,v} is the set of all possible linear combos of u & v. by defn span{u,v} is lin dep, so no way span{u,v} serves as a basis for V

[wintry steppe]

thank you

[gray dust]

you're welcome

[wintry steppe]

Now I kinda need help with 36.4

[gray dust]

try showing if $P\cup Q$ is closed under addition

[stoic python]

RokettoJanpu:

[wintry steppe]

what do you mean by closed under addition

[gray dust]

comes from defn of subspace

if you have a subset $S$ of a vector space $V$, $S$ is a subspace of $V$ iff for all $x,y$ in $S$, $x+y$ is in $S$, and for any scalar $c$, $cx$ is in $S$

[stoic python]

RokettoJanpu:

[wintry steppe]

yea I know that's the definition but I'm having trouble using that definition and applying it

[gray dust]

what do you mean by closed under addition
so wdym by this q if you know the defn of subspace

[cursive narwhal]

^They probably know the definition as they’ve stated it, just not as ‘Closed Under Addition’

[gray dust]

anyway, $P\cup Q$ is NOT a subspace of $\bR^3$ and you can show this by letting some nonzero vectors $x\in P$ and $y\in Q$, so from our choice we have $x,y\in P\cup Q$, then test if $x+y$ is in $P\cup Q$

[stoic python]

RokettoJanpu:

[wintry steppe]

Yea but how would I test if x+y is in P U Q

I understand that if X is in P and Y is in Q then X,Y must be in PUQ

And in the question I sent above it involves spam so i'm completely confused

[cursive narwhal]

Suppose x,y belong to P U Q. Then, x belongs to P or x belongs to Q. Similarly, y belongs to P or y belongs to Q. Now if x belongs to P and y belongs to Q, then x+y doesn’t belong to either of them necessarily. So you need extra conditions to guarantee that x+y belongs to P U Q

You can try proving this in greater generality. Suppose that V is a vector space and P & Q are subspaces of V. In general, is P n Q a vector subspace? In general, is P U Q a vector subspace? What are the conditions, if any, that guarantee that both are vector subspaces?

[wintry steppe]

Uhm I haven't dealt with proofs. My lin alg course is purely computational.

[gray dust]

you only need 1 example to show P cup Q isn't closed under addition. pick specific nonzero vectors x in P, y in Q. do some computations to show x+y is neither in P nor in Q and thus not in P cup Q

[wintry steppe]

but how does it relate to span in the question i sent above?

[gray dust]

no idea what your q means

[wintry steppe]

https://cdn.discordapp.com/attachments/540211747613704221/678716031908446349/Screen_Shot_2020-02-16_at_4.34.58_PM.png

like it has "span"

so what would I do?

nvm

[cursive narwhal]

Hmm still though, try doing proofs? Like try working through the proofs? It’s useful to do that

[gray dust]

My lin alg course is purely computational.
doubt they'll work the proofs

[wintry steppe]

I have no idea how to write proofs lol.

[gray dust]

that's why i suggested a computational way, providing a single example to show PUQ isn't a subspace

pick specific nonzero vectors x in P, y in Q. do some computations to show x+y is neither in P nor in Q and thus not in P cup Q

[south wadi]

if you have a free variable does that mean A is a linear combination of b

@vast torrent

what

if you solve this matrix

you get a free variable

does that mean it's a linear combination

since it has infinetly many solutions

[gray dust]

much of this is improperly phrased

sounds like you're finding vector(s) x such that Ax=B

[south wadi]

thats section 1.4

1.3 is vector equations

reviewing these sections for my test next week 1.3 - 1.9

and this question popped up in my head

[gray dust]

if you solve this matrix
so wdym by this

[south wadi]

you solve the matrix

find what x1, x2, x3 are

[gray dust]

what matrix?

[south wadi]

1 0 5
-2 1 -6
0 2 8

[gray dust]

no idea what you mean

there's nothing to "solve" in a matrix like that

the ONLY thing i can see is you're solving the system Ax=B

[south wadi]

bruh look

https://math.stackexchange.com/questions/931814/determine-if-a-is-a-linear-combination-of-b-when-a-free-variable-exists

this man had the same question

[gray dust]

ok look

[south wadi]

reduce the matrix

idk

[gray dust]

that guy is asking how to solve for a vector x such that Ax=B, this results in a SYSTEM OF EQUATIONS

which one can rewrite as an AUGMENTED MATRIX

[south wadi]

yeah Ax = b

they never mentioned it in that chapter tho

i mean section

only 1.4

[gray dust]

and there should be a vertical line between the 2nd to last & last column of the AUGMENTED MATRIX

[south wadi]

so is b a linear combination if theres a free variable

[gray dust]

and in this context

x is a vector given by x=(x1,x2,x3)

now that's all out of the way (please remember everything i said above) we can talk about free variables

at least 1 free var means infinitely many solutions to the system

[south wadi]

yo whats the difference between augmented matrix and matrix'

[gray dust]

a matrix is an array of numbers, that's all there is

an AUGMENTED matrix is a convenient way to write a (linear) system of equations

[south wadi]

ok whats with the caps lock buddy

alr

so is it a linear combination of b

cause theres infintely many solutions

it should be right

if u havbe a free variabl

e

[gray dust]

no we are looking for linear combos of A's column vectors that, when added, produce B

[south wadi]

so no

[gray dust]

your terminology is all over the place so i'm finding it hard to properly answer

[south wadi]

waht do you mean

[gray dust]

your phrasing of things so far is all over the place so i'm finding it hard to properly answer

[south wadi]

is b a linear combination of A if A has a free variable

[vast torrent]

that question doesn't make sense

[gray dust]

if the system of equations you're solving, represented by the augmented matrix, has at least 1 free variable, then there are infinitely many ways to take linear combinations of A's column vectors to produce B

[south wadi]

ok thats what i thought

yo rokabe

[gray dust]

[south wadi]

what year are u

[gray dust]

wdym

[south wadi]

1st year 2nd year 3rd year 4th year

[gray dust]

school?

[south wadi]

yes

[gray dust]

no year

[south wadi]

done?

u graduate?

[gray dust]

i'm out

[south wadi]

wht

what

im out

what

bro ur terminology isnt making sense

[nimble egret]

It means he's out

[gray dust]

it's up to you to interpret what out means

[south wadi]

so u graduated

?

[gray dust]

if that's your interpretation, so be it

[south wadi]

????

a;

f/????

yo rokabe

what dont you know

[gray dust]

how to speak french

[cursive narwhal]

Rokabe, why you prank everyone like this?

😄

[gray dust]

i derive pleasure from it

[thin bloom]

XD

[cursive narwhal]

Jesus rokabe

[gray dust]

that's right, i AM jesus

[boreal crescent]

can anyone help explain this solution

^that's the context to the question

[boreal crescent]

<@&286206848099549185>

[normal canyon]

whats your questions

do you know how to find the inverse of a 3x3 matrix

@boreal crescent

dude i'm helpig you

[boreal crescent]

using gaussian elimination yea

augment it with I3

[normal canyon]

ok bet

so once you augment that bitch

that whore of a matrix

you get the identity

and you know u+v+w = z

[boreal crescent]

yea

[normal canyon]

so once you reduce it to the identity

hold up let me think about this

give me like 2 minutes

then come back

alright cold man

[boreal crescent]

ok

[normal canyon]

ok i figured it out @boreal crescent

you better be thankful for my high IQ

lmao

(1,0,0)

(0,-1,1)

(0,1,-1)

are my three vectors

they form the basis for R^3

understand?????

@boreal crescent dude respond i'm helping you out

you should be giving me head for this

favor

[gray dust]

(0,-1,1)+(0,1,-1)=(0,0,0)

[boreal crescent]

what

i dont get it

ye

they dont form a basis for R3

lol

[normal canyon]

oh shit my b

on jahseh

[boreal crescent]

[normal canyon]

which class is this from? @boreal crescent

[boreal crescent]

intro to lin alg

[normal canyon]

whihc coollege

which college

dawg

[boreal crescent]

that doesnt matter sir

[normal canyon]

is it a T20 or not

[boreal crescent]

it is

why does that matter

[normal canyon]

i'm wondering how high your iQ is

[boreal crescent]

not very high

pls help with that question

[normal canyon]

can you send me the pdf iwth that question

so i can ask a couple of buddies

i'm actually curious to what the answer is

[boreal crescent]

i sent you the answer boss

its apparently [1,1,1]

[normal canyon]

@radiant prawn can u help with this one

please

[boreal crescent]

@gray dust could you help out sir

[radiant prawn]

Yeah. I've got you @normal canyon

First though, you have to tell me what university you go to

I want to know how high your IQ is

[normal canyon]

i go to georgia tech

@boreal crescent what topic is this a part of

maybe i'll use that idea

@sonic osprey r u mad atm e

mad

[slow scroll]

its given that $u + v + w = (1, 0, 0)$
This is the same as saying $$[u \quad v \quad w]\begin{bmatrix} 1 \ 1 \ 1\end{bmatrix} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}.$$ Note that $(1,0,0)^T$ is the first column of the identity, so $(1, 1, 1)^T$ must be the first column of the inverse matrix.

[stoic python]

kxrider:

kxrider:

[meager bolt]

Can someone help me with that one^

I tried to write it into an augmented matrix but cant seem to row reduce it to give me any meaningful answers

[slow scroll]

hint: equation 2 - equation 1 looks a lot like equation 3.

[meager bolt]

I am the dumbest row reducer in the west 😤

[fast pivot]

well, it's either that or computing a determinant, and solving for the "a" that makes that 0

[south wadi]

how do i do #11

[wintry steppe]

how did you do the other ones/what did you try/what didn't work

[south wadi]

What I did was solve the matrix

look at what the free variables were

and put it in parametric vector form

@wintry steppe

for #11 do i simplify to RREF then go from there

look at which variables are free

[wintry steppe]

I'm not sure where to get started for this question

[vast torrent]

is T even a subspace

convex hulls aren't generally subspaces

[fast pivot]

wtf is a "convex linear" combination? just a convex combination?

as in, the set of points inside the triangle with those three vertices?

[vast torrent]

I assume it means all vectors that can be generated by $t v_1 + (1-t)v_2$ for $t \in [0..1]$

[stoic python]

gfauxpas:

[dusky epoch]

[0, 1]

also that's two vectors only

[vast torrent]

I was using knuth's interval notation and I should have said for all v_i, v_j in the space

is what "all convex linear combinations" probably means

[wintry steppe]

so how would I approach doing this question

[south wadi]

What's the difference between R^N and R^M for Linear Transformation

Can someone say this in english

im on linear transformations

[gray dust]

One may colloquially refer to R^n as “n dimensional space”

[south wadi]

what is r^m

[thin bloom]

@south wadi wdym that is English

[south wadi]

yo also @gray dust is ur youtube channel mathispower4u

@thin bloom

what is r^m

[thin bloom]

Why don't you know what R^m means

[south wadi]

idk just dont

[gray dust]

R^m is m dimensional space. So T is a map from n dim to m dim space where n may not necessarily be equal to m

[cursive narwhal]

Strange, your book should’ve covered its meaning beforehand

R^m is the set of all ordered m-tuples with real entries

R^n, similarly, is the set of all ordered n-tuples with real entries. Perhaps this wording may be more familiar to you?

[south wadi]

nope

RokabeJintarou

is ur youtube channel mathispower4u

[gray dust]

No

[south wadi]

lies

u post videos on that channel

how do u do linear transformations

[thin bloom]

I don't get why you are reading this while not knowing R^n

[cursive narwhal]

Wait wait wait, can you show us what was covered prior to this?

[gray dust]

One may colloquially refer to R^n as “n dimensional space”
Have you at least seen this before

[south wadi]

yes

[gray dust]

R^m is m dimensional space. So T is a map from n dim to m dim space where n may not necessarily be equal to m
So you should at least get this too

[south wadi]

ive done sections 1.1, 1.2 , 1.3, 1.4, 1.5, 1.7 now im on 1.8

[gray dust]

Just confirm what I said

[south wadi]

ok

so ur transforming vectors from one space to another

in which it could be a different dimensional space

[gray dust]

Could be same dim, could be different dim. The book is just talking about maps between real vectors spaces in the most general sense

[south wadi]

ok

[gray dust]

If we declare a function T: R^2 —> R^2, this says T maps every vector in 2d space to some vector in 2d space

[south wadi]

what does the word map mean in this caps

case

[gray dust]

Map is a general word for “associate”, “send to”

In middle school algebra you talk about functions sending an input to an output, like f(x)=2x. It sends any input x to its output 2x

[south wadi]

so like how do i go about solving these problems

so for the first one

[gray dust]

I’m just explaining some terminology in the book and what I said

[south wadi]

yeah i understand

[gray dust]

Get that though before you work examples

[south wadi]

Ok

So I see some conditions

If a transformation is linear

T(u+v) = T(u) + T(v)

T(cu) = ct(u)

T(0) = 0

T(cu +dv) = cT(u) + dT(v)

[thin bloom]

Ye

[gray dust]

#prealg-and-algebra

[wintry steppe]

yeah

[gray dust]

iirc last time someone confused linalg for prealg was a month ago. nice to see it happen again

[wintry steppe]

lol

[north sierra]

hows it going all

i haven't been here in a while

hope all is well

[clear spoke]

$x \angle{135} = 5 + 5 \angle{90}$

[stoic python]

<3 Hoodie <3: