linear-algebra

[jagged saffron]

im not good at analysis

If you assume axiom of choice, these function is exists - you can construct this function to use Hamel basis of R over Q. - But if you does not assume axiom of choice, you can't show that nonlinear additive function - If you assume axiom of determinacy, every real function is measurable. And every measurable additive real function is linear.

[uncut forge]

The function f(t)=E(X(t)) where X(t) is a random process with stationary independent increments

But we can probably assume it is at least bounded

[jagged saffron]

oh

bounded is all you need

then

[uncut forge]

Thanks for finding the link it was an interesting theorem

[jagged saffron]

yea i actually remember doing this in my analysis class once

but i hate analysis

[uncut forge]

Yeah I think I did too for continous functions

And density arguments

Ah actually

Since X(t) is defined for all t it has to be bounded on every interval

[celest bridge]

Yea we actually covered that formula about the det wrt permutations, I just kinda glossed over that day and completely forgot about it.

I'm looking over my notes now and finally see it lol

Thanks again.

[clever cedar]

I cant wrap my head around this definition

What does it mean S can be extended tk a basis of U

To**

If we have a basis of U

Does it mean we can make S into that basis?

[worn crow]

yes

by adding vectors

[clever cedar]

Quick q

Does a subspace only have 1 basis

Or can it have multiple

[worn crow]

usually it has infinitely many

[clever cedar]

Oh

So we're trying to find one of the basis of U

Given those 2 vectors

[worn crow]

yes

[clever cedar]

Sick thank you

[uncut forge]

There's always infinitely many bases unless it has dimension 0

You can just scale one of the basis vectors and get a new basis

[clever cedar]

I thought since a basis is lin independant than its unique

I guess not

[uncut forge]

Nope because if for example u and v are linearly indep, then u and 2v is too

[clever cedar]

Oh shit ya

[uncut forge]

Try to think of it geometrically

If the dimension is 2 then all vectors not on a straight line are linearly independent

[clever cedar]

Oh and we can generate vectors not on a straight line by a sum of any two other vectors? (If that makes sense)

Nvm

Maybe not

[uncut forge]

Nope not in general

[clever cedar]

Alright. Well thank you very much

[uncut forge]

Np, btw a good youtube series about linear algebra is "Essence of linear algebra" by 3B1B

[clever cedar]

Ill definitely check that out

[clever cedar]

Im confused about the definition of image, if a kernel is the vector matrix that when multiplied by the matrix A equals 0, what is image

i cant understand notation

what is this saying

isnt that infinite?

thank you

another q

what is difference between kernel and nullspace

[blissful vault]

for this proof, is this enough?

if vector v = sum(k=1, n) a_k v_K but also = sum(k=1, n) b_k v_K

then sum(k=1, n) a_k v_K = sum(k=1, n) b_k v_K

and sum(k=1, n) a_k v_K - sum(k=1, n) b_k v_K = 0

therefore there is a unique solution only

[clever cedar]

unique basis u mean?

[celest bridge]

same thing

Just different word. Kernel sounds cooler, but null space makes much more obvious/intuitive sense.

[gray dust]

@clever cedar kernel's a more 'general' term for any function between two sets. kernel of a linear transform T on a vector space, ker(T), can also be referred to as null(T)

[clever cedar]

Thank you

[stoic elm]

how 2 do this

[gray dust]

parallel lines have direction vectors that point the same direction

[urban bough]

is the frobenius norm of a matrix the same as the 2 norm of a matrix?

and why or why not?

the two of them seem identical

[rancid sonnet]

https://cdn.discordapp.com/attachments/359052604149465088/641962102977855488/Capture.JPG

reposting because it went unnoticed probably the last time

The first two axes correspond to x and y

I'm guessing the last three has to stretch from minus to plus infinity?

And I could define the color to be the remainder when divided by 256?

[slow scroll]

@rancid sonnet can't help much since im going to bed, but note that nothing is being done to the position of the pixels in order to invert the colors. Lets say you have a color/position (x, y, a,b,c). Then according to the problem, the inverted color is (x, y, 255 - a, 255, - b, 255 - c). Think about what the linear transformation has to do to each coordinate to come up with a matrix representation.

[rancid sonnet]

yes I already noticed that

The problem is, just subtracting 255 isn't a linear transformation since it shifts the origin, that's why I thought I had to represent them with that remainder thing

[slow scroll]

Oh yea duh... hmm I know of way to do it by adding an extra dimension but that is probably unnecessary. Yea idk but I gtg

[versed topaz]

Dumb question: what does it mean for a quadratic form to be nonsingular? My prof said it meant that the associated matrix was nonsingular, where the matrix M associated to the form Q is the one satisfying Q(v) = v^T M v. But this doesn't seem well defined? If Q(x, y, z) = Ax^2 + By^ 2 + Cz^2 + Dxy + Eyz + Fzx, then we could pick M = [[A D F] [0 B E] [0 0 C]], which is nonsingular iff A≠0, B≠0, C≠0. But it seems like this would say xy + yz + zx is a singular. That sounds ridiculous, since this is just a hyperbola (there's no cusps or anything)

[wintry steppe]

when you're trying to find eigan values

and you're converting to an upper triangular matrix

how do you factor in the negative?

if you switch around the rows?

or do you just multiply by -1?

[lament atlas]

Hey guys can you help me pls. The question is about vectors. V.W1×W2 = 1 what is the result of W2×(W1-V).W1 ?

V, W1 and W2 are vectors

The answer is - 1?

[wintry steppe]

hey i need a sanity check

[lament atlas]

<@&286206848099549185>

Yeah, i did the same

Should be right

Thanks

[half ice]

@rancid sonnet
You could let the vectors be mod 255, then T(x) = -x which is linear. Problem is, you can't represent 255 seperately from 0 this way

[rancid sonnet]

Just let it be mod 256? @half ice

[half ice]

Then everything is 1 off

And my solution doesn't work so that's sad

[rancid sonnet]

how

[half ice]

I noticed that, if T(x) is the inversion of x:
T(x) + x = 255

Which makes mod 255 a natural choice

[rancid sonnet]

oh right

the transformation

hmm

so I'm still stuck

what if

I did it mod256 anyways, and defined the colors to be the remainder minus 1 ? @half ice

and defined 255 to be when the remainder is zero

[half ice]

Not linear anymore

I believe? Maybe check, but sounds ehh

[north cloak]

Anyone wanna help in B?

[quasi vale]

someone did reply

[clever cedar]

i dont understand the language of the example

does it mean describe the set that spans those two vectors?

for where u & v span(R)

where R is a set

[dusky epoch]

no

you're asked to describe the set SPANNED BY u and v.

[clever cedar]

Oh so one set simply has to be a linear combination of both vectors.

Thank you

[dusky epoch]

...

no, bad wording

[clever cedar]

sorry in my mind if a vector is an element of a span set than that vector can be derived as a linear combination of that set

is that wrong?

[dusky epoch]

that's less wrong than before

[clever cedar]

I guess ill need to re-study the formal definition

[clever cedar]

Oh i think see now, span is any linear combination of those two vectors. And from those two vectors I presume we generate the R^2 vector space

That makes more sense I believe

[dusky epoch]

...no not really

[clever cedar]

lol not really or no

[dusky epoch]

you're trying to say something but it inevitably ends up mangled bc of your unfamiliarity with mathematical language in general

[clever cedar]

I have no shame in my lack of mathematical parlance. If I fundamentally understand its nototion in my head than I'm happy regardless

[dusky epoch]

no offense but you cannot claim fundamental understanding if you can't word it coherently

[clever cedar]

non taken and thats fine. Hopfully as I read and study more my understanding will grow

"The span of a set of vectors is the set of all linear combinations of the vectors." @source https://homepages.rpi.edu/~mitchj/handouts/linalg/

so if we ask if v${\in}$span(U) than we're asking if v is a linear combination of the set

[stoic python]

Mike0x81:

[dusky epoch]

h bad tex

$v \in \mathrm{span}(U)$

[stoic python]

Ann:

[dusky epoch]

but yes

[clever cedar]

Incredible, thank you

[rancid sonnet]

what's a linear combination of a set?

all possible linear combinations of all the elements of the set?

[clever cedar]

Can u write down exactly what question / example is confusing u

[rancid sonnet]

No, my question was related to your statement

https://cdn.discordapp.com/attachments/540211747613704221/642444352966230046/360640011638669316.png

where you said "linear combination of set" @clever cedar

[clever cedar]

ohh

yes sir

i mean that the set contains some vectors (at least) that can be linearly combined to create the vector

[rancid sonnet]

oh

I thought you meant something else

First time I heard someone talking about linear combinations of sets

Is that a term used often? Sounds ambiguous

[charred stirrup]

for this question I solved it by finding the RREF and then finding row and col by inspection

and i did the null by writing pivots in term of free variables

but for this question the answer is kind of backwards

when I find the RREF, i need to transpose the original matrix, why is that?

so the answer to row(A) is a 5x2 instead of a 2x5

[clever cedar]

can someone tell me if this is a good answer for part 1

these vectors are not linearly independenant because there are 5 vectors and only 4 vectors are needed to the span the vector space R^4. So at least 1 vector is dependant

[burnt heath]

@charred stirrup u go to uofa ?

[charred stirrup]

ur boy is socal and why @ if no help

[clever cedar]

r u saying row(A) is incorrect?

[clever cedar]

i dont understand this question

assuming w and w_1 are in R^4 do i have to see if u is in the subspace of R^4 given those constraints?

is it asking me if u is a subset of R^4 given that w * u = 0 and w_1 * u = 0?

[gray dust]

asking if M is a subspace of R^4

[clever cedar]

what does the stuff after the colon do to the question

like what do i have to take into consideration

i shouldve asked with an easier example

[gray dust]

conditions that all elements in M have to satisfy

[clever cedar]

is M comprised of only one vector called u

or infinite vectors

[gray dust]

M contains all vectors in R^4 that satisfy the given conditions

[clever cedar]

and my job is to check if those vectors which satisfy the given conditions are also in the subspace of R^4?

sorry my professor didnt really explain these problems

[gray dust]

you really need to review how to read sets

[clever cedar]

i was never taught sets in the first place

[gray dust]

${x \in \bZ :x \geq 10}$

[stoic python]

RokettoJanpu:

[clever cedar]

x is an element of Z such that x >= 10

[gray dust]

no

[clever cedar]

i mean id be happy to pay u for tutoring if that would help

i am just eager to understand

[gray dust]

it's the set of ALL x's that are elements of the set of integers, such that x is greater than or equal to 10

[clever cedar]

Ohhh okay

I wasnt aware that the fancy Z meant integers

[gray dust]

there are many others

$\bR$ for the set of real numbers

[clever cedar]

that is the only one ive seen really

[stoic python]

RokettoJanpu:

[wintry steppe]

anyone knows how to use matlab or octave??

[clever cedar]

ask #computing-software

[wintry steppe]

okay

[half ice]

@clever cedar
u is a vector in R⁴ that, when dotted with w or w1, gives 0

[clever cedar]

When we say u we're generalizing for ANY vector in the R^4 space?

[half ice]

Well, any u that satisfies
u•w = 0, u•w1 = 0

[clever cedar]

Oh ok

[half ice]

There's a mix of languages here oop

"any u" or "some u that satisfies"

[clever cedar]

Ohhhh thats more clear

[half ice]

Any is best in this case

[clever cedar]

Roketto helped me realize that {} means some set of U

(i mean the curly braces)

[half ice]

If I said
A = {1, 2, 3}
That means "A is a set that contains 1, 2, and 3"

[clever cedar]

Ahh okay, theres just so much notation ive never seen in that problem that I couldnt break it down

[half ice]

In your above problem, M is a set that contains all of those vectors that follow those rules.

[clever cedar]

ohhhhhh thats 100000x more clearer

than i jsut use subspace test to see if that set is a subspace

[half ice]

Yus.

[clever cedar]

but am i assuming that im checking for if its a subspace of R^4?

since it doesnt clarify which space

[half ice]

It says u ∈ R⁴

[clever cedar]

god damn how can such few symbols say so much lol

crazy

[half ice]

I guess you're right that it's not saying which space it should be a subspace of, but R⁴ is a fair assumption

[clever cedar]

I see, thank you a million

if u have a donation page id be happy to donate

[half ice]

Lel nah sorry

I can't make a website. Not after last time

[clever cedar]

o what happened last time lol

[half ice]

Jk nothing that crazy

[clever cedar]

oh lmao

[wintry steppe]

How do I do this? Please explain as I want to prepare for a upcoming test

Sorry I'm only in the 7th grade

I'm new btw

[half ice]

A function passes the vertical line test

[wintry steppe]

So it would be c right? Because the x coordinate never repeats

[half ice]

There's only one y for any x, yes

[clever cedar]

it says M is a subspace in the bac of the book

wtf

how

[wintry steppe]

But what does it mean it asked the function of x

[half ice]

This isn't the right channel for you @wintry steppe, try #prealg-and-algebra

[wintry steppe]

O srry

[half ice]

Just a function where x is the dependent variable

@clever cedar
It passes the test!

[clever cedar]

if u1 = 0 then theres no way we can create a set for r^4

and thats the constraint

[half ice]

[ 0 ]
[u2]
[ 0 ]
[u4]
Is a member of R⁴

For any u2 or u4

[clever cedar]

but doesnt M have to be a set for all of R^4

[half ice]

M does not have to contain all of R⁴, no

Just the vectors that follow the rules given there

[clever cedar]

ffs why am i retard

[half ice]

This is notation you've never been taught, I get it

[clever cedar]

thank you

[clever cedar]

is my solution correct for this problem? If anyone oculd check

[half ice]

@clever cedar
"Passes the subspace test" isn't great reasoning. I have no idea how you'd check for the rules here. Instead, you noticed that S is the span of those three vectors, and a span is always a vector space

The logic is good on the basis and dimension

[clever cedar]

Yeah im still working on wrapping my head around subspace test. Its hard for me to think of a case when a subspace test may fail considering the infinite possibilities

Thank you for your response on the basis and dimension. That was a concern

[clever cedar]

is this a good example of proving the subspace test?

[half ice]

@clever cedar
M is the set of all vectors orthogonal to w

If I may use that notation.

1st subspace test:
Is the zero vector in M?

Remember, "to be in M" means "orthogonal to w" here. So, we're asking "is the zero vector orthogonal to w?"

[pallid swallow]

we just need M nonempty, addition and scalar multiplication closed in M

[clever cedar]

oh so not only does a zero vector have to exist but it has to be orthogonal to w

whatever ill just fail this part ive spent too many hours and its still over my head

[half ice]

"Existing" is "being in M"
and "being in M" is "orthogonal to w"

[clever cedar]

How am I suppose to know if the zero vector, being in m, is orthogonal to w, when w can be $\in$ of R^4

[half ice]

If you want to move on we can oop

[stoic python]

Mike0x81:
Compile Error! Click the reaction for details. (You may edit your message)

[clever cedar]

like the zero vector could be orthogonal to w = [1 0 0]

but what if w = [ 0 1 0]

there are literally infinite

posibbilities

[half ice]

No matter what w is, what happens if you dot it with [0,0,0,0]?

[clever cedar]

oh fuck

its 0

[half ice]

Exactly. The zero vector is orthogonal to anything. So, the zero vector is in M

[clever cedar]

ur mind is fucked that you're able to realize that without problem

[half ice]

Just lots of time with it

Plus some important theorems come from this idea

Fine, I'll give you spoilers

[clever cedar]

lol

[half ice]

The set of all vectors orthogonal to a list of vectors S is called the orthogonal subspace from S

It's a common way to generate subspaces

[clever cedar]

oh thats crazy

[half ice]

Want to move onto the second test?

[clever cedar]

please

[half ice]

You need a special property of the dot product. It distributes over addition
(a + b) • x = a•x + b•x

So let's say u•w = 0 and v•w = 0. That means both u and v are in M.

Is u + v in M?

[clever cedar]

as long as both u and v satisfy the constraint and are in R^4 then yes I believe so

[half ice]

That is, is (u + v)•w = 0?

[clever cedar]

Yeah because it satisfies the constraints

so logically it must be

[half ice]

(u + v)•w
= u•w + v•w
= 0 + 0
= 0

So if u and v are orthogonal, u+v is too

[clever cedar]

I gotta brush up on my knowledge of dot product and other theorem axioms

[half ice]

Yeah that one's important. Luckily there's a very similar one for the next test

k(u•v) = (ku)•v

Makes the third test similar to the second

[clever cedar]

oh wow so in order to do these tests I literally have to mathemtically prove them

for the given constraint

[half ice]

Maybe your teacher might want it easier on you, but this is proofs!

Like a pure mathy

[clever cedar]

Lol wow now I know to not take the proofs and conjectures class

too intimidating

[half ice]

Why not? You're already there

You're doing the best part of lin alg right now

[clever cedar]

😛 only with help for 80% of it

[half ice]

Actually, you're getting a lot better

[clever cedar]

That means a lot

[half ice]

The proofs are eh but you're actually stating what you need to prove, and that's the hard part

[clever cedar]

interesting, glad im not completely clueless

[charred stirrup]

what's the standard matrix for reflecting about line in y = -x ?

[quartz compass]

you can work out the columns of a matrix by looking at where your basis vectors go

[young pasture]

Stuck on 2a

S2 and S3 specifically

I dont think it is subspace

[charred stirrup]

@young pasture I wouldn't say that cant be the case, but the question by inspection heavily implies that all three of the rules hold for the question, otherwise I think it would ask you to prove it is not a subspace

[half ice]

@young pasture
I was just doing this exact question above with Mike

[prime needle]

45. Let f(x) = x and g(x) = x3 be vectors in C[0, 1]. (a) Find 〈f, g〉. (b) Find |g|. (c) Find d(f, g). (d) Orthonormalize the set B = {f, g}.

can someone help me with that?

[dusky epoch]

by x3 do you mean x^3

[prime needle]

yeah

x^3

[dusky epoch]

uhhhhhh

wait

what's your inner product

[prime needle]

I have the answers but I have no idea how to get it

[dusky epoch]

are you defining your inner product as $\left< f, g \right> = \int_0^1 f(x)g(x) \dd{x}$

[stoic python]

Ann:

[dusky epoch]

or something

[prime needle]

oh i think so

[dusky epoch]

wait no

hang on

ok no this is C[0,1] it's fine

[prime needle]

i saw that somewhere in my textbook but I thought it was just for a specific example instead of for all f and g

yeah i think that's it

thanks a lot

[prisma kestrel]

yo

[dusky epoch]

ok

so

as i was saying

[prisma kestrel]

your net wont matter too much i guess since i dont know if its your exercise to teach me lmao

[dusky epoch]

matrix multiplication is defined such that composition of linear transformations corresponds to multiplication of their matrices

[prisma kestrel]

can you explain the composition of linear transformations or linear transformations at all?

[dusky epoch]

...are you saying you don't know what those are

what the fuck

i was going to recommend the essence of linear algebra series by 3b1b, as it does a good job explaining certain bits of linear algebraic intuition

[prisma kestrel]

im not at a university for math so nope

so if you know this fact its probably not "what the fuck" anymore 😄

[dusky epoch]

[prisma kestrel]

🤣

[dusky epoch]

@slender yarrow i saw you typing, do you mind taking this over

what with you being discount me and all

[slender yarrow]

err i don't feel like i could explain that super great since i haven't really done lin alg for quite a while

but i could try

[dusky epoch]

rip

[gray glen]

translation: "I ain't tryna talk to these mfs smh"

[dusky epoch]

appuyez sur F pour payer des respects

[sonic osprey]

im here

[prisma kestrel]

well i've just looked up linear mapping but probably just the conditions under which a map is a linear map isnt everything which is worth noting about those

its pretty interesting how different the german and the english wiki articles are in contrast lmao

[dusky epoch]

,,,

[gray glen]

,,,

[prisma kestrel]

well i already guessed this kind of reaction occurs if somebody who hasnt got the opportunity to study at a university wants to learn something trys to achieve that 😉

[sonic osprey]

You're really projecting your own attitude onto their reaction

[prisma kestrel]

nah

[sonic osprey]

That's not what they were trying to say at all

[prisma kestrel]

...

this is just a reaction

like

are you serious

are you dumb

why

just why

[sonic osprey]

You're actually just projecting lmao

I know both of them well

[prisma kestrel]

then i'll be sorry

but as you might guess, its a bit hard for me to learn everything on my own

[sonic osprey]

Sure, you should be proud of doing tough things

What were you questioning anyways

[prisma kestrel]

not begging for compassion, just summarizing it

[gray glen]

what did you say first

[prisma kestrel]

well i had two initial questions

1. why are matrix multiplications defined the way they are
2. why do we need multiply a vector with a matrix when rotating the vector

[sonic osprey]

Well first thing about matrix multiplying a vector

[gray glen]

I'd argue 2 is just not a good question

even if you don't know anything

[sonic osprey]

If you think about all the vectors, matrix multiplication onto vectors is a function from vectors to vectors

[prisma kestrel]

so you wanna say that for A x B the function is mapping from $X^{m x n} x Y^n$ to $Y^n$ ?

[stoic python]

Marino:

[prisma kestrel]

have i wrote that even correctly? lol

[sonic osprey]

Nah, I'm talking about matrix times vector multiplication right now

So take a matrix and multiply it on the right by a vector

[prisma kestrel]

yeah

i just tried to formalize it

[sonic osprey]

The mapping is just Y^n to Y^n

[prisma kestrel]

and tried to say that matrix m x n multiplied with vector of dimension n results in vector of dimension n

well the mapping takes two arguments doesnt it? matrix and vector

[sonic osprey]

The function is just from Y^n to Y^n because you take vectors from Y^n and the matrix multiplication sends them to vectors of Y^n

No you just fix some matrix

[prisma kestrel]

oh so we imagine a static matrix?

[sonic osprey]

Every matrix gives rise to a function from Y^n to Y^n

[prisma kestrel]

well you mean that for every matrix we could define a function which maps the vector to its product (with the corresponding matrix)

[sonic osprey]

yes

[prisma kestrel]

yeah makes sense

[sonic osprey]

So every matrix is just a function

But now we have some idea of multiplying matrices

[prisma kestrel]

stop

[sonic osprey]

But if we take two matrices A,B and a vector v

[prisma kestrel]

So every matrix is just a function

well you mean that we map an index to its corresponding coefficient?

[sonic osprey]

uh what

This is what we've been talking about

"Every matrix gives rise to a function from Y^n to Y^n"

[prisma kestrel]

well for given vector element of Y^n

right?

[sonic osprey]

The matrix is a function that takes elements of Y^n and outputs other elements of Y^n

[prisma kestrel]

wait xDDD

we talk cross i think

can you give me an example to show me what is getting mapped to what?

because if we talking about matrix as a function i had thought we map an index to the coefficient, so I -> X where I is the index set and X the set which the coefficients lie in

[sonic osprey]

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

[stoic python]

Zopherus:

[prisma kestrel]

so for the static matrix (1,2,3,4) each vector gets mapped to another vector which is the result of the multiplication matrix * vector?

[sonic osprey]

yes

[prisma kestrel]

ok

[sonic osprey]

So now,

if we have two matrices A,B and a vector v

We want that $(AB)v = A(Bv)$

[stoic python]

Zopherus:

[sonic osprey]

In other words, if we multiply the matrices AB and then apply the function to the vector v

It would be the same as if we applied the matrix/function B to the vector v, then applied A to the new vector

[prisma kestrel]

well but do we know how the function has to look like? the Y² -> Y² one

i just used ² for readability reasons

[sonic osprey]

Yeah

[prisma kestrel]

then we should clarify that function design first

how does this function look and why?

the "map static matrix * vector to new vector" one

[sonic osprey]

How it looks is easy, the why is a bit more tricky

But starting with the why is more important

[prisma kestrel]

yeah sure

[sonic osprey]

The idea is that as you read, there are linear maps from Y^n to Y^n

These are vaguely like lines through the origin

[prisma kestrel]

that would be x ↦ a*x ?

[sonic osprey]

Yeah

Its basically this idea, but in higher dimensions

And skipping over some of the details here, it turns out that all linear maps from Y^n to Y^n can be represented as a square matrix

[prisma kestrel]

"it turns out"

werent we talking about the why? 😅

[sonic osprey]

Yeah, there are just a lot of details here that I don't really have the time to fill in

If you really want to learn this stuff, you should read a book

[prisma kestrel]

okay

but the square matrix you talked about

so do you wanna say this matrix is finite?

[sonic osprey]

The dimensions of the matrix are exactly n by n

[prisma kestrel]

but static matrix * every possible vector can be summarized in n*n values? o.O

[sonic osprey]

Because it's a linear function yes

[prisma kestrel]

and i guess this matrix is the matrix multiplication right? 🤣

[sonic osprey]

uh what

[prisma kestrel]

oh wait no

wrong

but how is it going on? which relation does the n*n matrix have to the matrix?

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

any idea for this problem?

$\begin{vmatrix}1+x_1&1+x_1^2&\cdots&1+x_1^n\1+x_2&1+x_2^2&\cdots&1+x_2^n\\cdots\1+x_n&1+x_n^2&\cdots&1+x_n^n\end{vmatrix}$

[stoic python]

Nguyễn Thành Trung:

[lost comet]

A particle is fired downwards at speed v from height z equals H. It then moves freely under gravity. At what time does it reach z equals 0?

hi guys i dont know how to start this question

could anyone help

[lost comet]

nvm i got it

nah i got it wrong lol

[surreal shuttle]

@lost comet wrong channel

[lost comet]

which channel

[surreal shuttle]

you should be in either #prealg-and-algebra or #precalculus

[lost comet]

ohhhhh

thanks

ye i dont mind

i think i got my sign wrong

[north sierra]

im having trouble finding the determinant of b

using co factor method

im doing something wrong

the first matrix is the original matrix

im supposed to be getting -2

but i get like -8 or something

or maybe this is right?

idk could someone check

[clever cedar]

@north sierra

ur using cofactor method but ur row reducing?

[north sierra]

lol mike i was reading your old messagees

i made the same mistake again

[clever cedar]

o which mistake

[north sierra]

the one you were telling me about last time

[clever cedar]

oh lol

[clever cedar]

happens lol

[north sierra]

c.

[clever cedar]

oh haha

[north sierra]

lol yeah

[clever cedar]

theres a alot of rules to remember

so its understandable

do u still need help with the above q?

[north sierra]

nah im good!

thanks though

i got the answer

[clever cedar]

oh nice

[keen thorn]

Can someone please explain to me what reduced cost means in layman terms in linear programming?

Everything I read online makes no sense to

[north sierra]

not sure how to do this

[sonic osprey]

try out some examples

see if you can find a pattern

[north sierra]

u talking to me?

[sonic osprey]

yes

[north sierra]

im not sure what r is

[sonic osprey]

So?

You don't know what A is either

That's the point of examples

[north sierra]

lol

true

what do you mean exampleS?

i checked the explanation online

but i still dont get it

not sure where they get the identity matrix from

[sonic osprey]

Pick some random r and A and see what happens

See if you notice a pattern

[north sierra]

ok

oh i get it now

lol

so n is just the size of the matrix right

r^n

[sonic osprey]

yes

[north sierra]

ok

making some examples helped yeah

thx

[clever cedar]

u havent asked a question

[topaz plover]

Hey anyone know what I should do to find a basis for col A

I know how to get col A

and I know how to get a basis of something

I know how to get something in col A*

[half ice]

@wintry steppe
A change of basis matrix (Not an identity matrix!) maps β to η.

Any linear transformation in the first basis, when composed with the change of basis, gives the same linear transformation in the second basis

I'd like to repeat, absolutely not an identity matrix

@topaz plover
Row reduce. Any column without a pivot represents a vector that's causing linear dependence. Cut them off, you have a basis

[mossy saffron]

what's the difference between the dot product and matrix multiplication?

[slow scroll]

A dot product is a binary operation on vectors to the target field, while matrix multiplication is the composition of linear transformations. The difference is pretty much... everything

[half ice]

Vector dot vector = real number
Matrix multiplied by matrix = matrix

[quartz compass]

matrix multiplication is the dot products of all the row vectors of the left matrix with all the column vectors of the right matrix

[mossy saffron]

I see

[topaz plover]

@half ice isnt that how you get a basis

but how do you get basis of col(a)

like what you said thats how you get a basis of a mtrix

is it the same for getting basis of col(a)

what about getting basis for nul(a)

[half ice]

@topaz plover
What you're calling "the basis of a matrix" is actually "the basis of col(a)"

For null(A), multiply by a general (a,b,c,d) vector, find out how to get the zero vector as an output. This is also done by row reducing

[north sierra]

why is that for a system to have uique solution, the determinant of coefficient matrix must not be zero?

unique*

[slow scroll]

because invertible matrices have non-zero determinant @north sierra

[north sierra]

true

and if the determinant is zero, it's linearly dependent right?

the columns of the matrix

[slow scroll]

yep. And if the columns of a matrix are linearly dependent, then the kernel is nontrivial, which means non-unique solutions

[north sierra]

kernal is another word for null space right?

kernel8

[slow scroll]

yea.

[half ice]

And the kfc guy

[north sierra]

lol

thank you

[slow scroll]

npnp

[half ice]

Property to know:
det(A^-1) = det(A)^-1

[topaz plover]

could someone help me with this plz

[half ice]

That's why a matrix with determinant zero has no inverse. If it did, it would have a determinant that divides by 0

@topaz plover
You've got to reduce!

Try to get the pivots in the same place

[north sierra]

could someone check my answer for this

just for x1

the back of the book says 7/(3p-3)

but i think they're wrong

i double checked my mark though

i let p = s

cause s looks like 5 and it gets confusing

[north sierra]

i dont know how to do this

[half ice]

You want to find a vector u in H, and a scalar c, such that the vector cu is not in H

[north sierra]

so i chose a random vector u

(3,7)

and scalar 2

and 2(3,7) = (6,14)

now?

[half ice]

Is (6,14) in H?

[north sierra]

how do i check?

i tried row reducing

but im having trouble

[half ice]

H is the set of all vectors of the form (3s, 2 + 5s)

[dusky epoch]

you're already overthinking it

there is no row-reduction to be done here

[north sierra]

oh

[dusky epoch]

does there exist a number s such that (6, 14) = (3s, 2+5s)?

if yes, then (6, 14) ∈ H. if not, then (6, 14) ∉ H.

[north sierra]

how to i figure out if it's an element of H?

[dusky epoch]

does there exist a number s such that (6, 14) = (3s, 2+5s)?
if yes, then (6, 14) ∈ H. if not, then (6, 14) ∉ H.

i just wrote it

[north sierra]

lol

[dusky epoch]

i know your brain's stuck in overthinking mode

[north sierra]

yeah

[dusky epoch]

but please read stuff people are saying

[north sierra]

it is lol

[dusky epoch]

snap out of it

[north sierra]

LOL

[dusky epoch]

does there exist a number s such that (6, 14) = (3s, 2+5s)?
if yes, then (6, 14) ∈ H. if not, then (6, 14) ∉ H.

[half ice]

You want
(3s, 2 + 5s) = (6, 14)

By looking at the first entry,
3s = 6
s = 2
But that gives me (6, 12), not (6, 14). So my attempt to shove (6, 14) into H has failed me.

[north sierra]

ah lol

my brain was sooooo over thinking this

thanks all

[dusky epoch]

you could've chosen your scalar to be 0

(0,0) is clearly not in H

[half ice]

Oh yeah, that's the easiest way to break it. But your random choice did too

[north sierra]

true

[topaz plover]

im still confused

on how to do this

like

how to get a,b,c,d,e,f

[dusky epoch]

i'd suggest working backward from the rref to get the right constants in your rows

[unborn rover]

Hello, i'm trying to create a spherecast against a triangle and stumbled upon a problem for which i need vector math help. I am trying to detect collision between a spherecast and a line segment, however, not just any point, but the closest to the spherecast's origin. I've transformed the logic of the problem to be: find the points on 2 defined rays: AB and CD which form a third ray EF that has a defined length. The point E on ray AB should be chosen to be closest possible to ray's AB origin point A. https://i.imgur.com/hgUNnTm.png

[pallid swallow]

ah but the planes are important

use planes that are parallel to both vectors

@unborn rover

if r is really big, it's possible, right?

then it has something to do with distance between planes

[topaz plover]

is there a way to do this besides reverese RREF

cuz that sounds hard

[steady fiber]

column dependence does not change from row operations

if column A is 2*column B - 3*column C in the original matrix

that will still be true after doing row operations

[topaz plover]

@steady fiber what do u mean

[winter siren]

Can you see if two vectors is linearly dependant or linearly independant based on the diagonal having leading entries when reducing
to row echelon form?

2 Vectors:
(6, -3),
(-4, 2)

in R^2

According to this website it seems like that, but I want to be sure:
http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?cmd=li;vdat=2;vdat=R2;m1_1=6;m2_1=-3;m1_2=-4;m2_2=2;curn=2;curn=2;submit21=Submit&hideops=0

[half ice]

Oh that's very smart
3 3 1
b 2 and f
0 d 1
Must be linearly independent @topaz plover

@winter siren
You can always take any set of vectors, and cut some out to create a linearly independent set. Put them into a matrix and row reduce, any column without a pivot represents any vectors you should cut out.

[north sierra]

so this is not closed under scalar multiplication cause of the second row

but what about the first row and third row?

[topaz plover]

@half ice ya ik but like

idk what to do

I have a a matrix with variables

and

a reduced matrix

[half ice]

Basically,
3 3 1
b 2 f
0 d 1
Has a non zero determinant

[steady fiber]

you also know that [3 b 0]^T and [a -8 0]^T

are linearly dependent

so they have to be multiples of each other

and so a = -24/b

that was the easiest dependence to see

but there's a ton of easy dependences in the vectors you can see

just use them to find the relationships between the variables

[winter siren]

"You can always take any set of vectors, and cut some out to create a linearly independent set. Put them into a matrix and row reduce, any column without a pivot represents any vectors you should cut out."

I'm sorry @half ice, but I don't quite get it.

pivot = A non-zero entry in the diagonal?
"... any column without a pivot represents any vectors you should cut out."?

[half ice]

@winter siren
I suppose the thing I'm getting at is to row reduce

You won't get the identity matrix

6 -4
-3 2

Does not reduce to identity, so the set is linearly dependent. Another easy way to see that is
-2/3 (6,-3) = (-4,2)

You can make one vector using the other, so the set is linearly dependent

[topaz plover]

how do I get the general solution

for this

[winter siren]

Ohh, okay.

Thanks @half ice

[topaz plover]

do I like

[half ice]

You know that if you multiply that vector by A, then you get b

No matter what s and t are

[topaz plover]

true

@half ice so how do I go towards getting the general solution

[half ice]

@topaz plover
A is a linear transformation. So:
T[(1,0,-3) + s(5,1,0) + t(-2,0,1)]

= T(1,0,3) + sT(5,1,0) + tT(-2,0,1)

= b + sT(5,1,0) + tT(-2,0,1)

But that should be b. So, the other two parts must be 0

The same thing without the (1,0,-3) bit is the nullspace

[vernal salmon]

If v is a vector of arbitrary dimension (finite) with complex number coordinates, 
does < v, v > = 1, imply <v complex conjugate, v> = 1

(dot product)

so far, I've got

<v,v> = complex conjugate of <v,v>   as c. conjugate of 1 is 1
so complex conjugate of <v,v>  = 1
i.e. conj( v1^2 + ... + vn^2 ) = 1   [if v is n-dimensional]
=> conj(v1)^2 + ... + conj(vn)^2 = 1

I'm trying to get

conj(v1)*v1 + ... + conj(vn)*vn = 1 which would give the result <v complex conjugate, v> = 1

[dusky epoch]

<v,v> = 1 means v is a unit vector

but also, for $v \in \bC^n$ you do not have $|v| = \sqrt{\sum_k v_k^2}$ in general

[stoic python]

Ann:

[dusky epoch]

actually, $\frac{1}{\sqrt{2}} (1, i)$ seems \textit{orthogonal} to its own conjugate, if anything.

[stoic python]

Ann:

[vernal salmon]

hmm but if I let v = 1/sqrt(2) [1, i], isn't <v,v> = 0 and not 1

I'm trying to start with <v,v> = 1 and arrive at <conj(v), v> = 1

though, I'm not sure if it's true / can't find a counter example

[dusky epoch]

no

$\left<v, w\right> = \sum_k v_k \overline{w_k}$

[stoic python]

Ann:

[dusky epoch]

think about it, it makes no sense for <v,v> to be 0 without v itself being 0

[vernal salmon]

oh crap... yeah, it wouldn't be an inner product otherwise

I had the wrong definition in mind x_x thanks

(i.e. I thought you do the same as if it were for real coordinates, just multiply and sum corresponding entries)

[dusky epoch]

yeah no you gotta conj

[vernal salmon]

mhmm, I see that now, thanks !

[vernal salmon]

Why is it true that

where U* is the conjugate transpose

[gleaming topaz]

Sorry I'll wait

[dusky epoch]

$U^* = \begin{pmatrix} u_1^* \ \vdots \ u_n^* \end{pmatrix}$

[stoic python]

Ann:

[vernal salmon]

Ohh thanks

I see

[hardy blaze]

if you're asked to find eigenvalues for a matrix to the 5th power

could you just get the diagonal numbers and put them all to the 5th power?

sorry if tahts a dumb q..

[clever cedar]

is column space the same thing as the image?

[paper egret]

i know this is obvious as hell

but specifically

how do you take determinants of matrices in a matrix

nani

[sonic osprey]

It's a block matrix

The 0's stand for n x n 0 matrices

[paper egret]

yes

i know that

like i know how to take determinants of a matrix, but how do you take the determiannt of matrices in a matrix

formally

[sonic osprey]

So if $A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$

[stoic python]

Zopherus:

[paper egret]

det(A) = -2

but how do you take the determinant of a matrix, in which its entries are NOT numbers, but matrices

[sonic osprey]

then $\begin{pmatrix} A & 0 \ 0 & I \end{pmatrix} = \begin{pmatrix} 1 & 2 & 0 & 0 \ 3 & 4 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix}$

[stoic python]

Zopherus:

[sonic osprey]

That's the thing

The entries are numbers

[paper egret]

yes i know

but is there any easy way to do it?

as in, you don't know how large A and I is

[sonic osprey]

So

[celest bridge]

like for some arbitrary A and I?

[paper egret]

yes

[celest bridge]

what is the context of this question?

[paper egret]

without writing all the entries of A and I itself

lmao just that

any easy way to do it?

[sonic osprey]

Think about what you know about determinants

Or ways to calculate determinants

And try to use those

The elements of this matrix are not matrices, it's just a shorthand for writing matrices of numbers

[paper egret]

oohh..

when you say it like that

hold on gimme a sec

thanks, i got an idea

i thought there was like a shortcut/easy way to do it

any hints on this one?

<@&286206848099549185>

Ive broken down the matrix, but can't seem an easy way to do this at all. I was able to do the previous 2 by breaking down the matrix and seeing that row reducing made it simpler

[celest bridge]

I know that the determinate of an upper triangular matrix is just the multiples of the diagonal

maybe that can help?

[paper egret]

i thought about it

[buoyant pike]

How much linear algebra is reasonable to learn within a month?

Trying to study for putnam

[feral grove]

depends on how deep an understanding you want

for putnam for the most part information about invertibility and determinants are most important, but learning them without prior structure might not be ideal

[celest bridge]

what is putnam?

[paper egret]

hardcore math competition

[feral grove]

also goon

[paper egret]

waddup

[feral grove]

it is useful to try to decompose your matrix into the product of two matrices

then note that det(AB)=det(A)det(B)

[paper egret]

decompose matrix into product of two matrices?

[feral grove]

correct

[paper egret]

mmmm any examples

[feral grove]

besides this one

[paper egret]

yea

wdym by decompose matrix into product of two matrices

[feral grove]

i mean in general like

i could right I=AA^-1

[paper egret]

yep

[feral grove]

and in the determinant we can use that to show that det(A)det(A^-1)=1

for example

[paper egret]

how does that help with my problem

i don't see the connection

[feral grove]

like

i'm slow at latex so i'm hesitant to really construct an example but

like see the previous problem you did

[paper egret]

nope doesn't help at all lol

[feral grove]

you proved the determinant of that matrix with and A and an I on the diagonal is det(A)

[sonic osprey]

how did you do part (a)

[feral grove]

yeah there's a reason a and b come before c

[paper egret]

part A

i let I be an n * n matrix

[feral grove]

you can write your determinants in C as the product of determinants in a and b

of matrices rather

than determinants

[paper egret]

and what I did was, determinant of that is equal to the same thing, but instead of an I matrix that is nn in size, i let it be equal to (n-1)(n-1) in size

in short, shrinking it

til it just becomes

det(a)

the second part

I did row reduction

until the diagonal entry B and C disappeared

part C is where i'm stuck at, can't row reduce, can't triangulate for any finite sized matrix

because I'm supposed to show this holds true for any finite sized matrix

I can't just use a 2x2 example

and prove it

[feral grove]

but

[sonic osprey]

why can't you row reduce?

[feral grove]

what's your second matrix in b times your matrix in a

or you can row reduce i guess

[paper egret]

I can row reduce, but it has to be any finite sized matrix, i have to make my argument convincing, think of this as a proof

[sonic osprey]

uh okay? You row reduced for part (b)

[paper egret]

because the Identity made it easier for me to row redce

row reduce/column reduce

[sonic osprey]

Why wouldn't the same thing work in part (c) then

[paper egret]

too many variables to work with, and I have to show its true for any finite size

[sonic osprey]

part (b) also had a lot of variables

[paper egret]

in part B

a lot lot lot less variables

[sonic osprey]

I also don't see why more variables would be a problem

[paper egret]

by allowing me to row reduce nicely

for part b, the one with the C matrix

I just argued that the identity matrix, the leading entry in that row will just cancel out with each entry in C

since it was identity, and the one next to Identity is 0

it has no effect on the matrix D

thus allowing me to make C "vanish"

for part c

any row reduction has some sort of impact on other matrices

in short the identity let me row/column reduce very VERY nicely

think of it as a proof

[sonic osprey]

It'll work

[feral grove]

but so much work

[paper egret]

that was what I was avoiding lol ^

[sonic osprey]

It'll turn out that the values of C won't matter in the end

The proof is actually the exact same

[paper egret]

I tried row reducing to make all the entries in A disappear, but that modifies matrix A

[feral grove]

but like use what you've already proven

[paper egret]

and doesn't get me anywhere

oh wait

[feral grove]

c directly follows from reducing to a and b

don't row reduce more than you have to lmao

[paper egret]

ok ima just give up and say this lmao

[feral grove]

goon pls

[paper egret]

[A 0 C D] = A*[I 0 C D]

done

i can factor a matrix out of a row

determinant of that

done

dunno how valid that sounds

but w/e

[feral grove]

[I 0 C D] *[A 0 0 I] = [A 0 C D] peppega

[paper egret]

oh

oh

bruh

[feral grove]

🙂

[paper egret]

but that looks wrong

lol

when u multiply it out

it becomes [A 0 AC D]

[feral grove]

shouldn't

[paper egret]

uh do the mat

[feral grove]

i did

[paper egret]

i did too

[feral grove]

how'd you get AC?

[paper egret]

uhhh

gimme a sec

[feral grove]

you shouldn't multiply by a row with A in it for that term

[paper egret]

nice?

[feral grove]

no A should be there

wait

[paper egret]

.-.

bro fuck lin alg

fuck box math

[feral grove]

no wait mb flip them

[paper egret]

this shit aids

flip which one

[feral grove]

i may have written them wrong

yeah i did

mb

[A 0 0 I] * [I 0 C D] = [A 0 C D]

[paper egret]

oh

welp shit the final boss has come

LOL

actually fuck this

[feral grove]

that's probably actually not too bad

[paper egret]

rip

nvm

lol thanks for help, that was cancer for me

[north sierra]

Would x3 be a free variable or x3=0?

[paper egret]

free

[native lodge]

the final line just says 0=0

so doesn't mean that x_3 must be 0

[paper egret]

0 = 0 tells you nothing

the only thing you know is

x1 = 1
x2 + 2x3 = 1

[native lodge]

it tells you that 0 is equal to 0

[paper egret]

x1 is fixed

x3 is free

that's the solution to your system

however

[north sierra]

Aight thx

[paper egret]

if in the final row

[native lodge]

help me with reimann sum longhand calculation, memes

[paper egret]

if you get 0 = 1

i believe you get no solutions at all

[north sierra]

Yup

[gray dust]

thought it's spelled reeman

[paper egret]

omfg

who tf is that

[native lodge]

if you get 0 = anything other than 0 then you have no solutions

[paper egret]

fuckin amphy

[north sierra]

Ya

I just forgot some old concepts

how do i figure out a?

[paper egret]

solve the system

RREF

[north sierra]

oh ok i did

and i got the matrix i posted

https://cdn.discordapp.com/attachments/540211747613704221/643235122325487637/image0.png

[paper egret]

the second part idfk lmao

[north sierra]

what

[paper egret]

i wanna say 3, but that sounds too obvious

[native lodge]

it's saying is w in the column space of the v's

does matrix formed from the v's and then set equal to w have a unique solution?

[north sierra]

nah

[native lodge]

so then w is not in the span of the v's

[north sierra]

ok

what about the second part?

would it be 3?

cause its not asking span

[native lodge]

that is saying how many unique basis vectors do you have?

[north sierra]

true yeah

so its linearly dependent so its less than 3 right

[native lodge]

since we have a free variable we know that there is some non zero combo of c_1(v_1)+c_2(v_2)+c_3(v_3)=0

[north sierra]

cause one vecter is a multiple of another

[native lodge]

yes

[north sierra]

true

[native lodge]

so there are only actually two vectors for this span

the third is just some combo of the other two

[north sierra]

yeah true

aright

thanks

[native lodge]

so can you see geometrically what is happening?

[north sierra]

nope

lol

[native lodge]

how about a quick vc to let you see it a different way then

[north sierra]

sure

amazing

thx bro

that made a lot of sense

[native lodge]

[north sierra]

@native lodge why should a matrix formed by the v's have to have a unique solution for w to be in {v1,v2,v3}?

[native lodge]

because then that means there is one and only one combo of the columns of v to give w. When that happens that means w is in the column space of the v's

so now suppose that w is the zero vector

[north sierra]

ok

[native lodge]

we now have two ways to arrive at the zero vector, that wasn't just setting all constants to zero

we are back to being linearly dependent

[north sierra]

wait i think i mixing things up

so when it says is w in {v1, v2, v3} does that mean span?

[native lodge]

yes

that is asking: is w in the span of the v's

another way: is w in the column space of the v's?

[north sierra]

oh okay

thank you

[silent dune]

How am I supposed to do b

[native lodge]

how much work do we need to provide?

we can answer this question by inspection actually

oh wait, we need to do the math to show it

but first let's just answer: independent or no? Can you tell right away?

[silent dune]

not sure kinda forgot how

[native lodge]

our vector space we are in is given by the number of components in the vectors

so we have three components, so we are in R^3

the most vectors we ever need for a basis of R^3 is how many?

massive red flag detected

[silent dune]

Uh

[native lodge]

but still how many vectors are needed to form a basis for R^3?

[clever cedar]

whats the dimension of R^3 is what he's asking

[silent dune]

3x1 ?

[gray dust]

visualize R^3 as 3d space