linear-algebra

[pastel harbor]

what is your time limit

[earnest juniper]

I have 2 seconds.

[pastel harbor]

this will take maybe 2 minutes in C++

I see

so then you need to handle all these cases

and get a linear solution

[earnest juniper]

Do you mean the elements that aren't included within the intersection?

[pastel harbor]

yes

because those will be in the second swap

[earnest juniper]

Do I also consider zero till x1?

[pastel harbor]

that won't be affected anyway

neither will the elements after y2

but there are more cases

note that you can have y1 < x1 < y2 < x2

you need to put 4 more if else on this too

[earnest juniper]

This is so complicated, I'll have around 15 cases to cover it all up.

[pastel harbor]

possibly more that I did not immediately see

[earnest juniper]

X2 > X1
Y2 > Y1

[pastel harbor]

who even sets retarded questions like this

[earnest juniper]

😂 Ikr.

[pastel harbor]

well you can write the general solution

for varying x1, x2 y1, y2

but that is probably not expected at this level

https://www.hackerearth.com/practice/data-structures/advanced-data-structures/segment-trees/practice-problems/algorithm/reversing-segments-3cd67101/

this is very similar to what you ask

this is the general version

but you don't even need sum

[earnest juniper]

The submissions are reviewed by people to ensure the code wasn't copied from somewhere else, is this like a public algorithm that you believe is fine to use or what exactly?

[pastel harbor]

a segment tree is well known

but your course likely expects a stupid solution with case work

[earnest juniper]

And there's no chance you can help me finish these cases?

[pastel harbor]

of course not

it is 4:15 AM

and I have to be up at 9 for work

[earnest juniper]

Ripz.

[pastel harbor]

I wish you the best of luck 🚬

but I would find this hard too

to write such cancerous case work

[earnest juniper]

Thank you for the help.

[south sedge]

Finding the matrix for linear transformation of mirroring in a plane x1 + x2 + x3 = 0, n is the normal vector to the plane. and u is an arbitrary vector. I dont understand the step between blue and red. Why is it the matrix subtraction possible when they are arent the same dimension?

[gray glen]

that's just a thic column vector

[tall galleon]

Anyone got good resources for a geometric understanding of linear algebra other than 3blue1brown, specifically on change of basis and linear transformations

[south sedge]

khanacademy?

[tall galleon]

I’ve been finding that helpful although I find their notation a little simple for the course I’m doing

[hot willow]

is there an algorithm for getting a non zero matrix that multiplies to a matrix with non zero entries to get a zero matrix?

like if we have a matrix

|6 -18|
|-4 12 |

let's called this c

is there an algorithm to find a matrix such that cd = zero matrix?

[slow scroll]

@hot willow sure, cd = 0 iff the columns of d are in the kernel of c.

[hot willow]

kernel?

[slow scroll]

null space?

[hot willow]

damn im pretty new to linear algebra

so don't know too much terminology

[slow scroll]

Null space of a matrix A is the set of vectors x such that Ax = 0

By the definition of matrix multiplication, if you have a matrix A and a matrix B = [v1, v2] where v1 and v2 are column vectors, the matrix
AB = [A(v1), A(v2)].

So v1, v2 \in Null(A) implies that AB = 0.

Since there is an algorithm to compute the null space, thats pretty much it

[hot willow]

oh ok

tht makes sense

[slow scroll]

so in your example, you could choose $v_1 = (3, 1)$ and $v_2 = (6, 2)$ and check that $$\begin{pmatrix} 6 & -18 \ -4 & 12\end{pmatrix} \begin{pmatrix} 3 & 6 \ 1 & 2\end{pmatrix} = 0 $$

[stoic python]

kxrider:

[coral tinsel]

@half ice I think I understand now

[half ice]

Understand waht

I'm a little slow let me know things

[coral tinsel]

the thing about the uninvertability of the matrix

[covert star]

Can one learn linear algebra without calc? Or do I need to learn calc first?

[limber sierra]

linear algebra is a different subject, and mostly independent

some lecturers might expect you to know basic calculus concepts

like "what a derivative is"

but even then, that's not necessary for the material

just a source of convenient examples

[covert star]

Alright, Thanks.

I didn't want to get knee deep into it then find I gotta back out.

[obsidian jackal]

i have this question that confuses me

i have matrix A

and the invertible S matrix

now i know how to get the values of S

but i dont know what D is

am i supposed to know what the values of the diagonal D have?

i meant that i can get S if i know what d is

[nimble egret]

This looks like an eigenvector/eigenvalue question is it not?

[obsidian jackal]

oh

that might be why i dont know it

we haven't gotten through eigenvalues yet

thanks @nimble egret

[shrewd slate]

Would it be fair to assume that if T: ℝ⁵ → ℝ⁵ ⇒ dim range T ∈ ℕ?

To be clear, I mean a linear mapping T.

[teal lintel]

Depends on what you mean by "ℕ".

Do you include 0 with that?

[shrewd slate]

Yea

[teal lintel]

Then yes.

[shrewd slate]

So, it can’t be fractional

[teal lintel]

Yes.

[shrewd slate]

Epic

[teal lintel]

In linear algebra, we don't deal with fractional dimension at all.

[shrewd slate]

Where do you?

Just out of curiosity

[teal lintel]

Topology.

And even then, only if you're explicitly dealing with dimension.

[shrewd slate]

Cool thanks

[half ice]

Looks good. Looks to be a bit on the applied side and does avoid some of the grittier theory.

Pay special attention to chapter 4, as that's where the powerful stuff is

@real plaza

[half ice]

Did you buy it? I mean you can libgen the 11th edition

I don't know what the 4th edition has haha

Yes, for the love of god don't pay for textbooks

There's a lot there

[idle echo]

How are algorithms to solve for determinants designed? Do they recursively solve for the determinants of smaller and smaller matrices until you get to 2x2 ones?

[sonic osprey]

That's one way

But that's slow

Faster is to row reduce until you get an upper triangular matrix

[half ice]

Sounds like you're talking about Laplace Expansion

@idle echo

[idle echo]

@half ice yes, I think so

[gloomy arrow]

Can anyone check for me?

9 times the length means basically the idenity matrix times 9

But then to rotate it pi/2 rads I would put a negative one on the upper right corner and a 1 on the bottom corner correct?

[half ice]

Yes

An extension times 9 is
[9 0]
[0 9]

The rotation is
[0 -1]
[1 0]

[gloomy arrow]

And then the addition of the two matrices would give the new transformation?

[half ice]

Nup. The multiplication

Basically, if you want a matrix A to happen, then the matrix B to happen, you can instead do that with BA

[gloomy arrow]

So when I multiplied them

I got

[0 -9]
[9 0]

[half ice]

,w matrix multiplication {{0,-1},{1,0}}{{9,0},{0,9}}

[stoic python]

Display results online and refine query

Requested by Kaynex#7520

[half ice]

Yes.

[gloomy arrow]

Alright cool 👌

And for this one im not too sure

So it reflects over x

and then back pi/2 radians

[half ice]

Same idea. (Rotate clockwise)(Reflects over x1)

[gloomy arrow]

So the two matrices im multiplying are:

[0 -1]
[1 0]

and

[1 0]
[0-1]

Im getting something in correct i think

Im getting -1 along the other diagonal

[south wadi]

yo

how do i know this transformation is linear

or not linear

[limber sierra]

do you know the definition of a linear transformation?

[south wadi]

R^n -> R^m

[limber sierra]

uhh

thats not the definition of a linear transformation

[south wadi]

isn't it just one vector space to another vector space

[limber sierra]

no, linear maps have to satisfy a couple other properties

$f(x+y) =f(x) + f(y)$ and $rf(x) = f(rx)$

[stoic python]

Namington:

[south wadi]

Yeah I've done that but different notation

[limber sierra]

for all scalars $r$ from the field ($\bR$) and vectors $x, y$ from the domain ($\bR^n$)

[stoic python]

Namington:

[limber sierra]

so if T(x, y) satisfies those properties

then it's linear

[south wadi]

Also T(0,0) right

is = 0

that's always true if theres not a constant right

[limber sierra]

that's a theorem, not part of the definition

but sure

that must hold necessarily

[south wadi]

ok so usually what our prof did

[limber sierra]

since $f(0x) = 0f(x) = 0$

[stoic python]

Namington:

[limber sierra]

and ofc 0x = 0

[south wadi]

where he would let u = (x1 + x2)

and v = (y1 + y2_

)

so it would be

$f(u+v) =f(u) + f(v)$ and $cf(u) = f(cu)$

[stoic python]

PeakThePenny:

[south wadi]

that's the notation my prof uses

so

[limber sierra]

sure, same thing

[south wadi]

if i let u = x 1 + x2

and v = y1 + y2

we have

$T(u+v) = y1 + y2, (x1+x2)^3

yeah i dont think this is gonna add up

$f(u+v) =f(u) + f(v)$ this won't be satisfied

[stoic python]

PeakThePenny:

[south wadi]

so it's not a linear transformation

right?

[limber sierra]

find a counterexample then

[south wadi]

to make it a linear transformation?

[limber sierra]

to show that it is not

if you think it isnt

[south wadi]

i mean if we look at the function here

$f(u+v) =f(u) + f(v)$

[stoic python]

PeakThePenny:

[south wadi]

and we chuck some numbers in this thing

say we let u = 3, 2

and v = 1, 1

[limber sierra]

yep, that's what i mean by "a counterexample"

and we can verify

clearly u+v = (4, 3) so f(u+v) = (3,64)

while f(u) + f(v) = (2, 27) + (1, 1) = (3, 28)

so this isnt linear

(in general, having exponents should be a big red flag that "it probably isnt linear")

[south wadi]

Alright thanks for the help

wait so what exactly is a linear transformation

[limber sierra]

well, i gave the definition above; a linear transformation is a function $f$ such that $f(u+v) = f(u) + f(v)$ and $cf(u) = f(cu)$

[stoic python]

Namington:

[limber sierra]

but intuitively, you can think of a linear transformation as a function that only "stretches" and "scales"

never changing the shape of a plane

(this is the intuitive explanation 3blue1brown gives, for example)

[south wadi]

who

[limber sierra]

alternatively, you can think of it as a "really nicely-behaved function"

in a certain sense

the derivative, for example, is a linear function

and this is incredibly useful

whereas something like exponentiation is not linear

it'd be really handy if we could just say $(a+b)^3 = a^3 + b^3$, but we cant

[stoic python]

Namington:

[limber sierra]

since raising to the power of 3 isnt linear

[magic light]

https://i.imgur.com/xbL87tE.png

what is this symbol?

the two < >

[cursive narwhal]

Inner product of two vectors v & w

[limber sierra]

(sometimes used to denote the dot product, which is a type of inner product)

[south wadi]

yeah thanks for the explanation

[magic light]

So, from reading it,

if v = { 1, 2, 3 } and w = { 1, 0, 0}
then <v, w> = 1x1 + 2x0 + 3x0 = 1?

[limber sierra]

yep.

[south wadi]

Namington

do you think everyone has the ability to master math

or phyiscs

or

[cursive narwhal]

That's the standard inner product in R^3

[limber sierra]

i'm not a pedagogy expert

naively, i'd say "most people but not everyone"

some people have legitimate learning disabilities

but i dont really know anyhting about this

[south wadi]

true

[limber sierra]

so im kinda talking out my ass

[south wadi]

yo that sounds like my psychology prof

pulling random shit out of his ass

[magic light]

https://i.imgur.com/6RO7sHC.png

So what's this symbol above the inner product

[gray dust]

if working in complex vector spaces, complex conjugate of inner product

[magic light]

It's in this context

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

[gray dust]

complex conjugate of inner product

[magic light]

right

I don't understand exactly, so from my reading an inner product space has this as a quality, that is

for each v, u, <u, v> = <v, u>

with the line

So what does this look like?

if v = {1, 2, 3}, u = {0, 1, 0}

what's the complex conjugate?

what does <v, u>(with the line) look like?

ah

I think I get it

[gray dust]

first off do you know what, if u,v are in C^n, <u,v> expanded looks like?

[magic light]

ah, good observation, I don't

I mean

I assume it works the same way?

(a1 + b1i) (a2 + b2i)?

if v = {a1 + b1i, 0, 0...}

and u is the same with 2

and then normal multiplication to get a number like that

but that is a guess

[gray dust]

let's say $x,y$ are in $\bC^2$, and say $x=(x_1,x_2)$ and $y=(y_1,y_2)$

[stoic python]

RokettoJanpu:

[gray dust]

use the standard inner product on $\bC^2$ to say $\lbr{x,y}=\cjg x_1y_1+\cjg x_2y_2$

[stoic python]

RokettoJanpu:

[magic light]

wait what

so

hold on

[gray dust]

have you learned what complex conjugate is before

[magic light]

a + bi is the conjugate of a - bi?

[gray dust]

for real a,b yeah

[magic light]

okay

I don't really understand why with x, y in C^2

we take the conjugate?

I mean, is that just by definition?

[gray dust]

this is the defn of standard inner product for a C^n space

[magic light]

right, counter intuitive

So

the conjugate of <x, y> is basically conjugating the entire solution to x1y1 + x2y2 where x1, x2 are conjugates?

[gray dust]

let's say $x,y$ are in $\bC^2$, and say $x=(x_1,x_2)$ and $y=(y_1,y_2)$\\use the standard inner product on $\bC^2$ to say $\lbr{x,y}=\cjg x_1y_1+\cjg x_2y_2$\\$\cjg{\lbr{x,y}}=\cjg{\cjg x_1y_1+\cjg x_2y_2}$

[magic light]

right that's what I thought

nice

let me write that down

[stoic python]

RokettoJanpu:

[magic light]

Is this correct?

[gray dust]

your u,v look too similar. pick other names

[clear otter]

If I take the tensor product of a row vector with itself, do I just get another row vector that’s the square of length of the original?

[gray dust]

k yea

[magic light]

thanks, I got it

[gray dust]

👍🏾 np

[south wadi]

let's say $x,y$ are in $\bC^2$, and say $x=(x_1,x_2)$ and $y=(y_1,y_2)$\use the standard inner product on $\bC^2$ to say $\lbr{x,y}=\cjg x_1y_1+\cjg x_2y_2$\$\cjg{\lbr{x,y}}=\cjg{\cjg x_1y_1+\cjg x_2y_2}$

[stoic python]

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

[south wadi]

let's say $x,y$ are in $\bC^2$, and say $x=(x_1,x_2)$ and $y=(y_1,y_2)$

[stoic python]

PeakThePenny:

[south wadi]

how to get black background?

[pallid rampart]

Type “ ,tex ––colour black “ in #bots

@south wadi

[south wadi]

didnt work

[gray dust]

,tex --color black

[stoic python]

Your colour scheme has been changed to black

[gray dust]

^ this

[gentle hedge]

is it possible to do this?
| 1 - 1 | | 1 | = | 1 | + | -3 |
| 6 -4 | | 3 | | 6 | | -12 |

[dusky epoch]

uh

what

[stoic python]

Ann:

[dusky epoch]

this? @gentle hedge

[gentle hedge]

thanks yeah this

[dusky epoch]

yeah, this is correct

[gentle hedge]

explanation to this?

could've used first column... Or can I?

[dusky epoch]

wdym

[gentle hedge]

I mean can I uses first column for that multiplication?

[dusky epoch]

...

what's that supposed to mean??

[gentle hedge]

nvm, I made a mistake in calculation

ok I am having problems here. How the calculation was done here?

[gray dust]

refer to the square matrix. the product can be computed by 1(first column)+3(second column)

[brittle orchid]

Does this just refer to the group (or set?) of real numbers excluding 0?
$\mathbb{R} - {0}$

[stoic python]

WhiteBerry:

[gray dust]

set of reals, excluding 0

[brittle orchid]

Thank you

[thorn prairie]

I need to solve the system

But by using the proper inverse

[dusky epoch]

what

[thorn prairie]

@dusky epoch ?

Thats what it says

[dusky epoch]

Solve the system by using the proper inverse. is that what it says word for word

[thorn prairie]

I translated from greek

System it means matrix i guess

So anyone knows what we have to do?

[dusky epoch]

well i have a hunch that they want you to recognize that this is in fact a linear system in xy, sqrt(y) and yz, and to solve for the values of those things first and then x, y and z themselves

so perhaps you could make a substitution

like u := xy, v := sqrt(y), w := yz

solve the system, which is linear in u, v and w

then go back to x, y and z

[thorn prairie]

Ok thanks

[brittle orchid]

Hi, I was wondering what the Groups $S_{3}$ and $S_{5}$ are referring to

[stoic python]

WhiteBerry:

[sonic osprey]

This should be in #groups-rings-fields

[brittle orchid]

Apologies, I'm studying it in my first year of university

[half ice]

S3 is the actions on permuting 3 objects

S5 is permuting 5

@brittle orchid

[limber sierra]

the keyword is "symmetric group"

[brittle orchid]

Thank you @limber sierra and @half ice

[limber sierra]

the symmetric group on 3 objects is the group of all permutations of {1, 2, 3}

so it'll have 3! = 6 elements

[brittle orchid]

yeah

[limber sierra]

similarly, S_5 will have 5! elements

[brittle orchid]

So it's a group of sets?

[half ice]

It's a group of "actions", per say

"Taking the second element and swapping it with the third" is one of the elements

Its inverse is swapping them back

[brittle orchid]

Right

Okay this goes straight over my head

I'm missing knowledge here

I suppose I need to catch up on mappings?

[half ice]

This one is a bit abstract, I'll admit. The elements of the group aren't actual real objects, but actions on something else

See cycle notation, which is a great way to bring it back into the concrete

[brittle orchid]

it's not "bringing it back" I have missed this entirely xD

[half ice]

I mean that "cycle notation" is good to visualize this

[brittle orchid]

I see

Do you think it's worth studying Mappings first?

I need to study for my finals eventually anyway as I missed it

Cause like, I'm clueless about surjectivity, injectivity and bijectivity

[half ice]

Oh lol yeah those are important

[brittle orchid]

yup x_x

[half ice]

Despite being a mapping itself, Sn is kinda unrelated and has its own thinking. You can know them independently

[brittle orchid]

Oh okay

So do I not need to know those concepts for that question? I've got a test on Friday (other subject) so I don't mind delaying this

[vast torrent]

You can learn about those three concepts in like an hour and a half on wikipedia

[brittle orchid]

Okay

[vast torrent]

Idk what's gonna be on your test

[brittle orchid]

No no no, my test is unrelated

Like I said, I don't mind delaying the learning of these concepts due to my test (other subject) on Friday

But hey, if I can understand the required concepts in an hour and a half I'm down.

[vast torrent]

Khan academy has videos on them in his linear algebra module

https://www.khanacademy.org/math/linear-algebra/matrix-transformations/inverse-transformations/v/linear-algebra-introduction-to-the-inverse-of-a-function

It's not linear algebra but that's the first time in many students' education that it becomes super important to define injective/surjective rigorously

Imo it should be taught with algebra ii

Before lin alg many curricula are sloppy in not distinguishing between codomain and image,a function and its graph, a function and its image

¯\_(ツ)_/¯

[smoky lagoon]

can someone help me figure out what this question is asking?

I don't understand what I'm supposed to even be doing for it

[vast torrent]

Assuming that this is a 2x2 mtx

Taking a vector in the plane and rotating it about the origin theta radians

Is a linear operator

Since it's a linear operator,it has a matrix representation

Find it

[smoky lagoon]

I still don't understand what you're saying

I don't know what Ra is

[vast torrent]

Ra is the matrix corresponding to the linear operator "rotate the vector a radians"

Still confused?

👻

[smoky lagoon]

yeah i'm trying to do another problem presently so i can deal with that one later

[vast torrent]

Then why did you ask for help

[smoky lagoon]

because I still don't understand so I went to do something else in the work so i can come back to that later

[vast torrent]

I began an explanation and you didn't understand it immediately from the first 2 or 3 sentences you give up?

Okay

[brittle orchid]

🤦‍♂️

[smoky lagoon]

not really giving up just thinking about something else

[vast torrent]

I didn't leave the party, i just decided i wanted to be somewhere not at the party

[brittle orchid]

imo if someone replies to you out of their own time to help you, the least you can do is give them your attention

[smoky lagoon]

Yeah I'm trying to look at it and I just don't understand what it is asking at all in any way

was hoping that the problem I pulled from the book would help but it did not

[half ice]

@smoky lagoon
This is important, it's a rotation by angle θ

[smoky lagoon]

figured it out

i just didnt know what a rotational matrix was

[wintry steppe]

Hi, i have a nice question:

when you want to find the eigenvalues / eigenvectors you always do this:

[stoic python]

nxue:

[wintry steppe]

But what happens when you do:

[stoic python]

nxue:

[wintry steppe]

and get the values, and then multiply them, you will get the determinant of A

Why does it happen?

[austere cedar]

For a 2x2 system:

To write An= lambdan n

An - lambda n = 0 vector

(A - lambda I) n = 0 vector

The I come sure to the lambda isn't a vector

Since you could have n = 0 vector as a solution which isn't very helpful

From cramers rule you can get infinite solutions when determinant of
(A - lambda I) = 0

Meaning both lines of the matrix are the same or a multiple of each other

So you can make any solution you want as long as one is a multiple of the other in n Eigen vector

Hopefully this makes sense, this is how I think of it anyway

[vast torrent]

I don't see how you answered the question stabulo

I've never seen that equation used anywhere, the

det(A-λI)=λ

@wintry steppe where did you see that expression?

[wintry steppe]

i made it up

[austere cedar]

I thought he was just proposing because he didn't know where the original came from

So I think I answered it :) 50:50

[wintry steppe]

but I dont understand how why when you multiply all the lambdas we get from the second equation i wrote, we get the original det

[austere cedar]

You don't get the second one?

Not multipled anything either

If you have 2 equations and 2 unknowns you can make one of the coefficients of one of the unknowns ewu as l and subtext the 2 equations , solving the simultaneous equations, cramers rule is as general method to do it

This method has

In the denominator of cramers rule formula for the matrix you're trying to solve here is the 2x2 matrix is the determinant of A- lambda I

[wintry steppe]

$|A - \lambda I| = \lambda
\text{You get:} \lambda_1 , \lambda_2, \lambda_3 , ... \lambda_n$

[stoic python]

nxue:

[wintry steppe]

and then multiply them

[austere cedar]

Oh for when you solve it?

The determinant

[wintry steppe]

again,

to find eigenvectors we do

det(A - xI) = 0

[austere cedar]

Why are you setting the determinant equal to lambda

[wintry steppe]

but I thought to myself, what if we do this: det(A-xI) = x

after you solve this ^

and get x1 x2 ... xn

[austere cedar]

Well it wouldn't do anything, since it would break the condition of cramers rule

Since it the determinant is non zero

Then cramers rule makes it solvable

[wintry steppe]

bro but im not talking about eigenvalues

im tlaking about something new

[austere cedar]

With 1 solution

[wintry steppe]

that does not exist

after you get x1 x2 x3 ...

and multiply them, you will get det(A)

[austere cedar]

One solution always exists for n = 0

[wintry steppe]

ok let me do an example i think it's confusing

[austere cedar]

I guess I don't know your question

[wintry steppe]

pick a 2x2 matrix plz

your choice

[austere cedar]

Don't know how to use texit so can't

And on phone can't even mash it together

[wintry steppe]

3 5
0 -1

sounds reasonable?

/ random enough

[austere cedar]

Yeah

[wintry steppe]

well call it A

det(A- xI) = x

(3-x)(-1-x) = x

[austere cedar]

Why

Why equal x

[wintry steppe]

it doesnt matter

[fervent spoke]

no idea why the hell u wud do that

[wintry steppe]

i dont try to find eigenvalues

[austere cedar]

It does

Ok you done something else

[fervent spoke]

But use ur example

[wintry steppe]

its something weird and different

[fervent spoke]

And for a 2x2 matrix

Solve the lambda vals

[wintry steppe]

dont pay attention to eigenvalues, its something different

ill continue:

(3-x)(-1-x) = x

[austere cedar]

Keep going then

[fervent spoke]

That make det(a - Ix) = x

You legit solve that with quadratic eq for 2x2 matrix and u get det formula lol

[wintry steppe]

multiply those together and you get the determinant of A
(-3)

@fervent spoke
can you explain it again? i dont understand

[austere cedar]

Determinant of A - 3?

[wintry steppe]

yup...
it was:

3 5
0 -1

-1 * 3 - 0 = -3

[austere cedar]

You've just made a random quadratic and solved it as far as I can see

[austere cedar]

?

[fervent spoke]

Whole point of orig formula is to find lambdas that make that eq = 0

[austere cedar]

I don't have as clue what is going on

[fervent spoke]

If u set it equal to lambda

Thats basically det(A - Ix) - x = 0

Replace x with lambdas lol

U do that and u get cancer equation to solve roots for

Even for 2x2 it kinda cancer

[wintry steppe]

if you replace x by lambda then what you get?

[fervent spoke]

But u get det(A) after solving for roots

[austere cedar]

It has to zero or you will get a single solution but to have the eigenvector formula which you have n on both sides so it will just be n = 0 which is why you must strive for infinite solutions which manages determinant of that matrix being zero, the n = 0 vector is not very interesting which is why you want the infinite solutions

I'd you had it equal to lambda not equal to zero I guess, then you will be able to solve the system of equations and have 1 solutions but you always know the n = 0 one will do this

An = lambda n

Either no solutions which can't be since always have n = 0

1 solution which would be the n = 0 trivial one

Or infinite

The determinant gives you the lambda values such that you get infinite solutions

To the best of my knowledge*

[boreal crescent]

what is the zero vector and basis for this

[austere cedar]

Looks very much like linear algebra

[boreal crescent]

yessir

[vast torrent]

Basis of what?

[boreal crescent]

R+ with addition and multiplication defined as they did

[vast torrent]

Let me see if that's a vector space at all

Sec

[boreal crescent]

it is, its given as a vector space

like a part of the question is to prove its a linear space

which i did with the 8 axioms

i think the zero vector is 1?

but idk how to find a basis for this

[vast torrent]

Cool

[fervent spoke]

@wintry steppe

[vast torrent]

If it's a 1d space then any vector is a basis

[fervent spoke]

Plug into quadratic formula and simplify

Its aids algebra but u get det(A) after it all

[boreal crescent]

whats the zero vector though

[vast torrent]

Tricky

[feral grove]

any vector

[boreal crescent]

what

[feral grove]

including the 0 vector 🤔

[vast torrent]

We want the solution to x+0=x

[boreal crescent]

0 isnt in R+ though

[feral grove]

he just said 1d space

[boreal crescent]

oh ok

[feral grove]

🙂

[vast torrent]

Solving for the unknown 0 i meant

"0"

With air quotes with your fingers

Let's use e for less confusion

x+e=x, solve for e

What's the equation x+e=x look like with your addition?

[boreal crescent]

ex = x

so e has to equal 1

[vast torrent]

"including the 0 vector" oh damn you got me

Any single vector except e spans the space

Indeed, e=1

[wintry steppe]

@fervent spoke
that does not tell you why the solutions of deltas, when multiplying them, gives the determinant

i have another question regarding this

why when you calculate

|A- xI| = x
you get a polynomial where the value without "x" is the det of A

[austere cedar]

...

[fervent spoke]

Wat

[austere cedar]

Are you setting it equal to x as some kind of experiment exercise?

[wintry steppe]

i know if you do:
|A- xI| = 0
and get a polynomial it has something with det in it (i dont think its the free value)

@austere cedar
bro im not talking about eigenvalues, leave it alone lol im just messing around with equations

[austere cedar]

What are your trying to do then?

[fervent spoke]

And were trying to tell u what ur doing is pointless

LMAO

[vast torrent]

nxue listen

[fervent spoke]

its like asking why is 2 + 2 = 4

[wintry steppe]

6 5 1
-1 1 3
0 3 -1
for example

A= ^

[vast torrent]

what are you claiming about the equation det(A-xI)=x

[austere cedar]

^

[vast torrent]

what do you think this equation does

[wintry steppe]

im giving an example it should clear it cus we got confused here

[vast torrent]

because it's not obvious to us why this equation is of any interest

[austere cedar]

^

[fervent spoke]

U want to do it for 3x3 matrix

[vast torrent]

any more than examining the solution to, for example

[fervent spoke]

Plug that shit in urself

[vast torrent]

det(A-xI)=5

[fervent spoke]

See how cancer that is

[wintry steppe]

bro be more mature im asking nicely

[fervent spoke]

Quadratic formula is alr cancer bc u have to manipulate it akgebraiically

im just saying

[wintry steppe]

and you're not explaining it either

[vast torrent]

maffs this has nothing to do with cardano's

nxue do you see what I'm asking you?

[austere cedar]

I'm mind boggled tbh

[wintry steppe]

yes, but i dont want it to be a specific value, im just messing around with this equation and want to see what properties it has

[vast torrent]

do you agree that det(A-xI)=5 is a wholly uninteresting equation?

[wintry steppe]

/it gives

[austere cedar]

No property

[vast torrent]

det(A-xI)=x is just as an uninteresting an equation

[wintry steppe]

why?

[vast torrent]

no, the burden of proof is on you to prove something is interesting

[austere cedar]

True

[wintry steppe]

im giving an example tho

[vast torrent]

by default making up a random equation is not interesting

[fervent spoke]

why

[vast torrent]

your equation is just as interesting as det(A-xI)=13

[wintry steppe]

thats not true, because im asking for what values we get that det(A-xI) equals x itself not just a random constant

thats not correct

your equation is just as interesting as det(A-xI)=13
@vast torrent

[vast torrent]

and I'm asking you, for what values do we get that det(A-xI) equals 13

[wintry steppe]

but it doesnt matter, the question matters

[austere cedar]

I think this guy is too far gone TFG

[wintry steppe]

TFG?

[austere cedar]

Too far gone

[wintry steppe]

can we start over please?

[austere cedar]

Like Alabama republican etc

[vast torrent]

yes

[wintry steppe]

i will give examples

[fervent spoke]

LMAOOO

[austere cedar]

Good luck

[wintry steppe]

can I ? or you're too mad at me for asking questions and not understanding your solutions ...

[vast torrent]

I'm not mad bro

[wintry steppe]

:(

[vast torrent]

you can try again

[wintry steppe]

thank you bro :)

all im saying is let's take the eigenvalues equation, and change it a little, so we dont =0 but =x (to see for what values the equation det(A-xI) gives the value itself)
when you calculate |A-xI| = x
you will eventually get a polynomial p(x) = x
all im aksing is why the value without x (i forgot whats the name of it, for example: 6x^2 + 4x -5 so -5 is the value without x's) is the determinant of A
that's all lol, im not here to make new rules, im just curious

[vast torrent]

the constant term it's called

[wintry steppe]

oops

[vast torrent]

okay so you're asking about the constant term of a polynomial

if p(x) is your polynomial

then p(x) + x^2 - 300x^11 has the same constant term, agree or disagree

[wintry steppe]

yes, but its not the standard poly, it's when you det(A-xI)=x and not =0

yes i agree

[vast torrent]

okay

so p(x) - x and p(x) have the same constant term

[wintry steppe]

yes

[vast torrent]

so your question can just as well ask

why is p(0) the determinant of A?

because you're asking

why is the constant term of p(x)-x the detemrinant of A?

[wintry steppe]

now i get it

(i had the big OHH)

[vast torrent]

yeeahhh there's that good light bulb moment

feeels good man

[wintry steppe]

thank to all of you guys

:)

[austere cedar]

Still confused

[vast torrent]

he we wasking why the constant term of the polynomial p(x)-x was the determinant of A

[austere cedar]

Constant term?

[wintry steppe]

i know that p(0) = det(A)
when you do |A-xI| =0

but when you p(0)-0
you still get the same constant term

[austere cedar]

Oh like the y intercept of a quadratic

[wintry steppe]

yes

[austere cedar]

Weird question I guess

[wintry steppe]

at least everything is clear now

lol

[vast torrent]

it's okay to ask weird questions

[austere cedar]

I don't even know what the question is still

Guess I'll never know

[wintry steppe]

nvm 😅

can i ask another question tho but now it will be more 'serious' because it was in my test

or i filled the questions capacity for the day

[vast torrent]

you can borrow a question from tomorrow

[wintry steppe]

lol

im going to use that card

i dont study in english, and im totally not sure to translate it, but given a ("multiplication space ????" i hope i translated it well)
f(u,w) = u^t Aw

which of these statements is 100% false:
1)A is symmetric
2) A-I is diagonlizable
3) AA^t is orthogonal
4) trace(A-A^t) =0

Inner product space, thats the term

[vast torrent]

quadratic form they're called

I see

I meant f is a quadratic form

[wintry steppe]

i dunno whats a quadratic form

it was 2 minutes to hand in the tests so i marked 3

[vast torrent]

a function of the form f(u,w) = u^t A w

[wintry steppe]

who knows tho

if its right

[vast torrent]

that's a quadratic form Im saying

[wintry steppe]

yes but what is the answerrr

;-;

[vast torrent]

hm

so by 100% false I assume they just mean false

[wintry steppe]

we call it here "American questions" (Multiple choice lol)

[vast torrent]

because 100% means 100/100 , 1, so I don't know what "1 false" means

I guess we deserve that

[wintry steppe]

so i made this logic: if there is only 1 solution right (and there is only 1 solution to mark)
if 1 is correct then 4 is correct, so i Xed 1

[vast torrent]

I dont understand how you can answer that quesiton unless there's a part you didnt translate correctly

we're not told anything about f

so how can we conclude anything at all about A

[wintry steppe]

its inner product space, takes 2 vectors and outputs this :

[stoic python]

nxue:

[vast torrent]

ooh it's saying

we know for sure that this expression

is an inner product

[wintry steppe]

standard inner product space,
<(3 , 6 ,1 ) , (1 , 1,2 )> = 3*1 + 6*1 + 1*2

but this one isnt standard

[vast torrent]

yes I See what it's asking

[wintry steppe]

the question is not standard inner ...(i keep forgetting the name)

[vast torrent]

inner product space

[wintry steppe]

lol

[vast torrent]

which of these statements is 100% false:
1)A is symmetric
2) A-I is diagonlizable
3) AA^t is orthogonal
4) trace(A-A^t) =0

it's confusing because

when they say symmetric, and transpose

do they mean symmetric in this product or symmetric in the usual product

[wintry steppe]

symmetric as A = A^t

[vast torrent]

it's worded weirdly

[wintry steppe]

ill tell you half the solution i got from someone

if its an inner product then f(u,w) = f(w,u)

(i forgot its R so no adjugated)

so A has to be symmetric

[vast torrent]

$u^\intercal A w = w^\intercal A u$

[stoic python]

gfauxpas:

[wintry steppe]

yup

if A has to be symmetric then ofc trace(A-A^t) =0 b/c trace(0 matrix) = 0

[vast torrent]

yes

[wintry steppe]

if A symmetric so A-I is diagonalizable (??? that i dont know)

[vast torrent]

100% true 😛

[wintry steppe]

😉

[vast torrent]

let me think about that for a sec

[wintry steppe]

ill tell you why i said "100% false" , b/c in the question they worded it like: which of these statements is necessarily false
(i guess because some of the others can be false sometimes, but not necessarily)

[vast torrent]

I don't see why that statement should be true

[wintry steppe]

which statement

[vast torrent]

if A is symmetric then A-I is diagonable

I don't see how A-I is of any relevance to the problem

[wintry steppe]

its just this annoying american question,
such as

A - 2B + AB = I
and 2 is an eigenvalue of B
which of the following is necessarily correct:
a) detA = 0
b) 1 is an eigenvalue of B
c) 1 is an eigenvalue of A
etc..

sorry, multiple choice question

😅

ok someone told me that if A is symmetric:
then A-I is symmetric
if its symmetric then its normal if its normal its diagonalizable

the problem is that ive never heard the term "normal matrix"

[vast torrent]

normal is usually called unitary I think? let me double check theyre the same

no, Im very wrong

normal is strictly weaker than unitary

normal means $AA^\dagger = A^\dagger A$

[stoic python]

gfauxpas:

[vast torrent]

unitary means $AA^\dagger = A^\dagger A = I$

[stoic python]

gfauxpas:

[fast pivot]

what is that dagger?

[pallid rampart]

Is that transpose?

[feral grove]

conjugate transpose

[pallid rampart]

Ah

I see

[south wadi]

help on 3

ping plz

[half ice]

@south wadi
It would have to be the case that
[a b] [1] = [3]
[c d] [3] [4]

and
[a b] [1] = [0]
[c d] [1] [2]

Multiply is out, that's four equations with four unknowns

[south wadi]

the answer is this

but i dont understand the fking first step

why is T(e_1) = T(e_2)

and like where it says we dudce

why is it pulling 1/2s out of nowhere

[gray dust]

it never said T(e_1) = T(e_2)

the matrix corresponding to a linear map T can be fully determined if you know where T sends the standard basis vectors

[boreal crescent]

can anyone help with part c

i proved part a true and part b false

[south wadi]

yo so rokabe

like i legit am losto n the first step

why is e1 the same as e2

and why are they jizzing out 1/2 and 3/2 like where this shit coming from

[nimble egret]

They're not the same

Perhaps should note the x and y in the first equation is not the same as the x and y in the second equation

We want to rewrite our standard basis vectors as a linear combination of the two vectors given

Each of the vector equations is just a system of 2 equations in 2 variables

Which you can solve however you want

[south wadi]

yeah i just dont see the -1/2 constant and 3/2 coming from

no idea

just spawned in

[nimble egret]

Well

Can you verify it works at least?

-1/2 + 3/2 = 1

• 3/2 + 3/2 = 0

[cursive narwhal]

jizzing out 1/2 and 3/2 😂

[nimble egret]

Which gives e_1

So even if you can't get them

You can verify these coefficients work

[south wadi]

why do we want the coefficient to be 1 on e1 and 0 on e2

[nimble egret]

??

[nimble egret]

I'm looking at e_1 only

If you look the right hand side of the equation

Simplifying gives [1, 0]

Which is e_1

[south wadi]

(1,0)

?

all i see is (3,4) and (0,2)

[nimble egret]

We're writing (1, 0) as a linear combination of (3, 4) and (0, 2)

[south wadi]

ok

why (1,0) though

holy fk dude this shit is so confusing

and useless

sorry im back now just had a tantrum right there

honestly im lost in this question

the first 2 were a breeze

So we are writing a linear transformation T

[gray dust]

the matrix corresponding to a linear map T can be fully determined if you know where T sends the standard basis vectors (which is to say, what T(e1) & T(e2) are). for a linear map T, we can say T(x)=Ax where we're finding the matrix A. what the previous sentence means is, the columns of A are T(e1) & T(e2) or T(1,0) & T(0,1)

[nimble egret]

And also recall the transformation is linear

[south wadi]

for T(1,3) = (3,4) and T(1,1) = (0,2)

[nimble egret]

This is why it's useful to write the standard basis vectors as a linear combination of known vectors

To utilise linear properties

[south wadi]

so we want e_1 to be (1,0)

and e_2 to be (0,1)

so why are we multiplying -1/2 by the x and 3/2 by the y

[gray dust]

no. in R^2, e1 is another name for (1,0)

[south wadi]

ok

[gray dust]

e1 can be expressed as a linear combo of (1,3) & (1,1), so e1=x(1,3)+y(1,1) for some unknown scalars x,y

[south wadi]

ok

[gray dust]

if we find what x,y are then we can compute T(e1) by taking T(x(1,3)+y(1,1))

which, because T is linear, is just xT(1,3)+yT(1,1)

[south wadi]

ok i think i get it now

[viscid vale]

if u call a variable "arbitrary" does that mean its an element of real numbers?

e.g. this textbook problem says: Let W be the set of all vectors of the form (some vector with b and c in it) where b and c are arbitrary

[dusky epoch]

if u call a variable "arbitrary" does that mean its an element of real numbers?
no

not in general

[viscid vale]

can we just assume that b, c ∈ R

oh

[dusky epoch]

well if it only makes sense with b and c as scalars

which it most likely does

then yes

[viscid vale]

the vector is:

(5b + 2c, b, c)

[dusky epoch]

yeah so

b and c are def real numbers

doesn't make sense to consider 'em as anything else

[viscid vale]

ok thanks

[undone lake]

hello. does someone knows why, here, composition is commutative ?
Let E be a K vector space, f an endomorphism, and f^0 = Id_E, f^k = f o f^(k-1)

i think that’s because of the properties of the linear endomorphism but i’m not sure

[dusky epoch]

f^k is the composition of k copies of f

the commutation of f with its powers boils down to the associativity of composition itself

[spiral sonnet]

So row echelon form is where a row's first non-zero number is atleast one column to the right of the previous row right?

[south wadi]

How do you know if a transformation maps onto its codomain

[cursive narwhal]

Let $T:V \to W$ be a linear transformation. If you want to check if it's surjective, then you just need to show that $T(V) = W$.

[stoic python]

Abhijeet Vats:

[spiral sonnet]

Is there a rule to row reducing?

[south wadi]

yea

[spiral sonnet]

What I'm having trouble with is reducing and finding the pivot columns

https://i.gyazo.com/806aaa93158c180c440f877ec171efab.png

In this example, the pivot columns are 1, 3 and 4 when reduced

[fervent spoke]

if that confuses u jst focus on making matrix look like this

[limber sierra]

uh

it wont look like that

[fervent spoke]

Iobv not all the time

point is

[spiral sonnet]

Yeah that's what I wanted focus on, but not all matrices will look like that

[fervent spoke]

Solve it how u want

[spiral sonnet]

so if it doesn't look like that I'm lost

[limber sierra]

in any case, do you know what operations are "valid" in row reduction?

[spiral sonnet]

yes

[limber sierra]

so try and reduce it best you can

to something similar in "spirit" to that

that is to say

a "pivot variable" is the leftmost variable in a row

you want every "pivot variable" to have all 0s below it

[limber sierra]

in this ccase, the pivot columns will be 1, 3, 4

so we'll expect something like

[fervent spoke]

ye that dude explaining better lmfao

[spiral sonnet]

might just have to grind out some problems

[fervent spoke]

if u have a 1 in a row there cant be 0s om the left of it

[spiral sonnet]

I get it when my professor goes through it, but I don't know if I can do it myself

[limber sierra]

the box region denotes the "we dont care" section

because its not below any pivots

also, in this case, you dont have to row reduce to 1 exactly

just to have it in this "shape"

since all you care about is the position of the pivots for this problem

[spiral sonnet]

doesn't there have to be 0's above and below a pivot variable?

[limber sierra]

for the purpose of this question? doesnt matter

you only care about where the pivots are

i..e columns 1, 3, 4

yes if you're "fully reducing" you'd have to reduce all the entries directly above the pivots to 0

[spiral sonnet]

oh okay

Not sure if I'm phrasing this correctly, but would I eventually reach that answer even if I don't do the optimal row operation?

[limber sierra]

oh yeah, you dont have to do row reduction totally optimally

just try to make progress every step

in fact, doing it optimally is very hard - computer programmers spend a lot of time thinking about how to efficiently implement row reduction

[spiral sonnet]

okay gotcha

gonna get some practice in

[spiral sonnet]

So for transformation matrices, specifically the rotation one. How would I convert a clockwise to counter clockwise? Like -pi/4 (clockwise) would be 45 degrees in the 4th quadrant. So would the counter clockwise be pi/4?

[south wadi]

Namington

how did u get (2,27) here

u mean (4,27) right

[pallid rampart]

So for this problem

I see how at each step of the gram schmidts, we can multiple the vector by -1

But I can’t prove that any orthonormal list of vector satisfying that span equality is of this form

[vast torrent]

"List"

What's a list @pallid rampart

[wintry steppe]

I'm not sure how to get started for this question.

[wind yacht]

Think about how to map each point from the first n to the second

Maybe consider 2 "separate" transformations first - one to deal with italics and one for the size

[wintry steppe]

but which points should i Use?

to map it out

[pallid rampart]

@vast torrent finite set

[vast torrent]

List I think means a finite sequence

But not sure

[pallid rampart]

Well this is what I meant

So for each vector produced by gram schmidts, I can multiply it by -1

And the final finite collection will still be an orthonormal collection of vectors

So for each vector we have 2 choices

Thus the 2^m in the question

But I can’t prove that any finite sequence of vectors satisfying that equation is of this form

[vague cedar]

does anyone know the derivation for this?

would love a ping if you know it

[vast torrent]

@pallid rampart something is very wrong

With the question

[tacit storm]

with my life

[vast torrent]

I was googl8ng things

If {v1,v2,...,vm} is an orthonormal basis

Then for any isometry A

{Av1,Av2,...,Avm} is an orthonormal basis for the same space

So there are infinitely many bases

[sonic osprey]

I made this same mistake

note how it says span{v_1, v_2, ... , v_j}

for all j

so the span of the first number of elements must be the same too

i.e. span(v_1) = span(e_1)

and span(v_1,v_2) = span(e_1,e_2)

etc etc

@vast torrent

[pallid rampart]

Bruh yes

The whole point of the question is because of that span equation

[remote elm]

Hello

I am having trouble understanding vector spaces

first of all, i don't entirely understand what a vector space is

[limber sierra]

do you have your definition of a vector space?

a vector space is a mathematical structure where we have some sense of "vector addition" and "scalar multiplication"

[remote elm]

a set of vectors from v1 ... to vk

[limber sierra]

that is not a vector space

[remote elm]

oh ok

[limber sierra]

a vector space is a set of vectors associated with two operators, vector addition and scalar multiplication

addition/multiplication has to follow some "rules" here

in order for them to "resemble" what we want addition and multiplication to look like

do you have a full definition of a vector space? somewhere in your notes/textbook?

[remote elm]

yes i do, i have a pdf with notes

one second

[limber sierra]

yeah, so the definition is given on the second page

the basic gist is, one thing mathematicians like to do is generalize

you mightve seen vector spaces like R^2 and R^3 before

and maybe, say, C^2 or similar

but one natural question to ask is, "is there something special about these spaces? is there a way we can define a general notion that tells us 'what we want' a collection of vectors to look like?"

a vector space is an "algebraic structure", which basically means a set with some additional operations on it

(a set of vectors, in this case)

in theory, those operations can do whatever they want - but we want vector spaces to follow some sensible "rules"

the rules we want vector addition to follow include things like:

a + b = b + a
a+(b+c) = (a+b)+c
there's some element 0 such that a+0 = a
every vector has a "negative" (so we can talk about subtraction) such that a + (-a) = 0

these are fairly sensible things to want from something we'd call "addition", right?

and it's nice because they give us some "rules" that unify vector spaces

they let us easily tell:
a. what is and isn't a vector space
b. what properties we can use when talking about vector spaces in general (rather than just talking about R^2 or C^n or whatever, we can talk about all vector spaces at once - that's powerful! - by just using these rules)

similarly, let s, t be scalars and a, b be vectors; we want scalar multiplication to satisfy:

s(a+b) = sa + sb
(s+t)a = sa + ta
s(ta) = (st)a
there exists some scalar 1 such that 1a = a

again, reasonable things for us to want something we call "scalar multiplication" to satisfy

[shrewd slate]

Vector space more like diet field amirite

[cursive narwhal]

[limber sierra]

one thing to note: we never define vector multiplication here

[cursive narwhal]

namington, you're way too nice lol

[limber sierra]

multiplying vectors isnt generally very useful or well-behaved

alright, so, we have two operations: vector addition and scalar multiplication

but it seems kind of... arbitrary

like we have one set with two operations

but those operations seem pretty unrelated

how can we tie those two operations together?

well, you're probably familiar from elementary algebra with the fact that x(y+z) = xy + xz

[remote elm]

right

[limber sierra]

so that's where s(a+b) = sa + sb and (s+t)a = sa + ta come from

anyway, thats a bit of a long-winded side-tangent

bringing the focus back to your specific question

in teh screenshot

[shrewd slate]

In one you add vectors and in another you add scalars, note that

[limber sierra]

we want to show that R^2 with this funky addition

[shrewd slate]

@ shahrik

[limber sierra]

is not a vector space

that means we want to show that it "breaks one of the rules"

to recap, these rules are:

a + b = b + a
a+(b+c) = (a+b)+c
there's some element 0 such that a+0 = a
every vector has a "negative" (so we can talk about subtraction) such that a + (-a) = 0
s(a+b) = sa + sb
(s+t)a = sa + ta
s(ta) = (st)a
there exists some scalar 1 such that 1a = a

also something I didn't mention: whenever you add vectors or multiply vectors by scalars, we expect to get a vector from our vector space back

it wouldnt make sense if, say, $\begin{pmatrix}1\0\end{pmatrix} \oplus \begin{pmatrix}3\6\end{pmatrix} = \text{goldfish}$

[stoic python]

Namington:

[limber sierra]

(we call this "closure")

anyway, focusing on this specific question

what "rule" do you think this would break?

obviously we're only dealing with addition (the + is written in a circle to distinguish it from regular addition), so we only need to look at those rules

[remote elm]

a1 + a2

b1 + 1

the second should be b1 + b2 ?

[limber sierra]

that's how addition is defined; what rule do you think it breaks?

dont think about what it "should be", think about what the problem is

[remote elm]

the problem is that it wouldnt work for all values of b2

[limber sierra]

why does this break our "rules for a vector space"?

[remote elm]

it wouldnt*

[limber sierra]

what do you mean?

again, we're looking for one of these rules that it violates

if you're in doubt, you might want to try some test vectors