linear-algebra

[wintry steppe]

first try to write vector (-4, 8) as linear combination of other two vector, (1, 3) and (3, 1)

[safe locust]

ok

done

solved for c1 and c2

now what

[wintry steppe]

ok so now just apply the linearity property

[safe locust]

wut ._.

[wintry steppe]

$T(\alpha v_1 + \beta v_2) = \alpha T(v_1) + \beta T(v_2)$

[stoic python]

η/rιμ

[safe locust]

so i have
-4 = c1(v1) + c2(v2)
8

what do i do?

c1=-7/2 c2=-5/2
v1= 1,3 v2= 3,1

[wintry steppe]

look at what i just wrote

[safe locust]

im confused ;-;

[wintry steppe]

$T(c_1 v_1 + c_2 v_2) = c_1 T(v_1) + c_2 T(v_2)$

[stoic python]

η/rιμ

[safe locust]

ok

so is that my final solution to this?

[wintry steppe]

What is T(v1)?

[safe locust]

1
3

[wintry steppe]

so then just substitute

T is transformation so I think it should be () brackets. So now just apply the linearity property

what i wrote above

[wintry steppe]

now apply this: #linear-algebra message

[safe locust]

what u mean?

im confused ;-;

[wintry steppe]

what are you confused about?

[safe locust]

what im supposed to do now?

[wintry steppe]

I just told you what to do, i said in the message "apply this" linking the next step

[safe locust]

did i not do it correctly in the picture?

like, idk how to apply that

[wintry steppe]

what is c1

[safe locust]

-7/2

[wintry steppe]

actually you sure its -7/2?

I think its supposed to be 7/2, not negative

check again

[safe locust]

I got -7/2

i checked again

just run w/ it

[wintry steppe]

what is v1

[safe locust]

1
3

This is my recent work

like, im not sure where to go from here

[wintry steppe]

c1 and c2 both cannot be negative, only 1 of them is negative. But besides that
$T(-\frac{7}{2} \begin{pmatrix} 1 \ 3 \end{pmatrix}^T) = -\frac{7}{2} T(\begin{pmatrix} 1 \ 3 \end{pmatrix}^T)$ This is by the second property of linearity

[stoic python]

η/rιμ

[wintry steppe]

you know what

just read this

https://math.libretexts.org/Bookshelves/Linear_Algebra/Book%3A_A_First_Course_in_Linear_Algebra_(Kuttler)/05%3A_Linear_Transformations/5.03%3A_Properties_of_Linear_Transformations

i don't think you understand what you are doing

if you don't understand now, you are going to have hard time later

so make sure you properly understand what you are doing

[safe locust]

Lol, ur not wrong.

My teacher sucks at explaining stuff

[wintry steppe]

ok

[prime drum]

It's not really teachers fault at the moment. Hybrid learning is not good for anyone.

[candid abyss]

hi i'm having a real hard time at quadratics right now

is this the right place?

[safe locust]

Could u help me solve it with 7/2 and ill go back and check that link after?

Due in a little

Sorry ;-;

[wintry steppe]

$T(-\frac{7}{2} \begin{pmatrix} 1 \ 3 \end{pmatrix}) = -\frac{7}{2} T(\begin{pmatrix} 1 \ 3 \end{pmatrix})$

[stoic python]

η/rιμ

[wintry steppe]

and you already know what T(1 3) is so just substitute that, then multiply it out

[safe locust]

Ok so i did it for both

[wintry steppe]

make sure that either c1 = 7/2, c2 = -5/2, OR c1 = -7/2, c2 = 5/2

now just add your two matrices

thats your final ans

[safe locust]

Oh

Ok

[safe locust]

Ty 😭😭

[wintry steppe]

what was your final ans?

[safe locust]

11 7
13 6

[wintry steppe]

check again

u made a mistake

[safe locust]

Wym

How

i used positive 7/2

not negative

[wintry steppe]

what was c2?

[safe locust]

-5/2

[wintry steppe]

ok maybe its correct then idk, i havent actually solved it

but just check your work again

[safe locust]

its ok

i submitted

https://tenor.com/view/dog-hug-bff-best-friend-friend-gif-9512793

@wintry steppe

[wintry steppe]

nice

[safe locust]

ty for the help

[wintry steppe]

How do I do this

the symbol just means max value in the vector

[plain saffron]

From

#❓how-to-get-help

Rule 3

Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[wintry steppe]

hey quick question
if i have something like A = DB + DC
can i move D to the left side by multiplying by inverse on the left?
all matrices

[dire thunder]

assuming D is invertible

[wintry steppe]

its a diagonal matrix so sweet

[native rampart]

Diagonal matrices can be non invertible tho

[dire thunder]

yes but have you considered e.g zero matrix

it is diagonal

[lavish jewel]

as people said several times already

[verbal vessel]

Suppose A and B are matrices such that
rank(AB) = rank(A)
show that if $x_1AB = x_2AB$ , then $x_1A=x_2A$

[stoic python]

Alireza

[verbal vessel]

I have to show B is invertible but I don't know how... can I have a hint?

[dusky epoch]

rank(AB) ≤ min(rank(A), rank(B))

[fervent elm]

are you able to get the determinant of an 8x8 matrix from their sub 4x4 matrix determinants

[dusky epoch]

not in general no

[lapis cosmos]

Not true

For example A=diag {0,I_n*n}, B_ij=1 if i=j+1 ,0 otherwise .

You can directly prove that N(A)=N(AB) (the former is a linear subspace of the latter, and they have the same dimension )

[wild pilot]

Can someone explain why MATLAB is giving me apparently spurious complex numbers when producing eigenvalues for the following matrix A =

 5     2
-2     1

eig(A)

ans =

3.0000 + 0.0000i
3.0000 - 0.0000i

[dire thunder]

,w eigenvalues {{5,2},{-2,1}}

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Commander Vimes#4407

[dire thunder]

aight

@wild pilot ig it just solves det (A - lambda I) over C

and gets 3+i0 and 3-i0 as two roots

[wild pilot]

It's probably just one of these things that distracts me unnecessarily, but I can even find these eigenvalues in my head - it's just x^2- 6x +9 where x is lambda resulting in eigenvalues of 3, multiplicity of 2 - and I can't work out how the issue of complex numbers is enlivened here at all.

[quartz compass]

well matlab isn't giving you the wrong answer

maybe it just happens to be when there's a double root it does some method that involves complex numbers or something

I don't think there's anything to worry about, it just has repeated eigenvalues

[wild pilot]

like when you add 2 and 2 and a computer says 4.000000000001 I guess due to some binary addition magic

[dusky epoch]

adding 2.0 and 2.0 will always give 4 even with floating point arithmetic

0.1 + 0.2 = 0.30000000000000004 though, because of rounding errors since computers use binary scientific notation rather than decimal

[novel hamlet]

Anyone know online calculator that gives ste by step solution on spectral decomposition

[lavish jewel]

i don't think any calculator uses a method you would do by hand

[novel hamlet]

tbh thats not a problem

[lavish jewel]

https://en.wikipedia.org/wiki/Givens_rotation

you would do that several times

[hoary osprey]

yea its just the cartesian product of V with itself n times

[uneven shore]

Are parallel vectors always opposite or equal vectors?

<@&286206848099549185>

[wintry steppe]

no

[uneven shore]

can you explain

pleaseee

lmaooo

[wintry steppe]

Tell me the definition of parallel vectors

[uneven shore]

They have the same direction?

[wintry steppe]

tell me the definition of direction then

[uneven shore]

The measure of the angle that it makes with a horizontal line

[wintry steppe]

that's far too complicated of a definition for this, and does not work in general vector spaces

two vectors are parallel if they are multiples of each other by a non-zero constant

[uneven shore]

Ooh okay

[wintry steppe]

so consider for example (1,0) and (69,0) in R²

they are neither opposite or equal

[uneven shore]

Oooh

Yeah

[unique tide]

I get confused on this

How can the entries of A be a scalar multiple if you're going to be multiplying every entry by a different number?

[stable kindle]

for certain values of a and b and c and d

very specific ones

it works out

[unique tide]

so like a = 0 and b = 1

could that work?

[stable kindle]

perhaps :3

let's see

[unique tide]

I think im getting confused on the scalar multiple definition

because I think that scalar multiple means you multiply every entry in the matrix by the same number

[stable kindle]

well if it's a scalar multiple of (1 0) then it's just gonna be some (k 0)

so when you transform the vector by the matrix you get, what, (a c)?

[unique tide]

yes

[stable kindle]

so what do a and c have to be

[unique tide]

I got that

number wise?

[stable kindle]

i mean just

[unique tide]

can't it be any number lol

[stable kindle]

what forms do they have to take

for (a c) to = (k 0)

[unique tide]

uh

vectors?

[stable kindle]

no, what do a and c have to be

ok well what does c have to be

[unique tide]

0

[stable kindle]

yep

and what does a have to be

[unique tide]

some value k?

[stable kindle]

ok but k can be any multiple of 1

so just anything, right

[unique tide]

yea it can be anything

I'm still confused on how you got (k, 0)

from (1,0)

[stable kindle]

so a scalar multiple of a vector is k * (m n) = (km kn)

[unique tide]

yes

[stable kindle]

k * (1 0) = (k 0)

yes?

[unique tide]

I understand that logic

but I feel like that just creates more questions than answers lmao

[stable kindle]

why

[unique tide]

like you can just multiply the column vector (1,0) by some number?

[stable kindle]

yes?

that's what a scalar multiple of it is

a scalar multiple just means you multiply it by some number

[unique tide]

right

[stable kindle]

the number is called a scalar

[unique tide]

so then why say [1,0], why not just use k alone

[stable kindle]

??

[unique tide]

like if matrix A is some entries abcd, why do we say it's a scalar multiple of [1,0] when we can multiple that vector by any number, why not just use the number itself as a scalar.

[stable kindle]

the matrix isn't a multiple of the vector

the vector after being transformed by the matrix is a scalar multiple of the vector

[unique tide]

ohh

[stable kindle]

where the scalar is anything, so i called it k

[unique tide]

ok that makes a lot of sense

that was tripping me up a bit

[stable kindle]

ok

[unique tide]

thank you

[stable kindle]

np

[blissful vault]

just making sure if i am on the right track

here i should find a basis for f and g first

then i find a basis for [result] and h

s that correct?

[nocturne jewel]

since {f,g,h} is a basis of $\mathbb{R}[t]_{\leq 2}$, the set you get after gram schmidt will be a basis and be orthonormal

[stoic python]

moshill1

[nocturne jewel]

so just go through orthonormalization and you're done

[blissful vault]

ohhh ok

thanks

[sweet vine]

which book a proof?

has*

[dreamy iron]

From Treil’s LADW page 17.

What are the dimensions of this matrix T?

I know that it is n+1 entries wide. But I can’t figure out how tall it is.

My guess is that this is a square matrix.... meaning it is also n+1 entire tall.

[limber sierra]

n+1 "entries tall", for the same reason it is n+1 "entries wide"

[dreamy iron]

(My justification is that the vector f’(t) is n+1 entries tall.)

[limber sierra]

youre counting 0, 1, 2, ... n as you go down a row or column

there are n+1 such elements

[dreamy iron]

Okay. So it is square. Tyty

What do you call a vector with non zero entries directly below the main diagonal?

[limber sierra]

"not upper triangular"

[dreamy iron]

lol

[limber sierra]

dont think theres a special term

[dreamy iron]

Okay

[limber sierra]

you can use a phrase like

"the n-by-n submatrix in the bottom-left corner is diagonal"

i guess

or "the n-by-n submatrix given by removing the top row and rightmost column is diagonal"

but i dont think theres a snappy word for it

[dreamy iron]

Lol. I can totally visualize that too. Imma use that phrasing. Tyty

[limber sierra]

these matrices come up in DEs but i havent heard a name before

[dreamy iron]

Okay. I can tell by the function definitions that T is not surjective.... but the matrix makes this obvious too....cuz it’s not “full rank.” I think that’s the right terminology.

[dusk sage]

i have a question about cholensky decomposition

I'm confused by the statement B here

so does it mean that A is positive definite if the matrix L with positive diagonal entries is unique?

or each diagonal entry of L is unique?

so if I had L equal to this would A be positive definite?

[wintry steppe]

Hey guys, Im new here. I have a question. Showing that the finite set of (Zn, Xn) is a group iff n is prime. Does it suffice to state that ' bc 0 isnt an element and u can get numbers in the set multiplying to reach 0, closure of group is not maintained thus it must be a prime'

No, (Zn, *n) is never a group. They are talking about Zn* not Zn.

Do you know what Zn* means, @wintry steppe

Yes. I do

What does it mean?

the set (1 2 3... n-1)

under mod n multiplication

no

Oh ok

Zn* is the group of units of Zn

[stoic python]

Carla_

[wintry steppe]

I see.

Thanks a lot @wintry steppe

Thanks for asking questions, i'm really bored and noone asking questions

So I hope it;'s cool if I ask further then 🙂

it is

So when we say Q*, we mean all the rationals which make a group under multn. Is that the same as saying, 'Q * is the group of units of rationals'

You know what a ring is?

A group with extra operation

New to this

maybe nvm then, and no thats not right, it is a group and has an extra operation but thats not enough

when we say Q* we mean all the invertible rationals (for multiplication)

when we say Z* we mean all the invertible integers...

etc

I seee

turns out that if R is a ring, then R* will always be a group

this because in rings, the second operation is associative and has identity. if we restrict to only the invertible ones, you hopefully can see that this is definition of a group

[wintry steppe]

This makes sense to me bc it removes the addition identity 0

im thinking cartesian atm tho. not generically

the identity of the group of units will be the multiplication identity 1

0 can't be there cuz cuz of distributivity it is not invertible except in the zero ring but otherwise no lol

yup 🙂 ty ty

now last question for now

why should n be a prime in (Zn*,Xn) for it to become a group?

Is it due to the lack of closure ofthe group bc zero is there if two numbers multiply

well if it isn't prime, there will be numbers that multiply to 0 so it can't be group

Okay

but it's stronger than that: there will always be an inverse of every non-zero element if it's prime

so you have to prove that

yeaa, im trying to but it's doing my head in

yea iirc it's a bit tricky to prove it

so far i said. well if its not prime then u get 2 numbers multiplying to get it. so it has to br prime xD

but ur right

i feel like it's stronger than that

thanks carla. ur god sent

[wintry steppe]

@wintry steppe this is where I learnt my definition from btw. Are they equivalent to yours or is there a difference i'm missing on?

[wintry steppe]

What are the invertible for multiplication elements of Q?

I see. Sorry

lol why sorry

[reef prism]

yo guys what is matrix (A^2)^-1, is it A again? 😅

[native rampart]

No

[lavish jewel]

without any other context, identity ftw

[edgy kelp]

https://i.imgur.com/Uwuz1Hm.png

Can anyone help me ith this

[wintry steppe]

have you heard of diagonalization?

[edgy kelp]

Yes

@wintry steppe

[wintry steppe]

tell me what you know about it

[edgy kelp]

matrix is diagnalizeable if there is a match that satisifes that above equation

P^-1AP

I think idk

[wintry steppe]

no

a matrix A is diagonalizable iff there is a matrix P and a diagonal matrix D such tha P^-1AP = D

[edgy kelp]

Okay

So that Matrix D is the [2;0;0;-1]

That is diagonalizable

[wintry steppe]

thats what you want to show

have you heard of eigenv* where * is either ectors or alues?

[edgy kelp]

this question i feel like is to hartd

for me

[wintry steppe]

I suppose you can just set up some linear equations after multiplying both sides by P on the left side.

[wintry steppe]

Can someone help

All hope is lost for us😞

[solid flower]

lol

chicken scratch

[nocturne jewel]

help.... with what?

you wrote 2 inequalities, congrats

[solid flower]

it looks like some martian language to me, idk how you figured they were inequalities

[gloomy arrow]

When you diagonalize a matrix PDP^-1, if P is unitary, does it matter which side the inverse is on?

[nocturne jewel]

Cause my chicken scratch is only slightly better

[wintry steppe]

no Cuck,its okay, inverses of unitaries are still unitary

[gloomy arrow]

Ty!

[wintry steppe]

but um you cant just swap them around

[wintry steppe]

so be careful, it matters what you put in each side, not what you call it

[gloomy arrow]

Wait wait

So if P is unitary then P^-1DP=/=PDP^-1?

[lilac stratus]

yeah

matrice multiplication isn't commutative

even just in the space of inversible matrices

[boreal crane]

Jup that matrix space that is isomorphic to C for example

[gloomy arrow]

Okay I am asking because in one of my homeworks my professor wrote find D=PAP^-1 where P is unitary but I was like the P^-1 should be in the front

[lilac stratus]

well

if you find D = PAP^-1

then with P' = P^-1

you have D = P'^-1AP'

The order doesn't matter, if you find P s.t D = PAP^-1 and you're looking for P' s.t D = P^-1AP, then you've won since D = P^-1A(P^-1)^-1, so P' = P^-1 is the matrix you're looking for

(but P won't be equal to P', so the order does matter in that sense)

not sure if that's clear

[gloomy arrow]

The P' part is confusing me because I think hes just looking for P

[lilac stratus]

Ok lemme rephrase it

[gloomy arrow]

I can also send the question if perhaps my own wording is unclear

[lilac stratus]

Looking for P s.t D = PAP^-1 or looking for P' s.t D = P'^-1AP' is the "same thing"

since P^-1 = P' (or equivalently P'^-1 = P)

[gloomy arrow]

But then you just get back to D=PAP^-1

[lilac stratus]

no you don't ?

I think my notation is confusing you

[gloomy arrow]

PAP^-1 and P'^-1AP'

[lilac stratus]

Looking for M s.t D = MAM^-1 or looking for M' s.t D = M'^-1AM' is the "same thing"

my "P" wasn't the "P" of your question

i'm saying that if you find either M or M', you got the P of your question

if you have M'

well then just let P be M'

and if you've found M

then let P be M^-1

is this clearer ?

[gloomy arrow]

Yeah I would need to take the inverse of M' to just find M

[lilac stratus]

yeah, that's what I was trying to say (sorry i'm not really fluent in english, my wording probably confused you )

[gloomy arrow]

Yeah I understand. But I dont think thats what my professor intended

because them M' would be the matrix of the eigenvectors put together but inverted

Any then I asked my friend and he told me that MAM^-1 is equal to M^1AM no inverse (') because M is unitary

So in regards to what my friend is saying, that would be incorrect

[lilac stratus]

I was just trying to say that it doesn't depend on whether you look for M or for M' in the process of solving the exercise, since in the end you get the P you're looking up to one last step of inversion

[gloomy arrow]

Right Right. I just want to know if my friend is correct in his statement

No primes/inversions

That you can swap them

because he said MAM^1=M^-1AM which I dont think is correct because you arent inverting the M. So like no M' and M'^-1. Just the same M

[lilac stratus]

here's an explicit counter example

[gloomy arrow]

Okay okay thank you thank you. Just because M is unitary would change that correct?

[lilac stratus]

I don't understand your question

[gloomy arrow]

Sorry wouldnt*

I mean to ask if M is unitary, that would not change anything correct

[lilac stratus]

Oh

So like if M is unitary, is this still false that D = M^-1AM implies D = MAM^-1, is what you are asking ?

[gloomy arrow]

yes that is what I am asking

[lilac stratus]

Then yeah, that's still false

[gloomy arrow]

Okay thank you just making triple sure

[crisp cloud]

Hey, could someone check my work for this homework problem I miss

Teacher doesn't explain very well, but if you guys figure out what I got wrong, I'd appreciate it.

[split basin]

Yo
Question
how do u use the degree symbol on keyboard

[solid flower]

[glad acorn]

may i ask why the nullspace of a linear mapping =0 ,then the linear mapping is one-to-one

[native rampart]

Because if the mapping is not one-one that implies,null space is non trivial

T(a)=T(b) implies T(a-b)=0

[nocturne oracle]

basis vectors are linearly independent, so im assuming since you wrote that they were dependent, you must have forgotten to write "not" a basis

[crisp cloud]

F

[nocturne oracle]

indeed

[glad acorn]

Is it iff?

[lilac stratus]

yes that's an equivalence

T(a) = T(b) <=> T(a) - T(b) = 0 <=> T(a-b) = 0

(the last equivalence comes from the fact that T is linear, more specifically comes from T(a+b) = T(a) + T(b))

[glad acorn]

Ok thanks

[jagged granite]

If $A\in M_n(\mathbb C)$ and $(Ax,x)=0 \ \forall x \in \mathbb C ^n$, does it follow that $A=0 $?

[stoic python]

bert

[native rampart]

Yes

However,if it were M_n(R) it wouldn't

[jagged granite]

How do I see this?

For M_n(R) I can see that it is skew symmetric. The trick is to look at (A(x+y),x+y)

[broken sun]

Hello. If the intersection of the kernel and the image of an operator is zero, then why is the finite-dimensional vector space the direct sum of the kernel and the image?

[wintry steppe]

Are you sure of that?

[broken sun]

If this is not correct, please give a counterexample.

[wintry steppe]

0 0
1 0

or do you mean to ask that in the case their intersection is zero that its a direct sum?

[slow scroll]

If f^2 = f then the space is a direct sum of the kernel and image of f

[spiral star]

yes, i didnt know there was a different interpretation lol

pretty sure the question was trivial intersection implies direct sum

[broken sun]

Yes, but I cannot understand your first interpretation of my question.

[slow scroll]

Wait umm I think I’m missing a condition on my statement

[broken sun]

I think my question has only one interpretation.

[spiral star]

yea

well anyway, you can just throw rank nullity at it and you get your answer

choose a basis for the kernel and a basis for the image. then combining them will give you a set of linearly independent vectors

and they will add up to the dimension of your space

[broken sun]

How can we show that such a union spans the whole space?

[spiral star]

like, the only thing you have to convince yourself of is that their direct sum is the entire space but that just follows from rank nullity

its just by dimension

dim V = dim (ker T) + dim (im T)

the direct sum will be a subspace of V with the same dimension as V

so it is V

[broken sun]

And since their intersection is zero, they must be independent.

[spiral star]

yea

[broken sun]

Ok. Thanks. Can you give a counterexample for infinite-dimensional case?

[spiral star]

the trivial intersection tells you already that their sum is a direct sum and their dimensions add up, and then rank nullity tells you that that the dimension of the direct sum is the same as that of the entire space

thats just all you need

the only thing that can "break" is rank-nullity in the infinite dimensional case

you cant use that argument like that anymore

[broken sun]

Ok. But, is there any counterexample?

[spiral star]

probably but i would have to think about it lol

[slow scroll]

There might not be. You’ll still have an equality between the cardinality of their dimensions, and that’s all you need

[spiral star]

thats not how that applies tho

cardinal arithmetic isnt the same

[slow scroll]

Explain?

[broken sun]

Rank-nullity is not true for infinite-dimensional case?

[spiral star]

when a + b = max{a,b} then you cant count in the same way as with naturals

rank nullity is still true

you just dont add up infinities like naturals

[broken sun]

So, why can we not use your already-mentioned argument for a proof for infinite-dimensional case?

[spiral star]

you can still follow the proof of the rank nullity theorem

[slow scroll]

im not saying we should add up infinities like naturals...

[spiral star]

you just cant use a counting argument

[broken sun]

What counting argument?

[spiral star]

that you span the entire space because the dimensions add up

this particular argument doesnt work

but you can try to follow the proof of the rank nullity theorem for arbitrary vector spaces

[broken sun]

Why?

[spiral star]

because cardinal arithmetic

[slow scroll]

that's exactly what im suggesting?

[spiral star]

i wasnt saying anything against that, only against the counting argument

[slow scroll]

okay, but by dim V = dim ker T + dim im T, i clearly mean there is a bijection between a basis for V and the disjoint union of the bases for ker T and im T

nevermind, im being dumb, this (rank nullity for infinite dimensions) is not enough

The only thing you can conclude is that the cardinality of the dimension of ker T oplus im T is the same as V, not that they are equal

Also, sorry I said “I clearly mean there is a bijection...” above.

I really wasn’t being clear about what I was talking about.

[spiral star]

i think i might have a candidate for a counter example

from a sequence space

so if i write (a_n) for a sequence and define a map T(a_n) = (0, a_n)

i think that should be linear

it just prepends 0 to the sequence

T is not surjective

but injective

so ker T = {0}

but since its not surjective, ker T + im T != V

@broken sun happy now? :)

[broken sun]

Thanks.

[spiral star]

you can get a counter example every time a linear map is injective but not surjective

that's something that can only happen in the infinite dimensional case

[broken sun]

Yes.

[spiral star]

(all in the context of endomorphisms)

[jagged granite]

my question got buried

[jagged granite]

Oh this is clear to me now. we do get (Ax,y)=0

[lilac stratus]

wait why ?

If forall x € R^n, Ax = 0 then in particular Ax is zero on the canonical basis of R^n, so it must be the zero matrix, no ?

[native rampart]

I think someone found an explicit counter example

[lilac stratus]

where ?

[native rampart]

It's not Ax=0

The statement is about (Ax|x) being 0

For all x

[lilac stratus]

Oh god

I'm dumb idk how to read

sorry

[native rampart]

#calculus

[bleak geode]

does functions not come under algebra?

[native rampart]

This comes under calculus,tho

functions in algebra are very very different

[bleak geode]

ok ill post there mb

[wary lily]

you've better chances by asking in server for blender

there are good ones

[north steeple]

Can someone help me understand inner product spaces

[gentle igloo]

whats ur question abt them

[north steeple]

i guess just

what it is

first

the definition i have is confusing

[gentle igloo]

are you comfortable with what a vector space is?

[north steeple]

yes I have been working with them

but still confuses me sometimes

[gentle igloo]

can u send the definition you have

[north steeple]

- positivity
- symetricity
-  additivity
- homogeneity

and then i have the definitions of those 4 htings

[gentle igloo]

yes so that is the definition of an inner product. and to boil it all down intuitively, it’s a function which gives a notion of “length” in a vector space. thus a vector space which has a notion of length in it (i.e. has an inner product defined over it) is called an inner product space

and the special function defining “length” must satisfy those properties in that definition in order for it to be an inner product

[north steeple]

what is this function though?

where do i make it or get it

[gentle igloo]

it totally depends on which vector space we are dealing with...

[north steeple]

ok i can give an example

[gentle igloo]

did you mean to say can i get an example?

[fresh obsidian]

Just out of curiosity, is multiplying by the determinant of a 3d transformation a valid way to compute a rotational solid along a non vertical or horizontal line?

[north steeple]

THe properties are those 4 things

[gentle igloo]

yea exactly

[ancient perch]

how were these quantities calculated?

[north steeple]

how would I start this?

positivity: (u,u) > 0, for u != 0, and (u,u) = 0 iff u = 0 ```

i have this

and im a little confused about the notation for the brackets

is that just two vectors

shoudlnt it be the dot product instead?

[gentle igloo]

the dot product is a specific example of an inner product.

[north steeple]

oh

[gentle igloo]

notice that in that little excerpt they didn’t mention which specific vector space we are talking about

it just says any inner product will satisfy that

[north steeple]

ok so for this question, lets say I have three vectors. u, v and w.

then i can assume u is not the zero vector

which means (u,u) > 0

is that correct so far?

[gentle igloo]

are you stating that u is not the zero vector?

[north steeple]

yes

[gentle igloo]

ok then yea

[north steeple]

and do i also have to satisfy (u,u) = 0, iff u = 0?

[limber sierra]

thats what the definition says, yes.

[north steeple]

how would I do it if I am assuming u !=0?

or can I now assume it is

[limber sierra]

these are two separate statements

[north steeple]

Ok right, so how do I satisfy both of those with the same vector?

[limber sierra]

you dont

[gentle igloo]

u don’t

[limber sierra]

the point is that its showing both cases

either u is equal to 0 or it isnt

if its equal to 0, then (u, u) = 0

if its not equal to 0, then (u, u) > 0

[north steeple]

ok so lets say we had vectors with actual values

[limber sierra]

and this holds with "u" representing ANY vector

[north steeple]

u = (0,0,0) in R^3

[limber sierra]

perhaps this rephrasing would make more sense to you:
for all vectors u, (u, u) ≥ 0; and (u, u) = 0 iff u = 0

[gentle igloo]

u can also think of it in terms of “length” if that’s more intuitive, that statement would then be saying if a vector has zero length then it’s the zero vector and vise versa

[north steeple]

ok so It only has to satisfy one of them

for positivity

well,

ok so what would be someting that doesnt satisfy it?

[limber sierra]

define (u, v) = 1 for all vectors u, v

then (0, 0) = 1 which contradicts (0, 0) = 0

in fact, if you come up with a totally random relation, it probably doesnt satisfy it

define (u, v) = x - y where x is the first entry of u and y is the first entry of v

this does satisfy (0, 0) = 0, but not (u, u) > 0 for u ≠ 0

indeed, (u, u) = 0 for all vectors u

(not just 0)

[north steeple]

its more confusing now

[limber sierra]

what dont you understand about those non-examples?

[gentle igloo]

remember that we’re just talking abt functions which take in pairs of vectors and spit out numbers. he’s just giving examples of such functions which don’t satisfy certain properties

[north steeple]

the set of vectors that equal something

is this meant to represent a function?

[limber sierra]

the inner product is a function, yes

[north steeple]

oh ok

[limber sierra]

a function that takes as input two vectors and outputs a real number

[north steeple]

only two vectors?

[limber sierra]

we often write inner products as (u, v) or u · v or similar

but really theyre functions

in the same way that, say, addition is a function

which takes as input two things and outputs their sum

[north steeple]

can you give me a super easy example again

[limber sierra]

well, the most common example of an inner product is the dot product

in the above case, the inputs of the function are a and b

and the output is the sum of the products of their coordinates

[north steeple]

ok and this satisfies the 4 properties?

[limber sierra]

yes.

[north steeple]

can you explain how for positivity

i dont get where to start on that

[limber sierra]

u · u = u_1 * u_1 + u_2 * u_2 + ... + u_n * u_n, right?

[north steeple]

yes

[limber sierra]

where each u_i is some entry in u

are you familiar with the fact that the square of a real number is always nonnegative?

[north steeple]

yes I am

[limber sierra]

there you go.

every term of the sum is a square

so every term is ≥ 0

if you add together a bunch of things that are ≥ 0, the sum is as well.

[north steeple]

ok that makes sense

now for symmetricity

this would mean

u_1 * v_1 = v_1 * u_1

which is true

[gentle igloo]

yup

[north steeple]

Additivity:

(u+v,w) = (u,w) + (v,w) for all u,v,w in V.```

so

this one confuses me

i know its a property ive done many times before

but idk how to write it out

[gentle igloo]

just plug in u+v and w into the dot product formula

[north steeple]

ok i guess

ya

[gentle igloo]

i’m not trying to rush you but just lmk when ur done cause i got a question to ask too

[north steeple]

oh

yes you can ask

thansk for all your help!

[gentle igloo]

sweet thx np

i’m reading numerical linear algebra by trefethen and bau and currently learning abt algorithms for least squares problems. One of these algorithms is to use a modified gram schmidt procedure on the augmented matrix. this does not make sense to me.

specifically where it says that Q*b = R(1:n, n+1)

i can’t seem to prove why this would be true

if anyone needs any more of the page/any other part of the book i can send it

[amber sierra]

I dont really understand what the rank and nullity of a matrix is.
How do I determine it for a matrix like
1 2 1
-1 0 3
1 5 7
please @ me when replying, thx

[north steeple]

two similar matrices have the exact same eigenvalue??

[gentle igloo]

yea they do

[north steeple]

ok and

im trying to find a basis for a eigenspace

i ended up with the zero vector

but I want to be sure

3 1 0
1 3 2
0 1 3```

lambda = 3

is there a way for someone to confirm so I know if i did it right

[gentle igloo]

first off all the zero vector can’t be part of a basis

[north steeple]

ohh

[gentle igloo]

but how did you get three for an eigenvalue

[north steeple]

i am given it

the vector and value

[gentle igloo]

oh ok

[north steeple]

i did

3I - A```

i being identity matrix, A being my matrix

then I just solved the system

[gentle igloo]

honestly i’m kinda rusty on this haha so i might not be the best for this

what system did you solve?

[north steeple]

oh no worries

0  -1  0 | x1
-1  0 -2 | x2 = 0
0  -1  0 | x3```

those are two matrices

not one augmented

and 0 meaning zero vector

[gentle igloo]

oh i see

what’s the point of solving that system cause didn’t you say you were already given an eigenvector in that corresponding eigenspace?

[north steeple]

ya

but i need to find the basis

for the eigenspace

[gentle igloo]

ohh

i honestly don’t think i ever learned how to do that but now i’m curious and giving it a try

[north steeple]

wait i think i know another way

lol

[gentle igloo]

ok so this might be wrong but

nevermind i’m dumb

what’s the vector given as an eigenvector

[north steeple]

there is none

[gentle igloo]

you said that they gave you one though

[gentle igloo]

dude ngl the wording of that problem confuses me so much... good luck

[north steeple]

i think its (-2,0,1)

idk how to verify that tho

[nocturne jewel]

left multiplying the vector by the matrix should be the same as scaling the vector by 3

(defintion of being an eigenvector with eigenvalue)

$Av=\lambda v$

[stoic python]

moshill1

[north steeple]

yes I got it thanks

how the heck do i do this

[nocturne jewel]

what's the question?

compute <f,g>?

if so what are they defining as the IP

[north steeple]

I just have to solve this

[nocturne jewel]

Ok but there's nothing to solve in your screenshot

[north steeple]

\left(2,\:2,\:1\right) 2t^5/2 dt

oh

f 2t^(5/2)dt

the f is some symbol

from calculus

[nocturne jewel]

???

[north steeple]

ive never used it

[nocturne jewel]

can you show me how they defined the inner product?

[north steeple]

they didnt

but

you see this

the f

what is that

[nocturne jewel]

the integral sign?

[north steeple]

ya

ive never used it before

but basically i need to solve that f thing withf 2t^(5/2)dt

[nocturne jewel]

Ok.. so you havent taken a calc course at your uni?

[north steeple]

idk what the dt means either

no lol

[nocturne jewel]

integral is the reverse of differentiation is the tl;dr

$\dv{x}\int_a^x f(t)\dd{t}=f(x)$

[stoic python]

moshill1

[nocturne jewel]

$\int_a^b f(x)\dd{x}=F(b)-F(a), F'=f$

[stoic python]

moshill1

[north steeple]

ok so

lets say I had a the inner product space V of continuous function [0,1]

where

u=2t, v=-t^2

how can I find lets say

d(u,v)

[nocturne jewel]

what's d(u,v)?

[gray dust]

metric. d(u,v)=||u-v||

[north steeple]

yes that

how do i do that then

[gray dust]

but the norm above is defined by the inner product, which itself is defined by an integral, so this is more computation that you've not learned enough material to do

[north steeple]

how would I do it since I dont know what t is

||2t - (-t^2)||

[gray dust]

again, you need to know how to compute an integral in order to compute the above

but to lay out the dependencies so you at least know the route of computation

the 'usual' inner product on $V$ is defined by
$$\ip{f}{g}=\int_0^1fg$$
which naturally defines a norm ('length' function) by
$$\norm{f}=\sqrt{\ip f}$$
which in turn naturally defines a metric ('distance' function) by
$$d(f,g)=\norm{f-g}$$

[stoic python]

RokettoJanpu

[north steeple]

Can somebody help me with this

i sent this here before

people told me I was correct

so I asked my teacher for a regrade, but she said she doesnt agree with me

a) should also be a basis

[nocturne jewel]

did you ask why it's not a basis?

[north steeple]

no, she just said she doesnt agree and I should check the solution

but it follows the definition of a basis

[nocturne jewel]

Ok so ask her why it isnt a basis

[north steeple]

ok i will

[nocturne jewel]

cause it is, d is just the more obvious basis

[a,b,a] = b[0,1,0]+a[1,0,1]

[a,b,a]=b[1,1,1]+(a-b)[1,0,1] if I did it right

[north steeple]

yes

thats her solution

oh

the top

here it is

[nocturne jewel]

Yeah, a is also a basis

d is just the more obvious one

[north steeple]

ok ill ask her to prove me wrong then lol

She asked me to solve it on paper and send it to her

@nocturne jewel is the thing your wrote above it?

[nocturne jewel]

what?

[north steeple]

[a,b,a]=b[1,1,1]+(a-b)[1,0,1]

[nocturne jewel]

Yeah, though check it for yourself, and you'd need to show the basis vectors are lin indep

[north steeple]

any chanc eyou can help me with that, i really dont remember anything now that im done with math

if not i understand

[nocturne jewel]

follow the teacher's process

show that 0=a[1,0,1]+b[1,1,1] is only true when a=b=0

[north steeple]

wouldnt it be sufficient to show that the set of vectorts are linearly dependent

and span R^3

oh wait, does it span?

[nocturne jewel]

it spans W

unless Im mistaken

[north steeple]

whats W?

[nocturne jewel]

the subspace in the question

reading your teacher's working

[north steeple]

oh i thought u had to show the set of vectors span R^3

for the basis

i alreeady wrote down the vectors being independent

[nocturne jewel]

d doesnt span R3 either

[north steeple]

oh ok

[nocturne jewel]

since R3 requires 3 vectors in the basis

by exteremal properties

[north steeple]

so how would I go about doing this then

[nocturne jewel]

doing what?

[north steeple]

writing down the solution to this

so far i am stuck

[nocturne jewel]

"this"

[north steeple]

a) being a basis

[nocturne jewel]

which part of it being a basis?

span or independence?

[north steeple]

ill fix that

[nocturne jewel]

,rotate

[north steeple]

omg

thats sick

[nocturne jewel]

yes that's right

so you have 2 linearly independent vectors, do they span W?

[north steeple]

idk how to check that

[nocturne jewel]

Take an arbitrary vector in W, [a,b,a] a,b in R

the only way you can get the 2nd entry to be b is make the scalar of [1,1,1, b

so [a,b,a]=X[1,0,1]+b[1,1,1]

for unknown X

a=X+b -> X=a-b

so $[a,b,a]=(a-b)[1,0,1]+b[1,1,1]$

[stoic python]

moshill1

[nocturne jewel]

since a and b are real, a-b is real so it's a valid scalar

[north steeple]

tbh i dont understand what you did, but if i write that down would that be sufficient ?

[gritty swift]

intuitively those vectors point in different directions, so they're independent

one is not a scaled version of the other, so they span a plane

[north steeple]

how can i show the span part

[gritty swift]

what he did is the way you solve it in the general case, he showed the only solution to Ax = 0 is x = 0

[north steeple]

ugh

i dont like this lol

i dont understand

[gritty swift]

these are the vectors ([1,1,1], [1,0,1]) you had

[north steeple]

ya

[gritty swift]

to be dependent one must be a combination of the other

but they're on different lines, so that can't be

watch the first few videos of https://www.3blue1brown.com/essence-of-linear-algebra-page/

the first 4

saying "one is a combination of the others" is the same as saying Ax = 0 has a solution for x not being zero

[north steeple]

I get the independence part

i dont get how to write down and show my teacher it spans

[gritty swift]

oh right lemme see

spans what? R3?

[north steeple]

im not sure

here

I literally just need to prove im right

and i get 85% instead of 71%

and this is my only math class i have to take, so I just need help with this one last thing and thats it

[gritty swift]

oh thats a confusion question, what they mean is do those two vectors span [a,b,a]

so its a plane in R3

just write [a,b,a] as a linear combination of your two vectors [1,0,1] and [1,1,1]

[north steeple]

that proves it spans it?

[gritty swift]

yeah

[north steeple]

how can I do that

im sorry

i know im asking so much

[gritty swift]

1 sec

you just notice

then you multiply by the right guys to get a,b

they wrote it in a kinda confusing way, i'd have just said [a,b,a]

[north steeple]

ok

so

i honestly have no idea what you did above

but its correct

then its correct

[gritty swift]

wdym

you know linear combinations?

its where you multiply constants times vectors and add vectors

[north steeple]

ok yes

[gritty swift]

this shows a way to combine your vectors to get [a,b,a]

so our vectors must span all [a,b,a]

since I just showed how you can get it as a combination

[north steeple]

ok

thanks for all your help

i really appreciate it

[gritty swift]

[north steeple]

i hope she will give me the mark

[smoky patio]

hey guys, if A and B are conjugate square matrices, do they have the same eigenvectors + eigenvalues?

found out its same eigenvalues but not vectors

[smoky patio]

Suppose $T: V \to V$ is a linear transformation on some 2-dimensional real vector space V with eigenvalues 2 and 3. Then the limit $\lim_{n\to\infty} |trace(T^n)|^{1/n} equals?$

[stoic python]

Delta Syndrome

[gritty swift]

we know $T^nx = \lambda^n x$ so we know each eigenvalue is raised to the nth power, and $\text{Trace}(T) = \lambda_1 + \lambda_2$
so $$\lim_{n\to \infty} \left((\lambda_1 + \lambda_2)^n\right)^{1/n} = \lambda_1 + \lambda_2$$

[stoic python]

uli

[gritty swift]

does that work @smoky patio ?

[smoky patio]

oooh

why is trace T = lambda 1 + lambda 2?

shouldn't it be all the diagonals?

wait

omg

sorry

[gritty swift]

lol

it was wierd when i first learned that too

[smoky patio]

wait so the answer is 5?

[gritty swift]

I think

[smoky patio]

wait I think n going to infinity is a bit weird

[gritty swift]

they just cancel out, no?

[smoky patio]

somethings weird

can you take n out like that?

sry I don't mean to be doubtful of you

just trying to understand

[gritty swift]

please be doubtful of me

[smoky patio]

:'))

[gritty swift]

oh yeah you're right

$\text{Trace}(T^n) = \lambda_1^n + \lambda_2^n$

not the other way

[stoic python]

uli

[smoky patio]

oh so infty?

[gritty swift]

how do you figure infty?

[smoky patio]

wow this is cal1 stuff why am i struggling

[gritty swift]

isn't $\lambda_1^n + \lambda_2^n < (\lambda_1 + \lambda_2)^n$

[smoky patio]

is it infty or 0 lool

[stoic python]

uli

[gritty swift]

i'm tired too lmao
$$\lim_{n \to \infty} (2^n + 3^n)^{1/n}$$

[stoic python]

uli

[gritty swift]

proof by calculator

it approaches 3

[smoky patio]

oooo

nice

[gritty swift]

euler is rolling over in his grave

[smoky patio]

AHAHA

[gritty swift]

now uh we should prolly show it analytically

is there a smart way to find limits like this i've forgotten

oh yeah can't you like put the limit inside the thingy sometimes

[smoky patio]

yeah I'm trying to remember how

[gritty swift]

no but the inside isn't defined at infty hm

oh

$$\lim_{x \to \infty} \left(2^{1/x} + 3^{1/x}\right)^x$$

looks kinda like the limit definition for e

but i'm not sure

[stoic python]

uli

[gritty swift]

oh wait

now i think you can do the limit of the inside, since its defined at infinity?

[smoky patio]

my reasoning is just that this term will be dominated by the higher one as n grows more and more

but that doesnt add anything for us

[gritty swift]

its super obvious i bet

[smoky patio]

hahah

ikr it feels that way

[gritty swift]

yeah you can put the limit inside i think

just remembering the details now cough cough watching a khan academy video

[smoky patio]

I think you only would be able to if there was no x outside the brackets but there is right?

or

LOOL

nice

[gritty swift]

its like $$\lim_{x \to a} f(g(x)) = f(\lim_{x \to a} g(x))$$ if f is defined there

[stoic python]

uli

[smoky patio]

true yeah I'm braindead

[gritty swift]

i'm just tired trying to remember what f and g are

[smoky patio]

found it on math exchange if ur interested

https://math.stackexchange.com/questions/142176/convergence-to-infinity-of-2n-3n-frac1n-as-n-to-infty

[gritty swift]

oh sandwich works cool

[smoky patio]

hahah

thanks for the brainstorming tho appreciate it

love harp seals btw

[gritty swift]

here's how symbollab does it https://www.symbolab.com/solver/step-by-step/\lim_{x\to\infty}\left(2^{x}%2B3^{x}\right)^{\frac{1}{x}}

the put exp(log()) around it then do the "limit chain rule" which is what i was trying to do earlier

oh and this is a good explaination https://math.stackexchange.com/a/503053

[soft burrow]

it's sort of misleading to call it a "limit chain rule"; it's a property of continuous functions. A comment under that mathSE answer gives an example for a discontinuous function for which it fails

(why is this not in #calculus anyway)

[waxen flume]

https://i.imgur.com/eYoZv4Q.png

can someone check this over?

[gray dust]

you should be able to check A=PCP^-1 yourself

[waxen flume]

i checked it just now

seems right

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

yep, thanks @gray dust

[novel hamlet]

How do I get exponent inside matrix?

I have {{1,2},{2,4}}^n

And I know

But I dont know how to do that

[lavish jewel]

that matrix should be pretty simple to diagonalize

you could diagonalize it and exponentiate the diagonal matrix of eigenvalues

[novel hamlet]

You mean make it to A= PDP^t

And then do D^n?

[lavish jewel]

yes

[novel hamlet]

I'll try that, tought there was some "cheat" to do it, like for limits there le hospital rule

[lavish jewel]

i mean, it's a rank 1 matrix. you can directly write it as an outer product and the nonzero eigenvalue is the 2-norm squared of the vector

so all you have to do is find the vector that has this outer product and exponentiate its norm

i think?

[novel hamlet]

I'm not sure what is 2-norm

[lavish jewel]

the length of the vector

[novel hamlet]

But I'll see that spectral decomposition

Should not be problem for that smal matrix

[waxen flume]

https://i.imgur.com/1TyIMDM.png

Am I just being asked to find the eigenvalues, then form a basis that spans the eigenvectors?

[quartz compass]

pretty much, well your eigenvectors can be used as your basis

[waxen flume]

Thank you.

[wintry vine]

is anyone else reading Linear Alg. by friedberg insel and spence?

[zealous junco]

can someone explain why that $T|_{\mathbb{C}v^\perp}$ is normal

Its probebaly very obvious and i just skim reading

and i cant understand suddenly 😢

[stoic python]

Anticipation

[native rampart]

$T|{\mathbb{C}v^\perp} T^|{\mathbb{C}v^\perp}=T^|{\mathbb{C}v^\perp}T|{\mathbb{C}v^\perp} \newline
T^|{\mathbb{C}v^\perp}= T|{\mathbb{C}v^\perp}^$ which implies the result

[zealous junco]

$T|{\mathbb{C}v^\perp} T^\star|{\mathbb{C}v^\perp}=T^\star|{\mathbb{C}v^\perp}T|{\mathbb{C}v^\perp}$

[stoic python]

Anticipation

[native rampart]

$(T|{\mathbb{C}v^\perp}) (T^|{\mathbb{C}v^\perp})=(T^|{\mathbb{C}v^\perp})(T|{\mathbb{C}v^\perp}) \newline
T^|{\mathbb{C}v^\perp}= (T|{\mathbb{C}v^\perp})^$ which implies the result

[zealous junco]

why would first line be true?

[stoic python]

Buncho Drunk

[native rampart]

Because TT*=T*T in general

So that's true for restrictions too

[zealous junco]

oh crap

i forgot an assumption

👍

[wide trail]

So I need to find the Col A and turn it into an Orthongonal basis. A = {[1,1,1,], [0,1,2]}, B = {[6,0,0]} I found that b is not in the Col A how would I turn A into an Orthongonal basis

I guess I can't get my head wrapped around it

[oblique rune]

Is there any difference between $|\vec{v}|$ and $||\vec{v}||$? Both are the length of a vector right?

[stoic python]

Exynouz

[oblique rune]

Or does the double absolute value mean norm of a normalised vector?

[native rampart]

|v| is not a thing

[oblique rune]

Huh I see

I think I've seen that notation before though.. maybe it's a typo, idk.

[lavish jewel]

wolfram uses it

[native rampart]

[lavish jewel]

you also find it in physics books

[native rampart]

Ok, It's the same thing,then

[lavish jewel]

it's just a notation thing

double bars is more common, but you usually see single bars the first time they show it

there is no mistaking it when you can put stuff in bold, but it's pretty helpful when writing stuff by hand to use double bars for clarity

[oblique rune]

I see

Where can I ask my question about LaTeX?

I am getting a nonsensical error message

[lavish jewel]

someone on the math pedagogy channel was sharing invites to a latex server. you could also try your luck in the discussion general and chill channels

[oblique rune]

Ok nvm, I found a LaTeX discord server, thanks though :)

[lavish jewel]

aight

[faint lintel]

I am very lost as to what JD = DJ has to do with that binomial expansion

[wintry steppe]

everything

[faint lintel]

Like I subbed in J = N + D and then N = J-D