linear-algebra

[shy atlas]

@tiny grove

this property is actually the motivation behind matrix multiplication

and the reason its defined like that

[sonic osprey]

I mean, T is a map from V to W, and phi is a map from W to F so phi o T is a map from V to F, aka a map in V'

[shy atlas]

yup made sense after a lot of beating around the bush

@tiny grove what exactly are you confused about ?

we can talk in here cuz no latex in dms bro : /

[tiny grove]

gotcha

um so

using the thing you uploaded above

where S is the composition of standard matrices A and B

would the answer just be AB? 🤔

[shy atlas]

like do you understand what i wrote above ?

[tiny grove]

i guess not

[shy atlas]

ok so for every linear transformation T we can define a matrix for it M(T)

now im saying the matrix for T o S is the same as matrix of T * matrix of S

we actually define matrix multiplication so that it fits this property

[tiny grove]

wait, is T=A and S=B?

[shy atlas]

uh no

[cursive narwhal]

no

[tiny grove]

huh

[shy atlas]

T is a linear transformation and A is the matrix for it

[cursive narwhal]

linear transformations are not the same thing as their standard matrices

[shy atlas]

so M(T) = A

[tiny grove]

ohhh

and M(S)=B?

[shy atlas]

mhm

[tiny grove]

alright, i was confused

[shy atlas]

well i had to spend 3 days on this topic

so i was definitely more confused than u when i was learning this

[tiny grove]

i misunderstood the question

and i doubt it

im dogshit in this course

[shy atlas]

it'll get better

study hard

[tiny grove]

i wish

@shy atlas oh also

then would the range of T o S be span of each column?

[shy atlas]

yeah

[tiny grove]

alright thanks

[shy atlas]

span of the columns of M(T o S)

[gray dust]

wdym each column

[tiny grove]

what soap said

[gray dust]

beware your wording

[eager kestrel]

How is (1) a taylor approximation of F(x,h), it isn't really obvious to me. Where did the f(x+3h) go?

[eager burrow]

In a taylor expansion, you get the coefficients by taking derivatives and then evaluating the derivatives in zero

Hence, what happened is that when you set h = 0 in the derivatives, you don't see the x+3h or the x-2h anymore

(and actually this isn't liner algebra, this is calc/analysis)

[dreamy iron]

I’m trying to prove this result:

I have access to this lemma.

And I use the lemma in part 1.

It feels like a cheat. (The proof online doesn’t mention this lemma. And just kinda proves it from principles.)

That’s it right? It doesn’t need to be more technical than that, yeah?

[eager kestrel]

(and actually this isn't liner algebra, this is calc/analysis)
@eager burrow
The derivative with respect to h? You get d(x-2h)/dh = -2?

If so, that is OK. How about the differential approximation? Where did f^(3)(x)h^2 + O(h3) come from?

[eager burrow]

Awright, so what's happening is that they're writing down a taylor expansion for the enumerator of that big quotient

The terms of the taylor expansion of f(x - 2h) in h = 0, for example, are:
in order zero: f(x),
in order one: f'(x) * (-2) * h
in order two: 1/2 f''(x) * (-2)² * h²

All you have to do is write down the formula for the taylor expansion for the enumerator, and you'll get what you want

I don't think it's useful for me to go into the details of the formula, tho, you really just need to apply it. If I write it down it won't give you much insight

Then basically terms like f^(3)(x) * h² come from the third term of the taylor expansion, where you have (something) * f^(3)(x) * h³. Then you still need to divide by the h in the denominator, and then you get that thing

[gray dust]

@dreamy iron invoking LD lemma then saying "i know u is a linear combo of the v's" is too much of a jump. (v1,...,vn,u) is LD, so LD lemma says there is at least 1 vector in the list that's a linear combo of the others. there should be a step where you exhaust every vector before u so you can actually conclude u is a linear combo of the v's

[shy atlas]

ye just becuz ur set is LD doesnt mean every vector can written as a linear combo of the others

• said true by ann 2020

[gray dust]

ann doesn't say ur or becuz

except in satire

[dusky epoch]

yeah the shortening of "because" that i usually go for is "bc"

[shy atlas]

this is a revolutionary moment for me

bc is just 2 letters

how did i never notice that

[dusky epoch]

[tiny grove]

Is the second part of the problem possible?

here, A is 2x2 matrix

[dusky epoch]

you don't even need to know that A is 2 by 2

if you've done part 1 you can do part 2 easily

$A = E_1 E_2 \dots E_n$ gives $A^{-1} = E_n^{-1} E_{n-1}^{-1} \dots E_2^{-1} E_1^{-1}$

[stoic python]

Ann:

[tiny grove]

uhh

so i have 3 elementary matrices for A

for inverse of A, would it be:

but in different order (from right to left)

nvm ignore wat i said

i got it

thanks @dusky epoch

[shy atlas]

just wanna check my work

so for a i did

$T'(\varphi_1) = \varphi _1 \circ T = 4x + 5y + 6z$

[stoic python]

soαρ:

[shy atlas]

cuz phi_1 just picks out the first component of the output of T

same thing for T'(phi_2)

for (b) i did

$ T'(\varphi _1) = T'(\varphi _1)(e_1) \psi _1 + T'(\varphi _1)(e_2) \psi _2 + T'(\varphi _1)(e_3) \psi _3 \ = 4\psi_1 + 5\psi_2 + 6\psi_3$

[stoic python]

soαρ:

[shy atlas]

<@&286206848099549185> p i n g

[shy atlas]

[hoary agate]

bro soap

how do you keep jumping topics every day

[shy atlas]

@hoary agate wdym jumping

ive been sticking to lin alg for like 5 days now

[dreamy iron]

<@&286206848099549185> can @shy atlas can some help please.

[shy atlas]

[dreamy iron]

pretty please, with sugar on top.

It's from Axler's chapter 3F

(i gotchu fam, don't worry, we'll beat axler together)

[shy atlas]

LADR gang

[half ice]

Yes and yes

It's worth keeping in mind that T'(phi1) is a functional on R3. T' will "pull back" functionals

[shy atlas]

what does this "pull back" thing mean

[wintry steppe]

pull out, i think

[wintry steppe]

if you have a linear map $T : V \to W$, then given a linear functional $\phi$ on $W$, the map $T'(\phi) := \phi \circ T$ is called the pullback of $\phi$ by $T$

[stoic python]

TTerra:

[wintry steppe]

like it's "pulling back" the functional on $W$ to one on $V$

[stoic python]

TTerra:

[shy atlas]

oh i see why its called that now

[wintry steppe]

(i think that's the notation axler uses)

it's also called the dual map in some places

[shy atlas]

axler uses dual map

[wintry steppe]

yeah

[shy atlas]

nice thnx for clearing that up

i dont really get the purpose of this dual map thingy tho

like whats even the point

axler doesnt motivate it at all

[wintry steppe]

for example, if you learn about alternating tensors and differential forms, you'll be doing pullbacks a lot

that's one example i can think of, but thats a bit far off from axler

[shy atlas]

[wintry steppe]

haha

lemme try to think of a better example

hmm i cant really think of much, guess i have to brush up on my linalg

someone else can probably give a good example though

[shy atlas]

anyways thnx for the help tho

[wintry steppe]

do you know that if you pick bases $\beta, \gamma$ of $V,W$, then the matrix of $T$ wrt $\beta,\gamma$ is the transpose of the matrix of the dual map wrt the dual bases of $\gamma, \beta$?

[stoic python]

TTerra:

[shy atlas]

ive seen something like that but its on the very end of 3F

[wintry steppe]

then you can try thinking of vectors as column vectors and linear functionals as row vectors, and see if theres any connection there

idk that might not be helpful

but i feel like you could say something nice about the dual map using stuff like that

[shy atlas]

hmm probably

ill have to finish the section to get a better understanding to dual maps

lemme check

oh wait

its the dot product duality or soemthing

[wintry steppe]

oh, and if $V$ is an inner product space then there's a theorem that says that for each linear functional $\phi$ on $V$ there is a unique vector $v \in V$ such that $\phi(\cdot) = \langle \cdot, v \rangle$

[stoic python]

TTerra:

[wintry steppe]

but i dont think axler introduces inner products until later, so that might be useless

you might be able to find out some nice connection between the dual map and the inner product structure if you use that

@shy atlas

[shy atlas]

ooo ill look out for that when i get to inner product spaces

[wintry steppe]

yeah

i feel like theres definitely something to be said there

[stoic python]

slimvesus:

[wintry steppe]

maybe you can do some stuff with adjoints

[shy atlas]

ill have to check this out a couple of days later when i get to chapter 6

[stoic python]

slimvesus:

[wintry steppe]

sweet

ill check this out later, itll be a good LA review

[fair latch]

How would I go about solving this, I am completely clueless

[stoic python]

urbiin:

urbiin:

[fair latch]

wrong channel

sorry

[ocean sequoia]

So this seems kinda neat so i wanted to make sure I understand it correctly... Its saying that if the only solution to x1 + x2+...+x3 = 0 we know the vector space is injective which allows us to know its isomorphic as well?

however off the top of my head the only example I can think of that would be like Ax = x right? So we are basically just showing that a vector space is isomorphic to itself?

[ocean sequoia]

Thanks i get it

sorry my language is a bit imprecise but

'It doesn't make sense to say that a vector space is injective' makes sense why that doesnt make sense 😛

[ocean sequoia]

No of course not b/c the identity map will show that its surjective and injective with itself which means it must be invertible thus isomorphic

[silver bison]

Is the multiple choice option incorrect?

I was leaning either A or E

so I guess v cannot span R^3

Reading this theorem, not quite sure

I meant definition, that H is closed under vector addition, that is, for each u.v in H, u+v is in H

b. H is closed under scalar multiplication, that is, for each u in H and scalar c, cu is in H

and plugging in 0,correct?

we are using those properties, correct?

[ocean sequoia]

think about linear independence

[silver bison]

3x3 right?

or I guess just 3 vectors

no, let me go look that up

okay

@wintry steppe yes

I am trying to look at the notes

[ocean sequoia]

Oh sorry I wasnt trying to give the whole answer cause that doesn’t answer the HW question I’m sorry honestly

I’ve gotten a lot of help in this discord and I’m trying to get better at helping others

[silver bison]

t= a/6, t=b/0, t=-c/5

[ocean sequoia]

yea exactly now do you see something that doesnt seem right?

[silver bison]

t cant be equal to three different things

[ocean sequoia]

take a closer look

[silver bison]

how do I go about doing that?

yeah

yes

[ocean sequoia]

he needs to see it from

t= a/6, t=b/0, t=-c/5
@silver bison

[silver bison]

okay but aside from 6, 0, -5, is there something else in the matrix?

[ocean sequoia]

look at and tell me if that makes sense?

besides t trying to be 3 things what else is t trying to be

there is something that violates a math principle there

[silver bison]

we dont know what a, b, and c are

[ocean sequoia]

you are overthinking it

[silver bison]

0

[ocean sequoia]

Why?

[silver bison]

anything multiplied by 0 is 0

[ocean sequoia]

ok so what does that mean about the vector?

[silver bison]

it has the zero vector?

[ocean sequoia]

what is the zero vector?

[silver bison]

yes

I believe so

okay, maybe I need to go back and reread this

[ocean sequoia]

sometimes it helps

ok @silver bison what does span mean?

because to reiterate we are trying to figure out if this spans r^3 right

👍

[silver bison]

Im just going to go back and reread the section because I cant even being to answer that question, I dont want to waste your time

[ocean sequoia]

you arent

we are almost there

do you know what span means?

[silver bison]

its all the combinations of vectors

[ocean sequoia]

yea more or less so what does it mean for H in this case to span a space?

[silver bison]

Question, does R^3 mean three vectors?

or does it mean something else?

[ocean sequoia]

R^3 means 3D

so x,y,z

So in order for H to span R^3 it every vector in R^3 must be able to be described as a vector from H right?

[silver bison]

yes

[ocean sequoia]

ok but can you get every vector in R^3 from H?

[silver bison]

so its asking can you get (0,0,0), (1,1,1),(2,2,2) etc?

[ocean sequoia]

literally any vector in R^3

[silver bison]

gotcha, so no you cant get all vectors in R^3 from H

[ocean sequoia]

give me an example

how do you know?

[silver bison]

it involves t correct?

so you cant get (7,2, 9)

[ocean sequoia]

why not

try and solve that

[silver bison]

So maybe this is where the mix up is coming

to find a vector that isnt in R^3 from H

am I multiplying something with the 6,0,-5?

[ocean sequoia]

lets just keep it super simple

could you ever solve 0t = 2

[silver bison]

no

[ocean sequoia]

ok

then we know it cant span r^3

[silver bison]

because there is nothing that can be multiplied by 0 to get 2

[ocean sequoia]

exactly

[silver bison]

would i be correct to say you cant multiply 6 by anything to get 7?

[ocean sequoia]

6t= 7

what does t equal?

[silver bison]

there is nothing to get that

[ocean sequoia]

really? I want you to think about that

6x = 7

[silver bison]

oh so we can use fractions

7/6

[ocean sequoia]

Do you know what R vs Z vs N means?

[silver bison]

no

[ocean sequoia]

like R^n vs Z^n etc..

[silver bison]

doesnt seem familiar, no

[ocean sequoia]

ok so that means the Reals which is the set off all numbers from {-inf,inf} which is the space we are talking about here

[silver bison]

oh okay

[ocean sequoia]

and that contains fractions right

[silver bison]

yeah Real , Natural, Integer

okay

[ocean sequoia]

ok so we know A is wrong then right?

because v doesnt span R^3?

[silver bison]

yes

thanks!

I know its not B, C, D, or F

as well as A

so that leaves E

[ocean sequoia]

How do you know its not those

give me a reason

thats probably better...

[silver bison]

Its E because we found that for 0, there is nothing that it can be multiplied to get another y, therefore its does not span the vectors 6,0,-5

F is a false statement, you need to be closed under addition and zero vector

D is also false for the same reason

[ocean sequoia]

Its E because we found that for 0, there is nothing that it can be multiplied to get another y, therefore its does not span the vectors 6,0,-5
@silver bison I dont want to be a pendant here but it doesnt span R^3

[silver bison]

now I am confused lol

[ocean sequoia]

Why is A wrong

[silver bison]

it doesnt span the vectors of the form 6t, 0, -5t

other words it doesnt span H

[ocean sequoia]

oh uh no it doesnt span R^3

If V spans W then it means that every vector W can be written as a linear combination of V

so a vector would have to span itself because every vector is obviously some scalar of itself

so H cant span R^3 because there is no linear combination of H that could get you the vector (x,y,z) where y is not 0

[silver bison]

wait how is what I said different then?

[ocean sequoia]

because you said H doesnt span H

therefore its does not span the vectors 6,0,-5
@silver bison

[silver bison]

misused a word I guess but thats what I meant

[ocean sequoia]

ok but i wanted to make sure

[silver bison]

yeah thanks

[ocean sequoia]

trust me i get it im still working on making my language more precise but hey its math right? its necessary

ok lets keep going

[silver bison]

@ocean sequoia you mind if I add you?

[ocean sequoia]

generally speaking i try to avoid that it makes me a bit uncomfortable however im in this server and in this specific channel alot

[silver bison]

ok...

[ocean sequoia]

if you want we can keep discussing the problem?

[silver bison]

I got to head out, but I appreciate the help @ocean sequoia and @wintry steppe

[ocean sequoia]

NP

[gaunt briar]

$$$

$u$

[stoic python]

☯Koͫzͤzͫmͤic☯:

[hexed steeple]

Whats the fastest way to find all the eigenvalues of a given 3x3 matrix?

[wintry steppe]

sometimes you can find them easily by inspection (e.g if the matrix is triangular)

Doing many examples so you understand how to do them, is the fastest way

[ocean sequoia]

id say using a computer

[rigid edge]

<@&286206848099549185> anyone familiar with matrices ?

idk what to do for this Q

[zinc copper]

dont need matrices for this

are you familiar with how to derive the cartesian equation of a plane based on its normal vector?

(and a point on the plane ofc)

[rigid edge]

umm no

[zinc copper]

okay si

so*

[rigid edge]

alright

[zinc copper]

say your plane passes through a point <x0,y0,z0>

(this is the position vector of the point)

and say it has a normal vector <a,b,c>

Now if you take any general point on the plane, with position vector <x,y,z>

you will notice that the vector between your general point and the point the plane passes through must be perpendicular to the normal

get me so far?

[rigid edge]

umm nope

we havent even done anything like this

I think professor wants us to do the longer way

[zinc copper]

as in?

[rigid edge]

is there any other way to do this Q

im certain we arent to use normals and stuff

[zinc copper]

i mean

im sure there is

but i dont know if it's as intuitive

better to understand what your doing than just proceed through an algorithm

[rigid edge]

yhhh i guess

im kinda on a time limit

need to submit in 3 hrs last q is this

[zinc copper]

so do you want me to show you my way?

[rigid edge]

alright surte

[zinc copper]

k so

do you know what a normal is?

[rigid edge]

yh

[zinc copper]

k cool

so you agree that if I take any vector along a plane, that vector will be perpendicular to the normal of that plane right?

[rigid edge]

right

[zinc copper]

k so

the vector that goes between a general point <x,y,z> and the point on intersection <x0,y0,z0> is one such vector right?

so the vector <x-x0,y-y0,z-z0>

[rigid edge]

yeh

[zinc copper]

now how do you know when two vectors are perpendicular?

[rigid edge]

no

[zinc copper]

?

[cursive narwhal]

Stop giving one word answers to narwhal's questions

[zinc copper]

wait let him think

[rigid edge]

so the 1st vector is 90* to the 2nd?

[zinc copper]

yeah

how can you check that?

just with the components fo the vector?

of*

ill give you a hint it has to do with a certain product

[rigid edge]

im not sure

[zinc copper]

what products that have to do with vectors do you know of?

[rigid edge]

dot

cross

scalar

[zinc copper]

okay gimme its definition, the dot product

[rigid edge]

idk the definition

[zinc copper]

wdym

[rigid edge]

i know how to use the formula

[zinc copper]

yeah that's the definition

give me its formula

[rigid edge]

u * v
multiply the 1st by the 2nd corresponding to the index its at

[zinc copper]

alright

[rigid edge]

so 1,2,3 * 4,5,6

4+10+18

[zinc copper]

and there's another formula

that has to do with the angle between the vectors

[rigid edge]

lenght of u * length of v * costheta = u*v

[zinc copper]

yes

now what happens when theta is 90 degrees?

[rigid edge]

0

so dot product of uv=0 is to see if they are normal

[zinc copper]

yes perfect

so if you take the dot product you get a(x-x0)+b(y-y0)+c(z-z0)=0

expanding

ax + by + cz = ax0 + by0 + cz0

let d = ax0 + by0 + cz0

then ax + by + cz = d

that's the cartesian equation

and it's entirely dependent on the normal and point of intersection

you already have a point of intersection

you just need to find the normal

[cursive narwhal]

@rigid edge

[zinc copper]

yeah that works

[rigid edge]

when writing in the form it asks for do I need to write 0y

[zinc copper]

i personally like the normal interpretation but it works

[rigid edge]

i was kind of doing that method but I stuffed it up

[zinc copper]

no dont write y at all

[cursive narwhal]

Now, that last step requires a bit more explanation. To see why $d \neq 0$, suppose that it was equal to $0$. Then, there exist $s,t \in \bR$ such that the following equalities hold:

$$1+2s = 0$$

$$1-s+t = 0$$

$$2+3s = 0$$

From the above, you can see that $2+4s = (2+3s) + s = 0 + s = 0$. So, $s = 0$. The issue, then, is that $1+2s = 1 \neq 0$. So that contradicts what we have above. Hence, such an $s$ and $t$ do not exist. It follows, then, that $d \neq 0$. So, that division by $d$ in the last step is absolutely justified.

[stoic python]

Abhijeet Vats:

[zinc copper]

it's definitely a valid method, however if you're short on time and need a reproducible method then id go for the normal interpretation

because it can be boiled down to an algorithm

[rigid edge]

@cursive narwhal why is b = 0?

[cursive narwhal]

Otherwise, d would depend on t

t is a parameter that varies across the reals so if you want to make sure that d is constant, you have to force bt to be 0. Only way that's possible is to have b = 0

In the same way that I forced the coefficient of s to be 0 because it being nonzero would result in d varying and that isn't what we want

[rigid edge]

oh ok

[fickle agate]

just to check

if a matrix A is unitary

then A* = A^-1 right?

[dusky epoch]

that's the definition of a unitary matrix yes

first off why the hell are you pinging me

second you're definitely not using the proper notation

[wintry steppe]

T(x,y,z) = (x-y, y-2z, x-2z) what is ToT?

[dusky epoch]

okay, so what's giving you trouble here?

at what point do you hit a wall in simplifying T(T(x,y,z))?

what are you even talking about

$T\big( T(x,y,z) \big) = T(x-y, y-2z, x-2z) = \ = \big( (x-y)-(y-2z), (y-2z) - 2(x-2z), (x-y) - 2(x-2z) \big)$

was this not obvious or was your goal just to have someone else do the substitution for you?

[stoic python]

Ann:

[wintry steppe]

Thank you. Ann.

[tame mural]

I notice that for Fields you can go from any number to any number under +, and the same is true for * excluding 0

What's the name of this property?

Or is this just a consequence of unique inverses?

[cursive narwhal]

A field is closed under multiplication and addition

@tame mural

[tame mural]

And that's enough to go from "anywhere to anywhere"?

[cursive narwhal]

No. "Anywhere to anywhere" doesn't have a meaning.

If you take two elements from your field and add them, you are guaranteed to get an element from the field as a result of adding them. That's what closure under addition means.

[tame mural]

yes

Or whether "terminal / initial" points exist

I see

[cursive narwhal]

Or whether "terminal / initial" points exist

[tame mural]

so it is a matter of inverses

[cursive narwhal]

"terminal" and "initial" reminds me very much of vectors and geometry. A field is more general than that so this is kind of a weird thing to ask.

[tame mural]

well like for the naturals under succession, 0 is an initial element

I guess I'm using the terms too loosely

yup

but c = b - a is enough to answer that question I think

ah I see

thanks

[half forge]

question, is gaussian elimination known as row echelon form ?

[wintry steppe]

gaussian elimination is the procedure (steps) to put a matrix in row echelon form

[muted holly]

yh

[wintry steppe]

all my boys hate proof based linear algebra

yuck yuck

[cursive narwhal]

don't use such language

[wintry steppe]

im blasian

[cursive narwhal]

<@&268886789983436800>

makes no difference

[stone drum]

Don't use slurs chopsticks.

[wintry steppe]

ok

[ripe hamlet]

$\begin{pmatrix}
1&-1&0\
-1&3&-1\
0&-1&1
\end{pmatrix}
\begin{pmatrix}
1\
x_2\
1
\end{pmatrix}=\begin{pmatrix}0\1\0\end{pmatrix}$

[stoic python]

trece:

[ripe hamlet]

to solve for x_2 i just form the vector equation like this, right? :

$-1 +3x_2 -1 =1$

[stoic python]

trece:

[cursive narwhal]

I mean

[torn silo]

yeah but the first line gives you your solution

[cursive narwhal]

you could just do it with the first row

[torn silo]

🤔

[ripe hamlet]

hmm?

[cursive narwhal]

so 1-x_2+0*1 = 1-x_2 = 0

[torn silo]

1 - x = 0

[cursive narwhal]

so x_2 = 1

[ripe hamlet]

my brain is kinda dead rn so

[torn silo]

what is x?

[ripe hamlet]

wait x_2 isnt 1 tho i think

wdym what is x

[cursive narwhal]

x_2 is, indeed, 1

[ripe hamlet]

$\begin{pmatrix}
1&-1&0\
-1&3&-1\
0&-1&1
\end{pmatrix}
\begin{pmatrix}
x_1\
x_2\
x_3
\end{pmatrix}=\begin{pmatrix}0\1\0\end{pmatrix}$

[stoic python]

trece:

[ripe hamlet]

what if we had this case

if i find the determinant of this 3x3 matrix,

[cursive narwhal]

Well, it would be the same thing as solving a system of linear equations for x_1,x_2,x_3

[ripe hamlet]

and then by using cramer's rule

$x_1-x_2=0\
-x_1+3x_2-x_3=1\
0-x_2+x_3=0$

[stoic python]

trece:

[ripe hamlet]

so like this?

[cursive narwhal]

Yea so x_1 = x_2 and x_2 = x_3. So, from the second equation, we get that x_2 = 1. So, all of them are 1.

[ripe hamlet]

ic

somehow

im supposed to get -1/3 for x_2

and onl if i would have an equation like

$-1+3x_2-1=1$

[stoic python]

trece:

[ripe hamlet]

i would get that

so im

[cursive narwhal]

my dude

[torn silo]

wut

🤔

[cursive narwhal]

-1+3x_2-1 = 1 => x_2 = 1

I don't know what kind of arithmetic you're doing but it ain't arithmetic on N

[torn silo]

have you tried sleeping?

[ripe hamlet]

$-2 + 3x_2 = 1$

[stoic python]

trece:

[ripe hamlet]

add 2 on both sides

yep

im

fucked up

sorry

ill go sleep or smth

[cursive narwhal]

go and sleep

no one does math when they're half awake

except for when they have an assignment due in 20 minutes

[median marlin]

Hello! :)
One of my coworkers had suggested to me at one point that it's possible to compute Levenshtein distance using linear algebra operations. Is this true?

I remember very little from the linear algebra course I had taken years ago, but this idea sounded very interesting to me

[limber sierra]

wagner-fischer?

[median marlin]

yes, that's the algorithm. But I'd like to understand how you might adapt it to use linear algebra operations

[fickle agate]

thats what i thought

but

lets say my lin combination is

px + qy + rz

[half ice]

What is the inner product on your space? Or, what space are you working with?

[fickle agate]

im having trouble computing it. bc i dont understand how to find the complex conjugate of px + qy + rz

and then multiplying that with x

i think its this

if thats what you mean

[half ice]

That's not the wikipedia one! Haha

Still, we can use this

[fickle agate]

hm ok

what does that mean.

hold on ill jsut write down the full q

sum of complex conjugates of px, qy, and rz?

hmm i see

im just confused bc i dont have actual numbers

like here's an example they had with numbers

no

[half ice]

Wondering if this isn't one?

Ah ic

[fickle agate]

if that even makes any sense

i see

well

i kind of guessed the idea

of splitting it as

[stoic python]

slimvesus:

[fickle agate]

yeh ive seen that idea

oh yeh true thats it

makes sense

what are you trying to say when you say "theres prob some conjugtes there"

because set A contains nonzero vectors

so magnitude is never 0

ok

i think i got it now ty

[wintry steppe]

Is there a way to find the characteristic polynomial of a matrix

Given only the eigenvectors and their respective eigenvalues?

Because I'm having a very hard time figuring this out

[sonic osprey]

the eigenvalues are the roots of the characteristic polynomial

So your characteristic polynomial will be (x - a)(x - b)

for example, if your eigenvalues are a and b

[wintry steppe]

@sonic osprey Yeah

But the matrix is a 3x3 matrix

So would it be (lambda - a)(lambda-c)(lambda-d)?

[sonic osprey]

yeah

[wintry steppe]

Got it, thanks!

[naive skiff]

https://cdn.discordapp.com/attachments/725194763598561412/725194772515389490/unknown.png

[quartz compass]

$v = c_1B_1 + c_2 B_2$

[stoic python]

Merosity:

[quartz compass]

$A'v = c_1 3 B_1 + c_2 4 B_2$

[stoic python]

Merosity:

[quartz compass]

$v = \begin{bmatrix} c_1 \ c_2 \end{bmatrix} $

[stoic python]

Merosity:

[quartz compass]

$B_1 = \begin{bmatrix} 1 \ 0\end{bmatrix} $

[stoic python]

Merosity:

[quartz compass]

$A' B_1 = 3 B_1$

[stoic python]

Merosity:

[quartz compass]

https://prnt.sc/t5e0j0

$A' = \begin{bmatrix} a & b\ c & d \end{bmatrix}$

[stoic python]

Merosity:

[quartz compass]

$A' B_1 = 3 B_1$

[stoic python]

Merosity:

[quartz compass]

$B_1 = \begin{bmatrix} 1 \ 0\end{bmatrix} $

[stoic python]

Merosity:

[untold citrus]

hello can someone help me with my homework question

[pallid rampart]

What's Theorem DRMM on page 365 of Breezer

[untold citrus]

give me a minute

DRMM therom

Theorem DRMMDeterminant Respects Matrix Multiplication

[pallid rampart]

$\det(AB)=\det(A)\det(B)$?

[stoic python]

Whoever:

[untold citrus]

yes

[pallid rampart]

Ah ok

We know that $\det(AA\inv)=\det(I)$, where $I$ is the identity matrix

Right?

Actually

Not by the theorem

[stoic python]

Whoever:

[pallid rampart]

You see that right?

[untold citrus]

yes

[pallid rampart]

K

What is the determinant of I?

[untold citrus]

tbh not sure

[pallid rampart]

Well try it out

Try to find the determinant of the 2x2 and 3x3 identity matrix

[untold citrus]

isnt it getting only first row?

[pallid rampart]

?

Anyways I'll spoil it, the determinant of the identity matrix is 1

[untold citrus]

right

so because the determinant is 1 so makes I = 1

[pallid rampart]

??

I is not 1

But the determinant of I is 1

So that means $\det(AA\inv)=\det(I)=1$

[stoic python]

Whoever:

[pallid rampart]

But by the theorem, we know $\det(AA\inv)=\det(A)\det(A\inv)=1$

[stoic python]

Whoever:

[pallid rampart]

So now we use division

[quartz compass]

you know how to evaluate a determinant?

[untold citrus]

i do not know but my professor went over it and i still dont understand

[tame mural]

Is there a name for taking the absolute value of every scalar inside a vector?

For example, [-1, 0, 3] -> [1, 0, 3]

[limber sierra]

"entry" is a better term than "scalar" here

for technical reasons

but no, i'm not aware of a name for that

i dont think that operation carries any real meaning, especially since its interpretation changes upon a change of basis

[tame mural]

I see, thanks

[limber sierra]

that said, if you say "vector generated by the entry-wise absolute value" or whatever

people will understand what you mean

[tame mural]

I was looking for clean ways to write my notes and formulas

but I didn't want to stray too far from conventional notation

[wintry steppe]

Using algebra u can solve it from:
(5sqrt(34))/34 = 10/sqrt(136)
arccos(10/sqrt(136)) = theta
10/sqrt(136) = cos(theta)
10 = sqrt(136) * cos
dot prod = |a||b|cos = sqrt(...) * cos
dot prod = +/- 10
The rest is easy, tho idk if there is a way while treating theta as a constant?
And without knowing the dot product. Finding A and B

[crystal oracle]

Hi. I am self-studying Axler's Linear algebra done right. There is a problem I can't solve.

Suppose V is a finite dimensional real or complex inner product space. Consider the metric induced by the norm induced by the inner product. Prove that this metric is complete (i.e., every Cauchy sequence converges).

The problem is, the chapter on orthonormal bases is next, so I am supposed to prove this without using an orthonormal basis.

My idea is as follows: I am given an infinite sequence x_1, x_2, x_3, ... of points in V. Consider a basis b_1, ..., b_n of V. Let α^n_i be the i-th coordinate of x_n in this basis. I want to prove that if for some i, the sequence of real or complex numbers α^1_i, α^2_i, ... is not Cauchy, then the sequence of vectors x_1, x_2, ... is not Cauchy. But I don't know how, and non-orthonormality of the basis doesn't help.

[sonic osprey]

hm

It's a fact that there exists some real number $c > 0$ such that $$||a_1 b_1 + \cdots a_n b_n|| \geq c(|a_1| + \cdots + |a_n|)$$

[stoic python]

Zopherus:

[half ice]

@wintry steppe
Heyo, saw your question and can provide a pretty simple answer. There's a few reasons we treasure the dot product:

• Very easy to calculate. If I give you two vectors, you'll have the dot product with minimal effort.

• Very nice algebraic properties. It distributes over addition like a product should. You can pull scalar multiples out of it.

• Relates to the collinearity of the two vectors. For example, if the dot product of two vectors are zero, those two vectors are orthogonal. This stems from the equation a•b = |a||b|cosθ

[wintry steppe]

And cross product?

[tame mural]

Cross product is entirely different, I believe

Given two vectors, find the third vector which is orthogonal to both of them shooting out in a direction

[opal osprey]

hello everyone. I'm having trouble with a simple exercise. Is it possible to have an orthogonal linear transformation such that f(3, 1) = (0,5)?

what I did was for any u = (x, y),

u*(3,1) = f(x, y) * (0,5),

and for the usual dot product,

u*(3,1) = 5*f(y)

so f(y) = (3x+y)/5

and so I have some information about an hypothetical orthogonal linear transformation

but how do I get the transformation for the first coordinate of the f(x, y)?

something like x - 3y would work, but this is a shot in the dark

[opal osprey]

okay - a transformation T is orthogonal iff T(e1), ..., T(en) is an orthonormal basis

but even if (3,1) is part of a basis, (0,5) could never be part of an orthonormal basis, right?

[dusky epoch]

neither (3,1) nor (0,5) are unit vectors and so neither can be part of an orthoNORMAL basis

[opal osprey]

(yes! unit vectors. I was missing the term in english)

thanks, @dusky epoch

that's enough justification for the original question, too, right?

[dusky epoch]

i think you overthought it

by a long shot

orthogonal linear transformations preserve length, but (3,1) and (0,5) have different lengths (sqrt(10) and 5 respectively). so the answer is no, it's not possible to have an orthogonal map send (3,1) to (0,5).

[opal osprey]

hmm

I see your point, which obviously makes more sense

[crystal oracle]

@sonic osprey Can you give me a suggestion on how to prove that? It's elementary for 1D space and I think it's pretty easy to show for 2D space, but what to do for 3D or higher dimensional, I don't know.
I think the problematic case is when Re⟨b_1, b_2⟩>0, Re⟨b_1, b_3)<0, Re⟨b_2, b_3⟩>0

[opal osprey]

(thanks, ann!)

[quaint heart]

Lol Phi you can actually show that any norm on R^n gives you the same open sets

The fact that zoph is invoking relies on some analysis facts

In particular you need to talk about compactness

Basically the proof is suppose you have 2 norms on R^n. Consider the map from R^n - 0 which sends an element to the ratio of the two norms. If you restrict to the unit sphere with respect to your norm, you can see that the image of the map on R^n - 0 is exactly the image of the map on the sphere. Then because the sphere is compact you get an upper and lower bound

This tells you that there exists some b so that $b \geq \frac{||v||_1}{||v_2||_2}\geq 1/b$

Which gives you that $b||v||_2 \geq ||v||_1\geq ||v||_2/b $

[stoic python]

Liquid:

[quaint heart]

This is what zoph is stating

But you need to know about compactness to prove this

@crystal oracle

[stoic python]

Liquid:

[quaint heart]

@sonic osprey was this the result you were invoking?

[spice storm]

It says y=x_y -61/55 but I got 89/55

Ping me please

x_2

Whyi got i

[wintry steppe]

What r u supposed to do

[spice storm]

I got it

you have to use cramers rule

[crystal oracle]

@quaint heart Thanks. It's late where I am, I'll try to understand what you wrote tomorrow.

[quaint heart]

Basically the idea is that the image of a map from the unit sphere to R has to be a closed interval

[ocean sequoia]

dim U = dim range(T) = dim V

so dim U = dim range(T) = dim range (S) = dim(W) so ST would have to be in the space

im not sure actually if that makes sense

[pallid rampart]

This doesn't show that ST is invertible

What does invertible mean?

[ocean sequoia]

oh wait

im dumb i misread the question

[pallid rampart]

Don't delete

It's fine

Now read the question correctly and do the problem

[ocean sequoia]

Let u1,u2,...,un, v1...vn,w1...wn be basis for U,V, and W. T(u1,u2,...,un) = (c1v1, c2v2,...,cnvn) = V, T(V) = (c1w1,c2w2,...,cnwn) = W which should show that both for S and T U and W are both injective and surjective the null space of S and T is 0 injective and because its a basis it spans W/U/V so its surjective so its an isomorphism which means its invertible

[wintry steppe]

does it say anywhere that your vector spaces are finite dimensional?

[pallid rampart]

Also T(U) = (c1v1, c2v2,...,cnvn) doesn't really make sense

[ocean sequoia]

yea this is only for finite dimensional vector spaces sorry

[wintry steppe]

(you dont need finite dimensionality here btw)

[ocean sequoia]

sorry i meant T(u1...un) = (c1v1, c2v2,...,cnvn)

[pallid rampart]

What's u1...un?

[ocean sequoia]

the basis for U

im editing that in the answer

[pallid rampart]

You can't multiply vectors tho

[half ice]

That's the components, I imagine

[ocean sequoia]

huh i mean u1,u2,...,un

[pallid rampart]

Anyways you can just show that ST is injective and surjective

Directly

[half ice]

Which is a scary thing, assuming your vector space can be written in terms of components

[ocean sequoia]

even tho its finite dimensional

[half ice]

Oh haha okay dorp. It's in terms of a basis

[ocean sequoia]

yea

[half ice]

And with that absolute brain-blast I'm out

[ocean sequoia]

LMAO

how would you show its injective and surjective directly? does what i edited too work?

[pallid rampart]

Really not sure what you mean by T(u1,u2,...,un)

So I'll just write what I will do for the problem

[ocean sequoia]

the transformation on the basis of U that takes it to the basis of V

[half ice]

The vector (u1,u2,...,un) I assume with a predetermined basis

Yaya it is

[pallid rampart]

Suppose S(T(u))=S(T(v)), then by the injectivity of S we have that T(u)=T(v), then by the injectivity of T we have that u=v. Therefore S(T(u))=S(T(v)) implies u=v which shows that ST is injective. Let w be in W, then since S is surjective there is an element v in V such that S(v)=w. Since T is surjective there is an element u in U such that T(u)=v. Therefore w=S(v)=S(T(u)), so for every element w in W there is an element u in U such that S(T(u))=w, so ST is surjective

[wintry steppe]

the composition of bijective (resp. linear) functions is bijective (resp. linear)

basically

[half ice]

Oh I should have read above lol. I didn't understand. You're expressing the vector in a component form, but you aren't using that component form anywhere. You don't need to put all that extra notation down

[ocean sequoia]

oh i was trying to be as explicit as possible

im pretty new to proofs in general so

[pallid rampart]

Basically linear transformations make the set of vector spaces into a category

just joking

[wintry steppe]

categories
new to proofs

dont scare him :(

[half ice]

Yeah! Don't scare me!

But basically don't summon something you aren't going to use. You can call a general vector u

[ocean sequoia]

I love that dont summon

makes math sound so much more badass but in all seriousness that makes sense

if thats an actual technical term i love that

[half ice]

Haha Math = YuGiOh confirmed

But no I just made that up

[ocean sequoia]

thanks guys

[pallid rampart]

Do you understand what I wrote?

[ocean sequoia]

yea I do

the only thing id ask is how you know S(T(u) = S(T(v)

the surjective part is 100%

oh we are supposing it

derp

derp i get it

[wintry steppe]

one way to do it in finite dimensional spaces is: if we have a basis of U, then T(basis) is a basis of V b/c T is an isomorphism, so ST(basis) is a basis of W b/c S is an isomorphism

which is what i think you were initially trying to do

[ocean sequoia]

yea that is

so you say the transformation is an isomorphism?

not that U and V are an isomorphism?

[wintry steppe]

a bijective linear map is called an isomorphism

[ocean sequoia]

ok

[wintry steppe]

and U and V are said to be isomorphic

[ocean sequoia]

got it

one way to do it in finite dimensional spaces is: if we have a basis of U, then T(basis) is a basis of V b/c T is an isomorphism, so ST(basis) is a basis of W b/c S is an isomorphism
@wintry steppe so to put this in super simple english terms it means that we are showing there is a one to one relationship between the basis of U and W with regards to ST

[wintry steppe]

careful point but not important: usually "isomorphism" requires that your inverse also have the nice property (here, linearity), but the inverse of a linear map is automatically linear so it doesnt matter

i guess that's a good way to put it, yeah

[ocean sequoia]

I like your way and whoever better but i want to make sure that i read it correctly

[hazy gull]

if a m×(n=<m) set of vectors is linearly independant, will their span just be the columns of the identity matrix ?

[limber sierra]

what does "m*n set of vectors" mean

like the number of vectors in the set is equal to the product m*n?

that depends on the dimension of the space and the values of m, n

er actually wait

"span just be the columns of the identity matrix" doesn't make sense

if the columns of the identity matrix are in your span, the entire space must be in your span

[ocean sequoia]

if you have a square matrix then it the indentity matrix will span it

[limber sierra]

since the columns of the identity matrix are the standard basis vectors, which span the space

[ocean sequoia]

but not every set of independent vectors has the standard basis

[limber sierra]

yes but my point is

its impossible for a span to just be the standard basis

[ocean sequoia]

oh im not disagreeing

sorry was trying to tack on

[hazy gull]

mxn is the size of the matrix that it produces combined

What would be an example of independent vectors that dont span the standard basis

[limber sierra]

so you have n vectors

uh if you have less vectors than the dimension of your space

like if you have 2 vectors from R^3

those span part of the standard basis (indeed the span of these two vectors will contain at most 2 standard basis vectors)

but not the entirety

by the dimension-basis theorem

[hazy gull]

O I see what your saying

[ocean sequoia]

correct me if im wrong but yea or wouldnt like {[0,0,1],[1,1,0]} are independent but dont span a standard basis

[limber sierra]

yep, the span of that set does not contain the vectors [1, 0, 0] or [0, 1, 0]

anyway this all relates to, like, one of the most important statements of linear algebra:

if a set of vectors is linearly independent and spans the entire vector space, it is a basis

and the amount of vectors in every basis is fixed as the dimension of the space

(i.e. every basis is the same size)

so if you have less vectors than the dimension, then you cant possibly span the space

or, if you have more vectors than the dimension, then they cant possibly be linearly independent

[hazy gull]

thats why I was saying m*(n<=m) , but if n<m then it would span R^n

wait no

idk im too sleep rn to fix it.

I get it though

[unique sluice]

Is this correct ?

[fresh acorn]

guys ı am in exam now , ı need solution fast can you help me ?

[shy atlas]

<@&268886789983436800>

this dude asking for help during a test

[fresh acorn]

online homework

ı dont have time

10 minutes left

[wintry steppe]

interesting i's

[fresh acorn]

👍

[crystal oracle]

@sonic osprey @quaint heart Alright, I am dropping the exercise I asked a question about, because (1) I am gonna get orthonormal bases in then ext chapter anyway, and I think it'll be easy to prove using them; (2) Proofs without using them seem too complicated, and I don't really know stuff about metric spaces, so it's difficult to prove the desired statement and not accidentally use something that depends on it.

[wintry steppe]

does anyone know how they would approach these three?

[dusky epoch]

try writing out some configurations of leading 1's in a matrix of appropriate size

[wintry steppe]

yeah, i mean i'm thinking that it could be an infinite amount because you could put any number is the place holders without the 0 or 1

but i don't think that's right

[dusky epoch]

no, it's not an infinite amount

you only care about the leading 1s

[simple hornet]

hello im not sure where to put this actually

however this part mentions linear algebra so maybe this is the right place

im having trouble understanding this

specifically the last part, the polynomial part

Shouldn't the sequence continue like the first term?

why is this part here

shouldn't the rest just look something like this?

[balmy kelp]

I have a question about singular value decomposition

does SVD give me a change of basis matrix?

[wintry steppe]

this might help

https://math.stackexchange.com/questions/2739820/singular-value-decomposition-in-terms-of-change-of-basis

[soft geyser]

wait lmao I just had a similar question

so you want to find a set of basis vectors which are all orthogonal to each other and have unit lengths, and you also want one of these vectors to be in the (1,1,1,...) direction
that's what singular value decomposition does when applied to (1,1,1,...)``` my professor said this

but I am having a hard time understanding it

[crystal oracle]

@sonic osprey @quaint heart Regarding the question I asked earlier. I emailed the textbook's author asking what the intended solution is. He responded that he included the problem in that section by mistake, and that it's supposed to be in a later section. So the mystery is solved.

[quaint heart]

Oh okay

[wintry steppe]

ik that cross products arent communitive

so do u go left to right?

when performing the computation

[quartz compass]

hmm technically it's the fact that they're not associative that is the problem here

without parenthesis it's not really well defined, but some information about u,v,w might be enough to reason out the answer independent of order

[wintry steppe]

well all i got are vectors defined in r3

ik this is just a simple computation problem, but im like kinda confused here whether to to go left to right

or whether it doesnt exist

[quartz compass]

not enough information

[wintry steppe]

alright, because the question does ask state it doesnt exist or isnt well defined

and then give a reason

ill just say because of the associative property im assuming?

[quartz compass]

reread my first message yeah

[wintry steppe]

alright for sure

th

x

[quartz compass]

you should probably make sure you see why, come up with some examples for yourself to see that associativity doesn't hold

[wintry steppe]

ok thx for ur help brother

[quartz compass]

👍

[tame mural]

this is a noob question but

in linear algebra, is it generally presumed that the Field isn't the rationals

because otherwise some inner products wouldn't be closed

[torn silo]

which field?

[tame mural]

a lot of times texts are vague and just say things like

F^n

so it seems like they're talking about all fields

but then because some inner products can be irrational, it makes me think they're excluding rationals

is that a decent reading of texts?

[torn silo]

that sounds weird generally you have to specify the field

unless you generalize to all fields

in which case whatever you're proving has to work for all of them

[tame mural]

so you're saying most texts don't mean to be so general

[sonic osprey]

I'm not sure why you think inner products wouldn't be closed?

[torn silo]

🤔

you have to read carefully is what I'm saying, if they just state F it has to work for F

[sonic osprey]

Inner products are defined as being functions to the base field

by definition, it has to be closed

Like, if you take the standard inner product on Q^2, where Q are the rationals, then (x,y) \cdot (a,b) = ax + by is still a rational number

[tame mural]

ah my writing was unclear, but I found out that the definition of normed vector space assumes we're dealing with R or C

[sonic osprey]

This is often done so you can use analytical things

but most of finite linear algebra works over any field

[karmic oracle]

be very careful with inner products over finite fields

they are definitely my least favorite very basic thing to think about because all the definitions and theorems have slightly different hypotheses.

[untold citrus]

quick question asking the subspace solution is a subspace of R^4 and what is R^4?

i am bit confused

[cursive narwhal]

$\bR^4$ is the set of 4-tuples $(a,b,c,d)$, where $a,b,c,d$ are all real numbers.

[stoic python]

Abhijeet Vats:

[untold citrus]

thank you

finding the subspace of 4-tuples of all real numbers of a b c d in this case

because i had to find the null space and coefficient of a matrix and then it says find solution set is a subspace of R^4

oh i have to determine solution set so from my answer if it is a subspace of R^4

[cursive narwhal]

ngl, this might just be my 6am brain failing me but that made almost no sense to me

[untold citrus]

Determine whether the solution set is a subspace of R^4

that the question

how to get the subspace?

[cursive narwhal]

Are you struggling with the problem? How about showing the full question?

[untold citrus]

that full question

Determine whether the solution set is a subspace of R^4

how you get the subspace?

[muted holly]

solution set ...

for what

[cursive narwhal]

my dude

you mentioned a matrix

and some other shit

send a picture of the problem

[muted holly]

not for me this one lol

[untold citrus]

i got a

just need help b

[umbral smelt]

Have you found it's solution set?

Oh nvm

[cursive narwhal]

Okay, so you're supposed to determine if the solution set is a subspace of R^4. All you really need to do is to check 3 things:

1. Does 0 belong to the solution set?

2. If I take any two vectors in the solution set and add them, is that in the solution set?

3. If I take a scalar and multiplied it with a vector in the solution set, is the scaled vector in the solution set?

If any of them fail, then it's not a subspace

[untold citrus]

ok thank you but i forgot how to check them but i figure it out

[cursive narwhal]

Uh note that when I said 0, I meant (0,0,0,0). So, gotta check if that's in there.

[untold citrus]

the solution is my null space? or the matrix after turning it into rref

rref is

[ 1 -2 0 -1 2]
[0 0 1 3 1]
[0 0 0 0 0]

[zinc tapir]

Prove that the eigenvalues of an upper triangular matrixMare thediagonal entries ofM.

any hints

[oak crater]

apply the definition of the characteristic polynomial

the determinant of an upper triangular matrix is very easy to calculate

it's just the product of the diagonal entries

[karmic oracle]

you could also write down the eigenvectors

[soft geyser]

can somebody help me understand how to do weighted least squares when I have standard deviation

so if I have the problem y = Ax

do I just subtract standard devs from both sides?

[untold citrus]

@cursive narwhal can you please give me example how to check those 3 questions you provided

or can someone else please give me example to understand it because i am stuck

[nocturne oracle]

its ok, it happens sometimes @pallid rampart

[cursive narwhal]

or can someone else please give me example to understand it because i am stuck
@untold citrus uh yea sure I’ll give an example, give me about 20 minutes and I’ll type it out. Sorry, just woke up

[obsidian nest]

hey guys, do you know if there's an interface where I can graph some functions in 3 dimensions?

[dusky epoch]

geogebra perhaps?

[obsidian nest]

I just want to check if I'm calculating some integrals correctly

I'll check it out

[dusky epoch]

oh

uh

hm

[dire thunder]

for integrals wolfram better

[crystal oracle]

Is Riesz representation theorem somehow related to kernel methods
in machine learning (e.g. kernel support vector machine)?

Riesz representation theorem: Suppose V is a finite dimensional
inner product space over a field 𝔽∈{ℝ,ℂ}. Suppose φ:V→𝔽 is a
linear functional. Then ∃! u∈V such that ∀v∈V φ(v)=⟨v,u⟩.

I suspect that this theorem allows us to work with such vector
spaces, where it's too computationally heavy to store points, but
calculating their inner products is cheap. But I can't figure out
how exactly they are related. And I think the last time I tried
to understand the theory behind kernel methods, my googling
stopped on something called Reproducing kernel Hilbert space and
Mercer's theorem. I couldn't figure out what that is because I
don't know functional analysis.

[sonic osprey]

They do seem related yeah

[ocean sequoia]

so Im reading a paper on algorithmic fairness that mentions "To
solve this “reconstruction problem,” procedures
have been proposed such as pre-processing the
data to orthogonalize the explanatory variables
(“inputs”) or outcomes to race'"
why would you orthogonalize the explanatory variables? To have the dot products be zero? So certain neurons just give zero? Thus making it less likely that product affects the activation?

[torn silo]

basically remove it altogether

[ocean sequoia]

yep

ok thatmakes sense

[zinc tapir]

can someone briefly explain why the above is true

isn't P_2(R) degree at most 3

[karmic oracle]

isn't that what this says?

[zinc tapir]

ya but the polynomial is of deg 3 lol

[gray dust]

deg=<2

[karmic oracle]

ahh, I see where the confusion is, good!

[zinc tapir]

i might be confusing myself afterall

what about this one

isn't the second line already an element of P_1

[zinc tapir]

ehh

wait a min

so V is given as P(R)

should f(x)=ax+b

since its deg at most 1

[shy atlas]

which one are you doing again ?

[zinc tapir]

b

[shy atlas]

consider $x^2 \in \mathcal{P}_2 (\bR)$. whats $T(x^2)$ ?

[stoic python]

soαρ:

[zinc tapir]

x^3

[shy atlas]

does that have degree less than or equal to 2 ?

[zinc tapir]

but they gave us P(R) not P_2(R)

[slow scroll]

since its deg at most 1
P(R) is not P_1(R)

[zinc tapir]

oh

[shy atlas]

^

[zinc tapir]

holy shit

my bad guys

[slow scroll]

its all good

[teal topaz]

Suppose we have a matrix $A\in\mathcal{M}{n\times n}(\mathbb{R})$ is similar to a diagonal matrix and has two distinct real eigenvalues and thus two eigenspaces $E{\lambda_1}$ and $E_{\lambda_2}$. If $\lambda_1=0$ and hence $E_{\lambda_1}=\mathrm{Null}(A)$, does it follow that $E_{\lambda_2}=\mathrm{Row}(A)$?

[stoic python]

Stract:

[teal topaz]

Yeah I just realized it's false, found a counterexample. Was assuming some things were orthogonal that aren't

[wintry steppe]

what's the counterexample?

[teal topaz]

$$\begin{bmatrix}0&1\0&1\end{bmatrix}$$

[stoic python]

Stract:

[teal topaz]

the rowspace is spanned by (0,1) but the second eigenspace is spanned by (1,1)

[wintry steppe]

nice, thanks

[teal topaz]

however @wintry steppe I think I can guarantee that Null(A)\oplus Row(A)=R^n. Doesn't that just follow from fundamental theorem of linear algebra

it's for for all m x n matrices

since $\mathrm{Row}(A)^\perp=\mathrm{Null}(A)$

[stoic python]

Stract:

slimvesus:

[teal topaz]

huh? no I'm sure this is a true fact

yes

https://math.stackexchange.com/questions/29072/how-is-the-column-space-of-a-matrix-a-orthogonal-to-its-nullspace/933276 here's a proof (the title is wrong but the answerer corrects it and proves the right thing)

ahh, np lol

[upbeat blaze]

I feel like the solution set should be empty since you cannot have 0=3, even if I were to do row operations, like swapping 3rd and 4th rows, it'll still have 0=3.. Does anyone have any tips on how to solve this?

[dusky epoch]

you are correct in that the solution set is empty

[upbeat blaze]

Thanks :D

[upbeat blaze]

Am I wrong to assume that since A is in R2, the columns of A are e1,e2,e3 and the matrix went from a 3x3 to a 3x2 matrix? Or am I missing something? Sorry for asking two questions in a row. Essentially all I need to do here is combine the given e's into a matrix and that'll give me the needed A ?

[slow scroll]

im not quite sure what you mean by "went from a 3x3 to a 3x2 matrix" but basically you have
T = [T(e1) T(e2) T(e3)]. This is a 2x3 matrix.

[upbeat blaze]

oops I mixed up the rowxcolum thing, my bad

[opal osprey]

hello everyone.

having trouble with orthogonal linear transformations

I want to define, by 'a general expression', the reflection wrt 3x-2y = 0

I am following a previous solution of this problem and here's what was done:

take vectors u = (2, 3) and v = (-3, 2); calculate their reflections, which are u' = (2, 3) and v' = (3, -2) respectively

A linear transformation T from Rn to Rn is orthogonal iff the vectorsT(~e1),T(~e2),. . .,T(~en)form an orthonormal basis of Rn

so at this point, I know I have an orthogonal transformation, but I haven't got the matrix yet

[dusky epoch]

ok so essentially you have an eigenbasis for your transformation is what you're saying

if you want, you can calculate the matrix of your transformation as $$T = \mat{2 & -3 \ 3 & 2} \mat{1 & 0 \ 0 & -1} \mat{2 & -3 \ 3 & 2}^{-1}$$

[stoic python]

Ann:

[opal osprey]

ann, thank you so much! let me think a little

ok, this is not what I was doing at all.

why eigenbasis?

[dusky epoch]

well you've got Tu = u and Tv = -v

so u and v are eigenvectors for T with eigenvalues 1 and -1 resp

[opal osprey]

I'd like to understand your process, but also let me add that the solution I am following states that after getting f(2, 3) and f(-3, 2), you can apply the linearity of functions to get f(1,0) and f(0, 1) and find the general expression, or the matrix of f wrt the standard basis

[dusky epoch]

oh sure you can do that too

doesnt make much of a difference

yours is a lower-level solution in terms of abstractions

[opal osprey]

so in your method you're diagonalising the matrix?