linear-algebra

[tacit storm]

then the isomorphism could be phi : p4(R)/p2(R) -> R2 phi(ax^4 + bx^3) ->(a,b)

[wintry steppe]

phi(ax^4 + bx^3 + P_2(R)) = (a, b) could work

just make sure it's well-defined and what not

or, if you know the first isomorphism theorem, this is immediate

[stoic python]

TTerra

[wintry steppe]

sry just wanted to see how convoluted i could make that

[tacit storm]

im so confused now

[wintry steppe]

.

your map works

i just proved its an iso

[tacit storm]

i did phi(ax^4 + bx^3) ->(a,b)

you did phi(ax^4 + bx^3 + P_2(R)) = (a, b)

[wintry steppe]

yes, because you're supposed to define a map from the quotient

not from P_4(R)

ax^4 + bx^3 is not an element of P_4(R)/P_2(R)

ax^4 + bx^3 + P_4(R) is

i assumed you made a typo, but i think this clarification should be made now

[tacit storm]

yes

thanks babe

but doesnt ax^4 + bx^3 + P_2(R) have dinmension 2

[wintry steppe]

well, sure

what's wrong with that?

the quotient is gonna be two dimensional

[silk jewel]

How do you prove a linear transformation is isomorphic?

[wintry steppe]

you mean is an isomorphism?

you show that it is injective and surjective

[silk jewel]

What would I need to do?

to show that

[wintry steppe]

for injectivity, you can show that if Tx = 0, then x = 0

[silk jewel]

because idk how to show every element is onto and one to one

[wintry steppe]

you mean the linear transformation is onto and one to one

[silk jewel]

yes

[wintry steppe]

in general, checking that the only x satisfying Tx = 0 is x = 0 is easy

now, if the spaces you're working with are finite dimensional and of the same dimension, this is actually sufficient

since in that case, the rank-nullity formula tells you that rank T = dim domain = dim codomain, and that implies that your linear transformation is onto

if you're working with infinite-dimensional spaces, though, you'll have to manually check that

which just boils down to the definition or surjectivity, there isn't really a nice trick involved

[silk jewel]

Alright that helps a bit, thanks.

[tacit storm]

i need help

pls

Is this right

[thorny furnace]

seems fine to me

[jade lance]

By the way my name is Skylar .
I graduated with Mathematics and computer Science majors.
I can help with maths and computer science if needed blush

[wintry turret]

I was wondering if anyone could help me how to prove the second part. I’m thinking proof by contradiction?

[limber sierra]

yeah, a contradictive argument is probably a good approach here

[wintry turret]

Seems kind of long, but I’ll give it a shot and come back with more questions

[limber sierra]

suppose (WLOG) that U_1 is not a subset of U_2, and U_2 is not a subset of U_1. this means we can take u in U_1 \ U_2, and v in U_2 \ U_1. what's u + v?

[here \ is the set difference, so U_1 \ U_2 is the set of all vectors in U_1 that arent in U_2]

hint: u + v cant be in U_1, since otherwise (u + v) - u = v would be in U_1

which is impossible

and by a similar argument u + v cant be in U_2

[wintry turret]

Is that only for 2 subsets

Subspaces*

[limber sierra]

yes, but think of how you might modify it for 3.

[a hint: either u+v is in U_3 or it isn't, but if ALL such u + v are in U_3, then U_1 and U_2 are contained in U_3. so when picking u, v, we should also make sure that they're not in U_3 - but can you justify why we can do that?]

[wintry turret]

Well, u+v would be in the union

Right?

[limber sierra]

what union?

okay let me rephrase my argument slightly

[wintry turret]

Of U1 and U2

[limber sierra]

u + v can't be in U_1 or U_2

do you understand that much?

[wintry turret]

Oh lol

[limber sierra]

but if we assume that the union of all three sets is a subspace

that means u + v must be in it

since subspaces are closed under vector addition

but u + v aint in U_1 or U_2, so it must be in U_3

what can you reason from there?

[you might want to consider, say, 2u + v as well]

[wintry turret]

If its in U3 then it must be in U1 union U2 union U3

[limber sierra]

well yes, we established that prior since u is in that union, and v is in that union

and if we're assuming (for contradiction) that it's a subspace

that means u + v must be in it as well

let me lay out what i've deduced so far:

[wintry turret]

Yeah I was scrolling back up to read sorry

I wish I can do these proofs as quickly as you guys

[limber sierra]

• We're assuming (WLOG) that U_1 is NOT contained in U_2 or U_3, and U_2 is NOT contained in U_1 or U_3.
• So take a u in U_1 that's not in U_2, which we can do because we know U_1 is not contained in U_2. Similarly, take a v in U_2 that's not in U_1.
• Now assume for contradiction that the union of the three subsets is a subspace. This means it's closed under vector addition, so u + v must be in the union. But:
  --- u + v can't be in U_1, since otherwise (u + v) - u would be in U_1, but we know v isn't in U_1.
  --- u + v can't be in U_2, since otherwise (u + v) - v would be in U_2, but we know u isn't in U_2.
• So we can conclude that u + v must be in U_3, since it's in the union but not in either of the other two sets.

there are a bunch of ways to proceed from here

one of them is to apply the same argument to 2u + v that we did to u + v

to reason that it must be in U_3, but not in U_1 or U_2

but then (2u + v) - (u + v) = u must be in U_3 as well

and we can repeat this argument for u + 2v instead to reason that v is in U_3 also

so U_3 contains u and v for any choice of u, v

which means U_1 and U_2 are contained in U_3

uh oh

contradiction

[wintry turret]

I see

[limber sierra]

since we were assuming that none of these sets were contained in the other

there is a slight catch here: its possible that the scalar "2" doesn't exist in your underlying scalar field

can you see a way to amend the argument in that case?

[wintry turret]

Can we just let it be an arbitrary scalar

[limber sierra]

not quite, unfortunately

what we can do is drop the scalar entirely

instead of 2u + v

just write u + u + v

and similarly write u + v + v instead of u + 2v

[wintry turret]

Ahhhh, I see where the little disclaimer under the problem comes from now

[limber sierra]

so yeah, i'd recommend replacing "2u + v" with "u + u + v" in my above argument

and similar for u + 2v -> u + v + v

do you see why it works now?

obviously the scalar field doesnt really matter anymore since we're not multiplying by scalars

but the algebra is the same

anyway, that argument can be polished up a bit

but hopefully it gives you the idea of whats going on

[wintry turret]

Yes, I think I’ll understand it better once I write down the proof and get a chance to think about it

[limber sierra]

there might be a cleaner approach but this is what came to mind first for me

[wintry turret]

Thank you

I’m always impressed at how fast everyone is with figuring this stuff out. Im in my undergrad and I feel like I know nothing when Im doing exercises in the book

[limber sierra]

well in this case it's kind of just "looking for where we might expect problems to arise"

which is admittedly partially an intuition thing

but like

the way i approached this was "in order to prove something isn't a subspace, we need to show its not closed under vector addition or under scalar multiplication"

"obviously doing this unioning wont mess with scalar multiplication, so we care about vector addition"

"and we want this to go wrong SPECIFICALLY WHEN the sets arent contained in each other"

"so lets say two sets aren't contained in each other. that means we can pick an element from one that isn't in the other, and vice versa. we care about what happens when we add vectors - so let's try adding them and seeing if we run into problems"

and then it kind of just follows from algebra on there

i wouldnt call this an "easy" process, but it's one that becomes more "natural" as you get more practice

[wintry turret]

Wow, thanks for that

I hope I can be as helpful as everyone else is here someday. But for now, I'll keep on working on my books.

[thorny hemlock]

i dont seem to understand how he deduced that atleast one T-lambdaI is not injective

[native rampart]

There

[thorny hemlock]

lmao

this book is really popular huh

[wintry steppe]

oh this proof

[dire thunder]

priyet tterra

[wintry steppe]

i like this proof

privyet

,ti

[stoic python]

The current time for TTerra is 08:42 AM (EST) on Tue, 12/01/2021.

[wintry steppe]

,ti vimes

[stoic python]

The current time for Commander Vimes is 07:42 PM (+06) on Tue, 12/01/2021.
TTerra is 11 hours behind, at 08:42 AM (EST) on Tue, 12/01/2021.

[wintry steppe]

[dire thunder]

@wintry steppe kak dela

[wintry steppe]

не плохо

i still have the russian keyboard on my phone despite not being in russian for two semesters

[dire thunder]

learn russian

[pseudo thicket]

Any idea how I should do this?

Was thinking of trying individual challenge subtract/add to other challenge until I get the binary vector value for ca and cb

Any hints would be greatly appreciated, have been stuck for hours

[acoustic zodiac]

help finding the kernel of $f(x,y,z,t) = (x-y+t, 2x+z+t, x+y+z)$?
i'm having trouble with it

[stoic python]

rcatalang

[acoustic zodiac]

it's an R4 to R3 map

[limpid fiber]

What's good team

How does multiplying a vector by D, keep it on the hyperbola xy = C

[acoustic path]

whats good team

[limpid fiber]

whats up my swag squad

[acoustic path]

whats up my linear algebra ppl

[limpid fiber]

whats up my mathematical muchachos

I think you're right lol

The first explanation made sense thank you

https://textbooks.math.gatech.edu/ila/diagonalization.html, it's an online textbook

Rigth after the geometry of diagonalizable matricies

first example in that section

Yeah its a pretty good textbook besides the C thing, the visualizations are nice

Diagonilization, or the multiplication by D is a hyperbola

Yeah its weird to see the hyperbola because everything so far has been linear

[acoustic zodiac]

yeah that's what i was doing but it ain't working

[limpid fiber]

Thank you I appreciate that! It is quite pretty to look at

[acoustic zodiac]

i simply tried solving it by substitution but i end up with the same equations

[spark umbra]

I have two computing problems I think could be solved nicely via linear algebra (since they're about matrices) but I'm not sure how I'd actually do it (I feel like some matrix multiplication magic but idk)

I'll post the first one below:

1.) find if a chain of 3 exists of any symbol in the tic tac toe grid

e.g.

[X][O][ ]
[O][X][O]
[ ][O][X]

this is a valid sequence since a chain of 3 exists (3 A's diagonal)

while this isn't

[X][X][ ]
[O][X][O]
[O][ ][O]

[quartz compass]

what's it mean to be a chain exactly

[spark umbra]

have you played tic tac toe before? it's any of the same symbol joined horizontally or diagonally

3 of them

so the top example the three x's form a chain

[limber sierra]

so youre just asked to determine if a given game of tic-tac-toe is won?

[spark umbra]

effectively yes

[tropic trail]

Quick question: if you have a matrix A which is symmetric, is every congruent matrix with A then also symmetric?

[spark umbra]

if this is at me to make me realize something I'm way below this level of maths lol

[limber sierra]

not sure there'd be a great linear algebraic approach there, its hard to express notions such as "diagonal" through LA operations

[spark umbra]

I am going to try squaring the matrix since that helps find relations but idk

[limber sierra]

at least without row-reducing

[quartz compass]

I was kind of thinking the determinant might have some properties we could use

[spark umbra]

yeah I thought not, it just seems like there should be some nice way to do it

[limber sierra]

like you could view it as a matrix in $(\mathbb{Z}/3\mathbb{Z})^{3\times 3}$

[stoic python]

Namington

[quartz compass]

if we imagine the x,0, or blank as separate primes

[limber sierra]

and then take the determinant yeah

[quartz compass]

then check the prime factorization

[limber sierra]

but im still not sure that gets you anywhere

[quartz compass]

like any prime in all the same row or column would factor out

[limber sierra]

ah hm

[quartz compass]

but that wouldn't get diagonals, but diagonals might still have some info

[limber sierra]

that still leaves problems with the diagonal though

[spark umbra]

if you give me a minute I could try programming that

[limber sierra]

you could do fixed-in-place row reduction i guess

[quartz compass]

how's a diagonal behave, we might have some kind of thing we can derive

[limber sierra]

but this seems way more common than just a couple foor loops

[quartz compass]

problem is we might get false positives too

[spark umbra]

for a diagonal to exist the middle must be containing the symbol if that helps

because it's only 3x3

it's the middle is a symbol and so are either of the corners

[limber sierra]

that's a useful observation to make from a computing perspective but im not so sure it helps linear algebraically

[spark umbra]

yeah that doesn't really help find an elegant solution

[tropic trail]

Oh okay, but a symmetric matrix has only real eigenvalues so every similar matrix too?

[spark umbra]

and that's the main goal

I could combine that method and the row checking

if you could explain that to me

as in it might be the best if I combine finding the determinant and that check its sorta a clean way

but idk how the prime determinant thing works

[quartz compass]

I don't know how useful the prime determinant trick is

[limber sierra]

yeah i dont really think it helps here

[quartz compass]

if there is a row or column of the same thing, then it is divisible by some specific prime used to denote that symbol

[limber sierra]

unfortunately

[spark umbra]

this is unfortunate

[quartz compass]

but it's not necessarily true the reverse direction, but maybe it could be done to work out primes that would make it for 3x3 since it's small

[limber sierra]

there is a more elegant method than just for-looping 6 times for whatever its worth

[spark umbra]

if it helps I could simplify the problem to 1 symbol at a time

I'd like to know what you mean namington

[quartz compass]

yeah I wouldn't use linear algebra at all lol

[limber sierra]

but i dont think theres a linear algebraic one

[spark umbra]

what solution would you suggest

fair enough, idk just seemed prime for some matrix tricks but guess not

[quartz compass]

well there are only 8 things to check

3 rows, 3 columns, 2 diagonals

[spark umbra]

yeah that's true

that's slightly cleaner

[quartz compass]

so for looping is gratuitous

[spark umbra]

I guess I'll go and think on it some more

[quartz compass]

at least in the crazy manner

[spark umbra]

thanks for the help guys

even if we couldn't find a solution

[limber sierra]

wdym mero

like my approach would be

say our board is an array board[][]

[quartz compass]

I think I misinterpreted what you meant

we're saing the same thing

[limber sierra]

the naive method is to loop through each row and each column

[spark umbra]

yeah I have it as vectors because cpp is angry about functions and arrays lol

[limber sierra]

in any case theres a cuter way to organize the data

convert the board state to a 9-bit binary number

there might be some bitwise operation you can do from there? let me think

[spark umbra]

I can easily make it a bit board

I thought abput that

as in if x = 1

[limber sierra]

well if youre gonna do that youd probably have 2 of them

[spark umbra]

and o = 2

[limber sierra]

one for each player

[spark umbra]

yeah I could

[limber sierra]

since you dont care about player 2 when looking at whether player 1 has won

[spark umbra]

because in that case I could div by 2 for the os

and mod 2 for xs

[limber sierra]

a player 2 entry is the same as "empty"

[spark umbra]

yeah

I agree that bit arrays might present some opportunities

and would be nice

idk if that really fixes the diagonal problem though

what would multiplying a bit matrix by itself do

[limber sierra]

well multiplying matrices is computationally kinda slow

theres a reason your GPU has an entire part of its board dedicated to it

i'd be very surprised if a solution involving matrix mult turns out faster than the naive one

[spark umbra]

I don't really mind about speed

I just want something nice

[limber sierra]

an observation to speed up computation: a win is only possible if (at least) one of these 5 "blue spots" is filled in by that player

[spark umbra]

if I was programming the fastest tic tac toe then I'd be really good now lol

[limber sierra]

in fact, of the 9 possible ways to win, 5 of them require the centre

if you really wanna multiply matrices though

hm

[spark umbra]

hmm

[limber sierra]

i guess we could view the matrix as from $((\bZ/2\bZ) \times (\bZ/2\bZ))^{3 \times 3}$

[stoic python]

Namington

[spark umbra]

maybe using the middle check is a good idea

[limber sierra]

eh thats clunkier than just viewing it as 2 matrices

[spark umbra]

is there a way to find just columns and rows via matrices

because if so I could do that cleanly

then just do a dirty corner check at the end

[limber sierra]

by multiplying by the right matrix, sure

but its much easier to do that with more standard programming methods

and cleaner

[spark umbra]

ah well at least I tried lol

thanks for the help

[nocturne jewel]

$\mathbb{T} = \mathbb{R} \cup {\infty} \ v \oplus w = \min(v,w) \ c \otimes v = c+v$

[stoic python]

moshill1

[nocturne jewel]

T isnt a R-vector space b/c: (cd) * v != c * (dv), not all elements have an additive identity, and (c+d) * v != cv + d*v

Is that right?

[gray dust]

@nocturne jewel we just need to show T fails one of the vector space axioms. any more is excessive. also give actual values of c,d,v that show T fails an axiom. just one set of values suffices. now T actually has an additive identity and you recited the additive identity axiom in an odd way. it doesn't say the elements can each have their own possibly different additive identities; T must have a single element that acts as such for ALL elements

[wintry steppe]

why is Ax=b equivalent to x=A^(-1)b

is it just when you a matrix to the other side it turns into its inverse?

[limber sierra]

Multiply both sides by $A^{-1}$ on the left:
\begin{align*}Ax &= b\ A^{-1}Ax &= A^{-1}b\end{align*}
but of course $A^{-1}Ax = Ix = x$, so $A^{-1}Ax = x = A^{-1}b$

[stoic python]

Namington

[limber sierra]

it's just algebra.

[nocturne jewel]

yeah, the question is show which properties fail if the set and operations isnt a vector space. and I meant additive inverse, since inf is the additive identity

[pseudo thicket]

anyone? it's fairly simple, but i dont see it

[tacit storm]

anyone got this

anything

i cant do this

they can be infinite

anything

i have no idea

not sure

im would say no

alright

no

R[x]

i just dont know an injective one

i mean sujective one that is not injective

then i do

i dont know

differentiate

this is surjective

its linear

its not injective becuase all constants go to zero

yes

thank you brother

wait

one more

This must be infinite

vector space

[acoustic path]

Since its not surjective

That means that it doesnt span every element of T(V)

[tacit storm]

yews

[acoustic path]

So its inf dimensional

[tacit storm]

yes

[wintry steppe]

didn't we just do this one the other day

[tacit storm]

cant remember lol

[wintry steppe]

[tacit storm]

yeah but i just need an example

[wintry steppe]

consider space of sequences

[tacit storm]

ok

i dont know

[wintry steppe]

what are some linear operators you can think of on the space of sequences?

just throw a few non trivial examples out there

maybe you'll get one

[tacit storm]

cant think of one for infinite vector space

[wintry steppe]

can you write down what the space of sequences is?

[tacit storm]

not sure what it is

[wintry steppe]

it's the space of sequences,
[
\bR^\bN = {(a_1,a_2,a_3,\dots,) : a_i \in \bR}
]

[stoic python]

TTerra

[wintry steppe]

this is, as you can check, a real vector space

[tacit storm]

yeah ok

but how does this help us find a linear map that is injective but not surjective

[wintry steppe]

because there's a particularly simple injective, non-surjective linear operator on this space

[tacit storm]

not sure mate

[wintry steppe]

can you come up with a few examples of non-trivial linear operators on this space?

maybe try to come up with some injective ones?

play around with it

[tacit storm]

i dont know

im so lost

[tacit storm]

im pretty sure just adding an extra component would do it

right

[wintry steppe]

what do you mean

[tacit storm]

adding on extra dimension component to the vector that is the sum of the other components

[wintry steppe]

can you write it out explicitly

i have a feeling you're on the right track

you do, in some sense, want to add on an extra dimension

but what does the sum of the other components mean when there are infinitely many?

if your linear map is gonna stick another component on, why not make it particularly simple?

[tacit storm]

idk

maybe make it zero then

im not sure

[wintry steppe]

yes!

[tacit storm]

so tahts it

thats

[wintry steppe]

just, i think you should write the map down explicitly

it should look like ||the sequence is being shifted over to the right by one index||

if you did it right

[tacit storm]

hold on

[wintry steppe]

perfect

[tacit storm]

will this do brother

[wintry steppe]

👍

yup, that's it

[tacit storm]

i love you

[wintry steppe]

it's good to have a few examples of infinite dimensional spaces in your pocket

[tacit storm]

i get the genral idea

now

[wintry steppe]

R[x], continuous functions on some set, sequence spaces, etc.

[gray dust]

@nocturne jewel if you must cite all axioms failed then fine. infty does serve as an additive identity, and it's true not every element has an additive inverse. but my point still stands for the other axioms you cited, we just need to pick one set of elements for which they don't hold

[pseudo thicket]

is A the second column and b the third column of the table?

[tropic trail]

I gues that they mean the form ( A | B) , where A are the first two colums and B only the last column

Can anyone get me started with the next proof: If A is a square matrix which is diagonalisable (in IR) and has only eigenvalues 1 and -1, then A^2 = I (the identity matrix).

[wintry steppe]

just diagonalize it

A is similar to a matrix whose square is I, so A^2 is similar to I, so A^2 = I

[tame mural]

A = [1 1; 1 2; 1 3; 1 4; 1 5]
y = [2, 4, 3, 5, 8]
x = A[:, 2]

weights = inv(transpose(A) * A) * transpose(A) * y

b, m = weights

predicted = m .* x .+ b

[tropic trail]

What do you mean with A is similar to a matrix whose square is I? Thats what you need to prove

[wintry steppe]

write out the definition of diagonalizability

for simplicity i'm going to take things to be in R^2 but the proof differs only by notation for the general case

[stoic python]

TTerra

[wintry steppe]

@tropic trail

[tropic trail]

Ohh thank you!

[hollow finch]

for real 2x2 matrices, are there any other cases than diagonalizable with two real eigenvalues, one real defective eigenvalue, and diagonalizable with complex eigenvalues?

just need to make sure my proof covers all cases

[stoic python]

pancakehammer

[quasi frigate]

This has no solutions right?

because column1 = -(column3*column2)?

so columns are not independent thus this have no solutions right?

[soft burrow]

not necessarily, the vector could be in the space spanned by the columns still

[hollow finch]

Ax=b has a solution if and only if b is in col(A). so even if A is not invertible (or square) the system can still have a solution

[quasi frigate]

So even though det(A) = 0, this can have a solution?

[hollow finch]

thats right

if A is not square the determinant function isnt even defined, but Ax=b can still have a solution

[quasi frigate]

Okay, if you don't mind I will send solution in just a sec

[soft burrow]

just now I was checking with wolframalpha over at #bots

in case you want to see that

[hollow finch]

yeah so b=column2-column3. that means right away we know (0,1,-1) is a solution by the column perspective of matrix multiplication

and because column1 is just a combination of columns 2 and 3, we can set a parameter such that column 1 cancels out with some other combination of columns2 and 3. specifically t*(1,1,1)

[quasi frigate]

IDK, I got solution that is dependent from z, is that right?

@hollow finch

[soft burrow]

yeah you should get a dimension 1 space of possible solutions

geometrically you can think of it as the intersection of two planes since the matrix's column space has dimension 2

[quasi frigate]

Okay, thanks

[hollow finch]

Looks like there is a mistake at the last line

The third column should have -1 / -1

[quasi frigate]

Yea, you're right

[hollow finch]

and going from the x= and y= to the vector is incorrect as well

it should be
x=1+z
y=2+z
z= 0+z
so the solution would be (1,2,0)+z(1,1,1)

there we go

[quasi frigate]

sorry, but I don't understand how -1,-2 became 1,2

[hollow finch]

one thing to note is that when a square matrix is not invertible, if there is one solution there will actually be infinitely many of them. so in a way noninvertible matrices can have more solutions than invertible ones. you can actually think of invertible matrices as ones which always have exactly one solution.

you have
1 -1 0 -1
0 -1 1 -2

you multiply the second row by -1 to get a pivot in the second column and add -1 of the second row to the first to get a zero in the 12 entry.
so the first row becomes (1,-1,0,-1)-(0,-1,1,-2)=(1,0,-1,1) and the second row becomes -1(0,-1,1,-2)=(0,1,-1,2)

[quasi frigate]

Ohh okay

I think I understand this now. Thanks a lot!

[hollow finch]

[nocturne jewel]

"Give an example of a nonempty subset U in R^2 st U is closed under scalar multiplication, but U is not a linear subspace of R^2"

Kinda just need a translation/explanation so I can give it an attempt.
Do I just need to find a set of 2D vectors where scalar multiplication maps to something in U, but adding 2 things in U doesnt?

[umbral smelt]

Yeah, I believe so

Since a subspace is both closed under scalar multiplication and closed under vector addition right?

[nocturne jewel]

or if it somehow doesnt have the 0 vector

[umbral smelt]

I'm not quite sure how if it doesn't have 0 vector would suffice the condition?

[nocturne jewel]

Naive Test fails?

[wintry steppe]

it would have to have the zero vector because it has to be closed under scalar multiplication

and 0 is a scalar

[nocturne jewel]

oh yeah

so if v,w are in U, then v+w cant be

[wintry steppe]

you just need that to fail for some v, w in U

yes

[nocturne jewel]

Oh right, cause the condition is for all

[wintry steppe]

mmmhmm

[pure tangle]

hi everyone could someone please help me with the following problem. I think I did the first part right but I'm stuck on converting it to a vector equation:

this is my best attempt at a solution for the second part

[pure tangle]

<@&286206848099549185>

[empty copper]

Your z looks like a 2

[pure tangle]

is the work correct @empty copper

[round coral]

@pure tangle the cartesian eqn is correct, can't see the vector equation though mentioned in your work

[hollow finch]

the vectors youre using for your parametric solution are the same. that doesnt work

For the second part, all you need are two vectors orthogonal to the normal vector. So (4,1,0) and (0,1,2) would work (dont think too hard about it).
Then you just do (5,1,3)+s(4,1,0)+t(0,1,2)

[storm pilot]

I have a question regarding the proof of lemma 4.2.4 in Hersteins topic in algebra

basically it says that the cardinality of any linearly independent set is less than that of any basis set (Finite)

It can be done using Steinitz exchange lemma, but Herstein's proof is shorter, but I'm failing to grasp one step in it

Suppose v1,...,vn is a basis set and w1,...,wm is an LI set, he claims without proof that wm,vi1,..,vik is a basis (k<n)

[native rampart]

Can you share the proof?

Feels like you are missing something

If w_m is not in span{v_i1,v_i2...v_ik} that's true

[storm pilot]

can you see the pic above?

I don't exactly get "Thus some proper subset of these .... forms a basis of V" part

[storm pilot]

Thats what he's assuming

[native rampart]

Take {(1,0),(0,1)} as a basis of R^2

Now ,{(1,0),(0,1),(2,1)} is a dependent set

But if we remove (1,0) from this list, we end up with an independent set, which forms a basis

[storm pilot]

yeah, even i thougt so, but it seems a little hard to grasp without justification.

[native rampart]

Let w=c1 e1+c2 e2... ck ek c_i all non zero {e1,e2...ek,...en} is a basis

[storm pilot]

i mean wm works as some vj which we're replacing right

[native rampart]

Then {w,e1,e2,...en} is a linearly dependent set which becomes LI and spans the same space as {e1,e2..en}(making it a basis) if we remove say e_1

[storm pilot]

we know that wm is not zero, and that part is sufficient to show that it has some nonzero term of some basis element

yeah, if c1≠0

thanks

[lucid fulcrum]

hello

can someone explain to me like i am 5

how my prof went from line 1 to 2 to 3

this is least squares btw

also i am very unfamiliar with the notation on line 2

someone said inner product, but i don't get it

[limber sierra]

are you not familiar with the euclidean norm/distance?

[lucid fulcrum]

i am not familiar @limber sierra

so suppose i ignored the argmin

i understand how line 1 evaluates to a scalar

but nothing else after

[limber sierra]

so youve never seen $\norm{\mathbf{x}}$ before?

[stoic python]

Namington

[limber sierra]

thats... a bit concerning

if these are the calculations youre being shown

[lucid fulcrum]

it's in my machine learning lecture .-.

[limber sierra]

the norm of a vector $\mathbf{x} = (x_1, x_2, \dots, x_n)$ is given by $\sqrt{x_1^2 + x_2^2 + \dots + x_n^2}$

[stoic python]

Namington

[wintry turret]

Is this correct?

[limber sierra]

its a generalization of 2-dimensional distance, which of course follows the pythagorean theorem sqrt(a^2 + b^2) = c

anyway, if you manually compare that sum with this, you'll notice they're the same

[lucid fulcrum]

so what does each of the 2's mean?

One is for 2-d I am assuming

[limber sierra]

one of the 2s is squared

the other one means specifically the 2-norm

[lucid fulcrum]

ah i see

[limber sierra]

there are more versions of the euclidean norm

which we call "p-norms"

but they're using the standard one here

at least, that's what i'd assume

maybe ML uses a different notation since we usually dont bother to write the subscript 2 when we work with the standard norm

[lucid fulcrum]

oh so is the computation inside the norm notation going to compute to a vector?

in which we take the norm then it is a scalar

[wintry turret]

Oh, sorry about that. I didn’t see someone was being helped, bc I just sent it from the notability app.

[limber sierra]

yes, y - Xw will be a vector

[lucid fulcrum]

ah so it is just the magnitude

i am so unfamiliar, ugh thanks for the help

[limber sierra]

yes

"norm" and "magnitude" are the same thing here

sorry, bunch of different terms for the same thing

:/

anyway its probably a good idea to "convince yourself" that what they do between lines 1 and 2 is true

at least with some examples

[faint lintel]

Hi so I'm taking a proof based lin alg course starting in like a week

I have exactly 0 linear algebra knowledge past how to do dot and cross products

So my question is what is some good prep to do to get an overview of the subject

I feel like I shouldn't go in completely cold

Especially since it's a proof based course

[slate fox]

I found axler's book really good

[native rampart]

I guess try friedberg-insel-spence's book

[shut silo]

Are there math courses that are not « proof-based » ?

Proof is the whole point of math

[nocturne jewel]

high school

[slate fox]

engineer moment

[faint lintel]

Yes there's like 2 other lin alg courses here which are all computation

[shut silo]

Yeah indeed, that's why everything you learn in HS will be obliterated in serious courses

[faint lintel]

One of them is based in python which is neat

But that's besides the point

I don't know what textbook (if there even is one) that we're using

[faint lintel]

ok tysm I'll start to take a look at this

[quiet adder]

I don't think it matters too much if you don't have any prior lin alg knowledge for your course, because you'll most likely just start from the beginnings in a proper way

but ofc it never hurts to look more into it beforehand

[quartz compass]

if all you know is the dot and cross product I'd say learn how to do RREF first and how to use that to solve a system of equations and invert a matrix

learn how to take a determinant, maybe learn a few examples with polynomial spaces, and learn how to get eigenvalues/eigenvectors to diagonalize a matrix

idk, depends on what 'proof based' really means here, but going into that with 0 knowledge sounds possibly bad to me

all this sorta stuff is covered in Lay, I second the recommendation

[faint lintel]

Ok cool

Yea that was my guess as well get a good computation background

I'll grab that book

[rapid prism]

so i'm taking like an applied linear algebra class (it's supposed to serve also as an introduction to linear algebra)

however i'm thinking of supplementing my learning with a more theoretical version, especially since this is my first time doing LA? any recs?

[native rampart]

Friedberg

[rapid prism]

how does Friedberg compare to Strang/Axler?

[native rampart]

It looks good

A lot better than Strang

Strang is just computations

Not sure about axler

[wintry steppe]

I have a question that i hope a seasoned linear algebra master could easily answer

So in my book m = rows and n = columns right. Then why is a vector with multiple rows and 1 column notated as a nx1 vector ? Wouldn't that be a 1xm vector? In my opinion that makes no sense. Why bother with notation if you're not going to use it correctly? Or am I missing something

[native rampart]

a x b usually refers to a rows and b columns

[wintry steppe]

right, then why wouldn't it be mx1 vector

m rows , 1 column

[native rampart]

[wintry steppe]

LOLLLLLLLLLLLLLLLLLLLLLLLLLLLLL

I have a feeling its one of those things where the mathematicians realized it made no sense later on but made everyone do it anyway because fuck it

[limber sierra]

there are a bunch of reasons why $n$ might be used instead, like if youre studying a map $\bR^m \to \bR^n$ and the resulting vector is from the image

[stoic python]

Namington

[limber sierra]

then itll certainly be n by 1

the symbol used doesnt really matter, what matters is what comes first and what comes second

[wintry steppe]

Then they should say that, note it , explain that m and n are just any representational variable and it should really be called a rowx1 vector not explicitly a nx1 vector

[limber sierra]

i've seen some conventions default to mxn, and some default to nxm

[wintry steppe]

Because they didn't explain that at all in the book and all the sudden its nx1 and 1xn which is really confusing when they just said that n is columns and m is rows

ya know what I mean

[limber sierra]

i mean, i cant see the book, but this seems more like a misunderstanding of how variables work than anything

[wintry steppe]

Bruh

If i just told you x =5 and then all the sudden y = 5 thats pretty confusing

That would not fly in programming

[limber sierra]

sure it would

x = 5 y = 5

[native rampart]

It 100% will

[limber sierra]

will compile in any langauge

(where equality is assignment)

[wintry steppe]

Nah , you're not understanding

lol

[native rampart]

Are you trying to construct a matrix?

[gleaming knot]

5 = x compiles in math but in no programming language

Boo programming languages

[wintry steppe]

If you have a macro defined as X = 5 and later on in the program it suddenly became Y = 5 something is wrong and im pretty sure thats not even possible unless you declared a new variable in memory denoted Y

[limber sierra]

anyway im not exactly sure what your textbook looks like

but it sounds like you saw a definition that used the variables m and n

and it said "where m denotes the number of rows and n denotes the number of columns"

[native rampart]

Think of a function prototype
Matrix a(int i,int j);

[limber sierra]

but what we call variables doesnt really matter

what symbols we use is irrelevant

there are some conventions

but the variables arent the part of the definition thats important

when the definition was (presumably) referring to m and n, it means in the scope of the specific definition

to give a programming analogy:

[wintry steppe]

But it does matter when you define certian areas of math. Try doing Trig but changing cos to sin and changing sin to cos. Thats going to get pretty confusing real fast. Sure they're just names we could have called it " big boobies" if we wanted but those names have mathematical meaning

[limber sierra]

func1()
    x = 5
    y = 6
    return x + y

func2()
    x = 6
    y = 5
    return x + y

theres nothing wrong with this

[well, besides the fact that these are stupid functions]

in one scope, x and y mean one thing

in another scope, x and y mean something else

similarly, in the scope of the definition, m meant the number of rows and n the number of columns

but when youre doing your work, the specific variable used might not matter - or it might be context sensitive

[wintry steppe]

Right, its not that it changes the math. But why change the scope of the naming convention? That just adds unnecessary confusion

[limber sierra]

for example, if $T$ was a linear map from $\bR^n \to \bR^m$, then we'd call elements of $T$'s domain $n \times 1$ while elements of $T$'s image are $m \times 1$

[stoic python]

Namington

[native rampart]

for(int i=0;i<6272;i++)
cout<<i;

vs

for(int j=0;j<6272;j++)
cout<<j;

[limber sierra]

ew

[native rampart]

Do you complain the second loop is different?

[limber sierra]

no im complaining about the fact that this seems to be C code but theres neither curly braces nor indentation

😠

[native rampart]

You don't need curly braces for one line blocks

[wintry steppe]

I don't know anymore. I mean if you guys are right its just variable names could be yx1 it doesn't matter. Just pretty shitty for a new person for them to change it randomly and not explain they just decided to name it nx1 and 1xn instead of mx1 like it should be

[limber sierra]

oh true

well i mean

if n is supposed to be the same

between n x 1 and 1 x n

theres not really anything that can be done?

i mean we could introduce a new variable m defined by m = n

i guess

but thats just an extra step for probably less clarity in the end

[wintry steppe]

It should be mx1 and 1xn technically based on their original definition of what rows and columns are

[limber sierra]

for example, if $A$ is an $1 \times n$ matrix, then $A^T$ is $n \times 1$

[wintry steppe]

they just decided not to use m for some random reason

[stoic python]

Namington

[limber sierra]

if we wrote m x 1 instead, that would no longer be correct unless we explicitly say m = n

[wintry steppe]

wut

[limber sierra]

which is just extra work for no gain

and a loss in clarity

[wintry steppe]

m is rows

m rows and 1 column

makes perfect sense

[limber sierra]

but... what is m?

in the above case

[wintry steppe]

the one you defined?

or the book

[limber sierra]

in my above example

[wintry steppe]

Yes that makes sense the way you defined it

A^t is just the reciprocal

I'm not talking about the way you defined it though

This does answer my question though

that basically, it doesn't matter and the person who wrote the book was just lazy and didn't care to explain that its rowsxcolumns and n represents either or

So I just realized they do explain it but not till after they already confused me later on in the chapter 😂 🙄

Then after numerous confusing examples and more definitions

THEY EXPLAIN IT JUST LIKE YOU DID

thank you 😂 @limber sierra apparently thats for nxn vectors only thats why I was so confused

[supple saddle]

Suppose $v$ is a norm on $R^n$, and A is an nxn matrix. How can I show that $v(Ax) $defines another norm on $R^n$

[stoic python]

Problematic

[soft burrow]

it doesn't unless $\ker A={0}$.

[stoic python]

derivada.schwarziana

[soft burrow]

assuming that as an additional hypothesis, it's easy to verify the axioms using the linearity of A

that is, A(x+y)=Ax+Ay for vectors x,y and A(ax)=aAx for an scalar a

ker(A)={0} is used to check that v(Ax)=0 iff x=0

[celest jetty]

im confused about how to get from blue graph to red graph

https://gyazo.com/db62a30f82ecbc04d441ba924b136ae8

[limber sierra]

this doesnt look like linear algebra.

[quartz compass]

looks like they're taking your blue graph and only looking at the positive x direction so they're putting f(|x|) then translating to the left by 1

[hollow finch]

If A has orthogonal but not necessarily orthonormal columns, will QA have orthogonal columns (where Q is an orthogonal matrix)?
What about AQ?

[wintry steppe]

if A has orthogonal...

rows?

[stiff frost]

If A has [orthogonal but not necessarily orthonormal] columns,

[wintry steppe]

ah mb

misread, thought it said orthogonal twice

QA has orthogonal columns if (QA)^T(QA) is diagonal, since each entry of the matrix product is the dot product of a row of (QA)^T with a column of QA

and the rows of (QA)^T are the columns of QA

now, writing this out...

[hollow finch]

Orthogonal matrix as in orthonormal columns. Q^TQ=QQ^T=I

Also that product would be I iff QA has orthonormal columns. Otherwise it'll just be diagonal

[wintry steppe]

ree

[hollow finch]

But the principle still stands

[wintry steppe]

ya

i'm too used to just normalizing everything

but then, yes, this simplifies to A^T A, and since the columns of A are orthogonal, this will be diagonal

as for AQ, if the columns of A are orthogonal, (AQ)^T(AQ) = Q^T A^T A Q = Q^T (diagonal matrix) Q, and it might get funky probably

[hollow finch]

So we know A^TA is diagonal. If we do (QA)^T(QA) we get A^TQ^TQA=A^TA which is again diagonal meaning the columns are indeed orthogonal got it.

Qx dot Qy= x dot y too so that makes sense as well since the columns of QA are Q times each column of A preserving orthogonality.

Hm yeah what happens with Q^TDQ

[wintry steppe]

it's nice if D is a scalar matrix, but if not, it could be kinda bad

let me think

[hollow finch]

It should be

$$\sum d_i;\vec{q}_i\vec{q}_i^T$$

[stoic python]

nix

[wintry steppe]

i believe so

that's not that bad

[hollow finch]

If all the d_i are equal it equals a scalar matrix so still diagonal but if they aren't the same no idea what'll happen

[wintry steppe]

maybe, some kind of "weighted sum" interpretation is possible

i don't know, just throwing shit out there lol

[hollow finch]

Ooh I like that

The $ij$ entry of the product $Q^TDQ$ will be

$q_i^TDq_j$

[stoic python]

nix

[hollow finch]

So a weighted sum inner product type thing

I don't see a reason for it to necessarily be zero if i≠j.

[tame mural]

I have a potentially bad proof for the independence of polynomials

$∑_{i=0}^n σ_i x^i$

[stoic python]

meow

[tame mural]

I want to know if a polynomial of some degree n must have 0 for all coefficients if it is the zero polynomial

[native rampart]

Yes,a polynomial is zero polynomial iff all coefficients are 0

[tame mural]

well one potentially bad proof is to say that

groups have unique identity

and since you have identified one zero polynomial

that is the unique one

is that bad?

another idea is to induct on n

[native rampart]

I don't think that's a bad proof

In the space of formal polynomials,there is a unique zero polynomial

[tame mural]

woots thx

[crisp dawn]

i don’t think that’s complete

You have to show that all polynomials with differeing coefficients correspond to different functions

Which is not very hard to show

[tame mural]

ah

hmm

[native rampart]

How do you do that?

Without taking $\sum{a_i x^i}=0 \implies a_i=0$ as an axiom

[stoic python]

MoonBears-D-

[crisp dawn]

Well tbh this just avoids the whole use of “group identity is unique”

You want to show nonzero coefficients==> nonzero function

i.e , given the set of coefficients, find some x s.t its nonzero

[native rampart]

Actually,you need to show nonzero coefficients==> Never zero polynomial

[crisp dawn]

You can use FTA but that might be circular

Never zero as in not vanishing for any x? Because i don’t agree with that

[native rampart]

Never zero as in if f(x) is a polynomial,there is a b such that f(b) is nonzero

[crisp dawn]

Yeah thats what i meant

You can just take x large enough

For any x above some value, it has to be nonzero, because the magnitude of the last term is larger than the magnitude of all earlier terms

[native rampart]

We are talking formal polynomials in general

[crisp dawn]

*sum of the magnitude

Really? they said degree n

[native rampart]

So,You may not have a notion of magnitude or partial order

[crisp dawn]

If the question was about formal polynomials, then I’m not sure it would make sense, cause there’s power series with radius of convergence 0

(Or, if we’re dealing with formal polynomials, then we’re identifying a polynomial with its sequence of coeffients, so its pretty much axiomatic)

[gleaming knot]

x^q - x over F_q says hi

[vocal isle]

I'm trying to solve a multi DOF damped system for U. [M] and [K] are symmetric nxn matrices, but [C] is not symmetric. Can I use the following approach?

Put in block matrix form
Solve the 1st order differential using exponential matrices
Is it true that because [C] is not symmetrical then it can't have distinct eigenvalues/eigenvectors, making it not possible to diagonalize? But even if that's true, e[A]t can always be found without diagonalizing, right?

[true tiger]

Hey, if i have line given with a system

How can i get the line equation from ti?

[tawny tulip]

@true tiger it will be easier if you send the question

[true tiger]

Like {2x + 3y + z = 0; x + y = 0}

[tawny tulip]

or an axample

[true tiger]

Something like this

Into one equation for a line

Should be something simple, but i can't remember

It's a line {2x + 3y + z = 0; x + y = 0} for exmaple

given with 2 systems

[tawny tulip]

are you familiar with Gaussian elimination?

[true tiger]

Yes

So?

@tawny tulip ?

[tawny tulip]

solve your 2 equations

[true tiger]

I'll get x y z

Id on't need the intersection point

I need the line equation

Or I can just solve it and use it as equation?

[tawny tulip]

I don't understand your question

can you explain? or give me an example you already solved

[true tiger]

Find distance between LINE:{2x + 3y + z = 0; x + y = 0} and PANE:{some random equation}

The line is given with a system

it should be just one line equation

[tawny tulip]

From what i remember this is not true, you don't find distance this way

[true tiger]

I just need...

A line

from the system

Some kind of equation

I forgot how you get it

{2x + 3y + z = 0; x + y = 0} can be written in like a normalequation or something

ah..

f me..

Why do I need equaton if point is sufficient for me anyways

ahh -.-

I just get the intersection point and do point to plane distance formula

[crimson pelican]

I have a question which is type true false

"Let A be an m × n matrix. If N(A)=0 (the null space of A) then A is invertible"

is this true ?

<@&286206848099549185>

[wintry steppe]

you've been here for how long and you're still pinging helpers immediately

[real stirrup]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[wintry steppe]

it's true if m = n.

[crimson pelican]

oops sorry

[wintry steppe]

but if m != n

can you even talk about invertibility?

are you @crimson pelican ?

regardless

yes

this is false in general, as you need m = n for it to follow

[crimson pelican]

we can take invertible even if not a square matrix

[wintry steppe]

no.

make sense of AA^{-1} = A^{-1}A = I for me when A is not square

[marble lance]

No, they are polynomial's alt

[wintry steppe]

been a while since i heard that name

invertible means has a left inverse and a right inverse

[crimson pelican]

my fault

but we can find left inverse right ? even if not square matrix ?

[wintry steppe]

depends on the matrix

if the matrix is injective then it's going to have a left inverse (compare with "a function is injective if and only if it has a left inverse")

i believe

but it need not have a right inverse

[tame mural]

hmmm i see

interesting

[wintry steppe]

see if that's true for me, i have a lecture starting in 3 minutes

[tame mural]

A matrix with full row rank has a right inverse

A matrix with full column rank has a left inverse @wintry steppe

[wintry steppe]

ya, and full column rank is equivalent to injectivity, by rank-nullity

okay, lecture starts

and full row rank is the same thing as injectivity of A^T, i.e. surjectivity of A (just use the properties of the dual map )

cool

thx for confirming my suspicions for me

[limber sierra]

please don't ping helpers before 15 minutes, as per #❓how-to-get-help

anyway, no, it is false

[tawny tulip]

if A is mxn then $\rho(A) \le min(m,n)$

[limber sierra]

easy counterexample: consider $\begin{pmatrix}1&0\1&0\end{pmatrix}$. then the row space is just the span of $\begin{pmatrix}1\0\end{pmatrix}$, which certainly does not span $\bR^2$

[stoic python]

Nyrre

Namington

[limber sierra]

m = n

if you really want them to be distinct, though, you could use $\begin{pmatrix}1&0\1&0\1&0\end{pmatrix}$ instead

[stoic python]

Namington

[limber sierra]

same idea

[crimson pelican]

i think it is true @sharp nest

[limber sierra]

you're misreading.

note that it says _sub_space

ie it may not be the whole R^n

and as my counterexamples show, there are situations where it is not the whole R^n

(this is related to the notion of rank btw)

[crimson pelican]

oh, you are right

mybad

[quasi frigate]

Hello, can somebody please check my solution?

[trail dirge]

@quasi frigate I will if I know the stuff

[quasi frigate]

If you want, that'd be great

[trail dirge]

can you plug in the values in the equation?

[quasi frigate]

Silly me, I just thought about that when I sent the message

[trail dirge]

but that's a weird technique tho

I mean I felt uncomfortable, how you were getting the echeleon form, don't we usually do it from left to right?

[acoustic path]

maybe hes from a country that reads right to left

[quasi frigate]

Well, I did not calculate that correctly

I'm using method to have only one 1 in column

idk how it's called in English

[acoustic path]

yeah gaussian elimination

[trail dirge]

@quasi frigate can you send the original equatiton?

[quasi frigate]

I was told to do that with method where you have a matrix with only values 1 in columns

[trail dirge]

Gauss Elimination

reducing to echeleon forms

check web for some guides

[quasi frigate]

Yea I’m gonna do that

[trail dirge]

@quasi frigate answer is x=-1,y=0,z=1, t=2

[native rampart]

Maybe

[trail dirge]

if v1,v2,v3 are not null spaces

I am kinda confused rn, but are those linearly independent?

I think I am leaning more towards truw

[wintry sphinx]

what? what does "no null spaces" mean?

that statement is completely false

[quasi frigate]

thanks, I got the same answers.

[wintry sphinx]

See if you can try a few examples

also, you should probably re-read and review the section you're going over

[trail dirge]

Sorry

[heavy thistle]

Let V be a vector space, and let v1,v2,v3∈V. Then any vector in the subspace W=Span(v1,v2,v3) can be written uniquely as a linear combination of v1,v2,v3. is it true?

[wintry sphinx]

nope

not uniquely

[heavy thistle]

why?

can u explane

[wintry sphinx]

why do you think a vector can be written uniquely?

[hollow finch]

Only if the vectors are linearly independent

[heavy thistle]

I felt incomplete by definition. thanks for your answers

[hollow finch]

You can prove this by contradiction if you suppose

$c_1v_1+\ldots+c_nv_n=k_1v_1+\ldots+k_nv_n$

Where $c_i\neq k_i$.

[stoic python]

nix

[exotic quail]

@quasi frigate this stuff used to give me nightmares...professor shoved it down our throats in Calc 3 and said we’d love him once we got to Linear Algebra and Differential Equations.

[wary lily]

and?

do you love him?

[quasi frigate]

yea tell me about it, in 2 weeks I got an exam from 2 math subjects at uni

[exotic quail]

Yes for being incredibly well prepared..No for the night terrors

[quasi frigate]

meh now I'm used to falling asleep at 5a.m. 🙂

[exotic quail]

😁

Sleeps the prize for correct calculations

[wary lily]

I used to read math and algorithm books to fall asleep

now, they keep me awake

it's insane

[quasi frigate]

you know how people tell you, good grades + sleep = no friends, friends + sleep = no good grades, good grades + friends = no sleep

I don't even have time for friends and sleep while studying CS wtff

24/7 working on projects and stuff

[exotic quail]

I switched my Major from CS this semester actually

[quasi frigate]

what do you study rn?

[exotic quail]

I’m studying pure Maths right now

Went in thinking I’d want to be a software designer and found the math to be far more captivating

[quasi frigate]

yeaa I got you

I'm not that much into strictly software engineering, but at my uni I have opportunity to study AI and robotic autonomy, which are a part of my course

[exotic quail]

That sounds a lot more exciting than my schools CS program!

[quasi frigate]

Well yea, It's a new program that started a year ago.

[faint lintel]

Ok lemme know if this is a question I should ask after actually getting into linear algebra

So I learned that the kernal of a function f is the equivalence relation such that x_1 ker(f) x_2 <=> f(x_1) = f(x_2)

so like 0 ker(sin) 2pi for example

I'm watching the 3b1b linear algebra series

[limber sierra]

thats a weird way to define kernel, usually the kernel is a set and then the equivalence relation youre referring to is called "quotienting by the kernel"

[faint lintel]

and he says that the null space, which is the set of all vectors that get mapped to the zero vector after a transformation, is also called the kernal

hmm maybe I'm misremember the definition

[limber sierra]

yes, thats the usual definition (for linear maps)

$x_1 \ker(f) x_2$ is unusual notation/language, but $x_1 \sim_{\ker(f)} x_2$ is seen sometimes

[stoic python]

Namington

[limber sierra]

huh, thats a bit nonstandard

[faint lintel]

Ok then what's the standard?

[limber sierra]

$\ker(f) = {x \mid f(x) = 0}$

[stoic python]

Namington

[faint lintel]

I see

so the kernel is the set of zeros of the function

[limber sierra]

and then the corresponding equivalence relation is $x_1 \sim x_2 \iff x_1 - x_2 \in \ker(f)$

[stoic python]

Namington

[faint lintel]

Ok ok I see

Ah so that's why the null space is called the kernal in this case

[limber sierra]

you can establish equivalence between this definition and the one you're using

[faint lintel]

cause you have the zero vectors rather than just zero

is that correct?

[limber sierra]

not sure what you mean by that?

its possible a linear map maps multiple distinct vectors to 0

for example, $T\begin{pmatrix}x_1\x_2\x_3\end{pmatrix} = \begin{pmatrix}x_1+x_2\0\0\end{pmatrix}$

[stoic python]

Namington

[limber sierra]

(0, 0, 0) and (0, 0, 5) are both in the kernel

as are (1, -1, 0) and (7, -7, 35918593458)

etc

[faint lintel]

So in the video he defines that the null space of a transformation is the set of all vectors that get mapped to the origin (aka become zero vectors) after the transformation

[limber sierra]

yes

the set {x | T(x) = 0 the zero vector}

[faint lintel]

and then he says that this is also called tke kernel

[limber sierra]

yes

[faint lintel]

that was my question, basically if that naming was related

ok cool

Also cool to see that other way of defining the kernel

[limber sierra]

anyway it seems the terminology your source is using follows a different convention

which i havent seen before

but theyre related in any case

[faint lintel]

which source, the video I'm talking about or the notes I posted

[limber sierra]

the notes

[faint lintel]

yea that's my teacher's notes he wrote up

I think I've asked a couple times about other concepts and people had similar remarks as to that was weird notation / weird definitions

[limber sierra]

yeah this one makes sense, like if someone said "the equivalence relation ker(f)"

i'd know exactly what they mean

but its still not really standard

[faint lintel]

gotcha

[limber sierra]

use it for your course though

[faint lintel]

Well I'm going to a new course in a couple weeks

new semester and all

so we shall see

[tame mural]

What kind of digital note taking system do you ya'll use?

I've been using markdown + latex + visual studio code

[rapid prism]

🤮

vim+latex+zathura #1

vs code is nice but after getting used to vim, i don't really use it anymore

[wintry steppe]

latex in vscode

that's it

works pretty well imo

[frosty vapor]

im gonna get an ipad soon...

i like the idea of sharing screen often to do math like this

it also means i don't have to lug my Gamer™️ laptop around

[tame mural]

A linear map can be between vector spaces over different fields, right?

$T: F_2^2 → F_3^2$

[stoic python]

meow

[wintry steppe]

no

T(cv) = cT(v). how do you make sense of c being in two different fields at once?

[tame mural]

because one is embedded in the other

same with (R, R) → C

[limber sierra]

in order for cv to make sense, c needs to be in the domain field

in order for cT(v) to make sense, c needs to be in the codomain field

[tame mural]

and it is

[wintry steppe]

ty nami

[tame mural]

such as my example, right above, right?

[limber sierra]

no?

the elements of F_2 and the elements of F_3 correspond to different things

they have differnet addition, multiplication, etc

just because they use the same symbol doesnt make them the same thing

[tame mural]

that's the same with R^2 → C

[limber sierra]

in fact, this is why youll sometimes see the notation $0_2$ or $0_3$ used when its ambiguous

[stoic python]

Namington

[tame mural]

or Q →R

[limber sierra]

sure, but R is a subfield of C

and Q a subfield of R

F_2 is not a subfield of F_3.

[tame mural]

Ok, then let's talk about Q → R

they are different fields

[limber sierra]

and yeah, we cant "naively" define a homomorphism in that case

but what we can do is "extend" Q into R

and then we can construct a linear map between our domain (now considered over R) and our codomain

alternatively, you could restrict the codomain to being a space over Q

[tame mural]

it's not in most definitions of linearity that I see

[limber sierra]

hm?

here's wikipedia's

which is canonical

wikipedia later specifies this

which is kinda what i was trying to get at

i'm curious what definitions dont state that (unless the course legitimately hasnt covered vector spaces over different fields yet)

[tame mural]

like in Axler

let me check other texts

[wintry steppe]

right before axler defines a linear map.

[tame mural]

but it doesn't say the same F, right

[wintry steppe]

[limber sierra]

theyre both over F the same field

when we say two things are the same variable

we mean they're the same

for example, if i said $3 + n = 2 \cdot n$, you'd assume both $n$s correspond to the same value

[stoic python]

Namington

[tame mural]

I see

[limber sierra]

same idea here; both Fs refer to the same field (interpreting that sentence as "V is over F and W is over F").

[hollow finch]

If $x$ is an eigenvector of $A$ associated with eigenvalue $\lambda$, where $A$ and $x$ are complex, what can we say about

$\overline{A}x$?

$\lambda$ is not necessarily complex but I'm not sure if that affects things anything.

[stoic python]

nix

nix

[hollow finch]

This is so sad

We can say that $\overline{A}\overline{x}=\overline{\lambda}\overline{x}$ though right?

[stoic python]

nix

[quartz compass]

yup