linear-algebra

[simple depot]

i guess column one can be made the sum of all columns

and then factor out that sum

then try to triangulate??????????

geez

[half storm]

What exactly do they want you to do?

[simple depot]

oh my bad

Find all values of x that satisfy the following equation

[half storm]

Portugese?

[simple depot]

with a1, a2, a3, a4 in real numbers

spanish lol

its not hard just annoying lol

[half storm]

This notation, is the zero on the right hand side actually the number 0 or the zero vector. Am I looking at the determinant of a matrix?

[simple depot]

yeah thats the determinant

[dusky epoch]

yes this is a determinant

[simple depot]

thats pretty awesome that people with high ranks in the server still hang around these more basic math channels

[half storm]

Well, they are basically helpers. And sometimes something be LA, but it's pretty high level LA stuff and it gets posted in here.

People post high-level stuff in calculus and multivar-calc DE too.

[simple depot]

ahhhh

still pretty nice haha

Sea means let

and i need to find x so that A is not inversible

so i think i should say det(A)=0

[half storm]

Yea

that's what you do

You basically want to expand on that top row and find the roots of the resulting polynomial.

[simple depot]

but how do I make that matrix easier to calculate the determinant for?

wait i do?

i thought its better to leave the top row as is

[half storm]

You expand on it.

I mean you can expand on any row because the determinant of a matrix is a unique number. So no matter what row you choose to do a cofactor expansion on, the result will be the same.

I don't really have any tricks though to make it easier to calculate i.e. a shortcut or anything. I just know what's going to happen eventually is that you end up with a polynomial in the third degree and have to find it's roots.

[simple depot]

yeah i figured the same thing, i tend to know what to do but i just mess up on the execution

[half storm]

Like I know how to do all these but don't know any of the cute tricks that make the stuff easier to calculate

yea same, well not always execution. I just don't know how to make the problems easier for myself sometimes.

[simple depot]

welp fuck

ill go give it a second shot

i just saw the light

thank you

[rough olive]

Hey

I'm given this system of equations:

And I'm asked to find solution to the system. I'm given this information as well:

I'm unsure of what they want me to do with the information above in respect to the system of equations? I know that A is the coefficientmatrix squared

But what do I do with the other information? This:

[half storm]

These are two separate problems yea?

Like finding the solution of that first system has nothing to do with the matrix right?

[rough olive]

No they're not separate, it's one problem. a is set to -1/2 and then I'm given the info above ^

I'm just not sure what they want me to do with the info above

Like.. usually I reduce the system of equations to RREF and then read what the values of x_1, x_2 and x_3 is

[half storm]

You'll do the same thing here.

[rough olive]

but here, for some reason, they give me the above values? what am I supposed to do with them

[half storm]

Take the first system of equations and turn it into an augmented matrix.

then solve for $x_1, x_2, x_3$

[stoic python]

JohntheDon:

[rough olive]

but why would they give me the above then? i.e. $[x_1, x_2, x_3] = [3, 1, 1]$

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

is it because they expect the values of x_1, x_2 and x_3 to be 3, 1, 1 once the system is solved?

[half storm]

Yea, I'm not sure of exactly what's going on. So it wants you to solve the first system of equations right? Then after that it wants you solve the matrix equation $A^2x = \langle 3 , 1 ,1 \rangle$ right?

[stoic python]

JohntheDon:

[rough olive]

If I try to solve this:

I get a weird result:

[half storm]

I'm finding it hard to follow the information that you're giving me.

So that matrix A is a separate matrix.

That's the coefficient matrix A in the matrix equation $A^2 x = \langle 3 , 1 , 1 \rangle$?

[rough olive]

Matrix A is this:

They ask me to insert -1/2 into a and then square the matrix

So it becomes this:

[stoic python]

JohntheDon:

[half storm]

O.k.

[half storm]

Good

now we're getting somehwere

[rough olive]

So far so good

[half storm]

So just solve for what x has to be right? $A^2(x) = \langle 3 , 1, 1 \rangle$

do that.

[stoic python]

JohntheDon:

[half storm]

Just put that in an augmented matrix and solve for what $ \langle x_1, x_2, x_3 \rangle$ has to be.

[stoic python]

JohntheDon:

[rough olive]

That's the thing though, I did that and it gave me this result:

Which simply can't be true since they expect us to do this by hand. There's no way you're gonna get that result by hand.

[half storm]

You can get that result by hand

You would need to get common denominators and such

Wait what equation did you solve?

You want to find the answer to $A^2(x) = \langle 3, 1, 1 \rangle$ right?

[stoic python]

JohntheDon:

[rough olive]

The equation I solved was this:

which is an augmented matrix using A^2 and [3, 1, 1]

[half storm]

Yea that's the right thing, but the matrix calculator I put that into does give a different answer.

than what you posted before

how did you get that answer up top

hold on

wait one sec.

[rough olive]

What answer did you get?

[half storm]

Well i put it in wrong

I didn't see a negative for -13/2

Yea i got the same answer

[rough olive]

I put it into a different calculator and got the following result:

[half storm]

as what you got the first time.

[rough olive]

this seems more plausible:

this is from maple ^

the other result is from an online calculator

[half storm]

lol that's what I got when i didn't put the negative for the -13/2

[rough olive]

oh fuck

[half storm]

but if you do that then you get what you got the first time.

[rough olive]

wait

it's not supposed to be negative

I made a typo it seems

that's why it gave me that fucked up answer

this is the correct answer

makes much more sense

[half storm]

cool

[rough olive]

While I have you here, do you mind helping me with a problem that involves linear transformations? I've always had trouble understanding linear transformations

I'll brb in 2 min

[half storm]

yea

[rough olive]

Alright so I have the following two linear transformations:

I'm asked to show that

Say we have a function F(x)=y, which maps an object x to y and then a function G(y), which maps a function from y to z. The composition of F and G means that objects are mapped from x to z directly, right? this is my understanding of composition

but i'm unsure of how to start solving the above

I'm unsure of where to begin

[half storm]

Well you want to show that $(S \circ T)(y) = y$ for all $y \in \mathbb{R}^2$. Remember that $$(S \circ T)(y) = S(T(y)) = S \begin{pmatrix} y_1 -2y_2 \ y_2 \ y_1 \end{pmatrix}$$

[stoic python]

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

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

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

[rough olive]

Why is S(T(y)) equal to S?

Oh yeah

[half storm]

It isn't equal to S

[rough olive]

that makes sense

[stoic python]

JohntheDon:

[rough olive]

So does this mean I have to put the y values into the x's place in:

[half storm]

I mean you can use x's it doesn't really matter they're just saying that you need to show if you take something in $\mathbb{R}^2$ and apply $ S \circ T $ to it, you get the same vector back out.

[stoic python]

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

[rough olive]

So they're asking me to show for all values of y in $\mathbb{R}^2$, the transformations return y?

[stoic python]

Resident Tracksuit Advisor:

[half storm]

yea

[rough olive]

But.. how am I going to show that? Isn't it obvious that all values of y = y?

[half storm]

That isn't what their asking you. They're asking you that if you take a $y \in \mathbb{R}^2$ and you apply $(S \circ T)(y) = y$. You need to show what you put in you get back out.

[stoic python]

JohntheDon:

[half storm]

They aren't asking you assert that y = y

They're asking you to assert that $(S \circ T)(y) = y$

[stoic python]

JohntheDon:

[half storm]

That's what you have to show.

[rough olive]

And the way to do that is by using an arbitrary matrix to map onto y_1, y_2 and y_3?

[half storm]

No matrices are needed here.

[rough olive]

man this is confusing to me :/

[half storm]

You need to just show that when you take a $y \in \mathbb{R}^2$ that $(S \circ T)(y) = y$.

[stoic python]

JohntheDon:

[half storm]

Here I'll walk you through a proof

Let $y \in \begin{pmatrix} y_1 \ y_2 \end{pmatrix} \in \mathbb{R} ^ 2$. Then
$$ (S \circ T)(y) := (S \circ T)\begin{pmatrix} y_1 \ y_2 \end{pmatrix}$$

[stoic python]

JohntheDon:

[half storm]

$$(S \circ T)\begin{pmatrix} y_1 \ y_2 \end{pmatrix} = S\left(T\begin{pmatrix} y_1 \ y_2 \end{pmatrix}\right)$$

[stoic python]

JohntheDon:

[half storm]

Now you need to use what you know about T right?

So what is T(y)

$$ T\begin{pmatrix}
y_1 \
y_2
\end{pmatrix}
= \begin{pmatrix}
y_1 - 2y_2 \
y_2 \
y_1
\end{pmatrix}

[rough olive]

What I know about T is that $$T\begin{pmatrix}
y_1
y_2
\end{pmatrix}

\begin{pmatrix}
y_1 - 2y_2
y_2
y_1
$$

[stoic python]

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

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

[half storm]

O.k. good

I can see that's what you were going for.

Ok so then take that and apply S right

$$
S\begin{pmatrix}
y_1 - 2y_2 \
y_2 \
y_1
\end{pmatrix}

\begin{pmatrix}
y_1 - 2y_2 + 2(y_2) \
y_1 - 2y_2 + 3(y_2) - y_1
\end{pmatrix}

=
\begin{pmatrix}
y_1 \
y_2
\end{pmatrix}
$$

[stoic python]

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

[rough olive]

so I did have to replace each x with the y values

[wintry steppe]

i think the align environment works with the bot if you want to make this display nicer john

[half storm]

Cool

Sure, It doesn't matter though. You're getting hung up on notation.

I could have said x's here instead of y's.

it doesn't matter.

I need to relaern how to use align enviroment.

[rough olive]

OK so after doing the above, I have to reduce it so that it ends up as $\begin{pmatrix}y_1 \ y_2\end{pmatrix}$

[stoic python]

Resident Tracksuit Advisor:

[half storm]

yea

[rough olive]

which then shows that

[half storm]

Yup

[rough olive]

Ok cool, thanks so much for the help. Yeah, I kind of got hung up on the notation, I don't know why. It's my number 1 issue with math...

[half storm]

I used to too kind of. Once you get more used about the language of how math is expressed in the form of symbols and notation and translating those things into english, you won't really have that problem anymore.

[rough olive]

Yeah it's all about practice I suppose

[gray dust]

$$S\m{y_1-2y_2\y_2\y_1}=\m{y_1-2y_2+2y_2\y_1-2y_2+3y_2-y_1}=\m{y_1\y_2}$$

[stoic python]

RokettoJanpu:

[gray dust]

that whatcha going for?

[half storm]

Yea lol

[rough olive]

wait you can just use \m for vectors? so much easier to write than pmatrix

thanks for that 👍

[gray dust]

\m is custom cmd

[rough olive]

oh okay

[gray dust]

ye it's short for begin matrix

[half storm]

I need to learn how to make my own custom macros.

[gray dust]

it has an optional 1st arg for matrix type ie i can write dets or bmatrix

$\m[v]{1&1\1&1}\times\m[b]{2\2}$

[stoic python]

RokettoJanpu:

[polar imp]

hi - im reading an article that's talking about 3x3 rotation matrices and it says "since R (the matrix) is orthogonal, only 3 of its 9 components are independent." can someone help me understand why this is the case? in particle, they single out R_2,3 R_3,1 and R_1,2...but im not sure why these entries were chosen as the "independent parameters" of the matrix

[rough olive]

@half storm if you're still available, could you also help me with the next question? I'm supposed to find a basis for $ker(T\circ S)$ (kernel of $T \circ S$)

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

The kernel of T is the set of all vectors x such that $T(x) = 0$, right?

[stoic python]

Resident Tracksuit Advisor:

[half storm]

Yea

[rough olive]

So that means that I have to find the set of all vectors $\textit{x}$ such that $T\begin{pmatrix}x_1+2x_2\x_1+3x_2-x_3)\end{pmatrix}= 0$?

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

Or rather $T\begin{pmatrix}x_1+2x_2-2(x_1+3x_2-x_3)\x_1+3x_2-x_3\x_1+2x_2)\end{pmatrix}= 0$, no?

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

since $y_1 = x_1+2x_2 \ y_2 = x_1+3x_2-x_3$

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

My initial instinct tells me to solve the system of equations above where the far-right column is all zeros

[half storm]

Yea that's right.

Yea that's what you want to do.

[rough olive]

Ok cool and that'll be it for this question? that's how i'll get my basis?

[half storm]

Yea

Well

That'll tell you what the vectors have to look like.

But it won't give you the basis exaclty

But you'll be able to get one out of that.

[rough olive]

OK hold on let me solve this system first and then i'll get back to you

[half storm]

o.k.

[zealous widget]

can anybody help prove to me that the maximum determinant of a $3 \times 3$ matrix of $1$s and $-1$s is $4$

[stoic python]

catfood:

[zealous widget]

@half storm maybe?

[wintry steppe]

john gets to experience the thrill of being @'d to answer someone's question

[zealous widget]

this question has been driving me insane for 4 hours now

:(

[half storm]

lol let's hope that I can answer it.

Give me a sec

[rough olive]

I get the following result when solving the system of equations:

doesn't the bottom row imply that it is inconsistent and therefore has no solution

[spice storm]

It means is linear Depenent. So you will have free varibles

[rough olive]

Oh okay

[half storm]

I'm not sure @zealous widget You may want to ask someone else sorry

[rough olive]

so x_3 is in this case a free variable?

[half storm]

I don't know enough about determinants yet and some of their properties.

Yea that's right.

[spice storm]

yea

[rough olive]

x_1 = -2x_3
x_2 = -1x_3?

or rather
x_1 = -2t
x_2 = -1t?

where t is the free variable?

[spice storm]

x_2= x3,

is not -1x_3

[rough olive]

but x_1 is correct?

[spice storm]

Yes

[rough olive]

x_1=-2x_3?

[spice storm]

you are solving for x2

so you switch the signs

to make it x2=x3

[rough olive]

Ok so the first vector of my basis is
$$\begin{pmatrix}
-2t \
t \
0
\end{pmatrix}$$?

[spice storm]

you will have x2=-x3 then by solving x2 you'll have it x2=x3

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

I thought a basis was only one vector?

But John said simply solving the system of equations won't give me the basis for $ker(T\circ S)$ and that i need to do more afterwards

[stoic python]

Resident Tracksuit Advisor:

[half storm]

A basis is not always just one vector. it can be many vectors depending on the linear transformation because the linear transformation determines what the kernel is.

In this case it is only 1 vector, and it's the vector $\begin{pmatrix} -2 \ 1 \ 0 \end{pmatrix}$

[stoic python]

JohntheDon:

[half storm]

A basis for a vector space means that every vector in the space can be expressed uniquely as a linear combination of the vectors in the basis.

[rough olive]

why 1 and -2 instead of -2t and t? I thought you had to include the free variable, which is indicated by the t

but is doing 1 and -2 the same as t and -2t notation wise?

[half storm]

Because a basis is necessarily a linearly independent set.

A basis is a set of linearly independent vectors.

So if you say that $\left { \begin{pmatrix} -2t \ t \ 0 \end{pmatrix} \right } $, then you're saying that the set of all vectors of the form $ \begin{pmatrix} -2t \ t \ 0 \end{pmatrix}$ is a basis for the kernel. And this is not true, because such a set of vectors is not linearly independent.

[stoic python]

JohntheDon:

[rough olive]

Ohhh okay

I see

Cool cool

And you know that the basis found above is the only basis because it is linearly dependent?

is that why you know there's only 1 basis and it's the one above?

[gray dust]

bases in general aren't unique

[half storm]

A basis for the kernel is $ \left { \begin{pmatrix} -2 \ 1 \ 0 \end{pmatrix} \right } $ is linearly independent necessarily.

And yea

a basis is generally not unique.

[stoic python]

JohntheDon:

[half storm]

Any set that is a real multiple of the vector $ \begin{pmatrix} -2 \ 1 \ 0 \end{pmatrix} $ - and the set must contain only one vector - is a basis for the kernel.

[stoic python]

JohntheDon:

[half storm]

Because of the necessity of linear independence.

And because any such vector spans the same space.

@wintry steppe lol I actually found in Friedberg that it does talk about how two sets, whether infinite or finite dimensional, must have the same cardinality. It' like the penultimate statement of that section.

[wintry steppe]

neat

[half storm]

Yea it just states it and gives a refernece to where you can read more about it.

[rough olive]

Alright so next question

For the next part I need to find $ran(T\circ S)$

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

the range is the set of all vectors $\textbf{b}$ such that $T(x) = b$ for some vector $x$ in $\mathbb{R}^n$

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

what is the vector b here?

[half storm]

b is just a vector in the codmain of T.

[rough olive]

how would I go about solving this part? it's not immediately clear to me like the previous one

in the previous one i had to set the values in the far-right column to 0 because a kernel is all vectors x such that T(x) = 0.

[half storm]

Yea you're not setting any vectors equal to anything on the right-hand side in this instance. You just need to find a general form for the vectos that $T \circ S$ maps to.

[stoic python]

JohntheDon:

[half storm]

And that's really it.

[rough olive]

hold on

yeah no i'm not sure

what do you mean by general form? the general form surely must be the thing i wrote earlier i.e.:

with the first row reduced to

[half storm]

Yea , drop the T and it's that.

There's no reason for T to be there because you've already applied T to the vector.

[rough olive]

Oh

Right

[half storm]

So that's literally it, the set of all vectors that have that form.

[rough olive]

i can't tell you how many times i've had assignments like this that simply seem too simple and it always makes me question myself. "Did I do something wrong? This seems too easy!"

[half storm]

Yep, that's it.

[spice storm]

Linear Algebra concepts seem hard but once you understand it is pretty easy

[rough olive]

some of the concepts in LinAlg seem wack as fuck man

we all learn differently though. i couldn't understand much when i read the book they gave us in this course

[spice storm]

Tell me about it. Is hard but ended up passing the class with A- somehow

[rough olive]

but actively asking for help and clarification like this helps me understand so much

[gray dust]

the defn of range isn't actually unique to linalg though, have you seen anything like it before?

[spice storm]

what book you use? there is a cool youtuber that helped me pass the course

[rough olive]

@gray dust believe it or not i haven't lol. or at least I don't remember.

@spice storm I use a book that my profs compiled using parts of two separate LinAlg books

[gray dust]

none of the words domain/codomain/range ring a bell from algebra/precalc?

[rough olive]

no but i'm from denmark and we use a different word for it. this is the first time i've had to use it in english. i probably have encountered it before.

[spice storm]

Do you go to university of copenhagen? I'm looking at the university for my masters in math

[rough olive]

I do yeah

i study computer science though

and my anxiety is at an all-time high right now

[spice storm]

Ahh, I was going to me message you if you know the math department

[gray dust]

do you have a not-so-formal but decent defn of a function?

[rough olive]

I have an oral exam tomorrow. 30 minutes, no notes. if I fail, i get thrown out of uni. There are 10 questions and I have to randomly pick one. I've done 9 out of 10 now. I'm gonna take a short 30 min break from all this BS, but if you guys are still here, i'd love to get some help for the last problem/question.

@spice storm the math department is right next to us and we do some of their courses in CS

I can tell you a bit about it, but you can also read more at the UCPH website

i can imagine you've already done so tho

[cursive narwhal]

Oh you're in the university of copenhagen?

[spice storm]

I have I just want to hear some students experiences

[rough olive]

yeah i am. it's fucking hard man.

[cursive narwhal]

Haha I'm 30 minutes away from you 😄

[rough olive]

no fucking way

[cursive narwhal]

UCPH is really good but really hard

[rough olive]

dude i don't know if i'm not uni material, but i find it sooo fucking hard

the programming parts of CS are okay and i'm good at that

but i almost always fail the math courses

[spice storm]

I hear they are hevaliy good in pure math. I checked and they don't have applied math what is something I want to do

[rough olive]

this linear algebra exam tomorrow is the third time i do the exam

:(

[cursive narwhal]

They do have applied math but it's super theoretical lmao

[spice storm]

How many tries do you have?

[rough olive]

THREE

[spice storm]

😦

[rough olive]

THEY WILL THROW ME OUT IF I DONT PASS TOMORROW

[cursive narwhal]

You'll be okay my dude

[spice storm]

You will do great man

keep your head up

[rough olive]

I hope man, i really want to finish this degree

[cursive narwhal]

Do your exam well and let's have a cup of coffee in malmo after that lmao

[rough olive]

sure thing man

[spice storm]

What time is the exam?

[wintry steppe]

abhi basically asking out tracksuit advisor on a date

😳

[rough olive]

let's all go out on a date

we can do math problems together

[cursive narwhal]

abhi basically asking out tracksuit advisor on a date
Essentially

[rough olive]

or maybe i can just be there for emotional support since i'm shit at math

[pallid rampart]

I wish that you pass your exam with all questions correct

[cursive narwhal]

Especially since my professors are generally pranking me all the way

[rough olive]

thanks @pallid rampart ❤️

i was considering doing a masters in software engineering, but after 1,5 years of doing a bachelor... yeah nope.

[spice storm]

We math commuity are here for you. you'll ace the exam

[cursive narwhal]

Dude for linear algebra it's nothing

Just t(x+y) = t(x) + t(y)

that's it

[wintry steppe]

good luck on your exam, remember to just take a breath and don't panic if a question seems hard

[spice storm]

True but linear Algebra is hard for some people

[cursive narwhal]

If you get stuck on a problem, then just start spam-defining functions between vector spaces

[rough olive]

thanks @wintry steppe! appreciate all the love ❤️ math has never been one of my strengths sadly.

anyway guys, i'm gonna take a 30 min break. I need food, water and an episode of some show. i'll be back:)

[rough olive]

I'm back

@half storm I tried using an online calculator for calculating the basis for the range

the calculator outputs the same result we got in the previous question (x_1 = -2, x_2 = 1), but then it also says this:

specifically this part

i thought the range was supposed to be the general form for a vector?

but here they have specific vectors

[wintry steppe]

If I have $$ a \cdot b = c \cdot d$$ where $$ b \neq d$$ can i conclude anything about a and c

[stoic python]

google:

[wintry steppe]

a, b, c, d are vectors

[pallid rampart]

Let b=0, c be perpendicular to b, a,d be nonzero, then a*b=c*d=0 while a≠c

[limber sierra]

you just added like 2 more data points

no, you cant conclude anything based off what you gave alone

[wintry steppe]

okay

[half storm]

@rough olive that's because those vectors form a basis for the range.

A basis for the range is different than the range itself.

[rough olive]

Oh damn my bad

I should've mentioned I had to find a basis, but I'll just follow this guide

can't I just use the results from the previous question? I mean in this I also have to put the augmented matrix in RREF

simply read the values

[half storm]

I don't really know what you're referring to as in "the previous question"?

[rough olive]

nevermind, I figured it out

i just confused myself as usual

[half storm]

If you want to find a basis for the range all you have to do is express $\begin{pmatrix} -x_1 -4x_2 + 2x_3 \ x_1 + 3x_2 - x_3 \ x_1 + 2x_2 \end{pmatrix}$ as a linear combination of a set of vectors.

[stoic python]

JohntheDon:

[half storm]

it's all good

[rough olive]

what I did was use the reduced augmented matrix from the previous part/assignment (where we had to find the kernel) and simply wrote that since x_3 is arbitrary, we can set it to 1 and use $c_1v_1 + c_2v_2 + x_3v_3 = 0$ to express $v_3$ as a linear combination of vectors ${v_1, v_2}$ as such:

[stoic python]

Resident Tracksuit Advisor:

what I did was use the reduced augmented matrix from the previous part/assignment (where we had to find the kernel) and simply wrote that since x_3 is arbitrary, we can set it to 1 and use $c_1v_1 + c_2v_2 + x_3v_3 = 0$ to express $v_3$ as a linear combination of vectors ${v_1, v_2}$ as such:
```Compile error! Output:

! Missing $ inserted.
<inserted text>
$
l.54 ...the kernel) and simply wrote that since x_
3 is arbitrary, we can set...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info: Calculating math sizes for size <14> on input line 54.
LaTeX Font Info: Try loading font information for U+msa on input line 54.
(/usr/local/texlive/2018/texmf-dist/tex/latex/amsfonts/umsa.fd

[rough olive]

which is the range

[floral thistle]

@rough olive

[rough olive]

what's up

[floral thistle]

What's the problem?

[rough olive]

I have a matrix A:

I'm being told that 0 and 5 are eigenvalues for A and that $\begin{pmatrix}1 \ 1\ 1\end{pmatrix}$ is an eigenvector that belongs to the eigenvalue 5

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

what I have to do is diagonalize A and find an invertible matrix P and a diagonalmatrix D such that $P^-1AP=D$

[stoic python]

Resident Tracksuit Advisor:

[floral thistle]

I have a matrix A:
@rough olive Damn, I don't remember eigenvalues 😕

[rough olive]

aww

[floral thistle]

I'm relearning linear algebra

aww
@rough olive
Sorry

Eigenvalues is literally programmed for tomorrow in my study schedule

[half ice]

I can help haha. Is that all of the eigenvalues?

[rough olive]

Yeah it is

In the previous part I had to find the characteristic polynomium of A

i don't know if you need that

for this

[half ice]

,w matrix {{3,-2,4},{3,-2,4},{3,-2,4}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[rough olive]

wtf

i wish i had known about this earlier...

[half ice]

Rofl well that is the answer in case you ever want to check

[rough olive]

yeah i got the same answer

god damn

[half ice]

WolframAlpha is the site, definitely worth going and playing with

[rough olive]

spent ages computing the determinant of the 3x3 matrix A

lmao

anyway, could you help with the next part?

[half ice]

Sure thing. I guess you don't need explained how the diagonalization is found?

[rough olive]

hold on brb 2 sec

alright i'm back

yes i'd actually appreciate an explanation as to how the diagonalization is found

if you don't mind

[half ice]

Yeah yeah. So the form is
SDS'
I'll use ' as inverse cuz lazy

[rough olive]

SDS?

[half ice]

You want your matrix to "factor" into
SDS'

Or whichever letters you guys use, of course the labels aren't important here

[rough olive]

what is SDS in this context though

[half ice]

D is a diagonal matrix is the only important bit, really

[rough olive]

Okay

Oh hold on

[half ice]

SDS'
Where D is a diagonal matrix
S can be anything

[rough olive]

Ok yeah in this problem P is used instead of S i guess

So SDS^-1 = PDP^-1

[half ice]

PDP' works I'll do that instead

PAP'
Now I've seen everything haha

[rough olive]

Meant PDP

oops

wait no

PAP^1 = D

that's what I meant

sorry i'm all over the place. tired from studying all day

[half ice]

It doesn't actually matter what label you use for these matricies of course

[rough olive]

yeah

true

[half ice]

So that. You can see that J is the diagonal matrix. It's made from the eigenvalues on the diagonal

[rough olive]

Oh okay

[half ice]

Sadly two of those eigenvalues are 0 so it's a little hard to see oop

That's your diagonal matrix

S is the "anything" matrix. It's made using the eigenvectors as the columns

[rough olive]

So if you want to diagonalize a matrix, you simply put zero everywhere except diagonally (where you put the eigenvalues)

[half ice]

That will get you the A in
D = PAP^-1

A is a matrix with only entries on the diagonal. Everything else is 0.

Wolfram calls it J but we mean it as A

[rough olive]

Yeah

one quick question tho

I'm given the info that 0 and 5 are two eigenvalues

but how do I know that the last eigenvalue is also 0?

[half ice]

You'd need to get the characteristic

And factor

[rough olive]

$-\lambda^3+5\lambda^2$

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

that's the characteristic

[half ice]

Giving you those two eigenvalues is kinda pointless haha

-λ²(λ - 5)

With that factorization you can see that λ = 0 is an eigenvalue of multiplicity 2

And if the matrix is diagonalizable you should expect there will be two eigenvectors attached to it

[rough olive]

alright

back to where we left off

J is the diagonal matrix

S is the "anything" matrix whatever that means

[half ice]

So if you took that S, and that J, and actually carried out this multiplication:
SJS^-1

You'd get your original matrix back

[gray dust]

S is a transition matrix leading to the basis wrt which A is diagonal

[half ice]

That's the point. We're factoring our matrix to this form, and this form is REALLY useful for some things

[rough olive]

How are the eigenvectors that the S matrix consists of found?

[half ice]

They are the eigenvectors of our matrix

Want to go over how to find eigenvectors of a matrix?

[rough olive]

yes please, I really need to refresh my memory on this

[half ice]

Yeah np at all.
So we care about eigenvectors v and eigenvalues λ because they satisfy this:
Av = λv

That is, v is a special vector, that when acted on A, just gives that vector back (but stretched/shrunk/reversed possibly)

A won't change that v's direction

Algebraically you can do this:
(A - λI)v = 0

[gray dust]

λI

[half ice]

So what does that mean? If we compute the matrix A - Iλ, then its nullspace is A's eigenvectors

[rough olive]

what does the I stand for

a lowercase L? or a 1? or a uppercase i?

[half ice]

Identity matrix

[rough olive]

Oh okay

[half ice]

Uppercase i

[rough olive]

so uppercase I

cool

[half ice]

You often compute det(A - Iλ) = 0 to get λ

Now we want the nullspace of A - Iλ to get v

[rough olive]

det(A-Iλ) is what I did to get the characteristic polynomium

just a random comment, continue explaining

[half ice]

Best way to do this, find A - Iλ by plugging in your known λ, set this equal to the zero vector. Row reduce and solve the system

[rough olive]

my known eigenvalues are 0 and 5

[half ice]

Let's work with 5 first it's simpler haha

[rough olive]

yeah heh

[half ice]

A - Iλ is:
-2 -2 4
3 -7 4
3 -2 -1

See how that works?

Basically just taking 5 off the diagonal

[rough olive]

yeah i was about to say

you took the 5 from the bottom right

and added it to the -2 in the middle?

why tho

[half ice]

I'll do an extra step:
A - Iλ
|3 -2 4| |1 0 0|
|3 -2 4| - 5|0 1 0|
|3 -2 4| |0 0 1|

[rough olive]

ohhhhh

[half ice]

Basically just algebraically simplify

[rough olive]

right

seems so obvious

[half ice]

Glad I could make it make sense!

A - Iλ is:
-2 -2 4
3 -7 4
3 -2 -1

But I care about the nullspace of this. That is,
|-2 -2 4| |0|
|3 -7 4| = |0|
|3 -2 -1| |0|

[rough olive]

right so you solve the system of equations above

[half ice]

Yaya and row reduction is the easiest way

,w row reduce {{-2,-2,4},{3,-7,4},{3,-2,-1}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[half ice]

Good, we didn't reduce to identity. If I got the identity matrix there, I would have been doing something wrong

So let's say this was put back into an equation format:
x - z = 0
y - z = 0

Just using (x,y,z) for the fun of it atm

I want a free variable. So I'll let z = t. Then I can express the solution completely with this:
x = t
y = t
z = t

The important thing is "how do I interpret this result?"

Take the matrix A - Iλ. We care about the vectors that, when multiplied by this matrix, give back 0. In this case, those vectors look like [t,t,t] or are the vectors with all of the values set the same.

Note this can be written as t[1,1,1]
Or that this is a vector space with basis [1,1,1]

We take [1,1,1] (or any multiple of it!) to be the eigenvalue

[rough olive]

right so that's the first eigenvector

but what about the two others? how do you compute those? are you supposed to use the other two eigenvalues 0 and 0?

| 3 -2 4| |1 0 0|
| 3 -2 4| - 0|0 1 0|
| 3 -2 4| |0 0 1|

[half ice]

Not equal but -

[rough olive]

oh shit right

[half ice]

A - λI is just A in this case haha

But same process. You want to set this to 0 and solve in order to find the nullspace

The trick here is that you'll have two free variables, not one. Using them, you can find the solution is a plane and has two basis vectors

These two basis vectors are your eigenvectors

[rough olive]

This is too hard to wrap my head around. I've been up since 06:00, so I'm going to give up on this one and get some sleep. Have to be up at 06:00 am tomorrow as well...

Thanks for the help though. You're not bad at explaining at all, I'm just tired. I probably would've understood this a couple hours ago.

Anyway, goodnight!

[winged iris]

🤔

[half ice]

No no I get it, this can be a lot especially for a first pass. Come back to it after a nap

Good night!

[polar imp]

hey sorry for the repost (this is the last time, i promise!) but i wasnt able to get any help this morning so wanted to try one more time on the off-chance that someone knows what's up...will try stackexchange or something if not!

im reading an article that's talking about 3x3 rotation matrices and it says "since R (the matrix) is orthogonal, only 3 of its 9 components are independent." can someone help me understand why this is the case? in particle, they single out R_2,3 R_3,1 and R_1,2...but im not sure why these entries were chosen as the "independent parameters" of the matrix

[wintry steppe]

can you post the article

[polar imp]

its a whole book lol, but i guess i can post the relevant passage.. although out of context i feel like it might be difficult to understand whats going on

i mean i understand the gist of why this is true - you can represent a rotation matrix in R3 with 3 values, the Euler angles

but i dont understand how that translates to - "lets pick these 3 specific numbers out of the matrix and call them the independent parameters"

@wintry steppe there it is, if thats at all helpful lol, i marked the footnote in red that im referring to

basically, the problem is, we have a rotation matrix that we are trying to constrain so that it becomes closer and closer to the identity matrix, so at the top of that page, the author is saying, "lets create some loss functions based on these 3 values of the rotation matrix", and the reason why we only need those 3 is because an orthogonal matrix only has 3 independent parameters... but my question is, why those 3 ?

no what you said is correct @wintry steppe , ive read that elsewhere as well

yeah ok

2.39 is just stating what i said, that the product of several rotation matrices together should be the identity matrix

maybe i should just write out the system of linear equations that would arise from the equation R^T R = I and see what happens if you set those 3 values to known constants or something

my guess is that it would simplify in such a way that you could determine the other 6 matrix entries

still leaves the question, why not any other 3 of the parameters, but i guess maybe it doesnt matter in the end

i remember reading something a while back about numerical optimization and using off-diagonal entries instead of diagonal entries or somesuch but cant recall

[celest slate]

Not sure if this is the right channel

oops are you guys doing something already?

[polar imp]

no its fine, we were just talking through some stuff... is there a more appropriate channel?

[celest slate]

I don’t know, I just want to clarify my understanding of what a tensor is

[polar imp]

oh i thought you were saying my question wasnt in the right channel lol

yea i think youre in the right spot

[celest slate]

I think a tensor is basically- well a scalar is a 0 dimensional sheet of numbers, a vector is 1 dimensional, a matrix is 2 dimensional, and a tensor is n-dimensional, where n is the rank of the tensor?

[polar imp]

yeah pretty much

[celest slate]

ok

how do you write down tensors of ranks 3+?

[polar imp]

like how do you notate it mathematically?

[celest slate]

yeah

[polar imp]

good question lol, hopefully someone knows

[celest slate]

secondary question

I don’t really understand what an inner product is

it looks like a dot product, and when I look up the difference I get things basically saying the inner product is a wide class of valid products of which the dot product is an example, but I also get a lot of results for calculating “the” inner product as though there is a preferred one

[wintry steppe]

an inner product is a generalization of the dot product on R^n to arbitrary vector spaces

[limber sierra]

thats a weird description since R^n can admit multiple different inner products

the inner product is a wide class of valid products of which the dot product is an example
this is true
I also get a lot of results for calculating “the” inner product
this is also true, there's a canonical ("standard") inner product on some spaces

on R^n this inner product is the dot product

so if it's not otherwise specified

they generally mean the dot product

[vague cedar]

@polar imp @celest slate you denote a tensor in component form (eg sigma_{ijk}) iirc

[celest slate]

???

[vague cedar]

kinda like how you would denote a matrix via its components when you're talking about matrix multiplication

[celest slate]

yeah but like

I could write down

[vague cedar]

$(AB){ij} = \sum{k=1}^m A_{ik}B_{kj}$

[stoic python]

jacob:

[vague cedar]

same idea

[celest slate]

|A B|
|C D|

[limber sierra]

people rarely care about specific tensors

[vague cedar]

well following the same idea you would write a tensor in a cube form but that's inefficient

[limber sierra]

at least in a mathematical context

so theres not much reason to write it down

but yeah if you really want to you can write it as a "cube"

or as a "list" of matrices

at least in "nice enough" contexts

[celest slate]

so how do you express the formula for which entries of a tensor correspond to what positions

[limber sierra]

jacob just did it

well, gave the matrix version

[vague cedar]

an example is the stress tensor:

[limber sierra]

wikipedia surely has a writeup on this if you want more explicit details

[vague cedar]

in good ole vector notation you would write something like

[limber sierra]

and there are alternate notations fwiw

einstein notation for instance

[vague cedar]

$\mathbf{T}=\mathbf{n}\cdot\mathbf{\sigma}$ but in tensor notation it would look like $T_j=\sigma_{ij}n_i$

[stoic python]

jacob:

[celest slate]

like let’s say I wanted a tensor of rank 3, where the coordinates are represented by (abc), such that if a = b and c, the entry is 1; if a =/= b and c the entry is 0; and if a = exactly one of b and c the entry is 1/2

How would I write that down

[limber sierra]

you just did

[celest slate]

formally I mean

[vague cedar]

piecewise notation seems like the plan here

$T_{abc} = \begin{cases} 1 & a=b=c \ 0 & a\neq b \land a \neq c \ 1/2 & a=b \lor a=c \end{cases}$

[stoic python]

jacob:

[limber sierra]

$(T)_{abc} = \begin{cases}1&a=b=c\0&a\neq b \land a \neq c\\frac12&\text{otherwise}\end{cases}$

yeah you beat me

[celest slate]

Hm

[vague cedar]

i think yours is more correct cuz the 1/2 condition is true for the 1 condition lol

[celest slate]

Okay I have a new question then

[stoic python]

Namington:

[limber sierra]

forgot brackets

theyre usually omitted but probably more familiar to someone who knows matrix notation

[celest slate]

If people usually don’t care about specific tensors.... how are tensors actually used? Is it just a few common tensors that correspond to certain transformations that are usually used, like elementary matrices?

[limber sierra]

theres some more "sophisticated" ways to write tensors

ie ricci calculus

as elements of tensor fields

typically.

[celest slate]

um

[limber sierra]

this is a very important notion in differential geometry

[celest slate]

like an algebra field?

[limber sierra]

no

[celest slate]

hm

[vague cedar]

lol speaking of writing shit in a cube

[limber sierra]

are you familiar with vector fields?

(not vector spaces)

they essentially "assign" vectors to each point

they're commonly used in calc 3/physics visualizations

[celest slate]

yes

[limber sierra]

picture that but you're assigning tensors to each point in a space

[celest slate]

ah

[limber sierra]

(usually a manifold)

these tensor fields have pretty rich structure

[celest slate]

how do you write a tensor field down

[limber sierra]

usually $TM$ or $T(M)$ for a manifold $M$

[celest slate]

I’m more disadvantaged here since I always saw vector fields pictured, but never actually saw an equation defining a vector field

[limber sierra]

but differential geometry is infamous for having a billion different pieces of notation for the same thing

[stoic python]

Namington:

[celest slate]

So again, that notation doesn’t look helpful for describing a specific tensor field

[limber sierra]

sure, they're generally explicitly described

same thing as most other mathematical structures

except, like, sometimes people give groups by their cayley table i guess

[celest slate]

wdym explicitly described?

[limber sierra]

or categories by a graph sometimes

like

described using a mixture of english words and formulas

eg https://en.wikipedia.org/wiki/Riemann_curvature_tensor

[celest slate]

But how do you do math on it if you can’t describe it in the languge of math?

[limber sierra]

"the language of math" is english

okay let me clarify

since thats kind of a cheeky comment

you could express this using fancy-prancy logic if you want

no one in their right mind would do this

even in linear algebra, you use terms like "basis" and "linearly independent" and stuff

and say "consider the vector space given by the basis" or whatever

these descriptions are entirely unambiguous (if used right)

you could express this by, instead of using words, listing a logical formula for every definition you invoke

but why would you do that

[celest slate]

I think an example would help

Can you give me an example of a simple tensor field?

[limber sierra]

probably not

[celest slate]

...?

[limber sierra]

this is the result of fairly sophisticated theory

i dont think theres an easy example available

i mean okay

technically scalar fields count as tensor fields

so just like

pick a number, assign it to every point in space

voila, tensor field (but admittedly a degenerate one)

you could do the same but return a tensor of your choice

if you prefer that

like you can view a tensor field as a function f(x, y, z, ...) = a tensor

[celest slate]

wait what dimension is the field

just any number large

[limber sierra]

(assuming your manifold can have coordinates specified as x, y, z, ... that is)

what do you mean by "dimension"?

[celest slate]

the picture you showed earlier of a vector field was two dimensional

[limber sierra]

uh

it was built on a two dimensional space yes

but you can define a vector field on any (nice) manifold of any dimension

same thing with a tensor field

it can be defined on a manifold of any dimension

this https://maths-people.anu.edu.au/~andrews/DG/DG_chap12.pdf gives the way a mathematician would typically be introduced to the concept.

[celest slate]

ok

[limber sierra]

the point is: when we need to invoke a specific tensor, we generally just give an "english" description of the tensor

rather than using fancy notation

(at least in mathematics, idk what physicists do)

fancy notation does exist

a whole bunch of it in fact

ricci calculus comes to mind

but in general, its far more common for us to just talk about arbitrary tensors from a tensor field

or maybe a specific subset of those tensors or w/e

[celest slate]

To be clear, you mean talk about arbitrary tensors in a specific tensor field, or arbitrary tensors in an arbitrary tensor field on a specific manifold, or arbitrary tensors in an arbitrary tensor field on an arbitrary manifold? Which one of those?

[limber sierra]

any of those, but usually the first

this is what physicists care about for instance

(such as the riemann curvature tensor, which is actually a tensor field because physicists are bad at naming things)

[celest slate]

Oh okay

I thought you meant the third, and I was really confused as to how that could be useful at all if you never cared about the first two

[polar imp]

does anyone know what the significance of "the orthogonal projection of a vector onto the null space of the jacobian of a matrix" is? i understand each of these terms independently (orthogonal projection, null space, jacobian) but i dont understand what the combined terms necessarily mean

or like, why you would do that

guess i would even start by asking, what the projection of a vector onto the null space of a matrix means

[half storm]

I'm guessing you can view the Jacobian of a matrix as a surjective linear operator from the set of functions $f \in C^1$ into $C^1$. I guess every single function that belongs in $C^1$ can be expressed as the direct sum of the set of the range of the Jacobian and the nullspace of the Jacobian. lol whatever that means. I'm now also curious. (Lol this is the completely wrong idea).

[stoic python]

JohntheDon:

[half storm]

Interesting question. I'm also familiar with basically all the terms but I have ot think about exactly what it means too.

[polar imp]

yea @half storm heres one sort of interesting explanation i found (first answer to this post): https://www.quora.com/What-does-projecting-into-nullspace-mean

think exact set of terms comes up a lot in robotics and kinematics, so maybe there is something i can find there

the problem im working on is related to that

[half storm]

Oh cool.

I can see why you might want the projection though for maybe like optimization because the projection of a vector onto the nullspace of the Jacobian matrix would you give you the part of the function where the first partial derivatives are zero

And those are usually the points where there are critical values.

If you see what I'm saying.

[polar imp]

yea thats exactly what this is for - an optimization routine.. so i guess projecting a vector onto this space gives you the part of the vector that doesnt change the function? idk

the matrix that i take the jacobian of contains all of our "errors" in the system that we are trying to minimize

[half storm]

Yea that's what I was thinking.

[ocean sequoia]

so I have a question regarding change of basis. if we have a linear transformation defined in R^2 say (2x,3y) when people say they have different basis they basically mean what they plug into x and y? So then a change of basis would just be the linear transformation that takes your vectors to mine or vice versa

[half storm]

I'm not sure what you're asking but if someone says that there is a change of basis matrix, all they're saying is that they have a matrix that takes vectors with respect to one basis into another. So if we have two bases $\beta$ and $\beta '$, the change of basis matrix from $\beta$ to $\beta '$ is the matrix whose columns are the cooridnate vectors of the basis vectors of $\beta$ with respect to $\beta '$

[ocean sequoia]

thanks

that helps

[half storm]

No problem

[stoic python]

JohntheDon:

[dapper epoch]

Anyone have any suggestions for the best online calculator to check answers for second year college linear algebra? Looking for something UI friendly and can calculate generally anything related to lin alg.

[median forum]

@dreamy iron a sorry. I just meant that R^R can be a natural way of viewing the set of functions form R to R, such that each function is something of the form {f(a) : a \in R}. it helps with some intuitions of the product topology

or rather, viewing it as a product is a good intuition for that set of functions embedded in the product topology

[rough olive]

I fucking did it!! I fucking passed my exams

I'm not getting thrown out of uni

[floral thistle]

LOL

Congrats!

Anyone have any suggestions for the best online calculator to check answers for second year college linear algebra? Looking for something UI friendly and can calculate generally anything related to lin alg.
@dapper epoch Numerical problems?

[wintry steppe]

@rough olive congratulations! you worked hard for that and you got it

you should be proud of yourself

[rough olive]

Thanks❤️ I'm over the top happy

And thanks to everyone here who helped me, big love 😘😘

[slow scroll]

@polar imp I answered this exact question from you five months ago?

[gray dust]

whatcha thonking tera?

[wintry steppe]

'tis interesting, that's all

[slow scroll]

idk why I instantly remembered lol

[wintry steppe]

maybe mike_r is a time traveler

they didn't get a satisfactory answer to their question just now so they went back and asked it

[slow scroll]

i could have worded my answer better ig

[prisma pier]

I don't understand why the order of indices is different in these two equations (19.1) and (19.2). I tried proving the second equation but I got indices that were in a different order. Is this a typo in the book or is there something wrong with what I wrote?

[prisma pier]

actually I'm starting to think that 19.1 is a typo and should be written with indices like in 19.2

because later on it uses:

[bold python]

When determining the set of all matrices a b c d, and show that it is a subspace of m2x2. Is there any otherway to show it instead of using subspace properties?

like in a video, she used that span{v1,v2..} ... is a subspace of V

and wrote the set H (which is the set of all matrices a b c d, abcd are real) as a span of the vectors and said that this was enough to prove that H is a subspace of m2x2

[subtle walrus]

this is using subspace properties?

[neat peak]

can every hilbert space be factorized into tensor products of other hilbert spaces?

[urban seal]

Guys is it called epsilon math when your trying to get a matrix to have 1 and 0 below the 1s? I forgot what that is called but I need to practice/figure out how to do it

[half storm]

It's row reduced echelon form

[urban seal]

Thanks

[zealous widget]

Does multiplying a matrix $A$ by the transpose of its cofactor matrix $C^T$ give a matrix where the sum of the elements of each row of $AC^T$ is equal to $det(A)$?

[stoic python]

catfood:

[gray dust]

yes

[zealous widget]

huh

thanks!

[gray dust]

if it helps recall A^-1=cof(A)^T/det(A), cof(A)^T=det(A)A^-1, Acof(A)^T=det(A)AA^-1=det(A)I

[plush mural]

If this is the way to find the distance of a point and a hyperplane

how come the formula for distance of a point and a plane is

where does the extra D come from? what am I missing? Should you always subtract the last constant when doing all hyperplanes?

[polar imp]

@slow scroll haha thats pretty funny...nice memory! yea i put this book down for a bit, and picked it back up this week. i do remember your response, but i think i just needed some more perspectives. when im trying to understand something i feel like i need it explained a couple different ways, but your answer is helpful for sure

glad my question left an impact on you for some reason lmao

[half storm]

@plush mural That formula is given by the fact that the distance between a point and a blame is the projection of the distance between a point on the plane and the another point onto the normal vector of the plane.

You know that the scalar equation of a plane is given by $Ax + By + Cz = D$ right?

[stoic python]

JohntheDon:

[half storm]

and the normal vector for such a plane is given by the coefficeints $A,B,C$ right?

[stoic python]

JohntheDon:

[plush mural]

Yes I see why both work, but when we are doing hyperplanes as per the first picture, why are we not subtracting the last constant? Going by the fist picture we would not have "-D" in the formula for the distance in the second picture

[half storm]

oh i see what you mean.

sorry.

[plush mural]

ah no problem, any ideas though?

[half storm]

Dunno, could be a typo

[old flame]

Can a linear map $T \in L(V,W)$ exists, if there exists a $v \in V$ such that $Tv \notin W$ ?

[stoic python]

Otoro:

[wintry steppe]

what's the definition of a function from V to W

[old flame]

so I guess the answer is no

@wintry steppe hello, I kept working on the problem from yesterday, lemme show you another proof

Suppose $T \in L(V,W)$ and $null T$ and $range T$ is finite dimensional. Let $(w_1,…,w_j)$ be the basis of $null T$ and $(u_1,…,u_m)$ be the basis of $range T$.

By definition of $range T$, there are vectors in $V$ $(v_1,…,v_m)$ such that $T(v_i)=u_i$ for $i=1,…,m$.

Claim : The vectors $(v_1,…,v_n)$ are linearly independent.
Proof : Consider the linear combination of 0, $\sum_{i=1}^{m} a_i v_i=0$, where $a_1,…,a_m \in F$ $\rightarrow$ $T(\sum_{i=1}^{m} a_i v_i)=\sum_{i=1}^{m} a_i Tv_i=T(0)=0$ $\rightarrow$ $\sum_{i=1}^{m} a_i u_i=0$ Since $(u_1,…,u_m)$ is a basis $\Rightarrow$ $a_1=…=a_m=0$, so concludes $(v_1,…,v_n)$ is a linearly independent list.

Extend the basis of $null T$ with the list $(v_1,…,v_n)$, resulting in the list of $(w_1,…,w_j,v_1,…,v_m)$.

Claim : Since this list contains all the vectors in V that maps to $range T$, it must span V.
Proof : Suppose there exist $b \in V$ such that $Tb \neq u, \forall u \in W$. Then T is not a linear map from $V$ to $W$, which contradicts the existence of a linear map presented in the question.

Therefore, since the spanning list of V is finite, V is finite dimensional

[stoic python]

Otoro:

[wintry steppe]

you've established that there are v_1, ..., v_m such that T(v_i) = u_i, so what does (v_1, ..., v_n) mean (i.e. what is n)?

i'm fairly certain that's not a typo

try looking at the list (w_1, ..., w_j, v_1, ..., v_m)

[gray dust]

@plush mural D!=0 matches an affine subspace (think shifted) of R^3. W being say a 2dim subspace of R^3, there's no "shift", ie a plane eqn describing W has D=0

[half storm]

@old flame so are you asking that can there exists a linear map between two vector spaces if it is the case that for all T functions from V into W, there exists a $v \in V$ s.t. $T(v) = W$? If that's what you're asking, then your question is more about an underlying function can there exist a function between the two sets at all.

But even if this what you're asking this seems weird to me because a function is just a set of ordered pairs i.e. a relation between two sets with specific properties.

It seems weird to me that there would exist a specific element in a set in which you cannot construct ANY function between between the two sets.

[plush mural]

@gray dust I dont know what an affine subspace is but so what you're saying is that in that example D=0? If we have D!=0 should we always subtract is as in the second picture? For hyperplanes in any space?

[old flame]

oh that is a typo sorry

[half storm]

This is more an underlying question about functions / set theory and relations. You can always construct a function between two sets.

And the way you've phrased your question is a not a logically sound statment; the premises are not true.

It is logically valid but not sound.

[gray dust]

@plush mural how familiar are you with subspace?

[wintry steppe]

"f : A -> B is a function" means (among other things) that for every a in A, f(a) is in B, so if T : V -> W is a linear map then you must have T(v) in W for all v in V @old flame (this is towards your first question, which is what i meant by "what's the definition of a function from V to W")

[stoic python]

JohntheDon:

[plush mural]

@gray dust I would say moderately familiar. But so I looked it up, a subspace without an origin etc?

[half storm]

@wintry steppe Like if the statement that he made was logically sound - then his conclusion would be true that there cannot be a linear map between the two spaces because there could not even be a function between the two sets V and W.

[wintry steppe]

Hi

I need some help

Let J be the induced almost complex structure on a complex manifold X, considered as a bundle endomorphism J : (T X )C → (T X )C . Let T X (1,0) denote the subbundle of (TX)C whose fibres are eigenspaces of J with eigenvalue i. Show that TX(1,0) naturally admits the structure of a holomorphic vector bundle.
(c) Suppose Y is a smooth analytic hypersurface of X. The normal bundle of Y in X is the holomorphic vector bundle NY/X on Y which is the cokernel of the inclusion T Y ( 1 , 0 ) ֒→ T X ( 1 , 0 ) | Y . P r o v e t h a t

Can anyone figure it out

[half storm]

@wintry steppe So his statement is logically valid - if the premises were true then you could say that there does not exist a linear map between the two spaces because there could not be a function between the two. But I'm pretty sure his premises aren't true because there can always be a function constructed between two sets. So such an element in V wouldn't exist.

At least that's what I think.

[wintry steppe]

I’m only in grade 8

bad troll

[half storm]

lol

[wintry steppe]

I want it to be analysed and explained to me lmao

I want to do this sort of maths

go to #point-set-topology then

if you're serious

Thats read only

type ,iam adv

But is that stuff grade 8 level

no

nowhere in the world

By far or by a little

far

Frick

i don't think there are 8th graders anywhere in the world doing complex manifolds lmao

Ok

[half storm]

lol Von Neumann wasn't doing doing analysis on manifolds in grade 8 probably.

[wintry steppe]

When did he do that

[half storm]

I have no idea lol.

[wintry steppe]

What is like grade 8 math

What topics

[half storm]

Elementary Algebra

[wintry steppe]

Ok

if you're completely serious mdriss8, there are very many prerequisites to go through before you can even understand the question you posted, and i don't know how much more anyone could say to an 8th grader about the problem beyond just saying which topics of math one must go through in order to answer (let alone understand) the question

[half storm]

lol that's pretty accurate.

[wintry steppe]

Ok

Is it possible to analyse manifolds at end of high school

if you cover a lot of things beforehand on your own, which would require a tremendous amount of effort

just going by usual highschool curriculums? probably not

[half storm]

Is it possible yes, likely no.

You would have to spend alot of time outside of class studying

[wintry steppe]

Ok

[half storm]

Go beyond everything that you learn in class.

[wintry steppe]

Ok

They do nothing in school

[half storm]

Why in particular do you want to do this kind of math is question?

[wintry steppe]

Interested

[half storm]

Right but why?

[wintry steppe]

It seems cool

i don't want to simply say it's impossible to be at the level to properly understand things about manifolds in highschool, because there are definitely bright minds out there who can do so. however, in order to do so there are just so many things you need to do beforehand, which would probably be overwhelming to even the most motivated highschooler

Ok

and this statement you posted looks quite complex

hehe

Lmao

[half storm]

Sure... but I think you're just looking at one problem that is drawing from a lot of different areas that you don't necessarliy under stand and are thinking "I want to do that" which may not be the best approach.

[wintry steppe]

I shouldn’t really try and do this it’s a bad idea..

[half storm]

I mean one day you might be able to

[gray dust]

@plush mural what i mean is if W is say a a 2dim subspace of R^3, W is a plane containing the origin described by Ax+By+Cz=0, D=0 necessarily

[half storm]

I'm not even saying that you can't but you probably want to have some other impetus other than "this kind of looks cool" y'know? But by the time you learn the neccessary topics then you'll probably realize whether you want to do that kind of math or not.

[wintry steppe]

Ye I understand lmao

Is that math needed for computer science

That sort of math

[half storm]

no lol.

I mean I'd ask someone who was doing CS why are they doing analysis on manifolds because that would be very interesting.

[plush mural]

@gray dust aha so hyperplane kind of implies that it is a subspace then?

[half storm]

Generally CS math is nothing like that.

[wintry steppe]

Ok

Would you say that’s phD level

[half storm]

Yea

[gray dust]

@plush mural no the book talking of W's orthogonal complement tips off W is a subspace

[half storm]

It's graduate level. A 2nd year masters student would have a handle on the concepts that it draws on or maybe an advanced graduate / undergrad.

[old flame]

@wintry steppe (w_1, ..., w_j, v_1, ..., v_m) is linearly independent too ? since each list themselves are linearly independent, consider 0 as a combination of this list, then all scalars would have to be zero

[plush mural]

@gray dust aha I see so not being a subspace means not having an orthogonal complement? So you cant say that the normal vector of any plane is its orthogonal complement?

[gray dust]

@plush mural the wording of your q's suggest you're not so familiar with the idea of orthogonal complement

[plush mural]

@gray dust yeah I guess, thank you for the help though

[spiral star]

@old flame yes they are linearly independent (and when your proof is complete it will imply that they form a basis)

but you are missing a correct argument for why those vectors span V

also: if you are really stuck with this exercise, i can show you how i would have solved it. but i leave that up to you

[old flame]

@spiral star it seems like I'm only missing one part that completes the entire proof, I'm not in a hurry, so I want to try thinking of the argument for the span. Thank you for your help though, I will get back to you when I have some thoughts 🙂 I would also really appreciate it if you can show proof, after I hopefully solve the question

[spiral star]

alright 😄

then keep working at it

maybe as some hint: at some point you should probably use the fact that you have a basis for the kernel

[gray dust]

@plush mural but to answer, from this excerpt W very much must be a subspace to speak of W's orthogonal complement

So you cant say that the normal vector of any plane is its orthogonal complement?
this doesn't make sense as one doesn't speak of a single vector as being a subspace's orthogonal complement except in the trivial case where the orthogonal complement of V wrt V is {0}. if you work through the defn, you find that the orthogonal complement of W is span{n} where n is any nonzero vector orthogonal to every vector in W

[floral thistle]

maybe as some hint: at some point you should probably use the fact that you have a basis for the kernel
@spiral star What's kernel? The null space?

[spiral star]

yes

[floral thistle]

Oh

[spiral star]

i guess people around here are more familiar with range and nullspace

but i never learned those terms :p

[wintry steppe]

they are both common

[spiral star]

idk, kernel and image makes more sense to me if you wanna talk about different algebraic structures as well

why should i say kernel and image in group or ring theory but use something different in linear algebra

[wintry steppe]

exactly lol

idk, maybe they're more intuitive terms to people learning linear algebra for the first time

"image" and "range" are straightforward but "kernel" and "null space" aren't exactly

[spiral star]

i guess null space is kinda descriptive for LA

[wintry steppe]

yeah

[floral thistle]

"image" and "range" are straightforward but "kernel" and "null space" aren't exactly
@wintry steppe nullspace is intuitive, tbh

[wintry steppe]

i should have worded that a little better

"image" and "range" are easily understood as the same thing and the terms are intuitive, whereas "kernel" is not easily understood to mean the same thing as "null space," which is a descriptive term and rather intuitive to a beginning LA student

i was halfway through a lecture so i didn't think about how my first message would be read lol

fuck i am too stupid to put what i mean into words

[spiral star]

[wintry steppe]

[floral thistle]

LOL

fuck i am too stupid to put what i mean into words
@wintry steppe Actually, what you said above is pretty understandable.

[wintry steppe]

yeah i edited it like four times

lmao

[gray dust]

@wintry steppe pay attention to lecture

[wintry steppe]

it was an optional lecture that got too confusing because the instructor started handwaving insanely hard

[gray dust]

ew

[wintry steppe]

and it was supposed to end at 3 but kept going so i left

[gray dust]

that's the worst

[floral thistle]

it was an optional lecture that got too confusing because the instructor started handwaving insanely hard
@wintry steppe I hate that

[devout void]

can someone briefly explain
what
x mod y
means

[wintry steppe]

the remainder left when x is divided by y

[half storm]

Wrong section

That's a discrete math / elementary number theory question

[devout void]

my book is called linear algebra and analytic geometry and they used the term here

we have abstract algebra and a lot other stuff

so i wasnt sure

[half storm]

That's cool just giving you a heads up

[devout void]

are complex numbers in linear algebra

idk the proper definition of linear algebra

[wintry steppe]

yes

[devout void]

alright cool

[wintry steppe]

linear algebra is the study of the algebraic properties of linear functions (those functions "respecting scalar multiplication and vector addition"), as well as of vector spaces (the domains of linear functions)

if you want a definition

[devout void]

@wintry steppe thanks btw

oh makes sense we got vectors after complex numbers

this book is weird tho

a lot of stuff mushed together

im Electronic engineering first year

i mean ill go second but will give last exams so studying

thanks for the help TTerra

[wintry steppe]

there is probably a better definition of "linear algebra" but i think the one i just gave is pretty comprehensive

[spiral star]

study of linear maps is a good definition lol

[devout void]

so i^n = (-i^-n) if n<0 ?

[floral thistle]

are complex numbers in linear algebra
@devout void Yes, and also polynomials

[half storm]

You'd probably do better to post it in complex variables section though lol.

[devout void]

wait so that is cause complex numbers can be expressed by vectors as well

[half storm]

yea

[floral thistle]

there is probably a better definition of "linear algebra" but i think the one i just gave is pretty comprehensive
@wintry steppe The study of linear combinations and structures associated with them?

[devout void]

kk makes sense

[floral thistle]

wait so that is cause complex numbers can be expressed by vectors as well
@devout void Yes

Look for something called Argand diagram

[devout void]

alrighht ill check complex variables

[floral thistle]

X axis is the real component, Y axis is the imaginary component

[half storm]

Linear algebra is decently described as the study of vector spaces and the mappings between them.

[floral thistle]

Linear algebra is probably best considered the study of vector spaces and the mappings between them.
@half storm I like this one

[devout void]

wait so is there any quick question to the "where do we even use complex numbers and their functions irl?"

[floral thistle]

wait so is there any quick question to the "where do we even use complex numbers and their functions irl?"
@devout void ???

I don't understand your question

[devout void]

why did mathematicians come up with i in the first place

where did it come handy

why do we learn it

[floral thistle]

To solve sqrt(-1)

[devout void]

any particular good use

solve it for what tho

like till today still seems imaginary to me

for the math game

idk

[floral thistle]

https://www.ukessays.com/essays/mathematics/application-of-complex-number-in-engineering.php

Read this

In EE is used a lot

[devout void]

imaginary numbers have essential concrete applications in a variety of sciences and related areas such as signal processing, control theory, electromagnetism, quantum mechanics, cartography, vibration analysis, and many others.

well

i know nothing of those yet

but i got the answer i was looking for 🙂

thanks @floral thistle

[floral thistle]

control theory is gonna make you cry a few semesters from now XD

[devout void]

wdym

ah

shit is it hard xd?

[floral thistle]

I always heard so from my EE friends

[devout void]

dang imagine how hard that is when my mind can comprehend basic linear algebra

what do you study max

[floral thistle]

I studied industrial engineering

Finished several years ago

Now self-studying to get into a data science grad school

[devout void]

well i hope you achieve your dreams