linear-algebra

[vast torrent]

Yes

Even if the matrix is real

[wintry steppe]

Ive never did something like that

Good to know, thank you very much guys! @vast torrent @sonic osprey

[gray dust]

@wintry steppe for future reference don't double post

[wintry steppe]

@gray dust
Sorry, i dont know whats the dofference between linearalgebra and the questions section...
Can i ask questions in #linear-algebra or is it just for discussions and all questions need to be posted here?

diff..*

[gray dust]

if you know what topic your question belongs to, you should post in the related channel (so yes linalg q's go to #linear-algebra). the question channels are for if you don't know where your q belongs or if the related channel is currently occupied

[wintry steppe]

@gray dust
Gotcha, won't do it again, thanks!

[gray dust]

yep

[boreal crescent]

where am i going wrong wtf

first column of the B matrix is the coordinate vector of T(v1), where v1 is the first element of the basis

T(t^2) = 4 - 6t + 3t^2

[dusky epoch]

your B is ordered

as t^2, t, 1

it's written that way

[boreal crescent]

ohhh

[dusky epoch]

3t^2 - 6t + 4

[boreal crescent]

yea got it

damn

[lilac kindle]

No idea what im doing

[dusky epoch]

$\pdv{u}{x} = \pdv{u}{\xi} \pdv{\xi}{x} + \pdv{u}{\eta} \pdv{\eta}{x} = \pdv{u}{\xi} + \pdv{u}{\eta}$

[stoic python]

Ann:

[dusky epoch]

$\pdv{u}{y} = \pdv{u}{\xi} \pdv{\xi}{y} + \pdv{u}{\eta} \pdv{\eta}{y} = a\pdv{u}{\xi} + b\pdv{u}{\eta}$

[stoic python]

Ann:

[lilac kindle]

I dont know how you got that

[dusky epoch]

multivar chain rule

[lilac kindle]

I thought x had to be dependent on E for chain rule to work

[dusky epoch]

what's E

[lilac kindle]

The weird E

[dusky epoch]

there's no E here

$\xi$ is "xi"

[stoic python]

Ann:

[dusky epoch]

$\eta$ is "eta"

[stoic python]

Ann:

[lilac kindle]

Ah, thanks i will try and solve it now

[dusky epoch]

u = u(xi, eta) = u(xi(x,y), eta(x,y))

[lilac kindle]

Wait, why do you have u = u(xi, eta)

I dont see that

[dusky epoch]

bc you're making a change of variables

so you want to express u as a function of xi and eta

your new vars

[lilac kindle]

So rearrange to x = xi - ay?

[dusky epoch]

no you don't need to do that

[lilac kindle]

I cant see how you merged them together

[dusky epoch]

,,,

forget it

[lilac kindle]

Ok, thank you

[wintry steppe]

given a subset - does it have to include the zero vector in order to be a subspace?
suppose i have:

$ \mathbb{W} \subseteq \mathbb{R}^4$
is $W = \{(a b c d)| a^2 + b^2 + c^2 + d^2 > 0 \}$ a subspace?

[stoic python]

nxue:

[wintry steppe]

(a b c d)* ...

[dusky epoch]

given a subset - does it have to include the zero vector in order to be a subspace?
yes it does, otherwise it'll fail to be closed under addition guaranteed

or under scaling

[wintry steppe]

because in this subset - i could do
0 0 0 0 + 0 0 0 0 and it would be = to 0 0 0 0 which is not in the subset because 0^2 + 0^2 ... is not > 0 , right? @dusky epoch

[dusky epoch]

the zero vector isn't in your subset

it's literally the only vector you're excluding

[wintry steppe]

yes i know so it lacks the zero vector and its not a subspace but i need to circle only one option- i have:
not a subspace because it lacks closure to addition /
scalar multiplication / both

@dusky epoch
so if it does not have the zero vector ---> it lacks closure to addition?

[dusky epoch]

if it doesn't have the zero vector it's always guaranteed to lack closure under scaling bc 0v = 0

in this particular case your subset also lacks closure under addition because you can take any nonzero vector v and write v + (-v) = 0 because -v isn't zero either and so is in your subset

[queen tundra]

Any good material on tensor product, tensor dot product and tensor double dot product?

[severe magnet]

Let $V$ be a $K$-vectorspace and $F:V \to V$ be a linear transformation with $F^2=-F$. Show that $V=Ker(F) \oplus Im(F)$. $\$ proof: "$\subseteq$" Let $x \in V$. We have two cases to look at: $\$ 1.case: $F(x)=0 \implies x \in Ker(F)$ and since we know that $0 \in Im(F)$ we know that $x = x+0 \in Ker(F) \oplus Im(F)$. $\$ 2.case: $F(x)=w \neq 0$ $\ \iff F^2(x)=F(w) \ \iff -F(x) = F(w) \ \iff F(w+x) = 0 \implies w+x \in Ker(F \circ F) \supseteq Ker(F)$ (proven in another exercise). We also know $w \in Im(F) \implies -w \in Im(F)$ (since Im(F) is a subspace of V). Thus we know $x \in Ker(F) \oplus Im(F)$ in all cases.

Is this correct so far?

[stoic python]

Nabil | GMT+1:

[lone flume]

I don't understand why you need to invoke the $\text{Ker}(F^2)$, I mean, you have that $F(w+x)=0$ so $w+x\in\text{Ker}(F)$ and since $w\in\text{Im}(F)$, we have that $x=x+w+(-w)\in\text{Ker}(F)\oplus\text{Im}(F)$.

[stoic python]

jnkena:

[severe magnet]

I just wrote that since I used F twice but yea it is trivial that is right

since we know $ker(F^2) = ker(F)$ we know that Im F $\cap$ Ker F = $\lbrace 0 \rbrace$

[stoic python]

Nabil | GMT+1:

[lone flume]

Your proof is correct as I see it

[severe magnet]

thank you for the input 🙂

[wintry steppe]

hi i need to prove/disprove this statement:
if L1 is an eigenvalue of matrix A and L2 is an eigenvalue of matrix B
then L1*L2 is an eigenvalue of at least one of these matrices: AB , BA
i dunno where to even start :(

[vast torrent]

Well you agree it's definitely true if AB or BA share an eigenvector of A and B, right?

[wintry steppe]

how do we know this?

[vast torrent]

Write out the definition of eigenvalue and eigenvector

[wintry steppe]

of I mean, yeah sure if they share an eigenvector, but what if not ?

[vast torrent]

I'm on medication rn that makes me sleepy so

I'm on low power mode

[wintry steppe]

we agreed that if AB and BA share an eigenvector with A and B it's true, but that's not the proof in fact i think it's not part of it

[vast torrent]

@wintry steppe i found online that the answer is false. You just need a counterexample where the eigenspaces don't share a common vector

So just experiment with 2x2s

I searched online bc i couldn't think of any reason why the statement should be true if the matrices considered don't share eigenvectors

And i didn't feel like trying to make a counterexample

[wintry steppe]

thank you!!

ill try and think of one

[vast torrent]

It shouldn't be too hard because

Most matrices don't share eigenvectors

So most ones you try should work

Btw my tutor taught me a trick in creating counterexamples

Want to hear

@wintry steppe he said

Fix most of the entries as constants, 0, 1,-1

And then you can vary one or two entries

By introducing a parameter t

And then after you matrix multiply

If you didn't find your counterexample

You can vary the parameter without having to multiply the matrices together each time

[pallid rampart]

Need help with this proof:
Let V be a complex vector space, p be a polynomial with complex coefficients, and T:V->V a linear transformation. Prove that a complex number a is an eigenvalue of p(T) then a=p(λ) for some eigenvalue λ of T

[dusky epoch]

hmm

let's see

so ker(p(T) - aI) is nontrivial

[pallid rampart]

Yeah

[dusky epoch]

hm.

[pallid rampart]

Um

If p=a_0-a+a_1x+...+a_nx^n, then we can find a root

then is this root an eigenvalue of T

just my wild guess

[dusky epoch]

[pallid rampart]

I think you'll have to do that, since the next exercise says

Show that this is not true of V is a real vector space

and all C is replaced with R

Got it @dusky epoch

[dusky epoch]

oh :?

do tell

[pallid rampart]

4.8 is just basic polynomial factorization over C

for some λ1, ..., λn

[pallid rampart]

Btw if you're interested ig, the example I gave was

V=R^2, T(x,y)=(-2y,x), p(z)=z^2+1, then -1 is an eigenvalue for p(T), but obviously no real value of λ satisfy p(λ)=-1

[thorn prairie]

Guys, on matrixes ( A + B ) ^-1 equals A^-1 + B^-1?

[quartz compass]

if they're equal, you can multiply both sides by (A+B) and get the identity

[thorn prairie]

What?

[quartz compass]

take this equation: ( A + B ) ^-1 = A^-1 + B^-1

multiply both sides by (A+B)

[thorn prairie]

But i asked if they are equal

Not that i want to prove they are

[quartz compass]

I'm saying assume it

for the sake of contradiction

there are other ways, take B=0

[quartz compass]

(A+B) is invertible but B isn't so it doesn't make sense

[thorn prairie]

basically i want to proove this

if i can multiply them seperately with -1

then its easy

left side is A + B

and right side is also A + B

also I^-1 equals I?

[torn hornet]

well verify it

[thorn prairie]

how?

i think it is

[torn hornet]

how do inverses work

[thorn prairie]

inverse is the ^-1?

[torn hornet]

yes, but what is its defination

[thorn prairie]

um A*A^-1

gives I

[torn hornet]

yeah

so using this, is the inverse of I I

[thorn prairie]

so basically I and I^-1 i guess are the same

right?

[torn hornet]

yes

since I*I=I

[thorn prairie]

ok so i found these equations

im struggling with what to do on this one

[torn hornet]

for this problem you need this information:

[thorn prairie]

it tells me that

A is a nxn matrix

with A^4 = 0

sorry these are new to me

[torn hornet]

yeah, its like power series

[thorn prairie]

and like no where on my notes i wrote about A^2 , A^3 etc

[torn hornet]

Suppose $A$ and $B$ are square matrices of the same dimention, and are invertible. Denote the inverse $A^{-1}$ and $B^{-1}$. what is the inverse $AB$

[stoic python]

JohnDoeSmith:

[thorn prairie]

what?

[quartz compass]

this is like your third problem you've posted

one at a time

[thorn prairie]

Im sorry i do the terms on greek so im trying to translate

i solved the others

[torn hornet]

yeah, answer what i asked

this will lead u to proving ur problem

[thorn prairie]

ok my only issue

is what denote means

[torn hornet]

umm, it just means im naming the inverses that

[thorn prairie]

if what you are asking is what AB^-1 gives its B^-1*A^-1

[torn hornet]

yeah

good

Now can you use this information on the problem u sent

the first one lol

[thorn prairie]

i solved that one haha

thanks

[torn hornet]

the one u posted like, 5 mins ago?

[thorn prairie]

i solved everything i sent besides the last picture

with the I - A ^-1

[torn hornet]

ok well then

[quartz compass]

use parenthesis

it's just annoying lol

[torn hornet]

yeah lol

anyhow

just use the defination of an inverse

i.e do they multiply to the identity

[thorn prairie]

identity is I?

[torn hornet]

yeah

[thorn prairie]

well the left part

will give me I - A^-1

[torn hornet]

paranthesis pls

[thorn prairie]

I - (A)^-1

haha what

[torn hornet]

ok so first

$(A+B)^{-1} \neq A^{-1}+B^{-1}$

[quartz compass]

I seriously doubt you solved those previous problems

[thorn prairie]

Its so funny whenever i have questions in math

[stoic python]

JohnDoeSmith:

[torn hornet]

which mero showed to u like

2 different ways

[thorn prairie]

@torn hornet You gotta be shittin me. those 2 are unequal?

[thorn robin]

yeah I'm positive he did something illegal to find the inverse of A^{-1} + B^{-1}

[thorn prairie]

I asked before if those are equal and i was getting philosophical responses so i guessed they are

[torn hornet]

yeah ok u didnt prove problem 1

no u got a proof

lmfao

[quartz compass]

wew

[thorn prairie]

yeah i guess its wrong

[quartz compass]

don't guess, prove

[torn hornet]

he did not give u philosophical responses mate, he told u the reasoning as to why its false

which you should pay attention to

[thorn prairie]

well tbh on my notes nowere i see those e2 being equal so yeah

[torn hornet]

can you tell my why these two arent equal

again if $B=A^{-1}$, then $AB=BA=I$

[stoic python]

JohnDoeSmith:

[torn hornet]

this fact should be how you do all problems

[thorn prairie]

but youre assuming

[torn hornet]

assuming what?

[thorn prairie]

if B = A^-1

[gray dust]

not even a nitpick

[torn hornet]

yeah idk how to respond to that except

this is how math works

lol

[thorn prairie]

Honestly you dont get how hard is to understand the math terms youre saying which i do them in another language

[torn hornet]

english aint my first langauge either

[gray dust]

listen. if $A$ is invertible, ie $A\inv$ exists, by definition $A\inv A=AA\inv=I$

[thorn prairie]

yes but i dont do english terms at all. What does that has to do with what im saying

@gray dust yes we are aware of what you sent

[stoic python]

RokettoJanpu:

[torn hornet]

ok but do you understand this defination

[thorn prairie]

no wait

[gray dust]

john just said "let B be the inverse of A" so that he didn't have to keep typing A^-1

[thorn prairie]

oh okay. I didnt understand thats what you meant

[torn hornet]

anyhow moving on

[thorn prairie]

Just so we are on the same page

for which problem youre talking rn?

[torn hornet]

apply this defination to prove things right

first i want you to understand this properly

[thorn prairie]

i understood it

[torn hornet]

so i said $(A+B)^{-1}\neq A^{-1}+B^{-1}$

[stoic python]

JohnDoeSmith:

[torn hornet]

right, can you prove this

using the defination

[thorn prairie]

ok

[gray dust]

defination?

[torn hornet]

defination of inverses that is

[gray dust]

definition

[torn hornet]

bruh

[thorn prairie]

ok so for this problem

[torn hornet]

Prove what i just said first

otherwise you wont get anywhere

[thorn prairie]

the unequality?

[torn hornet]

yes

[thorn prairie]

it makes sense since on the left part you add 2 matrixes then you find the inverse of that lets say C matrix where on the right side you find the inverse of both A and B then add them so it will be a different result

[torn hornet]

i mean intuition is good, but lemme just right this out for you

[gray dust]

write

[torn hornet]

If it were true, i.e $(A+B)^{-1}=A^{-1}+B^{-1}$, then by definition of inverse $(A+B)(A^{-1}+B^{-1})=I$, but this expands to $2I+AB^{-1}+BA^{-1}=I$, which is clearly not always true

[stoic python]

JohnDoeSmith:

[thorn prairie]

i see

[torn hornet]

in your problem, you have $(I-A)^{-1}=I+A+A^2+A^3$. can you use the definition of an inverse here

[stoic python]

JohnDoeSmith:

[thorn prairie]

umm

[torn hornet]

We are saying the inverse of I-A is I+A+A^2+A^3 yeah

[thorn prairie]

so i multiply left side with (I-A)?

[torn hornet]

yes

[thorn prairie]

and that gives me...

I?

[torn hornet]

I on left

what about right

[thorn prairie]

So if A^4 = 0

that means A = 0 i guess

[torn hornet]

no

[thorn prairie]

so i get I on right side as well

A^4 isnt AAA*A?

[torn hornet]

there are non-zero matrices whose powers can be zero

[thorn prairie]

oh i see

[torn hornet]

$A=\begin{pmatrix} 0 &0 \ 1 & 0 \end{pmatrix}$, whats $A^2$

[thorn prairie]

then how does A^4 = 0 help me

[stoic python]

JohnDoeSmith:

[thorn prairie]

its 0

[torn hornet]

yeah

anyhow, expand (I+A+A^2+A^3)(I-A)

[thorn prairie]

oh i multiply both sides

ok let me see

[torn hornet]

(matrices who hava a power that is 0 are called nilpotent matrices btw)

[thorn prairie]

ok so basically i get on the right side

I - A^4

so I

[torn hornet]

yep

[thorn prairie]

alright cool

one down

[torn hornet]

So since (I+A+A^2+A^3)(I-A) is I, I-A and the other thing are inverses

(note just muliplying both sides of the eqn wont prove it)

[thorn prairie]

why not

i say A part = that

then B part

[torn hornet]

$-x=x \implies (-x)^2=x^2\implies x^2=x^2 \implies 1=1$

[stoic python]

JohnDoeSmith:

[torn hornet]

does not tell us anything about the validitiy of the first identity does it

[thorn prairie]

whats this

[torn hornet]

False premises can lead to truths, so showing your identity implies I=I is not a proof

is my poiny

point

[thorn prairie]

oh i see

[torn hornet]

so instead say that (I-A)(I+A+A^2+A^3)=I, hence these are inverses

[thorn prairie]

so what was the point of telling my to multiply both sides with I - A

[torn hornet]

i just wanted you to multiply those two things lol

[thorn prairie]

so i know that the right side is inverse

to I - A

[torn hornet]

indeed

[thorn prairie]

so (I - A)^-1 = that right side

[torn hornet]

mhm

[thorn prairie]

ok one down finally

so now

what i did before was wrong

so i cant do anything on left part

[torn hornet]

lets do what we discussed, try showing multiplying those two is the identity

[thorn prairie]

ok so i multiply both wiith the inverse of the left side

so i get an I on the left side

and right side gives something that doesnt help me

[torn hornet]

so this one is a bit trickier

[thorn prairie]

wait

with the logic applied

your logic for the previous one is wrong

nvm

the logic on the previous one is correct

[torn hornet]

remember $(MN)^{-1}=N^{-1}M^{-1}$

[stoic python]

JohnDoeSmith:

[thorn prairie]

but we cant apply it here

[torn hornet]

this will significantly simplify your problem

[thorn prairie]

hm

um i dont have any matrix multiplication here

with the -1 on top of them

[torn hornet]

well, what im saying is

try inverting both sides

[thorn prairie]

ok so

left side is A^-1 + B^-1

[torn hornet]

ye

[thorn prairie]

um right side

A+B?

[torn hornet]

not quite

remember how to invert when matrices are being multiplied

do that here

[thorn prairie]

i cant see it

i multiply right side

with like (A+B)^-1?

sorry A+B

[torn hornet]

$[A(A+B)^{-1}B]^{-1}$

[stoic python]

JohnDoeSmith:

[torn hornet]

is the RHS right

now how do we invert product of matrices

[thorn prairie]

yeah we can multiply with that

its

B^-1*(A+B)*A^-1

[torn hornet]

yes

now this should be much easier to simplify yeah

[thorn prairie]

so its B^-1 + A^-1

[torn hornet]

exact;y

[thorn prairie]

so basically i prooved

that the inverse of the right side

is equal to the inverse of the left side

thus the normal ones equal as well?

[torn hornet]

mhm, and inverses are one-to-one so yeah

[thorn prairie]

ah i see. Lovely

i just realised

that probably most of these excersises im not supposed to do them

[torn hornet]

still good practice

[thorn prairie]

since its first time im seeing them

yeah. But i have like exercices to do with Cola rank nula nullity and stuff like that and then go to calculus to do sequences

anyways if youd be kind to help with the last one

[torn hornet]

yeah sure

[thorn prairie]

ok il send it but let me try it first

so i apply the same logic?

[torn hornet]

should be

[thorn prairie]

ok done

so inverses are equal with both being B^-1 * A + B^T

so normal ones equal

[torn hornet]

yep

[thorn prairie]

lovely

btw quick question

is it efficient to use det to find inverse of a table bigger than 2x2?

or should i do the A | I ~ I| A^-1

since they only taught us the det for 2x2 not bigger matrixes

for example for exercises where i just need to tell if a matrix is inverse or not and not to calculate it

it would be way easier to find the det

and see if its != of 0

[torn hornet]

theres several ways to find inverse, i would say its easier to do the A|I ~ I|A^[-1} by hand

[thorn prairie]

rather than trying to find the inverse table the other way until you find a 0 line

[torn hornet]

gaussian elimination is usually efficient

[thorn prairie]

somewhere i heard of Gaus i think

[gray dust]

gauss is very nice for inversion

though if i see a fair amount of zeros in a row or col then i don't mind taking the minors

[solar osprey]

If we want to prove the converse of this. Can we say that since for every w that belongs to B we have that w belongs to span then V is subset of span(B). Then since B is a subset of V then span(B) is a subspace of V hence span(B) is a subset of V. Hence the equality

[cursive narwhal]

You are trying to prove the following:

If each v in V can be uniquely expressed as a linear combination of vectors in beta, then beta is a basis for V.

Clearly, $L(\beta) \subset V$. Since $v \in V \implies v \in L(\beta)$, it follows that $V \subset L(\beta)$. This proves that $\beta$ spans $V$. Now, you just need to show linear independence.

Now, if we have $a_1u_1+a_2u_2 + .... +a_nu_n = 0$, then $a_1 = a_2 =...=a_n = 0$ certainly satisfies that equality. In fact, they are the unique scalars that satisfy that equality. This proves that $\beta$ is linearly independent. That shows that $\beta$ is a basis for $V$.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

@solar osprey

[solar osprey]

thank you dude

[empty copper]

Hotaro Oreki from Hyouka

[cursive narwhal]

Also, note that equivalences don't have converses. What you meant was the converse of the forward implication.

[thorn prairie]

Let F be a 5 × 5 table whose space produced by its columns is not R5. What can you say about its core?

Can someone help with this one?

[dusky epoch]

uh

did you... translate this from another language

[thorn prairie]

its matrix 5x5

yes..

[dusky epoch]

what did you translate it from

i'm pretty sure that you meant kernel rather than core

[thorn prairie]

greek

[dusky epoch]

okay nevermind idk why i asked that

w/e

it's kernel

[thorn prairie]

about tis core

i think it means

about the nullspace

the dimension of NulF

correct?

[dusky epoch]

ok if you wanna use the word nullspace so be it

so you have a 5 by 5 matrix whose column space isn't R^5

i.e. whose rank is less than 5

[wintry steppe]

is this channel taken? :v

[dusky epoch]

i'd err on the safe side and say yes

move to #❓how-to-get-help

[wintry steppe]

ok

[pallid rampart]

Translated from Greek

[thorn prairie]

@dusky epoch so what can i say about the kernel?

that its less than 5?

[dusky epoch]

no, you can say that the kernel is nontrivial

i.e. has something other than 0 in it

[wintry steppe]

i need to find orthogonal basis for this one

is it
-3
1
0

and

0
0
1
?

[pallid rampart]

That is an orthogonal basis

[wintry steppe]

thanks

[pallid rampart]

Now make it an orthonormal basis

[dusky epoch]

problem doesn't say to do that tho does it

[wintry steppe]

lol

no

in order to complete a set of vectors to an orthogonal basis
do i complete to a basis (with the standard basis) and then do gramshmidt ?

[dusky epoch]

that's one way of doing it ig

are the vectors you start with orthogonal?

[wintry steppe]

i start with one

lol

so basically ye, but what if they arent? i need to do gram shmidt before also

[shrewd slate]

If U is a subspace of V, but U ≠ V. Can it have the same number of dimensions; more specifically, can its basis be the same length?

I’m thinking no

Yeah nvm it’s very clearly no

[slender seal]

hey how do you define a line?

[pallid rampart]

If its finite dimensional

[agile gyro]

dim(U) = dim(V) <=> U = V

[quaint heart]

Iff it's finite dimensional

Not just if

[pallid rampart]

How do you define same dimension in infinite dimensional vector space?

The cardinality of the basis?

Also, is orthonormal basis kind of generalization of the standard basis?

Since the vectors are pairwise orthogonal and each has length 1

[quaint heart]

Yeah cardinality of the basis obviously

Lol saying a generalization of the standard basis is dumb

What does the standard basis for a k dimensional vector space V even mean @pallid rampart

[pallid rampart]

That was what I meant

It doesn't make sense to talk about standard basis

So we have orthonormal basis which is pretty nice to work with

[quaint heart]

I think it just let's you do geometry

[upbeat mauve]

I dont understand how this solution gets -3

Can someone explain please this is determinants

[dusky epoch]

laplace expansion along third row

do the chessboard pattern thingy and you'll find that the 3 in position (3,4) is on a minus square

[ + - + - + ]
[ - + - + - ]
[ + - + - + ]
[ - + - + - ]
[ + - + - + ]

[upbeat mauve]

Oh ty

[versed topaz]

If T has an eigenvalue of multiplicity > 1, does its minimal polynomial necessarily have a repeated root?

[dusky epoch]

no

[versed topaz]

Okay, it didn't sound true

Ty

[dusky epoch]

take T to be the identity transformation

its minpoly is x-1

[versed topaz]

Lol duh

[formal bridge]

Not sure how to prove this

[pallid rampart]

🤔 is this linear algebra?

maybe should be in #groups-rings-fields idk

[formal bridge]

yeah maybe, the class is abstract linear algebra so could very well be both

ill go there though, more on the abstract algebra side

[wintry steppe]

Not certain on how to approach this, a little guidance or hint would be helpful

[brittle juniper]

you're asked to find θ so you should try to at least make θ appear out of all this info

for example, you can have a look at ||u+v||²

[wintry steppe]

Does this somehow relate to the defenition of dot product?

[nimble egret]

yes

[wintry steppe]

Oh ok so i can square both sides of eq1 and expand dot product

[nimble egret]

try it and see if you get somewhere useful

[pallid rampart]

\|\|u\|\|

will work

[wintry steppe]

arcos(-2||u||/||v||)

i feel i might have missed something

[quartz compass]

|u| and |v| are related

so you can simplify that

@wintry steppe

[wintry steppe]

Oh, so then arcos(-1) = pi

Thank you guys, i appreciate it

[quartz compass]

yeah yw

geometrically it should hopefully be obvious to confirm

[wintry steppe]

Yeah now i notice for u+v to have same mag as u, v must travell back 2||u|| in opposite dir

[idle osprey]

Hi can someone please explain how one would find a basis for R3[x] when you have a basis A=2, B=3, C=5, D=-5 of a vector subspace of R3[x]
The vector subspace in this case is U={p(x) € R3[x] | p'(0) = p(1)} and we also know that R3[x] is a vector space of polynomials of max coefficients 3

[torn ermine]

Quick question: Is a matrix diagonalisable iff it is a normal matrix?

[short root]

Good question

[torn ermine]

... and the answe is? 😛

<@&286206848099549185>

[pallid rampart]

For this we only need to assume that (e_1,...,e_n) is an orthogonal basis right?

doesn't need to be orthonormal

Oh wait nvm

We need orthonormal

But how should I interpret this theorem?

What's the intuition for inner product?

[pallid rampart]

It's not R^2, it's the subset of R^2 with integer entries

[coral tinsel]

I'm writing a calculator for a factory game, and I need to figure out how I came to an answer so I can abstract it and put it in a program

starting data

The equation on the bottom is correct.
The matrix is an.... attempt.

[pallid rampart]

@real plaza yes

I was just correcting the fact that you said "since R^2 only takes integers"

which is not true

[half ice]

Oh fuq you're playing satisfactory

[coral tinsel]

yeap

[half ice]

What does the answer represent?

[coral tinsel]

what ingredients that iron ingot constitutes

[half ice]

How's update 3? I haven't had much time to play

[coral tinsel]

I haven't even build coal yet

down a rabbit hole with this calculator

[half ice]

Sadly representing this with a matrix is eh

A tree is the better way, if that makes any sense

[coral tinsel]

Can't

Directed cycles, also two-product recipes

[half ice]

... There's two product recipes now?

Oooh

[coral tinsel]

Everybody's math is screwed up

https://cdn.discordapp.com/attachments/559071846591365154/680652984702599183/unknown.png

[half ice]

Oh wow that's interesting

[half ice]

What's that on the right? Oil?

[coral tinsel]

Fuel

[half ice]

IC

[coral tinsel]

Pipes btw

[half ice]

What's your goal? Just to reduce everything down to most basic elements?

[coral tinsel]

Yup

[half ice]

Those cycles do screw everything up, as you can arbitrarily waste resources now. You might want to just ignore them

Or come up with a good rule on how they should be used

[coral tinsel]

Which I thought matrix operations could do

But I p much forgot how to matrix

tho this is Haskell so it won't be crazy to define everything from scratch

[half ice]

I mean you could have a massive vector of every product in the game, then creating a product is the same as multiplying that vector by some matrix

[coral tinsel]

sparse matrices

[half ice]

But you're looking to do this backwards, and these matricies won't have inverses lol

Or, they might actually. They'll all be mostly 1

[coral tinsel]

yup

[half ice]

Congrats on knowing Haskell

That's a skill

[coral tinsel]

*relearning Haskell after writing almost no code for 2 years

[half ice]

That inverse matrix idea might actually be pretty good, but it still won't really explain how to deal with cycles

Is there a "best strategy" on how to use those alternates?

[coral tinsel]

as in which ones to use and which ones not to?
some people have their rankings

but I was wanting to preserve the alternates data

[half ice]

@coral tinsel
The problem is you're introducing something that isn't invertible into the system. If I gave you my final product, you can't possibly tell me what I started with, because you can't know how many times I used the cyclic alternates

Even if you knew exactly which recipes I used

[coral tinsel]

They incur a fuel cost each iteration

I intend to add energy cost too....

It suddenly feels like the second law of thermodynamics and quantum non-determinacy are involved...

Did I mention that I was going to use RREF?

[gloomy arrow]

Can someone check this for me?

[nimble egret]

What's you justification for the last one

It might be linearly dependent

But I can't see the dependence immediately

And I don't have paper

[gloomy arrow]

I said that the last ones are linearly dependent because the amount of vectors is greater then the entries of the vectors

[nimble egret]

Oh lmoa

True

Yeah

Looks fine

Since the first two are linearly independent

They span R2 and so the third is necessarily in their span and stuff

Ok

Yeah

[gloomy arrow]

First 2?

I said the second one is lineraly dependent because in includes the zero vector

[nimble egret]

Yeah, that's fine

I mean first two vectors in e

Zero vector, you can adjust the coefficients

To satisfy the definition of linear dependence

[gray dust]

this all looks fine

[gloomy arrow]

OKay thank you. I might come back frequently to have my answers double checked

[gray dust]

but it never hurts to also show some work, just in case you get smth wrong and i need to check what you did

[gloomy arrow]

Oh yes I will show my work next time

[copper cobalt]

does nayone know min plus algebra

[gloomy arrow]

Let me also send the work

[nimble egret]

This one

Is pretty easy to check

Just do the multiplication with your result

And see if you get [6, 18]

[gloomy arrow]

Yup, I got it

Thanks

This would just be the idenity matrix multiplied by negative 8 right?

[gray dust]

don't forget we have rotation by an angle of pi too

[nimble egret]

The negative does that doesnt it?

[gloomy arrow]

Yeah the negative would account for the rotation

[gray dust]

you're right it does

[gloomy arrow]

Alright, this will be my final question. I was able to reduce this to RREF. I have all 0s in the last row, and ones on the diagonal up top

So its one to one because there is no free variable

And i forgot what onto is

[gray dust]

onto <-> surjective

[gloomy arrow]

I uh, dont know what that means

I dont think our professor used the term surjective

[gray dust]

that's interesting

take a look at the size of A. it tells me that T is a linear map from R^3 to R^4

[gloomy arrow]

Right

3 columns 4 rows

So it means that every image is a transformation of a vector?

[gray dust]

uhh not sure what you're getting at

[gloomy arrow]

Me neither

[gray dust]

i mean for any function T, we call T(x) the image of x under T

anyway your rref shows that the columns of A are linearly independent

so the span of A's column vectors which in this case is also the image of T is a 3d subspace in R^4

[gloomy arrow]

We havent gotten into spaces and subspaces yet

But I see what youre saying

[gray dust]

a 3d hyperplane in 4d space if you like. point is, the dimension of this plane is 1 less than the "ambient" space

[gloomy arrow]

Okay but how does that relate to onto

[gray dust]

first you'd have to recall what a function's domain and codomain are

as well as what injective/surjective functions are

[gloomy arrow]

I know what domain and codomain are

not injective or surjective

[gray dust]

and you know the difference between codomain & image?

[gloomy arrow]

Yeah

[gray dust]

say f is a function from X to Y. we say f is surjective aka onto if for all y in Y, there exists at least one x in X such that f(x)=y

[gloomy arrow]

But when would it not be equal

Example wise

[gray dust]

$\fxn{f}{\bR}{\bR}{x}{x^2}$

[stoic python]

RokettoJanpu:

[gray dust]

dom(f) is the set of reals, codom(f) is also the set of reals

we definitely have -1 in codom(f) but there doesn't exist anything in dom(f) that gets mapped to -1

[gloomy arrow]

ohhhh i see

[gray dust]

now let's say

$\fxn{g}{\bR}{[0,\infty)}{x}{x^2}$

[stoic python]

RokettoJanpu:

[gloomy arrow]

Thats onto because we dont have to worry about the negative numbers

[gray dust]

say f is a function from X to Y. we say f is surjective aka onto if for all y in Y, there exists at least one x in X such that f(x)=y
an easier to digest rephrasing of this is, every element in Y is the image (under f) of at least one element in X

[gloomy arrow]

But for the example with the matrix wouldnt every Y be an image? I cant really think of a counter example

[gray dust]

hold on what's dom(T) and codom(T)?

[gloomy arrow]

domain would be R4

co domain would be R3

Wait no

Other way around

Domain would be R3 and co domain would be R4

[gray dust]

ok and through RREF you showed that the columns of A are linearly independent

[gloomy arrow]

yes

[vital raft]

could somebody help me with a quick question

are you guys busy

?

[gray dust]

A has 3 lin indep columns so this leads to T's image being a 3d hyperplane in R4

[gloomy arrow]

Right

But not every thing in R4 would be something from R3

Right?

I mean what would be a quick way of checking

[gray dust]

sure, clearly there exist vectors in R4 that aren't the image of smth in R3 under T

for me it's enough to do the RREF to show A's cols are lin indep thus the dimension of T's image is 3 which is one less than the dimension of codom(T)

[gloomy arrow]

So therefore its not onto

[gray dust]

T isn't onto

[gloomy arrow]

Gotchu

Thanks

[gray dust]

you're welcome

[vital raft]

rokabe

could you help me real quick

?

[gray dust]

just post, don't ask for me specifically

[vital raft]

oh sorry

but anyways

is y=4x^2 (parabola) considered increasing faster than y=5x+10

?

[gray dust]

you should post that in #prealg-and-algebra instead or #precalculus

[vital raft]

my bad

[gray dust]

no worries

[gray dust]

@wintry steppe yeah no

[wintry steppe]

ok thanks

[sturdy raven]

if u say so

[limber sierra]

you mean

the columns of the matrix are linearly independent?

yes

[sturdy raven]

well if you were to find the zero matrix of a singular non-zero element that is any given position in any size matrix then you would see to be proven correct

[limber sierra]

write, say, $\begin{pmatrix}1&0&0\0&0&1\0&1&0\end{pmatrix} = \lambda_1\begin{pmatrix}1\0\0\end{pmatrix} + \lambda_2\begin{pmatrix}0\0\1\end{pmatrix} + \lambda_3\begin{pmatrix}0\1\0\end{pmatrix} = \begin{pmatrix}\lambda_1 \ 0 \ 0\end{pmatrix} + \begin{pmatrix}0\0\\lambda_2\end{pmatrix} + \begin{pmatrix}0\\lambda_3\0\end{pmatrix} = \begin{pmatrix}\lambda_1\\lambda_3\\lambda_2\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

linear independence should be obvious from here

yeah but this technique generalizes (at least to finite matrices)

just do induction or w/e

theres probably a better way than induction but

idk what you know

[wintry steppe]

I'm confused how to approach 38.2

I understand 38.1 because its the basis vectors which is easy

but I don't understand how to do it when i'm writing it in terms of the basis vectors

[limber sierra]

you need to solve $\lambda_1 \begin{bmatrix}2\1\end{bmatrix} + \lambda_2 \begin{bmatrix}5\3\end{bmatrix} = \begin{bmatrix}1\0\end{bmatrix}$

[stoic python]

Namington:

[limber sierra]

and similar for $= \begin{bmatrix}0\1\end{bmatrix}$ (but different values of $\lambda_1, \lambda_2$)

[stoic python]

Namington:

[cursive narwhal]

You have two equations with e_1 and e_2 as unknowns

[limber sierra]

huh

e_1 and e_2 are the standard basis

theyre not unknowns

[cursive narwhal]

Yea sure but you could treat them as 'unknowns' and work from the equations in part (i), no?

Like, eliminate e_1, find e_2, then find e_1 using the expression for e_2

[limber sierra]

oh sure

that definitely works too

in any case, you're just solving a system of linear equations

my technique would just have you write $\lambda_1 \begin{bmatrix}2\1\end{bmatrix} + \lambda_2 \begin{bmatrix}5\3\end{bmatrix} = \begin{bmatrix}2\lambda_1 + 5\lambda_2\\lambda_1 + 3\lambda_2\end{bmatrix} = \begin{bmatrix}1\0\end{bmatrix}$

[stoic python]

Namington:

[limber sierra]

and hence you need to solve the system $2\lambda_1 + 5\lambda_2 = 1, \lambda_1 + 3\lambda_2 = 0$

[stoic python]

Namington:

[wintry steppe]

Oh I see because the [1,0] was already known

makes sense

[limber sierra]

(and then do a similar thing but for [0,1])

[wintry steppe]

but how would I do it with @cursive narwhal method

[cursive narwhal]

There's no specific method over here

It's almost the same as namington's method

[limber sierra]

well i'd probably take $c_2 = 5e_1 + 3e_2$ and subtract $3c_1$ from both sides

[stoic python]

Namington:

[limber sierra]

$c_2 = c_2 - 3c_1 = 5e_1 + 3e_2 - 3c_1 = 5e_1 + 3e_2 - 3(2e_1 + e_2) = -e_1$

hence $e_1 = -c_2 + 3c_1$

and use that to solve the other equation

lmao ignore that

[stoic python]

Namington:

[wintry steppe]

well i'd probably take $c_2 = 5e_1 + 3e_2$ and subtract $3c_1$ from both sides
why do we subtract 3c1 from both sides

[stoic python]

hehexd lol:

Namington:

[limber sierra]

sorry just realized im an idiot

had to fix my work

uh i subtracted 3c_1 because it eliminates the 3e_2

do you know how to solve systems of linear equations?

this is just elimination

[wintry steppe]

oh it makes sense now

thank you

[wintry steppe]

how would I do 39.4

[restive shuttle]

im so lost at this conceptual problem

[normal canyon]

Is it not exist @restive shuttle

[restive shuttle]

i think i have a solution now

[normal canyon]

What’s the answer

@restive shuttle

[restive shuttle]

ima write it up first, 1 sec

Since A and B are invertible, they have nullity 0. The product of two invertible matrices is also invertible. So AB and BA also have nullity 0. Therefore by rank-nullity theorem, the rank of both AB and BA is n.

@normal canyon

[normal canyon]

So it’s false?

[restive shuttle]

yea

[pallid rampart]

Um

B is not invertible

[restive shuttle]

gg

[restive shuttle]

I don't know how to go forward on this. I got that the rank(A) = rank(B) = 2 by rank nullity theorem

and rank(AB) = 0, since AB = 0

[dusky epoch]

the nullspaces of A and B both have dimension 1 and their column spaces both have dimension 2 and for AB = 0 you need the col space of B to be contained in the nullspace of A

and that's impossible

[restive shuttle]

why does the col space of B have to be contained in the nullspace of A for AB = 0

[dusky epoch]

you want ABv = 0 for all v

Bv is an element of B's col space by defn

[coral ferry]

i've looked at a lot of linear algebra books like strang, lang and hoffman kunze

but i don't find them to be both comprehensive and also clear

do you guys have any recommendations?

im a first year uni student btw, i guess pure maths linear algebra is the flavour im learning

[sonic osprey]

These are pretty much the standards, what exactly didn't you like about them?

[coral ferry]

i tried hoffman kunze but i didnt find it to be that clear

strang was great but it wasn't very proof based

[sonic osprey]

Hm, from what I've read of it Hoffman Kunze is pretty clear

You just might not be super used to reading math textbooks?

[coral ferry]

yea that's probably it

very precise

[sonic osprey]

They're pretty different from reading anything else and things are never really clear at first

You have to struggle with and think about things before they become clear

[coral ferry]

yea

should i just keep at it with hoffman kunze?

[sonic osprey]

Take your pick

That's what I usually recommend

[coral ferry]

alright thank you

[bold blade]

okay sorry this is a dumb question but its like 4am and ive been doing linear for so long

how do u get from

here to

here

is it a property?

[dusky epoch]

i'm p sure this is false as stated

unless there's some missing context

[bold blade]

yea theres context ;ol ill send the full thing

[dusky epoch]

ah so {v1, v2, v3} is an orthonormal basis

[bold blade]

um.. sure

honestly dont rly understand this stuff conceptually which i know is bad but my schedule this quarter sucks and the teacher doesnt rly go deep in theory so i just memorize how to do stuff

okay wait it is . aproperty

[quartz compass]

the cross product takes two vectors and gives you a new vector perpendicular to both of those given by the right hand rule

the length of that new vector is the product of lengths of the input vectors, times the sine of the angle between them

if you know this, it's enough to get the direction and magnitude and answer your question, any of this new to you @bold blade ?

[bold blade]

yea i learned that stuff in calc 3 but im taking it at the same time

i didnt end up actually needing it for this question tho..

i just took column vectors to find the basis

i was just confused bc there were no numbers but its the same process i guess

[quartz compass]

if you didn't "need it for this question" then I don't know how you could have answered it

this isn't a property on its own

it's something that comes from the cross product with angle of 90 degrees between the vectors and the vectors having length 1

[lone plover]

Well I have a transformation of rotation followed by an enlargement

2sqrt3 -2
2 2sqrt3

and to find the SF, the mark scheme found the determinant, then said that SF is 4, and the determinant of that matrix is 16

and it didnt really explain the reasoning for the SF being 4

^Wondering if anyone can help clarify

[vast torrent]

What's "SF"?

[spiral sonnet]

What does R^4 mean? R is the real number symbol.

[pallid rampart]

R^4 means the set of tuples (x, y, z, w), where x,y,z,w are in R

[vast torrent]

Often implied with it is a way to add the tuples or multiply the tuples by numbers, component-wise

@spiral sonnet

But not always

[buoyant hemlock]

can some1 explain how they got from the first line to the second, like it makes sense when I look at it but thats only by intuition, is there a formalized way to make the transition of sums to vector/matrix representations

[lone plover]

What's "SF"?
@vast torrent scale factor

[rain lotus]

the geometric interpretation of the determinant is: if you take the square made by the standard basis (or hypercube, in n dimensions) it gets distorted when you apply the matrix, and the area/volume changes by some factor. that factor is the determinant. so when you have a rotation and a scaling, the determinant is the scaling factor of the area, which is the square of the scaling factor of a basis vector

[limber sierra]

ok?

[earnest juniper]

If I have a matrix and want to reverse 2 segments (parts) of that matrix . Is there a better approach to reverse only a smaller part if they intersect?
Both start and end are inclusive as shown in the example below.

Example:
We have the matrix [1, 2, 3, 4, 5] and want to reverse from position 1 > 4 ( becomes [4, 3, 2, 1, 5] ) and then 3 > 5 so it should become [4, 3, 5, 1, 2] as the final result.

[rain lotus]

@limber sierra i was trying to answer the question asked by @lone plover

[limber sierra]

ah, sure

@earnest juniper so you mean a permutation? not sure how matrices relate here

[lone plover]

Ah ok

[earnest juniper]

No idea, it's a list of numbers.

[limber sierra]

im not really sure what this means:

Is there a better approach to reverse only a smaller part if they intersect?

like, do you just want the programming method?

[earnest juniper]

The problem is that I cannot use the normal reverse methods as I'm limited in the time.

[limber sierra]

limited in time? like computational time?

[earnest juniper]

I'm learning competitive programming and I'm solving a couple of things now, so in competitive programming I have a time limit that I cannot exceed.

Yes, the computational time (time taken to execute operations).

[limber sierra]

ok, that clears up what youre asking

[earnest juniper]

So I'm looking for a better approach to these flips/reverses.

[limber sierra]

i recall there being an O(n) algorithm for exactly this, but i cant recall it off hand

gimme a minute to think

[earnest juniper]

Okie.

[limber sierra]

so for context i know python has a .reverse() list method, which is O(n) in the length of the sublist, but i'm unable to find good pseudocode for that

here's the source code:

static void
reverse_slice(PyObject **lo, PyObject **hi)
{
    assert(lo && hi);

    --hi;
    while (lo < hi) {
        PyObject *t = *lo;
        *lo = *hi;
        *hi = t;
        ++lo;
        --hi;
    }
}

static PyObject *
listreverse(PyListObject *self)
{
    if (Py_SIZE(self) > 1)
        reverse_slice(self->ob_item, self->ob_item + Py_SIZE(self));
    Py_RETURN_NONE;
}```

as far as i can tell, here's the explanation:

1. Keep track of 2 indices, initially set to the start and end of the list.
2. Swap the elements at these indices.
3. increment the left index, decrement the right index.
4. Repeat from 2 while the right index > the left index.```

in this case, you'd start these indices at the start and end of the subarray

but same gist

[earnest juniper]

I'll need to do 2 * 10^11 reverses, how much will that method take?

[limber sierra]

if i'm understanding it correctly, this algorithm takes n/2 iterations, so 10^11 iterations

i'm not aware of if there's a faster direct method

[earnest juniper]

Is that for each reverse or for all of these reverses?

[limber sierra]

to do 2*10^11 reverses

er wait

sorry misunderstood what you meant

these are separate reverses? hmmm

well, it still takes n/2 time for each reverse, so k*10^11 where k is the average length of a reversed subarray

roughly

[earnest juniper]

Array Size (1 ≤ N ≤ 100)
Reverses (1 ≤ K ≤ 2 * 10^9)

[limber sierra]

this doesnt seem particularly efficient, unfortunataly; there may be a better way

like i doubt theres a faster way than what python is implementing, since otherwise python would use it

oh wait hold on

do you know your arrays are always initialized in ascending order?

like do all your arrays start as 1 2 3 4 5 6 ...

[earnest juniper]

Yes, that is guaranteed.

[limber sierra]

ah, that makes this easier

[sonic osprey]

I think his question was more that, if we knew we had to do multiple reverses, is there some way we could save time rather than just doing each reverse after the other

[limber sierra]

zoph the answer to that question is "no if you don't know the original setup of the array at all"

but since the array is in natural ascending order like that, there's probably some tricks you can use

[earnest juniper]

I couldn't find these "tricks", so I was wondering maybe I could use another brain to help me 😂

[sonic osprey]

I'm not sure why the array being in ascending order matters here?

[limber sierra]

yeah, i'll think about it but it's been a while since i programmed anything tricky like this

[sonic osprey]

okay, things are easier to compute

[limber sierra]

zoph it means we dont need to keep track of each element's value

[sonic osprey]

yeah yeah okay

maybe just ping Horse tbh

[limber sierra]

i mean i want to look at this as a cycle composition now

lmao

thats probably not the best answer but

its what my mind jumps to

let me see if thats practical actually

using the cycle composition algorithm

hm, nah, the only way i can think of using the cycle composition algorithm here would probably be O(n^2) (or perhaps O(nlogn))

yeah, sorry, this is a bit above my programming paygrade

[earnest juniper]

Thanks for trying, is there possibly someone else that's more familiar with these things? (time, tricks, etc.)

[limber sierra]

well, as zopherus said

@pastel harbor

if you wouldnt mind ❤️

if you're wondering what my cycle idea was, it exploited the property that a "reversing" permutation from a to b can be written as the cycle starting with (a b ...); sadly the only way i could think of to make use of this property required nested for loops, which wouldnt work so fast

there's probably a cleaner way in haskell, of all things

[earnest juniper]

Alright, I'll provide as much details as possible that can help obtain the trick behind this question.

1: The starting array is sorted from 1 until 100
2: The reverse operation consists of 2 parts:
2.1: Reverse all elements in positions X1 .... X2
2.2: Then, reverse all elements in positions Y1 .... Y2

When repeating the reverse step you're repeating Step 2 in the same order with the same numbers every time.

Does that help? [@limber sierra]

The reverse operation is done on the same indices every single time.

[limber sierra]

oh interesting

[pastel harbor]

oh

hi

looks like I'm not too late to this

[limber sierra]

your services have been conscripted

lmao

[earnest juniper]

Yeah, awesome timing :P

[pastel harbor]

ok so the question you have

is to reverse twice in an array?

or are you trying to generalize

to k reverses

if the question is for k

[earnest juniper]

I basically repeat Step 2 K times.

[pastel harbor]

you can generalize with a fenwick/segment tree

yeah, then you can use a segment tree or fenwick tree

to keep track of the endpoints

and it reduces to a prefix sum problem

let me explain

notice that you only need to know the endpoints

[earnest juniper]

You mean X2 and Y2?

[pastel harbor]

I mean X1, X2, Y1, Y2

[earnest juniper]

Alright.

[pastel harbor]

so what you can do is

mark X1 with a +1

X2 with a -1

now the prefix sum indicates

whether an element is reversed or not

[0, 1, 0, 0, -1, 0]

prefix sum is [0, 1, 1, 1, 0, 0]

[earnest juniper]

Ok.

[pastel harbor]

so now you can extend this logic

you can apply all of your reverses in bulk

and then finally extract the resultant array

tbh

wait

[earnest juniper]

But at each step, I need to do the X first then Y after.

[pastel harbor]

yes

I think there might be a simpler way where you don't even need this, let me think if that is possible

[earnest juniper]

Oki.

[pastel harbor]

I mean the second reverse doesn't really matter

because the question is basically saying 2K reverses in total then

is that right?

[earnest juniper]

But the second reverse is reversing a different segment with different endpoints, doesn't that make a difference?

[pastel harbor]

are you saying

that X1 and X2

are the same in each step

[earnest juniper]

Yes, that's correct.

[pastel harbor]

oh

then this is much simpler

I was solving for different X1 X2 in each step

that is still solvable in n log n

[earnest juniper]

When repeating the reverse step you're repeating Step 2 in the same order with the same numbers every time.

[pastel harbor]

right, my bad

one sec, I should be able to devise something simple

[earnest juniper]

Alright.

[pastel harbor]

ok

you can just make cases

and do it in linear time

if they don't intersect

[earnest juniper]

I thought the first case can be if there's no intersection.

[pastel harbor]

and if they intersect

if they intersect

take section from the front

of length equal to intersection

if this exceeds half intersection point

then you will need to recursively repeat

I could write some code tomorrow when I'm up 👀

what language do you find easy to read?

[earnest juniper]

I need to hand over the thing now, can you just explain the last part one more time?

I'm working with C++.

[pastel harbor]

ok, so

let x1, x2 be 1, 5

[earnest juniper]

Alright.

[pastel harbor]

y1, y2 be 3, 7

3 to 5 is intersection

this means

after the first swap

you will have elements from

x1 to x1 + (intersection length)

inside that region

now, you will have x2 to y2 as the other points

note that in this case x2 to y2 is smaller

you will need to make different cases here as well

based on whether it is larger or smaller

this is what I meant where you do the x1 + intersection length as well

you will need to make two cases there

here the 3 will still come into the intersection

because intersection is larger

you must consider smaller intersection as separate case

anyway, if you handle all these cases

you can get a linear solution

but this is just pain tbh

for n = 100

no one cares,

you can just write the dumb n^2

and it will still run in under a second

[earnest juniper]

But K can be up to 10^9, doesn't that make a difference?

[pastel harbor]

oh