linear-algebra

[winter harbor]

the thing is that U is not right invertible

because again

right invertible linear map iff surjective linear map

but U can't be surjective

[fathom portal]

but if U is on right, we can do SU = I

[winter harbor]

its image are the vectors with first component 0

[winter harbor]

Yeah, SU = I

But US ≠ I

[fathom portal]

yeah, so U can't be left, so its not left invertible?

[winter harbor]

what? ok maybe I am getting confused with the left/right invertible thing

[fathom portal]

left invertible U means putting S on the left of U?

so if SU = I then U is left invertible

[winter harbor]

No, U being left invertible means that SU = I for some S.

and U is left invertible because it is injective

in fact

we explicitly constructed S

which is a left inverse for U

But U is not right invertible

i.e

US ≠ I

so if T=I

Then STU = I

[fathom portal]

why bring T?

[winter harbor]

but T^{-1} = I ≠ US

[winter harbor]

because that's what we want to disprove

[fathom portal]

i thought we proved it lol

[winter harbor]

We are working over an infinite dimensional vector space now

the shift operator doesn't even make sense over a finite dimensional vector space.

[fathom portal]

ah ok

[winter harbor]

We are woking over the space of sequences of real numbers.

So V = space of sequences of real numbers

which is infinite dimensional

and the result indeed doesn't work

Notice that finite dimensionality is really necessary here.

[fathom portal]

hmm i see

poggers

[winter harbor]

btw

still working with the shift operator on the space of sequences

try to find a counterexample

where T is not even invertible to begin with.

[fathom portal]

counter what?

[winter harbor]

counterexample?

typo

[fathom portal]

isnt that what we did?

[winter harbor]

I mean

We actually found one

Where T is invertible

we took T = I

but like

T could not even be invertible

and still satisfy STU = I for some S and U the shift operator.

[fathom portal]

in our example, US is not invertible, so we could not, but SU is right?
In this case we don't have (US)^-1

but we do have (SU)^-1

[winter harbor]

Yeah, SU = I.

So SU is invertible

you are correct

[winter harbor]

yup

[fathom portal]

poggers

ty

[winter harbor]

think a bit more about this problem

it is an interesting one

makes you think about why finite dimensionality is such a strong condition

try finding more counterexamples even

[fathom portal]

you can freely shift stuff around or cut out bits, and you can still end up with an infinite range
like, cut off decimals, or every 10th component, or whatever rocks your boat

this one was a mind boggler! i sat there for a long time thinking about it, but finally got it. felt a bit brute force-y

[fringe fjord]

Hmm, a straightforward plan would be to show that if E contains any nonzero matrix, then it also contains every matrix with 1 in a single entry and 0 in the rest.

[fathom portal]

Thats what I did

sorry, I don't know how mathheads write, this is just what I do

never been in a math module in university

[winter harbor]

First of all, you should start being more descriptive and using more words if you want to be better understood.

[fathom portal]

nobody sees my work

this isn't for uni

[winter harbor]

Yeah

[fathom portal]

i write it, learn it, and throw it

[winter harbor]

I mean

[fathom portal]

i just do math for fun

do you understand it?

[winter harbor]

In any case, I think it is a bad habit of people learning math for the first time to not use a lot of words in their proofs.

[winter harbor]

Yeah

The idea is that if E is not zero

[fathom portal]

yeah but idk how to write like mahthead do

[fringe fjord]

Well, you're showing it here, so we're nobody?

[fathom portal]

hmm

i got rekt

[winter harbor]

We can find a linear operator T such that Tv_s = v_c for some v_s and v_c (here you should be less ambigous, but I suppose what you mean is that v_c and v_s are both non zero). And for such v_c, we can always find another linear operator S such that Sv_c = v for any vector v we pick.

That's the first thing.

Then, by hypothesis TS is in E for each such S.

Ok

but the next step just seems plain false

why is it that TSv = v_c ?

again

This is really confusing

[fathom portal]

ahh yeah

something i didn't write

brb

[winter harbor]

anyways 1080p

being more descriptive in your proof is really helpful

both to like

yourself

and the people reading your proofs

Clear presentation is really important

[fathom portal]

Yeah

I usually dont do what i did there

Lol

...Anyway...

Ugh so im out of the cafe, that was weird lol

I was fully in the zone with lin alg

And the girl that sat not far was giving off some energy all that time

Then as i was showing you my solution, she started doing her makeup really slowly looking at me

So, i was a bit torn, i wanted to explain my solution lol, but idk if i should approach her

I felt like, she should just make a move if she wanted to

She loitered around as the cafe was closing, spraying perfume and stuff

(She could have left as the staff said they were closing but i had to pack)

Maybe should have spoke to her idk, but i jist left

[wintry steppe]

[fathom portal]

If im in the cafe, the only time someone makes makes a move or wants to talk, im in the zone 😩

[wintry steppe]

are you saying that lowmath cockblocked you?

[fathom portal]

Idk, feel a bit mixed here, like, if she wanted to approach she should have, but oth she did go half way

Lowmath?? This is linear algebra 😫

I just dont feel the urge, im ace, so this stuff feels more like effort to me. I was just being lazy, should start putting more effort in

[wintry steppe]

exactly

[vague crane]

all math is lowmath

[rapid ivy]

Hi question. "For any reflection matrix A and rotation matrix B, the equality AB = BA holds?

Wouldnt this be true?

I feel like im misunderstanding things

[quartz compass]

try weirder combinations of reflections and rotations maybe, I did a reflection by 45 degree plane then rotated 180 degrees and was able to get two different results

[rapid ivy]

crap, im an idiot

Yeah im looking too surface level

thank you

ill look into it further, I didnt realize this would be false

[quartz compass]

here's a pic of what I had in mind but maybe find your own still and play around a bit

the blue is the mirror plane, the yellow is the rotation, basically I figured I should try to set it up so that the rotation does nothing before I reflect but does something after reflection to get them to split apart

[rapid ivy]

got it

thank you for the clarification

[quartz compass]

you're welcome

[velvet moss]

im having trouble with this question

I tried writing out the definition of fg

then used the algebra of linear transformations

but im not really coming up with anything sufficient

[fringe fjord]

You'll need to do it in stages. A reasonable first stage would be to prove that L(1) must be either 0 or 1.

[wintry steppe]

https://gyazo.com/d529b051cac0b665a399a1d660fcea5c

what does this Q mean

[mental crane]

Let $A=\begin{bmatrix}a_1 a_2\end{bmatrix}$ so that
[a_1=\begin{bmatrix}3\-2\1\end{bmatrix},\quad a_2=\begin{bmatrix}-5\6\1\end{bmatrix}.]
The word "u is spanned by the columns of A" means that
(u) is a linear combination of (a_1) and (a_2).
That is, to say if (u) can be expressed as
[u=k_1a_1+k_2a_2]
for some real numbers (k_1) and (k_2).
The above equation can be reduced to
[\begin{bmatrix}0\4\4\end{bmatrix}=k_1\begin{bmatrix}3\-2\1\end{bmatrix}+k_2\begin{bmatrix}-5\6\1\end{bmatrix}]
and to
[\begin{cases}
3k_1-5k_2=0\
-2k_1+6k_2=4\
k_1+k_2=4.
\end{cases}]
If there is a solution for (k_1) and (k_2), then the sentence in quotation marks is true, else it is false.

[stoic python]

shyzealot

[dusky epoch]

OP posted this same question in #help-11 and it looks like they are away

[wintry steppe]

https://gyazo.com/c92aff8c387afc2a634377472635d687

How come W1 is not in the subspace of W?

My RREF is

https://gyazo.com/af7ea104f26673be26f1e28f2d9c1108

[native rampart]

w1 is in w

w_1=12(u_1)-3(u_2)-3(u_3)

[quiet salmon]

how can I show that the span of a list of vectors is a subspace?

not getting the idea just from the definition of span

[wintry steppe]

you mean vectorspace

[quiet salmon]

I mean I think I get the idea I just don't have no clue on what notation I should use I guess

[quiet salmon]

yeah I want to show that "The span of any list of vectors (v1, ..., vm) of a vector space V is a subspace of V"

[wintry steppe]

oh i see

[wintry steppe]

Proof:

Problem:

u just have to prove this upholds for the span

[quiet salmon]

yeah, nice

I was just confused like

how would I denote a linear combination of a vector v_i and also the linear combination of a vector u_i + v_i

idk if that was clear

but I get it now

thanks

[wintry steppe]

np

Is it trivial that if a vector space has at least 2 elements then the dimension is atleast 1?

[hard drum]

Yes, take a non-zero vector v. Then {v} is a linearly independent set.

[wintry steppe]

do i just say: $0 \in V$ and $0 \ne v \in V => dim V \geq 1$

[hard drum]

\ne

[stoic python]

Jester

[wintry steppe]

ah yes i see

why is it true that number * vector = vector * number?

[fringe fjord]

If it's true, it's just a notational convention.

[wintry steppe]

huh?

[fringe fjord]

A vector space comes with only one scalar multiplication.

[wintry steppe]

so only n*V is valid?

is defined*

[fringe fjord]

Yes, according to most books.

[wintry steppe]

ok makes sense

[fringe fjord]

If you want to define V·n to mean the same as n·V, you're free to do so, of course.

[quiet salmon]

guys

this will sound very stupid

but why is the span of an empty list () equal to {0}

why is it not like {}

bc if the list is empty, the set of all linear combinations of it should also be empty no?

[fringe fjord]

The sum of no vectors is 0.

[quiet salmon]

that sounds very weird

wait

it's just notation then?

oh no nevermind

ok I get it

lol

[wintry steppe]

is this how a proof should be written?

[quiet salmon]

u could try to use more words imo

try to make it more clear what argument you're trying to sustain

[wintry steppe]

hmm but then its hard to type words on a computer

also this proof is wrong

[gray dust]

setting the sum of an empty list to 0 is a handy/common convention

[leaden tide]

(likewise, the product over an empty set is usually taken to be 1)

[wintry steppe]

do nothing of multiplication

[wintry steppe]

bruh they expect me to use this to solve an exercise on a LA book

[dusky epoch]

is this a LA book or a DE book?

[restive raft]

span is also defined as the smallest subspace consisting of the set, so what is the smallest subspace/vector space that contains the empty set? Well dimension zero i.e. the set which is only the zero vector.

[leaden tide]

(the empty set is not a vector space because it has to be a group for addition)

[quaint frost]

Could someone help me with this?

Show that $p=\sum{i=0}^{n-1} x{i} t^{i} \in R[t]$ is a polynomial of degree $(p) \leq n -1$ and $x=\left[x{i-1}\right] \in R^{n, 1}$ is the matrix of the coefficients of the polynomial, then it holds

$$
V{n} x=\left[p\left(\alpha{i}\right)\right]=\left[\begin{array}{c}
p\left(\alpha{1}\right)
\vdots
p\left(\alpha_{n}\right)
\end{array}\right]
$$

[stoic python]

Levens

[quiet salmon]

oh my

[slender yarrow]

Could someone help me with this?
``
Show that $p=\sum_{i=0}^{n-1} x_{i} t^{i} \in R[t]$ is a polynomial of degree $(p) \leq n -1$ and $x=\left[x_{i-1}\right] \in R^{n, 1}$ is the matrix of the coefficients of the polynomial, then it holds

$$
V_{n} x=\left[p\left(\alpha_{i}\right)\right]=\left[\begin{array}{c}
p\left(\alpha_{1}\right)
\cdots
p\left(\alpha_{n}\right)
\end{array}\right]
$$

uh

[stoic python]

aPlatypus

[slender yarrow]

that's a little bit less weird at least

anyway it's being answered elsewhere already

@quiet salmon

[quiet salmon]

ah yes that's way better lol

[humble token]

I’m starting to learn linear algebra and as my first problem I decided to work on rotations, could someone tell me if my deriviation for the rotation matrix in 3 dimensions is correct, I have double checked my calculations https://media.discordapp.net/attachments/989250727643873320/993577975138222221/IMG_0094.jpg

Lemme define the angles

One sec

[wind shore]

any recommendation for advanced linear algebra books that are friendly to read, with good explanations but i guess rigorous enough, with good cover of overall theory?

[zinc timber]

hoffman kunge maybe

[wintry steppe]

friedberg insel spence, but i wouldn't exactly call it advanced

it fits all your other criteria, though

for a slightly unusual topic coverage, you could try reading the book by curtis and place. it's very short and covers some fun stuff, and i might consider placing it in "advanced" just for its topics

for a truly advanced book, see "advanced linear algebra" by roman

[wind shore]

ok i will look into those

[wind shore]

is it used for grad level advanced linear algebra courses?

[wintry steppe]

linear algebra isn't a graduate level topic

either FIS, hoffman/kunze, or axler are used for more sophisticated linear algebra courses. FIS is the better book of the three

i have never seen a course use roman's book

[hushed ocean]

Tiny definitional question: could you say that normalization is projecting onto a parallel vector of unit length?

its been several years since I've learned a definition of normalization, and the one i learned was a formula instead of a "axiomatic" definition

[wintry steppe]

define "parallel"

it makes sense in R^n, but i want a definition that works in any inner product space

[hushed ocean]

having the "same direction"
is there a good axiomatic, linalg way of defining "having the same direction"?

[velvet moss]

linear dependence maybe?

[hushed ocean]

I see, i think that would be beyond me, but thanks. Oh yeah and linear dependence totally makes sense for parallel 🙂

[velvet moss]

direction isnt very abstract way of thinking of it though

[wintry steppe]

if all you ask for is linear dependence, then the definition for normalization is wrong. (1, 0) and (-1, 0) are linearly dependent, but they are their own normalizations

vector divided by its length seems pretty straightforward as a definition

[velvet moss]

is this very abstract though

[wintry steppe]

no?

not really

[hushed ocean]

that does make sense, thank you. Sorry, it's been several years

🙂

[velvet moss]

thought he was asking for a more abstract def of normalization

[wintry steppe]

what do you mean by "abstract" then?

[restive raft]

it is a completely formal definition if you have a norm on a vector space

[wintry steppe]

to follow up on the last thing, you can make sense of angles in any inner product space, so direction should be fine to talk about. gets a little weird to think about when you go to infinite dimensions though lol

and normalization is fine in any normed space (maybe not an inner product space)

i wasn't intending to close out the conversation with the "vector divided by its length" message, because that didn't seem to be what you were asking for

[wind shore]

is there any reasons why courses don't use roman's book?

[wintry steppe]

i don't know

it's probably because it's written for a high level audience, when linear algebra is a low level topic

[wind shore]

oh

well thx for the recommendations, i appreciate your insights @wintry steppe

[winter pond]

Hey. How do you obtain a glide reflection matrix? For example translation of 5 units to the right and reflection by x axis for 3x3 matrix.

[wintry steppe]

translation is not a linear map, so you're not going to be able to represent it by matrix multiplication

this will be an affine transformation, something of the form x -> Ax + b

"5 units to the right" and "reflection by x axis" are a little vague in three dimensions, though. do you mean to work in R^2?

[fringe fjord]

In some application areas (computer graphics), it seems to be common to represent the plane as vectors of the form (x,y,1) such that the last column of a 3x3 matrix can create a translation.

(In more matemathecally dignified terms, that is moving to the projective plane and using projective transformations).

[winter pond]

This is one of my task

[fringe fjord]

The final row of 0 0 1 is what you get in the representation I described.

[winter pond]

If u see at the bottom I have to apparently talk about what is the glide reflection used matrix and how can I mathematically obtain it.

[winter pond]

Okay so how do I represent it as matrix moved by 5 units and reflected by x axis?

[fringe fjord]

If you write out the matrix multiplication, the output is (ax + by + c·1, dx + ey + f·1, 0 + 0 + 1·1). What should a, b, c, d, e, f be such that this is always the same as (x+5, -y, 1)?

[winter pond]

a = 1
b = 0
c = 5
d = 0
e = -1
f = 0

Is it?

So the glide reflection matrix is [(1, 0, 5), (0, -1, 0), (0, 0, 1)]?

[fringe fjord]

Yes.

[wintry steppe]

How would I go about constructing an isomorphism between $C^0[0,1]$ and $C^1[0,1]$?

[stoic python]

Color, Hermes

[wintry steppe]

I really have no idea how I should think about preserving structures and accounting for the functions who aren't continuous after differentiated

isomorphism of vector spaces or of normed spaces?

vector spaces

i would imagine something like C^1 -> C^0, f -> f'

[orchid remnant]

Can I get opinion on my solution?

[stoic python]

AuHasard

[vague crane]

gods

[fringe fjord]

It has a fatal lack of words to explain what you're doing, or even aiming to achieve.

[vague crane]

^

[orchid remnant]

ok, so the first row is just the system of equations that we want to solve right? and then i represented it in an augmented matrixx

second row is just saying that i'm adding row 3 in the matrix with 2*row 1

is this good so far?

last step is where i get the solution for my variables

[wintry steppe]

i don't want to read the whole thing, but i can tell you that the final result is correct

[orchid remnant]

is there a good rule of thumb for when you write a solution like this

like how you should write out the steps?

[wintry steppe]

if you truly insist on writing out a bunch of row operations, do it all in one go instead of rewriting the matrices a bunch of times

[orchid remnant]

wouldn't that omit some steps?

like wouldn't teachers dock points for that

[wintry steppe]

no

you'd still be writing out the same thing, just with less repetition

[orchid remnant]

could you give an example please

[wintry steppe]

here's an example of a shitty row reduction question from my first year linear algebra course

it's all just in one go

(i may have purposely made it bad to spite the graders)

[orchid remnant]

by ”in one go”, do you mean that I should skip all the steps between first and last step?

[wintry steppe]

that's not what i meant

i meant without breaking lines to rewrite matrices

you're wasting a lot of space rewriting the same matrix over and over again, it makes it annoying to read

[orchid remnant]

oh i see, that makes sense!

i assume you used overleaf

[wintry steppe]

i hate overleaf

if the internet goes down or there's a problem with the website, then you cannot work

[orchid remnant]

oh lol, any advice where you can write the solution so it looks better?

because that picture is from texit

[wintry steppe]

i write my latex in vs code

there is an add on or something to it that makes that work out

[fringe fjord]

I find it strange that you switch back to equations halfway through instead of doing row operations all the way to RREF.

[orchid remnant]

is it ineffective to do it that way?

bear in mind that i'm studying this as a HS course, so i've not studied this extensively

[wintry steppe]

it's not really ineffective, just odd

stick with one or the other

[orchid remnant]

what'd I have to do here to make it RREF?

[wintry steppe]

you should figure it out yourself

[orchid remnant]

The first non-zero number in the first row (the leading entry) is the number 1.
✅
The second row also starts with the number 1, which is further to the right than the leading entry in the first row. For every subsequent row, the number 1 must be further to the right.
✅
The leading entry in each row must be the only non-zero number in its column.
So the second column in the first row, and third column in the first and second row?

[wintry steppe]

the last condition is saying that those three entries must become zero for you to be in rref

more row operations are needed

[orchid remnant]

yeah, those numbers in blue must be 0, right?

[wintry steppe]

right

here's a starting point if you're really stuck: use the 1 in the last row to kill the entries above it with row operations

[wintry steppe]

$dim W= 1,$ where $W = {x(t) + y(t) = 0, x,y \in \mathbb{R}}$, right?

[stoic python]

Color, Hermes

[wintry steppe]

I'm a bit confused about x and y being functions here

Wouldnt the set of all solutions just be (x, -x) w/ x being in R

this doesn't make sense

we share the same confusion

set of solutions to what?

what is the full context?

also, what even are the elements of W? is it supposed to be something like the set of (x, y) with x(t) + y(t) = 0? that's my best guess

[dusky epoch]

@wintry steppe where did you see this?

[wintry steppe]

It's asking whether the two spaces are isomorphic

vector spaces

Ah

the variable would be t not the functions, right?

So wouldn't it be dependent on the two functions?

In which case, I'm not sure what the dimension would be

contingent on the funcs

[dusky epoch]

what two functions

i think you might be speaking nonsense

[wintry steppe]

$x_1(t) + x_2(t) = 0$

[stoic python]

Color, Hermes

[dusky epoch]

??

[wintry steppe]

x1 an x2 would be functions here

and the dimension would just be = the nullspace of the two functions

[dusky epoch]

????

what are you talking about?

do you think that elements of W look like pairs of functions?

[wintry steppe]

They take in a value, t, and output another value.

That is my working definition of a function

[dusky epoch]

that is not what i was asking you.

let me try to ask again more clearly

according to you, what do elements of W look like?

are elements of W functions? pairs of functions? lists of numbers? arrows? something else?

[wintry steppe]

for future reference please make sure you post the full question to the server when you ask a question

elements of W would be the nullspace of the two functions $x_1$ + $x_2$, which are undefined

[stoic python]

Color, Hermes

[keen sierra]

how does the parenthetical remark relate to the problem statement? seems like there is missing context

[wintry steppe]

Oh

Geez

Here

[keen sierra]

oh i see, it applies to the subpart you pasted earlier

[dusky epoch]

this does not make any sense

i think you are stuck in a haze of linear-algebraic terminology and the words are not coming out of your mouth right.

[wintry steppe]

lol

that's a funny way of putting it

it really does make no sense though, you should use the words right

[zinc timber]

scientists in science fiction movies

[winter pond]

Hey so I have a doubt

Well I know that to get from blue to red it’s rotation 90° so the homegenous coordinate would be

$$\begin{pmatrix} 0 & 1 & 0\ -1 & 0 & 0\ 0 & 0 & 1\end{pmatrix}$$

However it doesn't end there. Apparently the translation matrix is:

$$\begin{pmatrix} 0 & 1 & 3\ -1 & 0 & 5\ 0 & 0 & 1\end{pmatrix}$$

How does it move 3 units to the right and 5 units up?

[stoic python]

Tony Chiba

[winter pond]

Is (3, 5) center of rotation?

[slender yarrow]

nah (3,5) represents how much you translate your points

try it

take some point $\begin{pmatrix} x\y\1\end{pmatrix}$ and compute $$\begin{bmatrix}1& 0& 3 \ 0&1&5 \ 0&0&1\end{bmatrix} \begin{pmatrix} x\y\1\end{pmatrix}$$

[stoic python]

aPlatypus

[slender yarrow]

@winter pond

[winter pond]

Yeah I'm trying hold up

Okay I took a point x = 0 and y = 3 from the blue coordinates

The translation result I get is $\begin{pmatrix} 4 \ 8 \ 1 \end{pmatrix}$

[stoic python]

Tony Chiba

[slender yarrow]

yeah

if you take a general point (x, y, 1), multiplying by this matrix should give you (x+3, y+5, 1)

[winter pond]

Okay

But here we are just talking about translation

But can u explain with the 90 degree rotation so I'd understand better

Because its rotated 90 degrees along with 3 units right and 5 units up

[winter pond]

Also can you please tell me how do I obtain the last translation matrix shown here because I have to make a report on it lmao. All I know is the 90 degrees rotation from the blue to red but I dont know how to show I got 3 and 5.

[winter pond]

If you could teach me this I could do for the rest of transformation shown here from blue to other colors

[brazen drift]

Would anyone be so kind as to try and see where I went wrong here?

[winter pond]

@slender yarrow

[slender yarrow]

well do you know how to represent rotations in plain 2D (non-homogeneous) coords?

[winter pond]

Yes

$\begin{pmatrix} 0 & 1\ -1 & 0 \end{pmatrix}$

[stoic python]

Tony Chiba

[winter pond]

I also do know with homogenous as well

[slender yarrow]

so why do you ask me how to do it then ?

[winter pond]

But just the 3 and 5 is one thing I dont know how to get by looking at the graph

[winter pond]

Oh I was asking for this

Because the transformation doesnt just end with rotation of 90 degrees

There is more and I want to know how to find that transformation

[vale ore]

Can someone help with this plz

[slender yarrow]

#❓how-to-get-help

it's not really suited here

and there's already 3 million ppl in this channel anyway

[winter pond]

Talking to me?

[slender yarrow]

to grant

[winter pond]

Oh my bad

[slender yarrow]

well if you look at "homogeneous matrix" with translation that you gave at the beginning

can you see in which order rotation and translation are done ?

because the order matters here

[winter pond]

Is it rotated first and translated?

[slender yarrow]

rotating then translating vs. translating then rotating don't necessarily give the same result

[winter pond]

Thats how I see it

[slender yarrow]

yea indeed, if you put a general rotation and translation in your homogeneous matrix you could prove that

then if you just rotate your stickman 90 degrees from its original position, you're one translation away from the transformed stickman right ?

[winter pond]

Right yeah

So do I find the difference between the rotated point and the translated point in the image

Which could eventually give translation vector

[slender yarrow]

well rotated vs rotated+translated yeah

yup yup

[winter pond]

Thank you so much.

Bless u

[brazen drift]

I'm stuck on a linear algebra problem in #help-9 if anyone would be kind enough to help. 🙏

[ivory orbit]

help with this 5^x+2=3^y

[wintry steppe]

what are you asking? what do you even want to do with this equation?

this probably isn't the right channel for it. it doesn't seem relevant to linear algebra

[rapid ivy]

Im struggling to understnad how dot product is different from matrix multiplication

seems like the exact same thing

yes they are "different objects"

but its still the same thing no? or am i thinking about this wrong

[wintry steppe]

how in the world do they seem like the same thing?

matrix multiplication takes two matrices and returns a matrix, the dot product takes two vectors (which could be matrices) and returns a scalar

they can't be the same thing if they give different results

it may be worth remarking, though, that the entries of a matrix product are dot products of the rows and columns of your matrices

i.e. the (i, j) entry of AB is the dot product of the ith row of A and the jth column of B

[rapid ivy]

i guess im confused how a vector isnt necessarily a matrice

do u have any resources so i can look further into that

[wintry steppe]

a (column) vector with n entries is an n x 1 matrix; the dot product of two vectors v and w is (after identifying scalars and 1 x 1 matrices) the matrix product v^Tw (superscript T meaning transpose)

that might be what you're looking for, i was just about to type it lol

so the dot product is really a special case of matrix multiplication after this

so maybe instead of my first message, i should say: the dot product is a special case of matrix multiplication

[rapid ivy]

interesting, ok thank you

any youtube videos you might recommend that helped you grasp this concept of dot product?

Also the idea of vector != matrix

[wintry steppe]

i have never watched a youtube video for math

but 3 blue 1 brown seems to be a common recommendation

[rapid ivy]

thanks

[wintry steppe]

it's usually good to think of vectors as matrices when you're doing things related to the dot product (and inner products in general)

lets you write some formulas neatly and what not (like the dot product one above)

[keen sierra]

it's best to distinguish between vectors and linear maps on the one hand, and the matrices that represent them on the other hand, because a single vector or a single linear map can have multiple matrix representations when expressed with respect to different bases

[rapid ivy]

thank you for clarification

also

for something liek this

would i just plug vectors v and u into a matrix

and reduce

[copper pelican]

basically yeah

[rapid ivy]

am i missing a step?

[copper pelican]

oh wait no

[rapid ivy]

😳

what was i doing wrong?

[stoic python]

ab

[rapid ivy]

oh

crap

[copper pelican]

yeah so you need to do some more stuff

[rapid ivy]

wouldnt i add each vector? to create a "new" vector?

[copper pelican]

wdym by add

[rapid ivy]

wait i think im jut confusing things

like [6,8,10,12]

[copper pelican]

ah no that just gives you another vector in W

[stoic python]

ab

[rapid ivy]

perpendicular to the vector?

[copper pelican]

sort of, except this time we're comparing subspaces

i.e. $W^\bot$ is the subspace of vectors that are each perpendicular to every vector $w \in W$

[stoic python]

ab

[rapid ivy]

ahhhhh

yes im following

so then how do i construct that?

[copper pelican]

hm, so first do you know how to find one vector that is orthogonal to both?

[rapid ivy]

jesus christ i have alot to learn

ill look into that

since no i dont think i know how to find that

[copper pelican]

do you know how to find the cross product?

[rapid ivy]

wait, wouldnt it just mean the dot product is 0?

or wrong thing? 😳

[copper pelican]

yeah that's the definition of orthogonality

i.e. two vectors are orthogonal if their dot product is 0

oh in this case actually you may need to do that manually, since the cross product (a separate thing) is only defined in 3 dimensions

so find a vector that has a dot product of zero with both, then another that has a dot product of zero with all three

[rapid ivy]

wait huh?

so i get if dot product = 0

but where would i apply that

to the span () portion?

[copper pelican]

yeah you at least need two linearly independent vectors that are orthogonal to the spanning vectors in W

a bit of a cop-out, but https://math.stackexchange.com/questions/2307669/find-a-basis-for-orthogonal-complement-in-r⁴ shows a nice way to find these vectors

[vocal bone]

Hello, can someone help with this question pls

I know we need to use the definition of even no. of negative slopes giving positive and odd no. of negative slopes giving negative, however idk how to use it in this case

[wintry steppe]

slopes? what's the definition in sec 1.3?

[vocal bone]

from the shilov textbook

[tired flint]

Why is the ROC different?

a is a constant for |a|<1

In picture one, the ROC is |z|>|a|

in picture 2, the z-transform is equal, but has an ROC of |z|<|a|

[wintry steppe]

this is definitely not linear algebra

try #real-complex-analysis

[tired flint]

ok, thank you

[primal swift]

Solve det this way takes more time

This way also its sarrus but simplified

When you get use to it you manage to solve 3 order dets like 2+2

[quiet salmon]

you could just use the normal way to solve determinants

which is like

going column by column

instead of this confusing arrow bs

[primal swift]

It looks confusing, but is more simple that it looks

Life saver in low time tests

Im telling u, once you get the practice u manage to solve 3x3 dets by head

[echo elm]

I know this is probably very easy to most of you but I just cannot figure this out. I'm trying to get the inverse function for f(x) = 2√(3x+1) + 4
The four is outside of the square root which is what most online calculators don't read
I have the answer but not the steps on how to get there
Some help would be very much appreciated

[wintry steppe]

not linear algebra

also not linear algebra

did you mean the channel above this one?

[wintry steppe]

Can someone give me some tips on the second part of this question?

I'm not sure how I would get matrices from considering the bases

I was told to look at the restrictions that the domain imposes on the function, but the domain is V, so I'm not sure what I would do with that.

[wintry steppe]

do you know the definition of "matrix representing T"?

[wintry steppe]

Yeah, I mean I don't know how I would get the three matrices which compose the block matrix

The matrix T relative to basis B would just be the vectors which B consists of. I could apply a transformation to get that same matrix relative to gamma, but I don't see how I'd turn that into the block matrix

you don't need to find out what those three matrices are

you just need to show that some of the entries of the whole block matrix vanish

I see

[toxic herald]

Sorry for the brash wording its hard to explain with text, but does anyone know how I would go about finding the location of a vector/point on a sphere after an Euler transformation? Best example I can give is say you had a beach ball with the valve facing up/skyward and rotated the ball so that the valve was pointing to the right/horizon. How would you describe the translation/offset from initial to starting in world vector space?

[keen sierra]

there are infinitely many rigid transformations that can move the valve from "up" to "right" - the ball can be rotated about the axis that connects the valve to the center of the ball

[white lantern]

is there a way to calculate what direction rotation one vector is of another

[keen sierra]

in R^3 you can use the cross product

[quiet salmon]

how can I check if a list of polynomials is linearly independent

here's what I wrote

doe sit make sense or am i tripping

this is weird

[wet stratus]

part of it makes sense. but the other part? what exactly do you even want to show? that 1, z, ..., z^n are linearly independent?

[quiet salmon]

yes

[wet stratus]

how did you define polynomials?

as sequences that are 0 after a certain index?

[quiet salmon]

one moment

it's just a notation

so my guess is that hes using z to denote p(z)

but yeah confusing to me as well

[wet stratus]

well z is a polynomial. and p(z) is (probably a different) polynomial

[quiet salmon]

so weird

[wet stratus]

hmm your definition is kinda weird

because it defines polynomials as functions

[quiet salmon]

yup

[wet stratus]

even though they aren't technically

and over finite fields we can have that some polynomials describe the same function

(for example x and x^2 are the same function over F_2)

[quiet salmon]

[wet stratus]

so yeah it's a bit handwavy

[quiet salmon]

wait

[wet stratus]

with this definition you can prove that 1,z,...,z^n etc are linearly independent over say the reals

[quiet salmon]

theres a part i didnt show

might clarify things

[wet stratus]

it's skipping over details there

what it means: let $p(z)=a_0+a_1z+\ldots a_mz^m=0$ the zero function. assume that $a_i\neq 0$ for at least one index $0\leq i\leq m$. Then we know that $p(z)$ has at most $m$ roots. However the zero function has infinitely many roots over $\bR$ (or take another infinite field here) and this gives a contradiction

[stoic python]

Denascite

[quiet salmon]

wow

ok that makes sense

[wet stratus]

and we see where it breaks down over finite fields

[quiet salmon]

it gives a contradiction bc the field in question is infinite

so there cant be a list that is linearly independent over said field?

[wet stratus]

you mean over a finite field?

yes there can if for example m is less than the number of elements in the field

[quiet salmon]

we are over F = R right

my list has length m so finite

i dont get the contradiction part why does it lean towards a contradiction

ooohh I guess because of the definition of the 0 function?

[wet stratus]

so we started with the assumption that a_0+a_1z+...+a_mz^m=0 is the zero function with a_i != 0 for one index i. that is the assumption that 1,z,...,z^m are linearly dependent. then from that we get a contradiction. so 1,z,...z^m are linearly independent over R

[quiet salmon]

alright!! ok

that makes sense yeah

okay

thanks man

can you help me verifying another assumption?

[wet stratus]

yes

well depends on the assumption

[quiet salmon]

any list of vectors containing the zero vector is linearly dependent

[wet stratus]

yes

[quiet salmon]

i can also prove this using a contradiction right

supposing the list is linearly indepdent

[wet stratus]

you can also show it directly

which is easier

[quiet salmon]

how so

[wet stratus]

just show that there exists a nontrivial linear combination of the vectors which equals 0

[quiet salmon]

ok lemme see

[wet stratus]

maybe as a follow up exercise, show that in general if A is a linearly dependent set and A is a subset of B, then B is also linearly dependent

your claim is A={0}

[quiet salmon]

holy shit this concept is confusing

[quiet salmon]

i cant see why this makes sense

[wet stratus]

yeah it's confusing at first but really nice after you understand it

[quiet salmon]

I saw something

in stack overflow that gave me some light

look

tried to write it in my own words kind of

but the actual post is this

deos it make any remote sense

[wet stratus]

I see what you are going for but what you wrote is not what you (hopefully) mean

[quiet salmon]

so

at least one of the bk is not zero right

[wet stratus]

yes but it is important which bk is not zero

you can't just pick one of them

[quiet salmon]

fuck

ok but im almost there right

[wet stratus]

like the stackoverflow post says, it's important that a is not 0.

if a=0 then b=c=0 aswell

but if a not zero then you get a nontrivial combination

[quiet salmon]

ok

the ak not equal 0 must be the one

eait wait

wait wait

OK SO

the ak not equal 0 must be the one accompanying the 0 vector

this is the condition

[wet stratus]

yes

but why

[quiet salmon]

bc then we have a non trivial combination that equals the 0 vector

choosing one ak != 0 and all the other coefficients equal 0

[wet stratus]

yes

[quiet salmon]

alrighttt

[wet stratus]

good

[quiet salmon]

NICE

dude thank you so much

[wet stratus]

can you now do this

[quiet salmon]

right i'll try

hmm

not really getting the idea here tbh

wrote a bunch of nonsense so far

what does it mean for a set to be linearly dependent?

a set, not a list

[wet stratus]

we know that $A$ is a linearly independent set. that means there exists vectors $v_1, \ldots, v_n \in A$ and scalars $(a_1, \ldots, a_n)\neq (0, \ldots, 0)$ such that $a_1v_1 + \ldots + a_nv_n=0$

[stoic python]

Denascite

[wet stratus]

a set is linearly dependent if it contains a linearly dependent list

[quiet salmon]

?

[wet stratus]

now can you find vectors $y_1, \ldots, y_m\in B$ (with $A\subseteq B$) such that there are scalars $(b_1, \ldots, b_m)\neq (0, \ldots, 0)$ with $b_1y_1+\ldots b_my_m=0$ ?

[stoic python]

Denascite

[quiet salmon]

ok so first guess

are these vectors of the form (supposing that the $b_{j}$th term isn't equal to 0): \

$$y_{j} = \frac{-b_{1}}{b_{j}}y_{1} + \dots + \frac{-b_{m}}{b_{j}}y_{m}$$

[wet stratus]

on the left is a vector y_j and on the right is a number

so that can't be true

you are thinking too complicated

[quiet salmon]

oh derp its because i forgot to multiply the vectors

one moment

[stoic python]

texaspb

[wet stratus]

while this rearrangement is true (assuming in the ... you are ignoring y_j), what does this offer you

review what you know from the assumption on A and what you have to show about B

[quiet salmon]

well I have to show that theres a nontrivial linear comb of vectors in B such that they equal the 0 vector aint it?

[wet stratus]

yes

[quiet salmon]

ok so for the same reasoning, we take the bjth not equal 0 and all the rest be equal to 0 thats a non trivial comb

that it?

[wet stratus]

but what is b_j. and which vectors do you pick

[quiet salmon]

oooh

good question lol

but is it just the same thing as we did earlier

this here

hm

the previous example

[wet stratus]

well it's a similar idea

[quiet salmon]

but there we had assumed that 0 is in the list of vectors

well lol I dont know

which vector does it have to be?

[wet stratus]

take a step back

what do we know

[quiet salmon]

that A = {0}

[wet stratus]

no no no

that was the previous thing we did

we know A is a linearly dependent set

[quiet salmon]

ok

im lost

lol sorry

[wet stratus]

ok lets start new

$A$ is a linearly dependent set and $A\subseteq B$. Show that $B$ is also a linearly dependent set

[stoic python]

Denascite

[wet stratus]

$A$ being linearly dependent means that there exists vectors $v_1, \ldots, v_n \in A$ and scalars $(a_1, \ldots, a_n)\neq (0, \ldots, 0)$ such that $a_1v_1 + \ldots + a_nv_n=0$

we need to show that there exists vectors $y_1, \ldots, y_m\in B$ (with $A\subseteq B$) and scalars $(b_1, \ldots, b_m)\neq (0, \ldots, 0)$ with $b_1y_1+\ldots b_my_m=0$

[quiet salmon]

did you really mean linearly independent here?

[wet stratus]

nope

[stoic python]

Denascite

Denascite

[quiet salmon]

ayo

wait a minute

shouldn't this already follow from the definition you gave me of linearly dependent set? 🤔 because given that A \subseteq B, then B already contains a linearly dependent list of vectors, therefore B is linearly dependent

lol

[wet stratus]

which list of linearly dependent vectors does it contain?

[quiet salmon]

THE...

zero vector

?

fuck...

[wet stratus]

no. we don't know whether B contains the 0 vector

[quiet salmon]

is there a hint

[wet stratus]

we know that v1, ..., vn is a list of linearly dependent vectors in A. and we know A is a subset of B

[quiet salmon]

$(v_{1}, \dots, v_{n}) \in B$

[stoic python]

texaspb

[wet stratus]

$(v_{1}, \dots, v_{n}) \in B^n$ or $v_1, \ldots, v_n \in B$ but yes

[stoic python]

Denascite

[quiet salmon]

wow

intereesting!!!

[wet stratus]

ok. next problem. again essentially same idea

[quiet salmon]

damn you really helped me learn this stuff, I appreciate it

[wet stratus]

let $A=(x_1, \ldots, x_n)$ be linearly dependent and $B=(x_1, \ldots, x_n, y_1, \ldots, y_m)$. Show that $B$ is linearly dependent

[stoic python]

Denascite

[quiet salmon]

okie

one sec

it's like the same exercise but we are working with lists now

which makes things a little less easier

hm

[wet stratus]

what do we know from A being linearly dependent

always make sure to check exactly what you know and what you need to show

always step 1 in any proof

[quiet salmon]

alright

so

A being linearly dependent means that there exists vectors v1, ..., vn in A and scalars a1, ..., an != (0, ..., 0) such that a1v1 + ... + anvn = 0

ok so this is what we know

AND ALSO

[wet stratus]

no this was for a set A. now A=(v1, ..., vn) is a list

[quiet salmon]

wait

hm

HAHAHHA

so A = (x1, ...., xn) being linearly dependent means that there exists scalars a1, ..., an with at least one ak != 0 such that a1x1 + ... + anxn = 0...?

[wet stratus]

yes

[quiet salmon]

okay

so far what is noticeable about B is that each x1, ...., xn in A is also in B

so thats something

[wet stratus]

what do we need to show?

[quiet salmon]

that there exists scalars b1, ..., bm with at least one bj != 0 such that b1x1 + ... + bnxn + ... + b1y1 + ... + bmym = 0?

idk if this expression makes sense

but like there is a linear combination

that contains the already known linear combination for A

I'll write this down

[wet stratus]

I'm gonna rewrite it: there exist scalars $b_1, \ldots, b_n, c_1, \ldots, c_m$ not all 0 such that $b_1x_1+\ldots +b_nx_n + c_1y_1+\ldots+ c_my_m=0$

[quiet salmon]

yeah

[stoic python]

Denascite

[quiet salmon]

and we know that the terms with b form a linearly dependent combination, therefore the whole thing is equal to 0

which yields the other thing also being equal to 0?

[wet stratus]

well we haven't chosen the scalars b_i and c_i yet

[quiet salmon]

ok

[wet stratus]

as what do you choose them

[quiet salmon]

if the vectors x's are linearly dependent

shouldn't it be the case that the terms b_kx_k form a linearly dependent comb

[wet stratus]

that will be our goal. but how do we choose the b_k

[quiet salmon]

at least one of them not equal to 0

and the rest equal 0?

[wet stratus]

no

lets again review what we know. we know that A=(x1, ..., xn) is linearly dependent. what does that mean

[quiet salmon]

there exists a1, ..., an st a1x1 + ... + anxn = 0

with at least one ak != 0

[wet stratus]

yes

good. so knowing these a1,...,an. how can we choose b1,..,bn and c1...cm such that b1x1+...+bnxn+c1y1+...+cmym=0

and that at least one of b_k or c_k is not 0

[quiet salmon]

we could take each bj be equal to each aj?

[wet stratus]

yes

and the c_i ?

[quiet salmon]

they all must be equal?

[wet stratus]

to ?

[quiet salmon]

b?

or a

either

[wet stratus]

no

[quiet salmon]

to 0

???

[wet stratus]

yes!

[quiet salmon]

OMG

🤯

I would actually never have thought of that alone

[wet stratus]

then $b_1x_1+\ldots +b_nx_n + c_1y_1+\ldots+ c_my_m=a_1x_1+\ldots+a_nx_n + 0y_1+\ldots +0y_m = 0$

[stoic python]

Denascite

[quiet salmon]

wow

that's just amazing

tbh

such a weird concept

I'll have to practice doing some problems on my own now

[wet stratus]

when we set those c_i= 0 we are "ignoring" those y_i. this is in a sense the same thing we did earlier when A was a subset of B. we ignored all the extra vectors that are now in B and only consider those from A that we already know

[quiet salmon]

makes sense

[wet stratus]

and again the same thing we did much earlier when we did the same thing with 0 being included in the list

we ignored every vector that isn't 0

and then only cared about the 0 vector because that was what we knew something about

[quiet salmon]

that is just amazing

really

thank you so much

[wet stratus]

you're welcome

[quiet salmon]

I'll write a pretty proof here and send it

1 moment

is this good? I really wanna improve my proof writing so hopefully this is ok

[wet stratus]

"$A\subseteq B = {x: x\in A \implies x\in B}$" is wrong. you can just write $A\subseteq B$, therefore $x\in A \implies x\in B$.

[stoic python]

Denascite

[wet stratus]

when defining $b_i$ and $c_i$ you should write $b_i=a_i$ for $i=1, \ldots, n$ and $c_i=0$ for $i=1, \ldots, m$ instead of this imprecise "each"

[stoic python]

Denascite

[wet stratus]

and then you put a couple of extra dots between bnxn and c1y1 which aren't there

but that's it

rest is fine

[quiet salmon]

oh ok!

thought the "each" would suffice this i=1,...,n thing

[wet stratus]

well it's obvious what you mean

but generally just be precise

[quiet salmon]

alright

[wet stratus]

it takes like 2 seconds longer

and half a line

but you have both enough space and enough time for that

[quiet salmon]

got it

anyways, thanks again. I'll do some exercises on my own now and move onto basis and dimension

I got the general idea of dependence

I think

[wet stratus]

yeah do a few more exercises. that never hurts

and always remember. check exactly what you know and what you need to show

[quiet salmon]

yep really important

[wintry steppe]

[torn goblet]

This is probably a really easy question but “T or F: Let A be an n x n matrix and S is an n x n invertible matrix. If x is a solution to the system $$(S^-1AS)x=b$$ then Sx is a solution to the system Ay=Sb”

[stoic python]

chriz

[torn goblet]

I get that’s it’s true but I just don’t understand what exactly is true

[fringe fjord]

You know S^{-1}ASx=b. Multiply both sides of that equation by S from the left.

[torn goblet]

Right, that creates the identity on the left leaving ASx=Sb. So does Y just represent Sx? I feel that that’s way too trivial to worth being a homework problem

I guess I’m just over thinking it. Thanks

[wintry steppe]

which book are you following?

[torn goblet]

Elementary Linear Algebra by Anton

Well, I’m sorta self-learning it with a previous syllabus. Taking Lin alg this year as a freshman so I’m trying to learn some beforehand

[wintry steppe]

same i was learning LA by lang but its too long ig

[torn goblet]

Ah I heard of that one, but yea that’s understandable. Hard to stay motivated sometimes when it feels endless

[wintry steppe]

if any two elements are linearly independent then is the whole thing linearly independent?

[hard drum]

Not necessarily no

Consider (1,0), (0,1), (1,1) in R^2 for example

[quiet salmon]

how to check if a list of vectors spans a vector space?

ex: I want to show that the list ((1,2),(3,5),(4,7)) spans R^2

[cyan zenith]

Just show that the list contains two lin. ind. vectors

[quiet salmon]

could you please elaborate

[cyan zenith]

The dimension of the vector space spanned by a finite system of vectors is the maximum number of linearly independent vectors in the system, and then we call the system of those independent vectors a basis for the vector space

As such, if you have any two independent vectors in R^2, they span R^2

[quiet salmon]

is it valid to check if a vector in my list is a linear combination of the others?

[cyan zenith]

that would be linear dependence

[fringe fjord]

You can check that (and the answer would be that every vector in your list is a linear combination of the two others). But what will you use that information for? It doesn't have much to do with whether the set spans R².

[quiet salmon]

hm

ok

I saw a method, put the vectors in a matrix and row reduce it

check if the resulting matrix is the identity matrix

[fringe fjord]

The resulting matrix can never be an identity matrix: It is 2×3 an identity matrices are square.

[quiet salmon]

yeah

[fringe fjord]

The systematic by-the-book method would be to row reduce and then see whether the matrix has maximum rank.

[quiet salmon]

I don't understand

the book I'm using didn't cover that

[fringe fjord]

Rank is just the number of rows that are nonzero in the row echelon form.

It gives the dimension of the column space (and of the row space) of the original matrix.

[alpine ferry]

is there a way to say that something is an element of a vector or matrix

[vague crane]

"x_i is an element of matrix M", i would guess

[alpine ferry]

no notation?

bc ik u can't use \in

unless you create a set of the elements of a matrix

but there doesn't seem to be a direct way

[vague crane]

index notation, maybe?

[alpine ferry]

I guess it's not useful then?

I guess

just curious

[fringe fjord]

I think one usually says "entries" rather than "elements", to avoid confusion with the "element" concept from set theory. There's no short general notation for "is one of the entries of" that I'm aware of.

[alpine ferry]

oh ok

interesting

[formal sparrow]

Could someone tell explain to mee what <a|b|c> is when a,b,c are vectors? Is it about inner product?

[hard drum]

Could you show us it in context?

I've seen minor variants of that in context for things like the triple product

[fringe fjord]

That looks like bra-ket notation.

In <a|B|c>, the B is a linear operator, and the whole notation stands for the (Hermitian) inner product of a with B(c).

Physicists often use it in contexts where B is self-adjoint, in which case <a|B|c> is also the inner product of B(a) with c.

The point of the notation is to make more explicit this symmetry about which side B works on.

[formal sparrow]

@hard drum and @fringe fjord Thank you.! Also @fringe fjord , if I may ask, I was reading about the Hermetian inner product and it said that the diff between that and normal product is that the Hermetian inner product is F(a,b) = conjugate(F(b,a)). The normal inner product is represented by transpose(U)*V, so like that what is Hermentian inner product presented by? Thank you

[fringe fjord]

(complex conjugate of the transpose of u) times v

This is according to the physics convention where a Hermitian product is linear in the right input and conjugate linear in the left input.
Texts written by mathematicians often use the opposite convention where the conjugate-linear input is the right one. But they then very rarely use bra-ket notation for it.

[formal sparrow]

Thanks

[velvet moss]

What is meant by A(i,k_i)

here

[wintry steppe]

(i, k_i)th entry

the entry in the i'th row and k_i'th column

[velvet moss]

oh I’m dumb thanks

[robust owl]

I have a conventions question. Friedberg defines the associates incidence matrix of a relation on the set {1,...,n} to be the matrix A such that A_{ij}=1 whenever i relates to j and A_{ij}=0 otherwise. Before giving this definition Friedberg says for convenience an incidence matrix has all 0's for diagonal entries. Should I take this as intending to exclude relations that have some entries that relate to themself?

Because if not then sometimes you could have an incidence relation with an associated matrix A such that A_{ii}=0 even though i relates to itself.

[wintry steppe]

where is this

[robust owl]

It's sec 2.3 (the one about compositions of linear xforms and matrix multiplication) in the 5th edition iirc.

Near the end under the applications part

First couple para

[wintry steppe]

ok found it

yeah, i don't really see any reason to make the diagonal entries zero. maybe friedberg just wants to focus on a special case?

i would think of an incidence matrix as encoding a graph's vertices and edges (put a 1 if vertices are connected, 0 otherwise); no reason you can't have an edge from a vertex to itself

i dunno, honestly

[robust owl]

The reason I was thinking of it is q20 showing A+A^2 (for A a dominance relation) has a row with all positive entries except the diagonal. They claim (roughly) you can consider the row number that dominates the max amount of entries (on pg 96 in the para above the example).

So, I let i be that row value and assume it has some zero entry not along the diagonal which i called column j. I kept getting all turned around considering the diagonal entries along row i and row j though.

Mmm, like I think they want you to show row j will end up having at least as many positive entries as j in {1,...,n}-{i} and then that since A_{ij}=0 that A_{ji}=1 so that row j has strictly more positive entries than row i (or I guess that j dominates more things than i).

[gentle igloo]

The following screenshots are from pgs 338 and 339 of Roman’s Advanced Linear Algebra. I’m putting this question here since it’s from a linear algebra book which people here might have read but it’s not necessarily linear algebra. Question placement advise is welcome.

The only way i can resolve these questions is if i assume that what he calls “function composition” is actually some arbitrary group operation on H. Function composition between elements of F and H could also be viewed as some sort of left group action on F. He never clarifies any of this and i feel like my answer is a long shot.

If anyone wants more context I can send more pages from the book

[wintry steppe]

weird, 338 and 339 is hilbert space stuff in my copy

this is what it says in my pdf @gentle igloo. what edition of the book are you using?

this is from the third edition

[gentle igloo]

ah yea i was using the second edition

thanks that clears my confusion for the first two questions but i’m still confused about why the diagram shows a set of functions which doesn’t satisfy property 3

or could i assume that he means “if t is in H and f is in f, and they can be composed, then their composition is in F”

[thin star]

I'm writing a game engine and have some trouble wrapping my head around tetrahedralization of piecewise linear complexes (meshes) which can sometimes be concave. I don't think I need delaunay tetrahedralization and no algorithms on that (with code) have been published anyway. I want the tetrahedralization to make do with the vertices it gets. I need the tetras to calculate properties like volume and center of gravity. I will also be using the tetras for collision testing since they are nice and convex.

Then there's a second tetrahedralization I need, where it's just a point cloud. Basically I'd need the convex hull, and tetrahedralize it based on the leftover point cloud inside it.

Does anybody know any good approaches for this? There is very little material on the net about tetrahedralization. I can only find stuff about triangulation but that's not what I want, I want the 3d stuff.

I've tried to read some papers about 3d delaunay triangulation but not only are the descriptions of the algorithms completely opaque, they are missing a lot of implementation details which I need, I don't have the creativity to "fix up" their poor writing. Another problem of delaunay triangulation is that it needs to insert extra points/vertices sometimes which I don't want.

[serene solstice]

What's the line above det(X +i*I_2)? The conjugate notation?

[thin star]

I just realized this is impossible: if you take a triangle faced prism and rotate one of the faces, you can't tetrahedralize it without extra points. fml.

[zinc timber]

yes

complex conjugate

[wintry steppe]

wth does this mean?

yeah?

[hard drum]

I was responding to what you deleted lol

[wintry steppe]

ah ok lol

[hard drum]

But yes I mean idk the domain so let's call it D (some set of matrices over a field F I assume) and they're defining a map $\langle -, -\rangle: D \times D \to F$ by $\langle A,B\rangle = \text{tr}(AB)$

[stoic python]

potato

[delicate flame]

How do people create simple matrices on a computer?

IS ther a website that allows you to build them. At the moment i write down on a peice of paper

[slender yarrow]

you mean how to display a matrix ?

[delicate flame]

yeah

[slender yarrow]

well there's LaTeX

which the person just above you used

there's quite a learning curve, but it's really good

$$
\begin{bmatrix}
1 & 2 & 3  \\
4 & 5 & 6  \\
7 & 8 & 9 
\end{bmatrix}
$$

[stoic python]

aPlatypus

[slender yarrow]

as an example

[delicate flame]

ok, i could just use visual studio code to display these or a jupyter notebook.

[slender yarrow]

I think LaTeX (or at least a close equivalent) can be used in jupyter notebooks yeah

[delicate flame]

What do you use for your own personal use?

[prisma socket]

if A and B are matrices by the size of nxn and x belongs to R^n then (A+cB)x = Ax+ (cB)x ?

[slender yarrow]

Most of the time I use overleaf (an online latex editor), but you have to write the whole document in latex there, which is a pain for a beginner

using notebooks is a good compromise if you know them

[delicate flame]

i have a overleaf account for a paper i helped with. notebooks i use regular so ill probably stick with those.

[slender yarrow]

is there anything that makes you doubt that statement ?

[prisma socket]

I know it is true but for some reason I need to make sure it is true and I didn't find any statement about that in my book

cuz A and B are matrices

[slender yarrow]

they probably should have mentioned something like that in your book hmm

at least in the form (A+B)x = Ax + Bx

[slender yarrow]

but anyway, if you're paranoid, you can also just prove it from scratch

using the definition of matrix multiplication (among other stuff)

[prisma socket]

yeah

thanks!

[clever totem]

if the group G operates regularly on M meaning that for all m,n im M there exists exactly one g in G so that g.m=n
what can be said about the cardinality of G and M?

[dusky epoch]

well one would think that M and G are in bijection, no? fix some element m_0 ∈ M and map m ∈ M to g ∈ G such that m = gm_0

[torpid moat]

How does one define the adjoint $m^*\colon{M_n(\mathbb{C})}\to{M_n(\mathbb{C})\otimes{M_n(\mathbb{C})}}$ of the multiplication map $m\colon{M_n(\mathbb{C})\otimes{M_n(\mathbb{C})}}\to{M_n(\mathbb{C})}$?

[stoic python]

thestonethatrolled

[zinc timber]

you can use orbit stabilizer for that by noting that id_G * x =x is the only element of the Stab_x(G)

[zinc timber]

what?

[fathom portal]

im a bit bamboozled here

[zinc timber]

hint: V must be infinite dimensional

like what's the first infinite dimensional space comes to your mind

[fringe fjord]

Hmm, what does × even mean in that question?

[wintry steppe]

[fringe fjord]

Ah, probably something like: forget that U1 and U2 are subspaces of V, but view them as unrelated vector spaces and give their cartesian product the obvious vector space structure.

[leaden tide]

probably that

[dusky epoch]

something something full polynomial vector space

[odd glade]

ight so I got a couple of questions on determinant functions and shit

so my first question is

the determinant function supposed to be a function on a single row of matrix and shit

hold the other rows fix right

then how come in the example above

its being used on multiple rows

like

they have D(A_11 * e_1 + A_12 * e_2, A_21 * e_1 + A_22 * e_2)

turns into

D(A) = A_11 * D(e_1, blah blah) + A_12 * D(e_2, blah blah)

and thats fine

but then they break it up again

on the second row

[keen sierra]

it’s a function of the entire matrix and shit, it’s a linear function if all rows except one are fixed

[odd glade]

ok thanks and shit

[graceful mauve]

Suppose we're working over C. If T is an endomorphism of a finite-dimensional space V, and a subspace W is stable under T (that is, T(W) is contained in W), then T has an eigenvector in W

To show this, can I just write T restricted to W in terms of a basis on W and then apply FTA to the characteristic polynomial?

[fringe fjord]

Yeah.

[graceful mauve]

bet, thanks

[fringe fjord]

Unless W is the trivial subspace, of course.

[graceful mauve]

ah, yeah that's a good point

forgor some linear algebra while reading Lie theory 💀