linear-algebra

[steady fiber]

lol works

[icy osprey]

So in your case, lamba and D are the same thing

my prof just uses fancy letters

noice

[steady fiber]

yes, most people use P and D

in diagonalization

[icy osprey]

that makes much more sense

so D is just the rhs

and S is an identity?

[steady fiber]

no

D is the diagonal matrix up there with the -1-i and -1+i

[icy osprey]

no I'm talking in your case

[steady fiber]

I dont have any other case

I only ever mentioned diagonalization

someone else talked about using identity to solve it trivially

which is pretty smart tbh

[icy osprey]

ahh, yea that was sonja

lol

[smoky lagoon]

I know that lambda = 2 with multiplicity of 2, but there's only one eigenvector

what am i supposed to do here

[ionic dust]

Can someone describe what is meant in the last sentence of this paragraph that states "E2 whose elements are the isometries of the plane which send a given pattern to itself."

Here is figure 25.1 which is what they refer to

and the shading portion of 25.2

[ocean sequoia]

First suppose U1 +...+ Um is a direct sum. Then the definition of direct sum implies that the only way to write 0 as a sum u1+...+ um, where each uj is inUj, is by taking each uj equal to 0.

im kinda confused as to what this is saying

what does it mean for each each uj is inUj, is by taking each uj equal to 0? Does that mean each element of the subspace U is the 0 vector? That cant be right cause then its not a direct sum right?

Or is it saying if you sum up the parts for the elements of Uj you get 0 this U1 +...+Um is 0

[odd kite]

it's saying that once you pick one nonzero vector from one of the subspaces, there's no way to completely cancel it with any sum of vectors from the other subspaces

@ionic dust E2 is the set of all rotations, translations, and reflections. They are talking about the subset of these transformations which leaves the lattice looking the same

(2 stands for 2 dimensions)

[ocean sequoia]

So the only way to get zero is a unique combination which is what we want?

[odd kite]

well it's a property of a direct sum

[ocean sequoia]

First suppose that U + W is a direct sum. If v is in the intersection of U and W then 0 = v + (-v) where v is in U and -v is in W thus U + W is a direct sum by By the unique representation of 0 as the sum of a vector in U and a vector in W,we have v = 0. Thus the intersection of U and W is {0}.

Im confused as to how/why we know -v is in W? Why couldnt it be in U?

Also how do we know v is 0? Why couldnt it be 1,-1?

[odd kite]

because elements of a direct sum are sums of one vector from each subspace. Since v is in U, the other vector has to be in W.

it has to be the zero vector because if it wasn't then U+W wouldn't be a direct sum

because there would be v from U and w from W such that v+w=0 for nonzero v or w

which is against the defn. of direct sum

[ocean sequoia]

because elements of a direct sum are sums of one vector from each subspace. Since v is in U, the other vector has to be in W.
@odd kite thank you that makes sense!

it has to be the zero vector because if it wasn't then U+W wouldn't be a direct sum

because there would be v from U and w from W such that v+w=0 for nonzero v or w

I think i might be confused on the definition of direct sum. The way i understood was that if you add two subspaces each sum could only be expressed in a unique way

even if v is 0 couldnt we also have some other vector j where j is in the intersection of U and W so that j + -j = 0?

of no cause then we are against the definition of direct sum

[spiral sonnet]

would this be considered a subspace of R^3?

[limber sierra]

check the axioms

[spiral sonnet]

yeah I got it!

[limber sierra]

or rather, the required properties

i.e. closure under addition and scalar multiplication

[also existence of 0 vector but thats obvious here]

wdym "I got it"?

[spiral sonnet]

Yeah I forgot that all axioms have to be true in order for it to be a subspace, but I know how to do it from there

Well nevermind guess I have no idea what I'm damn doing

If a set S1 consists of all vectors [a, b, c] such that a, b, c are integers. Shouldn't this be a subspace?

So it has the zero vector. If I choose any two vectors in S1, it will still add up to a vector that's in S1. And if I choose any scalar and multiply it with a vector in S1 I'll still get a vector in S1

[limber sierra]

will you?

what if i choose the scalar -1

[spiral sonnet]

I'm also not talking about the a,b,c >= 0 problem anymore. This is the problem I'm on now If a set S1 consists of all vectors [a, b, c] such that a, b, c are integers. Shouldn't this be a subspace? But for the problem I'm talking about it should

[limber sierra]

ah okay

but i'd still check scalar multiplication there

hint: you have "more" possibilities for scalars than possibilities for vector entries

that is, your scalars take possibilities from R, but your entries are all from Z

can you use this in any way?

[spiral sonnet]

so a scalar can be any real number. so fractions o.o

I thought it was just integers since that's what I've been using

[limber sierra]

well, youre talking about being a subspace of R^3 right?

[spiral thistle]

Generally this is specified as the underlying field of the vector space in the problem.

[limber sierra]

the scalar field of R^3 is the real numbers

if your scalar field is restricted (e.g. you're taking R^3 over Z) that'll usually be specifically stated

[spiral sonnet]

okay cool!

[spiral thistle]

You don't generally do a ton of linear algebra over integers because integers aren't algebraically closed over multiplication/division

[limber sierra]

[oh yeah, i shouldnt use the word "field" here]

[habit, sorry]

[indeed in such a setting they'd actually say "the module R^3 over Z" but... beyond the scope of this convo]

[or just "the Z-module" or whatever]

the real chad move is to say "the abelian group" instead of "the Z-module"

[spiral thistle]

If we're not talking about fields in their mathematical context, then I find it useful to sometimes use the layman definition of the word field like you did. 🙂

[spiral sonnet]

I'm convinced I'll never be able to fully understand math even though I want to. Too many terms that are way too confusing

[limber sierra]

don't worry, as i said, beyond the scope of this convo

just addressing potential nitpickers

[not robert]

[spiral thistle]

@spiral sonnet you'd be surprised at how things just click after awhile.

[spiral sonnet]

I just had my 3rd exam today and I can't even do addition and subtraction correctly with a calculator 😦

[limber sierra]

honestly, I don't think I've ever "understood" math while i was learning it

but like 2-3 months after

it's second nature

[spiral sonnet]

math makes my blood boil!

[limber sierra]

it just somehow mentally coalesces into a coherent picture

(with enough practice)

if that makes sense

[spiral thistle]

I think one thing that gets in the way is the inaccessible nature of all of the notation and jargon

[limber sierra]

jargon is mostly intimidation tbh

i think that's why it continues to be so widespread, like

once you've actually had a chance to work with it, everything sort of fits together

so people writing textbooks/teaching courses are like

"yeah it's a big jargony but it's not too bad, really the terms are simple"

because they've forgotten how it was like before it all "fit"

once it does click, the intimidation factor is gone

i've fallen guilty to this before

[and continue to, regularly]

[spiral sonnet]

In what case will a set of vectors not span R^3?

[limber sierra]

consider the vector $\begin{pmatrix}1\1\1\end{pmatrix}$; does this span $\bR^3$?

[stoic python]

Namington:

[spiral sonnet]

uhhh you can scale it so can you say it's a line that spans R^3?

[limber sierra]

can i get $\begin{pmatrix}2\1\1\end{pmatrix}$ using linear multiples of that vector?

[stoic python]

Namington:

[spiral sonnet]

nope

[limber sierra]

so it doesnt span R^3

but it DOES span the subspace of $\bR^3$ that is all vectors of the form $\begin{pmatrix}a\a\a\end{pmatrix}$ for real $a$

[stoic python]

Namington:

[limber sierra]

there's a slick condition for whether a set spans a space R^n: a set of n linearly independent vectors spans R^n

so if you have a set of vectors from R^3, and a size 3 subset is linearly independent, then the set spans R^3

[spiral thistle]

@spiral sonnet I like to think about this graphically. You need at least n linearly independent vectors to span a space of n dimensions.

To span 2D space, you need two non-colinear vectors.

To span 3D space, you need three non-coplanar vectors.

[spiral sonnet]

So if you want a set to span R^3, would you need atleast 3 vectors in the set?

[limber sierra]

yes

at least 3 vectors, and you need linear independence

[spiral thistle]

And they would have to be independent of each other

🙂

[limber sierra]

for example, the set $\left{\begin{pmatrix}1\1\1\end{pmatrix}, \begin{pmatrix}2\2\2\end{pmatrix}, \begin{pmatrix}5\5\5\end{pmatrix}\right}$ would not span $\bR^3$

[stoic python]

Namington:

[limber sierra]

in fact, its span is the same as the span of $\left{\begin{pmatrix}1\1\1\end{pmatrix}\right}$

[stoic python]

Namington:

[spiral sonnet]

okay got it!

[spiral sonnet]

What does P_d<-b mean? I know P is the change of coordinates matrix

[left hill]

um I have a question

_<

how does that equal 3/2? I keep getting 189/441

[steady fiber]

I get 3/2 as well

it's 3/2 bc the math just works that way ig

[left hill]

I was squaring the wrong vector lmao

sorry

I got it too

[daring solstice]

The answer given was to check if the dot product is 0 when checking each of them with each other but could I also just check if one is the cross product of the other two?

It seems standard way is to check dot products becasue it is easier to compute by hand

[steady fiber]

1. doing cross product of 2 vectors giving the third vector does not imply they are orthogonal. You know nothing about the two vectors that you cross product-ed with each other. You would need to do at least 2 cross products, which is definitely not easier than 3 dot products
2. the cross product is almost definitely not going to be the third vector, even if they are orthogonal. Even if they're orthogonal, you'll get a vector that is a constant multiple of the third vector, so you have to do a harder check as well

2 cross products + 2 harder checks

or 3 dot products and checking if it's 0

I'd say the dot product method is far easier

[daring solstice]

Ah that makes sense thanks for clarifying

[steady fiber]

also the dot product method can be generalized to n-dimensions

there is no cross product in most dimensions

[daring solstice]

That too. That makes a lot of sense

[steady fiber]

there is an exterior product but like it's weird

[daring solstice]

(u and v are vectors)
I don't quite understand how to expand (u + v) dot (u + v) here

[spiral sonnet]

anyone know how I would do this?

I setup an augmented matrix with [b1 b2 b3 | d1] then used d2 then d3

[daring solstice]

oh oh yea that makes sense that I wouldnt understand how to do it as I haven't touched inner products in depth yet

it kinda makes sense now but coming from scalar mutiplication it does take a bit of getting used to intuitively

wait no it's the same as scalar mutiplication i overthink iy

I think. I just didn't know I can 'expand' a dot like that cumulatively same as (a + b) * (a + b)

[smoky lagoon]

aaaah

only for linear functions ?

nope, only for constant functions

[eager locust]

Sorry to answer this late, but for those still interested, this is how this inequality is proven:
[By the way, this is the C2 [a, b] integral norm.]

[smoky lagoon]

you're not late at all

the thing that was tripping me up was that everything I was reading was using another y(x) as a second integral but it didn't include the b-a

and then I realized that b-a is the bohring integral so y(x) = 1 and since f(x)=ky(x) whjere k is a constant, that f(x) can equal any constant term

but your way looks more like the way they wanted me to approach it

[eager locust]

Sometimes, simple functions and concepts have surprising implications, I guess.
It did take me a bit of trial-and-error (and to see your answer) to get to the form they probably wanted, but I assume that this is what Math is all about.

[smoky lagoon]

unfortunately vectors never have and probably never will make sense in my chimp brain

[eager locust]

Don't worry, I bet they will sooner than you think.

[smoky lagoon]

meh I'm not all that interested in vectors at this point

i just want my baby infinite series back

[eager locust]

Honestly, one of the things that quite surprised me about Linear Algebra is how it could be applied to sets that aren't just ordered co-ordinates, but more generally, like with integrals, and how you can use such concepts in analysis.

And even if it's "just co-ordinates," there are still norms defined for infinite series, and the CS inequality applies for them as well.

[smoky lagoon]

I'm really excited for analysis

seems like I'll get a kick out of it

[eager locust]

You should be. It is a lovely topic, and it seems to be the most "intuitively useful" in Math.

[smoky lagoon]

I've heard it's really hard as well so hopefully that doesn't hurt me too much

I'm sort of struggling in multivariable calc but I bullshit through and get Bs on everthing but I don't really get whats happening

i have no clue what a line integral

even is

[eager locust]

Google Images is God's gift, I tell you.
I genuinely learned a lot just from Googling stuff.
I also use Google Images for my notes I make in class.

[smoky lagoon]

Yeah I know I look at that

but my brain just doesn't work in 3 dimensions

[golden cloak]

lets say that l: V --> R (l is a linear functional on V). Im just trying to "see" one of them exactly.
So, we define that l(x) = a_1 x_1 + ... + a_n x_n. RHS is so called "linear form" of that functional, and functional and its linear form are not the same, but are isomorphic. Ok. So what I want to know is the following:
Say l: R^3 --> R. So l = {(x, k) | x in R^3, k in R} = {(x, k) in R^3 x R}, does that mean that elements of R^3' look like ((1, 3, 2), 6))?

[dusky epoch]

RHS is so called "linear form" of that functional, and functional and its linear form are not the same, but are isomorphic.
sounds like a stupid distinction to make if you ask me

[golden cloak]

well I thought so, but think not so stupid

but ok, I get your point

because, elements of that set Im asking about are sure different animals from sum of some numbers

but nvm, that is not so important

am I right with construction of that set and its elements, thats the main question?

[dusky epoch]

not really

$l = {(x, l(x)) \in \bR^3 \times \bR \mid x \in \bR^3}$ i guess if you wanna go formal

[stoic python]

Ann:

[golden cloak]

ok ok right

ty

((1, 3, 2), 6)) is element of l not V'

ok, nice

[idle osprey]

is reflection over line x=y=z the same as reflection over origin (0,0,0)?

[dusky epoch]

no

[quartz compass]

in 3d space those don't seem to make much sense

you can rotate around a line by an angle

you can reflect through a plane

you can have an inversion through a point

[dusky epoch]

you can reflect around a line just fine

it's the same as rotating by 180° around it

[quartz compass]

I said you can rotate by an angle, but I wouldn't consider rotating by 180 degrees to be a reflection generically speaking

I'm not my own reflection when I rotate by 180 degrees

[golden cloak]

actually you are right

[idle osprey]

oh okay thanks

[golden cloak]

yeah, if you have a line in 3D and say a square which side is perpendicular to that line, you won't be reflecting over that line, but inverting through the point on that line

[spiral sonnet]

Anyone know how I'm suppose to solve this problem?

So I tried [B | D1], [B | D2], [B | D3] to find the columns of the change of coordinates matrix, but that didn't seem to work

[lilac iron]

its B

[spiral sonnet]

Isn't B just the change of coordinates matrix from B to a standard basis?

[gray dust]

the cols of $P_{\mathcal D\leftarrow\mathcal B}$ are the coordinates of $\mathcal B$'s vectors wrt $\mathcal D$

[stoic python]

RokettoJanpu:

[spiral sonnet]

maybe I did the calculations wrong or had a number incorrect. I did that

oh wait

I think I mixed up the matrices

[gritty sorrel]

does dim V = dim null T imply that T is the zero map?

[broken hawk]

try proving it

[gritty sorrel]

well I know from this that dim range T = 0 which means that range T = {0}

still I don't know why this implies T=0

[broken hawk]

what does T=0 mean?

like, unpack that notation

[gritty sorrel]

T(v)=0 for all v in V if T : V-->W

[broken hawk]

yea. so if range T is {0}, then what is T(v) for some v∈V?

[gritty sorrel]

what if for example T was the differentiation of all constants?

T(v) = 0 for arbitrary v yes

[broken hawk]

so you’re done

what if for example T was the differentiation of all constants?
what are the vector spaces here?

[gritty sorrel]

can I take V to be R?

and W to be R too

[broken hawk]

you can identify it with that but strictly speaking it’s a different space

V here would be the space of constant functions ℝ→ℝ

but you can obviously just identify those with the actual constants

but then the notion of “derivative” doesn’t make sense

you can’t differentiate a number

[gritty sorrel]

oh ok got it

[broken hawk]

but yes on this space the derivative is just the 0-function

[low plank]

How would I do this question?

[steady fiber]

find determinant

find what k make the determinant 0

those k don't work

all the other k work

[smoky lagoon]

basically, what ks gives you a free variable or zero row

and your answer is all k other than those

[wintry steppe]

what book do you rec for linear algebra? Lang, Hoffman Kunze or ?

[smoky lagoon]

my class uses Linear Algebra and its Applications by David Lay which I have liked for what I've needed it for

granted I'm not using as a method of learning but a resource but it's still pretty good and clear about things

[dusky epoch]

we don't do that here

also wrong channel

[mild pike]

So where can I post it ?

[dusky epoch]

#❓how-to-get-help

one of the 10 channels

which isn't occupied rn

[golden cloak]

Can anyone give me some fine explanation of second dual? I really can't grasp it rn

[wintry steppe]

silly and not rigorous explanation:
in finite dimensions taking the dual gives you the transpose of a vector, acting on vectors by left multiplication
taking the dual twice gives you the original vector, acting on functionals
how should a vector act on a functional? evaluation!

that's how i usually think of it, although the double dual hasn't really come up so i haven't had to think about it much

[golden cloak]

Is it like, in V'' are functionals that vectors in V can hire like "go through functionals in V' and spit its value for me?

ohhhhhhhhhh every vector can be thought of as a linear transformation itself

ty!

[wintry steppe]

for a more formal version of what i said, lemme find a pdf of my first year linalg textbook

i think it explained the finite dimensional dual and double dual well

[golden cloak]

nono I have a formal def

but why not

if you can find it

cool, ty

[wintry steppe]

https://ulissesgtz.files.wordpress.com/2019/02/stephen-h.-friedberg-arnold-j.-insel-lawrence-e.-spence-linear-algebra-pearson-2014.pdf

[golden cloak]

a fat one 😄 just as I like it 😄

[wintry steppe]

page 122

it kind of just powers through the definitions to give you the canonical isomorphism between V and V^**

[golden cloak]

yeah I'm going through it rn in my book

[wintry steppe]

but it kind of makes rigorous the silly expanation (which may or may not be helpful) i gave earlier

what book?

[golden cloak]

oh im from Serbia so... I don't think it will matter to you 😄

but that proffesor (Ljubisha Kochinac)

he was really well known world wide

https://www.youtube.com/watch?v=sxAtIQnqukQ

this guy

when he got old, he went to countryside, planting herbs, fishing and doing math

only now and then he will go worldwide to give lectures

really fine prof.

[sharp merlin]

wait none of these are in row echelon form right

bc leadering entrys arent 1

nvm

[limber sierra]

row echelon form does not require leading entries of 1

reduced row echelon form does

[sharp merlin]

yup

@limber sierra why is the third one not in form then

[limber sierra]

no clue

it is

[sharp merlin]

ok

[celest bridge]

god my lin alg is so rusty. how do I go about finding B here? I don't even know how the info about the trace and det are going to help me here.

or how I'm supposed to "use" eigenvectors to find it

[celest bridge]

I get that A and U are supposed to be similar matrices, so there exists P s.t. P^-1 A P = U, but I can't find how you can find P anywhere in my notes or looking online.

[wintry steppe]

diagonalizing

oh wait nvm that doesn't work methinks

[hollow finch]

Is there a term for the minor of a minor? So if I have a 4x4 matrix, taking the determinant of the matrix where I removed column/row 1 and column/row 3

Is it like a hyperminor? lol

[left hill]

um

😳

am i able to scale a matrix by e^3?

or is that illegal

[brittle juniper]

why would you think it would be illegal? why would you think it would be legal?

what kind of matrices are you working with?

[left hill]

I'm not sure
I heard someone say something about real numbers only but maybe I misunderstood
And this is the type of problem I'm working on

😳👉👈

[hollow finch]

e and its (real) powers are real

but you could scale it by 'i' i guess. not sure why you would though.

from what you sent, you could multiply both rows by e^3 and it would simplify things greatly

Although hold up

$\begin{bmatrix}1&3\1&3 \end{bmatrix}\begin{bmatrix}u_1\u_2 \end{bmatrix}=\begin{bmatrix}-3\1 \end{bmatrix}$

[stoic python]

nix:

[hollow finch]

That is an inconsistent system

[left hill]

it's what I got for my repeated eigenvalue

[hollow finch]

@left hill What was the original matrix, eigenvalue, and eigenvector?

[left hill]

@hollow finch

[hollow finch]

(-3,1) is the correct eigenvector

The system you're looking for is
$(A+\frac{3}{2}I)\vec{v}_1=\vec{v}_2$

[stoic python]

nix:

[hollow finch]

You simplified the matrix before you set up the system

So I believe it would be
$$\begin{bmatrix}\frac{1}{2}&\frac{3}{2}\-\frac{1}{6}&-\frac{1}{2} \end{bmatrix}\begin{bmatrix}u_1\u_2 \end{bmatrix}=\begin{bmatrix}-3\1 \end{bmatrix}$$

[stoic python]

nix:

[hollow finch]

The system will always be consistent if you set it up correctly

So if you get something like $$\begin{bmatrix}1&3\0&0\end{bmatrix}\begin{bmatrix}u_1\u_2 \end{bmatrix}=\begin{bmatrix}0\1 \end{bmatrix}$$

[stoic python]

nix:

[hollow finch]

which says 0=1

then you should check back for a mistake

🙂

[left hill]

waitttt

are you saying I made a mistake?

did I do the multiplicity part wrong?

[hollow finch]

You set up the system incorrectly

[left hill]

is my mistake in the second column?
like the very first line? or after that

[sharp merlin]

Anyone understand this

i dont get why the first two on the left dont work if the * are 0's

[left hill]

[hollow finch]

@left hill
$(A-\lambda I)\vec{v}_1=\vec{v}_2$

[stoic python]

nix:

[left hill]

oh

one sec

So like this?

[storm python]

uh yeh you forgot to minus the lambda I

[hollow finch]

In your case, $A=\begin{bmatrix}-1&3/2\-1/6&-2\end{bmatrix}$

[stoic python]

nix:

[hollow finch]

$\lambda=-\frac{3}{2}$

[stoic python]

nix:

[hollow finch]

$\vec{v}_1=\begin{bmatrix}-3\1\end{bmatrix}$

[stoic python]

nix:

[left hill]

I'm lost

[hollow finch]

Oops

[left hill]

???

that's what I had

[hollow finch]

Okay first things first, what is $A-\lambda I$

[stoic python]

nix:

[left hill]

$A=\begin{bmatrix}1&3/2\1/3&-2\end{bmatrix}$

[stoic python]

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

[left hill]

1 3
1 3

or I guess

[hollow finch]

$\begin{bmatrix}-1&3/2\-1/6&-2\end{bmatrix}-\left(-\frac{3}{2}\right)\begin{bmatrix} 1&0\0&1\end{bmatrix}$

[stoic python]

nix:

[left hill]

which becomes

1 3
1 3

right?

[hollow finch]

I've never seen 3/2 just turn into 3 before

[left hill]

😐

but but but

isn't the next step to combine them?

and simplify

[hollow finch]

To combine them yes

What do you mean by simplify

[left hill]

find x1 and x2

so I can make the eigenvector

[hollow finch]

If you're trying to solve Ax=b, and you change A before you start solving the system, you aren't going to get the correct answer.

First do the matrix addition, then solve the system

[left hill]

gotcha gotcha
that makes sense

is it still v1 on the left hand side?

[hollow finch]

No that was my bad

Your first eigenvector goes on the right hand side of the equation

[left hill]

Is it like this?

[storm python]

yes but you should probably put an equal sign

[left hill]

ty

[storm python]

cute.

[left hill]

This paper is dying

No u

U3U

[storm python]

[left hill]

👉👈

Um

Hehe

[hollow finch]

Yeah very nice. Notice that this system has a solution 😛

[left hill]

I got V1 + 3V2 = - 6

😳

I'm starting to think I don't know linear

[hollow finch]

Haha it's okay

Just use what you had up there

$v_1=-3v_2-6$

[stoic python]

nix:

[hollow finch]

So then $(v_1,v_2)=(-3v_2-6,v_2)$

[stoic python]

nix:

[hollow finch]

Have you ever solved a system of equations with a free variable?

[left hill]

oooo not yet

sorry this is for DEs and we kinda just got thrown into it

[hollow finch]

Aha I see

The idea is to separate the vectors and factor out the free variable.
So $\begin{bmatrix}-3v_2-6\v_2\end{bmatrix}=\begin{bmatrix}-6\0\end{bmatrix}+\begin{bmatrix}-3v_2\v_2\end{bmatrix}$

[stoic python]

nix:

[hollow finch]

$=\begin{bmatrix}-6\0\end{bmatrix}+v_2\begin{bmatrix}-3\1\end{bmatrix}$

[stoic python]

nix:

[hollow finch]

Notice that second vector is your original eigenvector

[left hill]

ohhhhh yes yes

does that mean something?

like
confirmation that we did it right?

[hollow finch]

It is a great sign

In fact, you've found your second solution for your system of DE's. You just need to put it into the right form (which is a bit tricky at least imo)

I would definitely look back at your notes

[left hill]

is it the y = c1v1e + c2(v2e + v1te) thingy?

I left out the exponents

[hollow finch]

Yes, exactly. $\vec{y}_2=\vec{v}_1te^{\lambda t}+\vec{v}_2e^{\lambda t}$

[stoic python]

nix:

[hollow finch]

Very nice 🙂

[left hill]

one sec
I am using the last of my bren energy

Looks like this right?

[storm python]

looks gud

[left hill]

so then find the solution thingy

[hollow finch]

Yikes. that's a rough system but it should give you the right answer.

[left hill]

I can scale everything by e^3 right?

[hollow finch]

Yes

[left hill]

Uwu

[hollow finch]

I wonder if instead you could write your solution instead as
$$\vec{y}=c_1\begin{bmatrix}-3\1\end{bmatrix}e^{-\frac{3}{2}(t-2)}+c_2\left(\begin{bmatrix}-3\1\end{bmatrix}(t-2)+\begin{bmatrix}-6\0\end{bmatrix}\right)e^{-\frac{3}{2}(t-2)}$$

[stoic python]

nix:

[hollow finch]

Because then c1 and c2 would be very easy to find

[left hill]

this is getting gross

[hollow finch]

But plug in $t=2$ to that and you get
$\begin{bmatrix}1\0\end{bmatrix}=c_1\begin{bmatrix}-3\1\end{bmatrix}+c_2\begin{bmatrix}-6\0\end{bmatrix}$

[stoic python]

nix:

[hollow finch]

Which is solvable by inspection

c1=0, c2=-1/6

Not 100% sure you could do that, but I don't see anything wrong immediately

[left hill]

I got this
My head hurts

Probably wrong

[hollow finch]

It's -1/6 e^3 but otherwise yes

[left hill]

pinky promise? 😳 👉 👈

[hollow finch]

Yeah I put it into a calculator. And it's also equivalent to the solution I got to my system above.

[left hill]

Final answer? 😳

[hollow finch]

@left hill You're missing the e term on the second part

[left hill]

Oh yeah
-e^3/6 right?

[wintry steppe]

is anyone able to help me with this ?

It is in the linear algebra by cambridge textbok exercise 10.16

[solar osprey]

I was thinking adding -1 times the first row to the second row

then adding -1 time the last row to the before last row

then adding -1 times before last row to the second row

we obtain a row full of zeros

thus det = 0

[dusky epoch]

yup

[solar osprey]

idk I feel like this chapter is all about smart matrix manipulation

[solar osprey]

ok this one is giving me a headache

I thought maybe turn it into upper tri

but its a tedious matter

[broken hawk]

oh that’s a classic
I don’t actually remember how you do it tho
but it’s a useful result

[solar osprey]

then I tried to see if we could

maybe split it

[broken hawk]

(well really the useful result is knowing that it’s nonzero if the a_i are all distinct)

[solar osprey]

into two matrices multiplcation

Dont spoil the answer

I just wanna know if Im on the right track

lol

[broken hawk]

the answer is “how to do it” not what comes out in the end, imo

[solar osprey]

mmmm

let me give it some extra thought and see what happens

will try looking for patterns or smt with 2x2 and 3x3

[broken hawk]

the class I’m tutoring had it as an exercise last semester apparently. in their exercise they got told what the determinant will be and just had to prove it

[solar osprey]

that makes it 70% easier

imo

[broken hawk]

looking for patterns is definitely a good approach

looking at the solution, it’s actually not that hard once you know what it should be

okay well it does require a trick of sorts I guess

[solar osprey]

mmm

im expecting some kind of summation

[broken hawk]

I’ll put a tip in spoilers here if you wanna glance at it ||figure out the pattern, then prove it by induction on the size of the matrix||

[solar osprey]

ill open it when im desperate lol

thank you !

[broken hawk]

I have the solution here so if you need more ping me

[solar osprey]

sure

@broken hawk I have an idea

but I have a question first

Does Type 3 column operation preserve determinant too? My book only mentions rows, and cofactor expansions along rows only?

[broken hawk]

which are type 3?

adding multiple of one column to another column?

if so, yes

they do

[solar osprey]

yes

[broken hawk]

(proof: det(A) = det(Aᵀ))

[solar osprey]

yes yes

is the answer 0 by any chance? @broken hawk

My approach yields to that result

[broken hawk]

if and only if some of the aᵢ are identical

what pattern did you find in dimensions 2/3?

[solar osprey]

some sort of power expansion

and alternating indices

I thought maybe

if we add

for each column j

we add all other columns to it

we get a geometric series

in each entry

[broken hawk]

you can’t actually do that simultaneously, mind

(with elementary operations I mean)

[solar osprey]

ahhh

Yes

I missed that

u are right

[torn silo]

Stupid question but the normal matrix addition + mutliplication defines an associative algebra?

[broken hawk]

yes

and also a (noncommutative) ring, for that matter

[torn silo]

okay thank you very much

[solar osprey]

@broken hawk I FINALLY FOUND the pattern

its basically the product of every goddamn pair

but like im trying to think of a straighforward proof

[broken hawk]

well, of the differences of pairs

Π(xⱼ - xᵢ) for i<j

[solar osprey]

yes

[flat depot]

Okay, so I just proved that set of all real functions is direct sum of set of odd real functions and even real function. That implies that every real function can be decomposed as sum of odd and even functions. But I have no idea how that decomposition would look like. Is there any kind of procedure for such decomposition?

[cursive narwhal]

$f(x) = \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

The first one is even, the second one is odd

@flat depot

[flat depot]

Okay, so I can make even/odd function out of any given function with some combinations. That's interesting.

[cursive narwhal]

Indeed

[torn hornet]

this is p useful

its been a while since i have done QM, but essentially for odd/even potentials in 1d

the shroedinger eqn is p trivial

[steady fiber]

Fourier also technically decomposes functions into odd and even

[torn hornet]

so doing this can make solving shroedingers eqns in 1d p easy

[steady fiber]

if you only look at the sine terms and cosine terms separately (or if you only look at the real and imaginary parts separately for complex Fourier)

[dusky epoch]

if you do this to the function e^x, you get cosh and sinh

[flat depot]

There is also another thing. I kinda tried to research on my own if set of period functions is subspace of vectors space of all real-valued functions. My results were inconclusive, but I get the feeling that entire set is not vector space, but any "class" of functions with fixed period p is vector space. Is my intuition correct?

[dusky epoch]

yes

[flat depot]

Great, thanks for confirmation.

[golden cloak]

is this line of reasoning ok?

[dusky epoch]

no

[golden cloak]

why?

[dusky epoch]

what even is that first step

{a ∈ P | x-a ∈ P} is either all of P or empty depending on whether x itself is in P

and why not use words instead of this formal mumbo jumbo

[golden cloak]

its a ∈ V, my mistake

typo

hm that typo ruined my "proof" 😦

its not formal mumbo jumbo its just formal

whats wrong with trying to be formal?

[dusky epoch]

you're obscuring your point

[flat depot]

My teacher used to say that intuition is more important than being formal, and using words is intuitive

[dusky epoch]

there's a healthy mix of prose and formality that makes for good proofs that are enjoyable to read

[flat depot]

You can refine intuitive reasoning with formality, but it's hard to start from formality, at least in the beginning of being a mathematician

[torn hornet]

using too much set notation when it can be said in words just make ur writing hard for ppl to read

including yourself lol. Its better to write things out in english so that its more clear/immediate

[dusky epoch]

using too much set notation when it can be said in words just make ur writing hard for ppl to read
yeah that was my point kinda

[torn hornet]

like a complex statement could be absolutely disgusting using logical things, with multiple nested stuff, but easy enough to read with words

i dont have an example readily avalaible for this, but im sure u can imagine this lol

[golden cloak]

ok, makes sense

I'll listen to you 🙂

[pale shell]

If a linear map has only the trivial solution for the zero vector, does that imply that it is injective?

[torn hornet]

yes, try proving it

[cursive narwhal]

Why not just try proving it lmao

[gray dust]

you're talking injectivity? not long ago you didn't know what a codomain was. very nice, keep it up

[cursive narwhal]

Legit?

[pale shell]

Thanks Rokabe!

[pale shell]

Im basically trying to assume there is only the trivial solution and that it isn’t injective

[half ice]

Best way to go about the proof is by the contrapositive.

Assume the linear map is not injective. Show that f(something) = 0, for a non-zero something

.
You may not know what the contrapositive is, now that I think about it.

If you want to prove something of the form A → B, you can instead prove ~B → ~A, and that's the same thing. This strategy works well here

[pale shell]

I know about a contrapositive

So basically if T not injective then there is not only the trivial solution

[half ice]

If T is not injective, then Ker(T) is not trivial

[pale shell]

Sure

[gray dust]

it's actually not absolutely needed to prove the contrapositive

[cursive narwhal]

You can show it directly

T(x) = T(y) so T(x-y) = 0. x-y belongs to the kernel of T and since that is just the zero subspace, x=y

[gray dust]

let $A$ be a matrix and let $T$ be a linear map defined by
$$T(x)=Ax$$
suppose $T(x)=Ax=0\implies x=0$. we can show $T$ is injective
$$T(x_1)=T(x_2)$$
$$Ax_1=Ax_2$$
$$Ax_1-Ax_2=0$$
$$A(x_1-x_2)=0$$
we now use $T(x)=Ax=0\implies x=0$ to say that
$$x_1-x_2=0$$
$$x_1=x_2$$
therefore, $T$ is injective $\qed$

[stoic python]

RokettoJanpu:

[cursive narwhal]

^of course, just fix the argument above with the relevant terminology etc

[torn hornet]

to add here this is true for homomorphisms in general algebraic structures, that kernal=0 implies injective. proof follows in much the same way

(linear maps are homomorphisms of vector spaces)

[cursive narwhal]

Jintarou, you don’t need matrices for this. You just need the linearity of the map.

[gray dust]

it's for earl, not you

if you really want this w/o A then ok

let $T$ be a linear map. suppose $T(x)=0\implies x=0$
$$T(x_1)=T(x_2)\implies T(x_1)-T(x_2)=T(x_1-x_2)=0$$
we now use $T(x)=0\implies x=0$ to say that
$$x_1-x_2=0\implies x_1=x_2$$
therefore, $T$ is injective $\qed$

[stoic python]

RokettoJanpu:

[cursive narwhal]

I mean, i gave it earlier so w/e lol

[gray dust]

mb

[sharp merlin]

how do u do this

[odd kite]

if $A \begin{bmatrix}x1 \ x2\end{bmatrix} = \begin{bmatrix}b1 \ b2\end{bmatrix}$ and A is invertible, then $\begin{bmatrix}x1 \ x2\end{bmatrix} = A^{-1} \begin{bmatrix}b1 \ b2\end{bmatrix}$

[stoic python]

Timon:

[sharp merlin]

oh thanks

[ocean sequoia]

Verify whether this is a subspace of F^3

yes right? It contains the zero vector, and its closed under both addition and multiplication?

(f + g )(x) = 0 = f(x) + g(x) = 0

f(cx) = 0 =cf(x)

[dusky epoch]

uh

(f + g )(x) = 0 = f(x) + g(x) = 0
f(cx) = 0 =cf(x)

what are f and g

and what's x

your set is in fact not closed under addition

(1,1,0) and (0,0,1) are both in the set, but their sum (1,1,1) is not

[ocean sequoia]

Oh i think im misunderstanding how you were suppose to add them

hang on I thought that f(x) = x1 * x2 * x3

and that g(x) = x1 * x2 * x3

[dusky epoch]

uh

...

ok so you've given two different names to the same function

dunno why you'd even consider it

[ocean sequoia]

oh shit

im a moron

wow

[dusky epoch]

like

[ocean sequoia]

no

[dusky epoch]

you overthought this

massively

[ocean sequoia]

yea

fuck.

thanks

ugh i got so caught up in trying figuring it had to be provable i did mental gymnastics

[elder robin]

I though I knew how to do this, but the way I solved it makes it seem like any h would make it linear dependent. I start by writing the augmented matrix with {v_1, v_2, v_3, 0}, then get it into RREF which gets me {{1, -2, 0, 0}. {0, 0, 1, 0}, {0, 0, 0, 0}.

Which leaves x_3 as a free variable, and h is nowhere in the solution set, so any h would work. I must be doing something wrong, right?

[half ice]

@elder robin
In your steps, pay attention to when you divide. Remember that you can't divide by 0

Wait why an augmented matrix? Lol

[elder robin]

I used wolfram alpha to row reduce tbh

[half ice]

Wolfram will probably give you a condition on h

[elder robin]

Augmented matrix because v_1(x_2) + v_2(x_2) + ... = 0 ?

Maybe I wrote that wrong

[sharp merlin]

How does this make any sense

what if A was D ^-1

[half ice]

That's true haha, we just normally drop the augmented part @elder robin

[elder robin]

c_1(v_1) + ... + c_n(v_n) = 0 is linear dependence, as long as it isn't trivial solution

Oh ok

[half ice]

Why is Wolfram failing me

,w determinant {{1,-2,2},{-3,6,-5},{-5,10,h}}

[stoic python]

Display results online and refine query

Requested by Kaynex#7520

[half ice]

Oh huh. It's always dependent

[elder robin]

Maybe it's just a weird question?

[half ice]

That's neat

Yeah that's an odd case. Any h you put in is still dependent

[elder robin]

Nice, ty

[sharp merlin]

does anyone understand this

[half ice]

@sharp merlin
If AD = I
Then D is A's inverse.

Then finally, AD = DA

[sharp merlin]

Yeah i got that one

[half ice]

So C = D in that case

[sharp merlin]

mhm

[half ice]

You're correct with D

A square matrix will always map Rⁿ to Rⁿ

But that map won't be injective/surjective unless the matrix is invertible

[sharp merlin]

ok'

i didnt learn about injective and surjective yet

[half ice]

@sharp merlin
The answer is using the word "onto" is means the same thing as surjective.

[sharp merlin]

oh ok

[half ice]

Worth knowing is the difference between a map "into" or a map "onto" Rn

[sharp merlin]

yeah

Also what does linear combination of columns have to do with making a matrix linear dependent

@half ice

[half ice]

A set of vectors is independent/dependent
A matrix is not

Now, we sometimes just take the columns of a matrix to be the set of vectors.

[sharp merlin]

how do we just the columsn to tell if its dependent

i see that if there is a linear combination with atleast 2 columns it becomes linear dependent

why is that

oh wait i see now

nvm got it

[ocean sequoia]

ok hopefully i didnt over think this one

it does contain the zero vector

if f(x) = x1 + 2x2 + 3x3 and g(x) = x4 + 2x5 +3x6

then f(x) + g(x) = 0 (f + g)(x)

because if f(x) is zero and g(x) is zero than a combination of them must be zero because addition is associative

and its closed under multiplication for sure because if f(x) is 0 then cf(x) = 0 = f(cx)

[subtle walrus]

what even is your notation

like what is (f + g)(x) supposed to be

[ocean sequoia]

honestly thats the notation ive seen online

lmao

and in the book

[subtle walrus]

and what does it mean

[errant imp]

Maybe it's the function that is equivalent to f(x)+g(x)

[subtle walrus]

and what is x

[ocean sequoia]

BAM is right

and its any x's that satisfy the equation right

[subtle walrus]

so an element of F^3?

[ocean sequoia]

yes

[subtle walrus]

why does it have x_1, x_2, x_3, x_4, x_5 and x_6 components

because otherwise f and g make no sense

[ocean sequoia]

wouldnt f and g have to have different elements in the function?

[subtle walrus]

what does that mean

ok so

i can see what you are trying to do

f and g are the same function

so i will just use f

and you are trying to define a function f: F^3 -> F, such that x is in the subspace iff f(x) = 0

which is fine

and the argument for being closed under scalar multiplication is then fine as well (although i doubt you understand that)

for closed under addition you would have to show that f(x+y) = 0 if f(x) = 0 = f(y)

but tbh there is not much reason to even introduce a function like that

[ocean sequoia]

Ok but because we know that f(x) = 0 and f(y) = 0 then by the associate property doesnt f(x+y) = 0?

[subtle walrus]

it does not follow from the associative property

[sonic osprey]

if f(x) = x^2 - 3x + 2, then f(1) = 0 and f(2) = 0, but f(3) = 2

[subtle walrus]

just write it down explicitly

compute the sum of 2 arbitrary vectors

and plug that into f

[ocean sequoia]

so instead do <x1,x2,x3> + <y1,y2,y3> = <x1 + y1, x2+y2, x3+y3> ?

[subtle walrus]

yes

[ocean sequoia]

i think this is where im getting confused

So I have this <x1 + y1, x2+y2, x3+y3> and i know that both of these function individually are 0

[subtle walrus]

<x1 + y1, x2+y2, x3+y3> is not a function

[ocean sequoia]

err sorry

yea

[subtle walrus]

its a vector

[ocean sequoia]

vectors

yea

[subtle walrus]

ok, so you want to check if this vector is in your subspace

[ocean sequoia]

yea

[subtle walrus]

it might be easier if you do something like

z_1 = x_1 + y_1 and so on

then your vector becomes z = (z_1, z_2, z_3)

now can you check if z is in the subset?

[ocean sequoia]

so f(z) = x1 + y1 + x2 + y2 + x3 + y3?

[subtle walrus]

no

forget about the function f for now

just look at the definition of the subset

[ocean sequoia]

so all the members are elements of F^3? where thier sum is 0?

[subtle walrus]

well

an element of F^3 is in the subset

if and only if it satisfies

so just plug in z instead

[ocean sequoia]

z1 + 2z2 + 3z3

[subtle walrus]

yeah

[ocean sequoia]

which is really just

x1 + y1 + 2(x2 + y2) + 3(x3 + y3)?

[subtle walrus]

exactly

[ocean sequoia]

and thats zero

because

x1 + y1 + 2(x2 + y2) + 3(x3 + y3) = x1 + 2x2 + 3x3 + y1 + 2y2 + 3y3 = 0

[subtle walrus]

i mean yeah

[ocean sequoia]

do i need to show that

sorry ive really new to proofs

i want to get better

[subtle walrus]

i dont know, i am not grading your homework

you should understand why its true

i.e. you can rearrange x1 + 2x2 + 3x3 + y1 + 2y2 + 3y3

to

wait nvm

you did that

so yeah, this is fine imo

i would mention that both x and y are in the subset and thus satisfy the equation, so 0 + 0 = 0

[ocean sequoia]

ok and if you have a minute i think you were right about

and the argument for being closed under scalar multiplication is then fine as well (although i doubt you understand that)

the doubting i understand it part

[subtle walrus]

(notice that you used the commutative law, not the associative law)

[ocean sequoia]

yep

i shouldve said communtative not associative

[subtle walrus]

you can do exactly the same thing you just did but with scalar multiplication

[ocean sequoia]

ok cool

[subtle walrus]

i.e. take a vector x, multiply by scalar a

(set the new vector to z, so ax=z if that helps)

[ocean sequoia]

so basically c(x1 + 2x2 + 3x3) = cx1 + c2x2 + c3x3 = 0

and obviously if f(x) = 0 then cf(x) = 0

[subtle walrus]

you still have to check the equation that defines the subset

[ocean sequoia]

i keep typing to in wrong im not sure why

[subtle walrus]

which is x_1 + 2x_2 + 3x_3

you dropped the coefficients

[ocean sequoia]

yea im not sure why i keep doing that

but is this ok c(x_1 + 2x_2 + 3x3) = cx_1 + c2x_2 + c3x_3 = 0

[subtle walrus]

you are still kinda trying to do this too fast

you would compute c(x_1, x_2, x_3) = (cx_1, cx_2, cx_3)

and then check if cx_1 + 2cx_2 + 3cx_3 = 0

[ocean sequoia]

ohhhh i see

[subtle walrus]

(which still has to be explained why)

here you can use another law

[ocean sequoia]

thats because the distrbutive property

i can pull out the c

[subtle walrus]

yes

[ocean sequoia]

ahh ok

so im skipping that first step

[subtle walrus]

you always want to compute arbitrary sums / scalar products first

and then check if they are in the subset

[ocean sequoia]

cool thanks man i really appreciate all the help

[ocean sequoia]

ok so i dont even need to check closure under addition or multiplication because it doesnt contain the zero vector

therefore its not a subspace of F^3

[pallid rampart]

Yes

[molten mica]

does anyone know how to do this?

[wintry steppe]

this is not linear algebra #prealg-and-algebra

[molten mica]

oh is it pre alg?

[steady fiber]

yes

[cold topaz]

which one is correct? [T1][T2] = T1 circle T2 or
[T1][T2] = T2 circle T1?

[eternal finch]

The former.

[ocean sequoia]

Ok so the trick on this problem is to show the sum of a differentiable functions is also differentiable

Ok so the trick on this problem is to show the sum of a differentiable functions is also differentiable

the f'(-1) = 3f(2) is "unimportant"

[half ice]

Did you have any questions about it? You seem to have a good idea

@ocean sequoia

[ocean sequoia]

nope just double checking sorry

this is like the first proofs ive done on my own and im trying to avoid mental misteps and i made a big one earlier thanks for asking

[wintry steppe]

hey, guys, just wondering why and how is that for finite groups of order "prime number" we only get just the cyclic group of that prime number?

[dusky epoch]

for any finite group G and for any g ∈ G we have that ord(g) divides |G|

if |G| is a prime p then ord(g) can only ever be 1 or p

[wintry steppe]

thanks, is there a way to count number of groups for order of even numbers and non-prime odd?

[sonic osprey]

no its hard

in general

[wintry steppe]

@sonic osprey i guess there must be some relation to prime factorization in some way🤔

[sonic osprey]

There's really not

Unless you're only working with abelian groups

Maybe as an example, there are 1,774,274,116,992,170 groups of order 2048

[wintry steppe]

jesus😲

[sonic osprey]

There really hasn't been an easy way to calculate the number of groups of a certain order

This is easy when you're only working with abelian groups but

[wintry steppe]

and then you see this monster group thing, what that all about?🤔 like how do they even find these lol

[sonic osprey]

I'm actually doing research on the monster group right now

Usually big groups are constructed as automorphism groups of certain objects

[wintry steppe]

oh, i just realized i was asking q's in the wrong chat lol

@sonic osprey you mean like the symmetries of some weird n-dimensional shapes for example?

[sonic osprey]

eh, you can take the symmetries of things that aren't always geometric

like you can take the symmetries of graphs, or the automorphism groups of groups

or the automorphism groups of vector spaces

[wintry steppe]

might seem like a stupid question but is there a notion for like a continuous number of elements in there group? like for infinite groups, is the order of group equal to aleph-null or aleph-one?🤔

would there be any difference anyway?

[sonic osprey]

I mean, you have infinite groups?

And you have infinite groups of different cardinalities?

And they can't be isomorphic?

I'm not really sure what you're trying to ask

[wintry steppe]

sorry, i think i meant to ask whether there exists things like Lie groups, where there is sort of continuous symmetry

[sonic osprey]

Ah yeah that's a bit different

The first step to Lie groups is to consider topological groups

Where basically the operation of multiplication of your group, and the operation of inverses have to be continuous functions

and you can use topology things to say certain things about your group

The unit circle S^1 is the classic example of a topological group (and also a Lie group)

where the elements are e^{ix} and the operation is multiplication

[subtle walrus]

(in a more geometrical fashion this group is actually SO(2), the rotations of the plane)

[viscid kernel]

What does the adjoint of a matrix mean ( geometrically )

[quartz compass]

I think I have an answer for that but I'm too lazy to work it out right now so I could be wrong stating these two things from memory

[gray dust]

it represents the inverse linear map, except off by a scaling factor

[quartz compass]

I think it's the derivative of the determinant with respect to a specific entry of the matrix

[sonic osprey]

Wait, adjoint here is just the hermitian right

[quartz compass]

or can be seen as like components of the vectors that are orthogonal

to the original matrix

I'm thinking adjugate maybe

[sonic osprey]

I think both of you are

[viscid kernel]

Hmm Im now starting to get an intuition., but Im still not 100%. Cuz Im actuallt trying to understand the formula of the inverse of a martrix. Inverse(A) = adjoint(A)/determinant

[sonic osprey]

Oh nope that's the adjugate alright

[steady fiber]

(very) few people call it the adjoint, usually the classical adjoint for some reason

[viscid kernel]

@gray glen oh Im sorry I mean that one. I mean adjugate, not adjoint

[steady fiber]

but ya, call it the adjugate

[quartz compass]

the second thing I said is sort of the geometric idea I have in mind

if I remember correctly it's like basically the entries are like looking at cramer's rule

and dividing by the determinant is like normalizing the vectors

I think the main thing to focus on is what the inverse looks like geometrically compared to the original matrix

it's taking column vectors of A and making a matrix of row vectors which have a dot product with them that is either 1 or 0

that being the identity matrix when multiplied

maybe I'm being too hand wavy and lazy to be helpful if you're not willing to try to play around with this yourself but

kind of like how the cross product is constructed, and how it's normal to the other vectors in the determinant

this is either too vague or not, to see why the determinant would normalize this

thinking of a determinant with a single row removed but filled with the basis vectors in the spots turns it into a kind of determinant operator waiting for a single vector

given that they're linearly independent, there's only a single possible vector that can fill this spot when dotted with it and when it does it becomes the determinant itself

it's probably more insightful if I draw pictures of the matrices lol

[viscid kernel]

@quartz compass wow. Thanks for the explanation. Im now understanding it a bit.

[quartz compass]

cool

probably a good thing to focus on as well is just the dot product, like projections in particular since they are kind of a building block for making linear transformations

[ocean sequoia]

I have a quick question regarding subspaces... if have a subspace of R^2 does the subspace have to be able to describe every point in R^2

[dusky epoch]

wdym by "able to describe every point in R^2"

[ocean sequoia]

can every vector in R^2 can be described by the function?

like is that a requirement for it to be a subspace of R^2

[dusky epoch]

what function

[ocean sequoia]

I guess another way to ask it is could R^2 contain a vector that is not in the subspace of R^2?

[dusky epoch]

you're not making much sense rn sorry

[ocean sequoia]

Yea i agree

lol I think i got my definition of a subspace a bit twisted

[gray dust]

you may not know the definition of subspace

[ocean sequoia]

and im trying to straigten it out

The way I understand subspaces is that its closed under addition/ scalar multiplication, contains the zero vector

and has an additive identity

but some how i got the idea in my head that if U is a subspace of V every point in V also must be in U?

which might be a better way to say it?

[dusky epoch]

that idea is wrong

[gray dust]

a subspace is a subset of the given vector space that is...

closed under addition/ scalar multiplication, contains the zero vector

[dusky epoch]

if U is a subspace of V then U ⊆ V by definition

and if you also require V ⊆ U then U ends up being equal to V

[ocean sequoia]

Thanks Ann that makes sense

[viscid kernel]

What does transposing a 2x2 matrix geometrically mean ?

I know how you transpose it but. The problem is that Im now trying to prove that. Det(A) = det(A^t)

[ocean sequoia]

you can just show it A = [A,B|C,D] the det(A) is 1/ad - bc when you take the transpose of a matrix all you are doing is turning columns into rows so you get [AC|BD] detA^t = 1/ad-cb = 1/ad-bc by the commutative property

[hoary agate]

yeah, algebraically thats all

[ocean sequoia]

and that kinda makes sense too if you think about the fact that a matrix is a linear transformation and det(A) is a scalar factor for the parallelipiped when you apply the matrix

so if you like flipped Length and Width for a Rectangle you would still have the same area from an intutive perspective

hopefully someone can make sure that is right for you

[viscid kernel]

Im still confused. 😦 so lets say you are transposing the identity matrix does that mean your i-hat becomes j-hat and j-hat becomes i-hat

[ocean sequoia]

you flip the rows and columns

[viscid kernel]

Yeah but I dont get the example you have given with the paraleliped. I know that the determinent of your three by three matrix is actually the width of the paraleliped, but then I got confused.

[ocean sequoia]

someone might be able to correct me here but i dont think its the width of the paraleliped its the scale factor by which the volume would change

[viscid kernel]

Hmm alright, still thanks for your explanation tho. It means a lot.

[glacial grove]

basic question: how is a vector u - some real number defined in linear algebra

[dusky epoch]

what

[glacial grove]

like v - 3

[dusky epoch]

that isn't defined

can you show where you're seeing this

[glacial grove]

yea that's what i thought, just wanted to make sure I want blanking

[dusky epoch]

i'm curious now tho. where did you see it

[glacial grove]

was thinking how R does element-wise execution on some vector - 1 and was trying to remember if that was defined in linear algebra

[half ice]

If your vectors are real numbers, then it can happen

But your vectors are probably not real numbers

[glacial grove]

yea

[nocturne arch]

hey

anyone experienced with sage math?

[glass harness]

hello

to calculate the adjugate matrix of a matrix, do you find the determinant of the adjugate matrix associated with each element?

and assign the resultant determinant of the adjugate matrix associated with each element as the new element for the adjugate matrix of the matrix?

[gray dust]

the roadmap is matrix of minors -> matrix of cofactors -> adjugate, which are the first 3 steps given at https://www.mathsisfun.com/algebra/matrix-inverse-minors-cofactors-adjugate.html

[glass harness]

youre a god

[vital swallow]

does anyone know if the concept https://en.wikipedia.org/wiki/Frame_bundle is related to https://en.wikipedia.org/wiki/Frame_(linear_algebra) ? I need to learn about the first, but don't know where to go (and have found an accessible book for the second)

[hollow finch]

If a nonsquare matrix transformation has a kernel of only the zero vector, then is the transformation injective?

[nocturne arch]

how to find the basis of the null space of a matrix modulo n, idk how sage math is doing it. help please 🙏

[hollow finch]

The matrix is not square so im pretty sure its definitely not bijective

$Ax_1=Ax_2 \implies A(x_1-x_2)=0 \implies x_1-x_2\in null(A) \implies x_1-x_2=0 \implies x_1=x_2$

[stoic python]

nix:

[hollow finch]

injective means one to one

yeah its cool

[vital swallow]

kernel = 0 <==> injective. doesn't matter if matrix is square or not

[sharp merlin]

is there a way to check is something in the span without having to think about it

like can u do matrices

to figure out the coefficients

[nocturne arch]

one way

for the example u have

is to notice that 3 non scalar multiples of each other 2d vectors are guaranteed to span all of 2d space

[gray dust]

you can set up a matrix A whose columns are v1,v2,v3 and solve Ax=(-3,-8) if you want smth more "mindless", though perhaps a bit more tedious work

[sharp merlin]

@nocturne arch but how do u find coefficients

[nocturne arch]

oh

[west spade]

That just requires the grunt work

[nocturne arch]

i just thought u wanted to know if it was in the span

[gray dust]

is provided a way to mindlessly obtain coefficients
ignores it

[nocturne arch]

also there are many coeffiecent that would solve it

do u want all solutions or only 1?

@sharp merlin

[sharp merlin]

1

[gray dust]

it's sufficient to show there exists at least one set of coeffs

[sharp merlin]

@gray dust but its a 2x3

i would have a free variable

[west spade]

So you could have infinitely many solutions

[sharp merlin]

1 0 5

2 -1 5

[gray dust]

AT LEAST one solution means the vector lies in the span, you realize that?

[sharp merlin]

yeah hold on

[gray dust]

infinitely many solutions doesn't change that one bit

[sharp merlin]

how do i even solve this

1 0 5 -3

0 -1 -5 14

and then

1 0 5 -3

0 1 5 -14

what do i do from here?

to get rid of 5

[nocturne arch]

u can't

[sharp merlin]

what do i from that step then

[nocturne arch]

u are already in row echelon form

[sharp merlin]

oh

[nocturne arch]

so u can write the solution in vector form

[sharp merlin]

so do i just make c3 = 0

[nocturne arch]

u can write ur infite solutions as a parameterized vector function

[gray dust]

look, you just need to show a solution exists. you're not there to get em all

[nocturne arch]

https://textbooks.math.gatech.edu/ila/parametric-form.html

[sharp merlin]

x1 + 5x3 = -3 and 1x2 + 5x3 = -14

[nocturne arch]

@sharp merlin

[sharp merlin]

isnt using free variable easier @nocturne arch

@gray dust yeah ik

[gray dust]

set the free var to whatever you like, then you got a set of fine coefficients

[sharp merlin]

nvm got it

[nocturne arch]

how to find the basis of the null space of a matrix modulo n, n can be a non prime. help please 🙏

[pale shell]

Does anyone know a resource

With linear algebra practice problems

Including proofs and computation

[slow scroll]

I think there used to be a website like that for algebra, but i can't remember the name. Just pick problems from a book or something.

[cursive narwhal]

https://yutsumura.com/category/linear-algebra/

^Try this, perhaps?

The problems aren't very difficult but they're good as practice.

@pale shell

[pale shell]

Thank you

I don’t care too much about difficulty at this stage

Just conceptual understnzifng

[safe garnet]

if vectors X Y Z are linearly dependent

are the vectors X, aX + Y, bY + Z linearly dependent for any a, b in R