linear-algebra

[dusky epoch]

what's a w/o subscript and what is a^2 supposed to mean

[undone garnet]

I do it in one row

the others are the same

using Vandermonde

for calculate determinant

[dusky epoch]

then your very first line already makes no sense!

[undone garnet]

make no sense?

why?

[dusky epoch]

a is a scalar so what you have there is a 1 by n matrix

matrices that aren't square don't have determinants

[undone garnet]

well

I mean

the others row will do like the very first row

not just do each row

and add them all

[dusky epoch]

write out the first step of what you are trying to do

in full

[undone garnet]

$\begin{vmatrix}a_1^{k_1}&a_1^{k_2}&a_1^{k_3}&\cdots&a_1^{k_n}\a_2^{k_1}&a_2^{k_2}&a_2^{k_3}&\cdots&a_2^{k_n}\\cdots\a_n^{k_1}&a_n^{k_2}&a_n^{k_3}&\cdots&a_n^{k_n}\\end{vmatrix}$\
$\prod_{i=1}^na_i^{k_1}\begin{vmatrix}1&a_1^{k_2-k_1}&a_1^{k_3-k_1}&\cdots&a_1^{k_n-k_1}\1&a_2^{k_2-k_1}&a_2^{k_3-k_1}&\cdots&a_2^{k_n-k_1}\\cdots\1&a_n^{k_2-k_1}&a_n^{k_3-k_1}&\cdots&a_n^{k_n-k_1}\\end{vmatrix}$\
$\prod_{i=1}^na_i^{k_1}.a_i^{k_2-k_1-1}\begin{vmatrix}1&a_1&a_1^{k_3-k_2+1}&\cdots&a_1^{k_n-k_2+1}\1&a_2&a_2^{k_3-k_2+1}&\cdots&a_2^{k_n-k_2+1}\\cdots\1&a_n&a_n^{k_3-k_2+1}&\cdots&a_n^{k_n-k_2+1}\\end{vmatrix}$\
$\prod_{i=1}^na_i^{k_1}.a_i^{k_2-k_1-1}.a_i^{k_3-k_2-1}\begin{vmatrix}1&a_1&a_1^2&\cdots&a_1^{k_n-k_3+2}\1&a_2&a_2^2&\cdots&a_2^{k_n-k_3+2}\\cdots\1&a_n&a_n^2&\cdots&a_n^{k_n-k_3+2}\\end{vmatrix}$\
...\
$\prod_{i=1}^na_i^{k_1}.a_i^{k_2-k_1-1}.a_i^{k_3-k_2-1}\cdots.a_i^{kn-k{n-1}-1}\begin{vmatrix}1&a_1&a_1^2&\cdots&a_1^{n-1}\1&a_2&a_2^2&\cdots&a_2^{n-1}\\cdots\1&a_n&a_n^2&\cdots&a_n^{n-1}\\end{vmatrix}$\
$\prod_{i=1}^na_i^{k_n-(n-1)}\begin{vmatrix}1&a_1&a_1^2&\cdots&a_1^{n-1}\1&a_2&a_2^2&\cdots&a_2^{n-1}\\cdots\1&a_n&a_n^2&\cdots&a_n^{n-1}\\end{vmatrix}$\
$\prod_{i=1}^na_i^{k_n-(n-1)}\prod_{1\le i < j \le n}(a_j-a_i) > 0$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

step by step to easy for lookin

[stoic python]

Nguyễn Thành Trung:

Nguyễn Thành Trung:

Nguyễn Thành Trung:

Nguyễn Thành Trung:

Nguyễn Thành Trung:

Nguyễn Thành Trung:

Nguyễn Thành Trung:

[dusky epoch]

ok yeah that makes no sense

otherwise you could "pull out" all the fucking entries like that, NOT CARING ABOUT WHAT THE ENTRIES ACTUALLY ARE IN ANY CAPACITY, and then conclude that the deteminant of an all-1s matrix is zero, which is CLEARLY BULLSHIT

[undone garnet]

?

I don't do that?

no

[dusky epoch]

that's EXACTLY what you did though.

[undone garnet]

no

[dusky epoch]

yes.

[undone garnet]

a^(k_2-k_1)

[dusky epoch]

oh that.

ok i'll give you that

now the first step makes sense

but the second no longer does

you can't just ignore those 1's now

you tried to pull out a multiplier of a_i^{k_3 - k_2} from each row

but miraculously $1/a_i^{k_3 - k_2}$ turned out to be 1!

[stoic python]

Ann:

[dusky epoch]

@undone garnet

[undone garnet]

huh?

you missomething

[dusky epoch]

i am talking specifically about the second transformation that you did to the det

but if you so insist

[undone garnet]

huh?

is there anything wrong?

[dusky epoch]

you tried to pull out a multiplier of $a_i^{k_2 - k_1 - 1}$ from each row, but then the first column somehow went completely unaffected by it!

[stoic python]

Ann:

[dusky epoch]

SO YES, THERE IS A LOT WRONG WITH WHAT YOU DID!

[undone garnet]

😮

I see it

thank you

[tame root]

How can you tell if the columns of a m x n matrix (where m != n) are linearly independent from reduced row echelon form?

for example, with a 6 x 8 matrix

should there be exactly 6 pivots for the columns to be linearly independent?

[dusky epoch]

a 6 by 8 matrix has 8 columns which are all vectors in R^6 so they are linearly independent precisely NEVER

[tame root]

So is the rule that you need n pivots for the columns of an m x n matrix to be linearly independent?

[dusky epoch]

well

as many pivots as columns so i guess yes

[tame root]

okay thanks

[low steppe]

the dimension of a homogenous linear equation is the number of basis it consists of, right?

and the basis of the vector is whichever row/column that carries a pivot variable, right?

[dusky epoch]

...what

ok let's just say
your question doesn't make any sense but if it did the answer would probably be no

[brittle juniper]

,rotate

[low steppe]

so based on that, this is the simplification

am i getting it right

[simple vector]

$dim\left(Nul\left(A\right)\right)=:dim\left(Nul\left(A^{:T}\right)\right)$

[stoic python]

mehmet:

[simple vector]

Is this statement always true? If so, what is that because

$Rank\left(A\right)=:Rank\left(A^{:T}\right)$ ?

[stoic python]

mehmet:

[slow scroll]

Rank(A^t)=Rank(A) but nullity(A^t) != nullity(A) in general @simple vector

[simple vector]

But is there a link between nullity(A^T) and nullity(A) or do I have to determine both Nul(A) and Nul(A^T) first?

[slow scroll]

Well if you know the number of rows of A, call it m, then nullity(A^t) = m-rank(A)

So for a mxn matrix, the link is nullity(A^t) + n = nullity(A) + m

Do u see why?

[simple vector]

Yes that was it! Was looking for something my prof told me yesterday during the lecture but forget it lol

Thank youu 😄

[slow scroll]

Npnp

[simple vector]

One follow up question: if I have a n x n matrix than nullity(A^t) = nullity(A)?

[slow scroll]

Yea

Basically this all comes from rank-nullity theorem

[simple vector]

Wow thanks I'm gonna read some wikipedia now 🙂

[tulip saddle]

Dont use wikipedia

[slow scroll]

If you haven’t learned it rank-nullity (in matrix language) just says that

rank(A)+nullity(A)= columns of A

This is because rank corresponds to the number of pivot columns in rref(A) and every column that does not have a pivot is “free”, corresponding to a dimension in the null space of A

[simple vector]

yes, I know Rank(A) = total pivots in rref and Dim(A) = total free variables in rref. This makes everything fall together 😄

[tulip saddle]

goodluck on your exam

[simple vector]

Thanks 🙂

[tulip saddle]

geen probleem

[wintry steppe]

,$ \left(
\begin{array}{ccc|c}
\beta & 2\beta + 2 & \beta + 1 & 0 \
1 & 2 & 1 & 1 \
-\alpha & -2\alpha & \beta + 1 & 0
\end{array}
\right)

[stoic python]

yami:

[wintry steppe]

so i have this augmented matrix, what are the steps to find values for alpha and beta so that the system of equations has no, a single one, or infinite solutions?

[slow scroll]

There are some things to notice: 2beta+2 = 2(beta+1) which sort of corresponds to the coefficient of alpha. Think about how you can restrict alpha/beta so that the first and third rows look the same (linearly dependent).

Similarly for no solutions, think about how you can make the system inconsistent (hint use second row to compare to first or third rows)

[jagged saffron]

how do i show that every real matrix is similar to a matrix in rational canonical form over R?

[tranquil junco]

heres a proof of an exercise im gunna post here so ppl can point out inevitable errors

ok wow zoph mean

[stoic python]

hegel:

[subtle burrow]

I mean if T is 3x3 then it has rank 2 else it is 3x2 it is nonsquare thus non invertable

Qed

[half ice]

Assuming U exists

Oh cool I didn't know you could do that

[stoic python]

hegel:

hegel:

hegel:

hegel:

[tranquil junco]

@subtle burrow ik but those theorems arent in this book yet

[half ice]

UT may be square

[tranquil junco]

[half ice]

But it's def not invertible

[tranquil junco]

o ye

ic

[subtle burrow]

how do you get a square matrix operating on a R2 vector that spits out a R3 vector

[tranquil junco]

wut

hes talking about UT

not U

[subtle burrow]

ohh u right u right

[tranquil junco]

this buk didnt even improve T is invertible iff T is non-singular for V and W of different dimension

veri cruel

made me confos for a moment

[half ice]

I like your proof, it works

[tranquil junco]

thanc

[half ice]

You can, with minor adjustment, show that a linear transformation can't make a basis larger

[tranquil junco]

does this apply to any UT where U and T are transformations between spaces of different dimension?

with T from V to W and U from W to V

[subtle burrow]

In general Rank(A*B) ≤ Rank(A)

[half ice]

Yes, that's what I'm struggling to say lol

You can't get the dimension of the output higher than the dimension of the input

So the image of U can't be R³

[subtle burrow]

Well no, you can Rank + Nullity = Dim

But the important thing is your Higher Dim stuff will still be in a lower dim Subspace in the Higher dim

This is talking about the higher dim from lower not the U stuff

[hexed sandal]

can someone help me with this

[paper egret]

#prealg-and-algebra

[subtle burrow]

I believe in you

[paper egret]

what defines a vector space?

as long as two operations are valid, vector addition and scalar multiplication right?

[half ice]

Copy/Pasting this as I've used it already
A vector space is a set that contains:

• Vectors, that you can add and subtract
• Scalars, that you can add, subtract, multiply, divide.
• A scalar multiplication, that takes a vector and a scalar, and returns a vector

As well, for vectors v, u, and scalars a, b, you get:

• Double distributive property:
  v(a + b) = av + bv
  a(v + u) = av + au
• Associative scalar multiplication:
  a(bv) = (ab)v

[paper egret]

vector spaces are ass

ah gotcha

is it right to say that this is like the cartesian coordinate equivalent of functions,

functions in cartesian coordinate

transformations in vector spaces

[half ice]

There's a proper definition that comes in 10 axioms

[paper egret]

tf

because the way i view vector spaces is like a very special kind of place where you can do vector math

hope that isn't a completely wrong way of htinking about it

[half ice]

That's fair, I imagine most people think of it that way. Ultimately, we generalize this idea so you can use the results of linear algebra on things that you might not see as "vectors" at first

[gray dust]

not all vector fields are made of pointy arrows

[half ice]

Polynomials fit the definition of a vector space, so we can get linear algebra results on them

[paper egret]

yes

so many interpretations of vectors

but the math version of vectors breaks it up to like the most basic and fundamental of it, it's nuts

[gray dust]

math version of vectors?

[paper egret]

i.e

[wintry steppe]

vectors are sets

[paper egret]

the physics version of it is udsually denoted by a bunch of arrows, the CS version of it is usually denoted by arrays

[half ice]

We kept what was necessary to get the important results. All vector spaces have a basis, a dimension

[quartz compass]

vectors are elements of a vector space !!

dont "vectors are sets" me

[half ice]

They have eigens too!

[subtle burrow]

vectors are n-tuples, prove me wrong

[paper egret]

in my notes,

H is a subspace of V if
0 is in H
H is closed under vector addition
H is closed under scalar multiplication

I get this, but then on the bottom, it says that these three statements are equivalent too

H is non empty
same
same

why is

0 is in H and H is no nempty

the same

[quartz compass]

vectors are not n-tuples

[paper egret]

or equivalent

[quartz compass]

prove yourself wrong and study more

[paper egret]

why are the statements 0 is in H, and H is non empty equivalent

H subspace of V

[half ice]

They're not, they're axioms

Oh, wait nvm

[paper egret]

welp fuck

o

[half ice]

I misunderstood

You can see that it's necessary for a vector space to have a zero vector, before we can actually call it a vector space

[paper egret]

why tho

because it seems like

[half ice]

A subspace is just a vector space that fits in the vector space

[paper egret]

for a vector space to be a vector space, you just need 2 things, vector additoin and scalar multiplication

[half ice]

So it also has that zero vector

[subtle burrow]

vector spaces are just weak fields, prove me wrong

[paper egret]

a subspace is a subset of a vector space that has one additional property, 0 is in the subspace

but how come vector spaces alone don't need the 0 vector?

subspaces do?

[half ice]

They all do

[paper egret]

why don't we have to explicitly state it for vector spaces?

[half ice]

That picture is the definition of a vector space

We do have to

[paper egret]

shit rly?

[subtle burrow]

the addative identity must be in a vector space so 0 vector must be there

it is D

[half ice]

I gave the crappy sparknotes version of a vector space earlier because I didn't think you were working in this much detail oop

[paper egret]

oh no im just curious, studying for an exam tmrw

[subtle burrow]

it is bullet point D

[paper egret]

oh in vector spaces, is it just kinda implied

[half ice]

All vector spaces need a zero vector. It's one of the rules that a vector space needs to follow

[paper egret]

then how come, for example, when you have to show something is a vector space, you only show 2 things, but when you attempt to show a subspace, you have to show 3 things?

1. vector addition
2. scalar multiplication
3. 0 is in the space

[half ice]

You might have had a shortcut version. It's very easy to show something is a vector space if you're using regular addition, regular multiplication

[paper egret]

ah

that explains it then

[half ice]

You can have much more exotic spaces if you mess with the limits though

Like a vector space where the scalars are the integers mod 3, this should work

[paper egret]

i think i just need a bit of help with terminology, gimme a sec

Let S = {v1, v2, .., vk} be a subset of V. Then Span(S) = {c1v1 + c2v2 + ... + ckvk | c1...ck is in F} is a subset of V

wtf does this really mean

[half ice]

I like to think of a "set" as a jar that can contain things

[paper egret]

uh huh

[half ice]

S is the name of the jar. In it, you have some vectors

[paper egret]

what does it mean when it says S is a subset of V

is V a vector space?

[half ice]

V is another jar, it also has vectors

[paper egret]

oooh

[half ice]

S is a subset of V, if every element in the S jar is also an element somewhere in the V jar

[paper egret]

but how come Span(S) is a subset of V, not Span(S) is a subset of Span(V)

[half ice]

One might say that V has everything that S has, plus more.

[paper egret]

i'm not understanding osmething here

like how is that logically possible

Span(S) is a subset of V

[wintry steppe]

name an element of Span(S)

[paper egret]

not Span(S) is a subset of Span(V)

[wintry steppe]

and it's in V

[paper egret]

how though

it doesn't sound like it's guaranteed

[wintry steppe]

Let's say that S = {[0, 1]}

[paper egret]

yup

[wintry steppe]

name an element of span(S)

Let V = R^2

[paper egret]

[0, k] where k can literally be any number

[wintry steppe]

is [0, k] in R^2?

[paper egret]

but then V = {[0,1], [1, 0]}

[wintry steppe]

no

V = R^2

[paper egret]

V is a space?

not a set of vectors?

[wintry steppe]

V = {(a, b): a \in \mathbb R, b \in \mathbb R}

V is a set of vectors

[half ice]

Every space is also a set of vectors

[paper egret]

holy shit my brain this is confusing

[wintry steppe]

it is the set of all two-dimensional real vectors

[paper egret]

oh so V is the set of all two dimensional real vectors

[half ice]

A set is a collection of vectors that don't need to follow any rules

A space is a collection of vectors that also follow those rules I put above

[wintry steppe]

(or the set of all functions from {0, 1} to R lmao)

[paper egret]

V = Span(E), where E is the standard basis im assuming

[wintry steppe]

yes

[paper egret]

boom

goteem

gotchu

thanks

[slow scroll]

i mean it doesn't have to be the standard basis tho xd

[paper egret]

oh yea ofc

but i get it

another question

uhhhh

[half ice]

If V contains a set of vectors S
Then V contains Span(S)

Which is to say Span can't leave vector spaces

[paper egret]

gotcha

this is actually helping a ton

holy shit

[half ice]

Even better, Span(S) is a vector space itself.

So it's a subspace of V

[paper egret]

waaaait

wtf

how

[half ice]

Span is the least number of rules that you'd need to have a vector space

[paper egret]

but

how is the 0 vector guaranteed?

wait nv

m

that was stupid

all coefficeicnts = 0

duh

bro lin alg is big brain math

[half ice]

Math: "How do I turn this set into a space doing as little work as possible?"

Nobody:

Math: "Oh yeah, use the span"

[paper egret]

Let W be a subspace of V
Let v0 be a fixed vector in V

Then v0 + W = {v0 + w | w in W} is an affine subspace of V parallel to W

i want to say i get it, but i don't

cuz idk how to put this into my own words

[half ice]

I'm not too sure what affine means in this context, help someone

[paper egret]

ah rip

[slow scroll]

im pretty sure affine subspace is just a 'coset' of the subspace.

[half ice]

Oh, like + a scalar?

Lemme Google

[slow scroll]

like let U be a subspace of V
let v be in V
an affine space would be the coset v + U.
I think ive heard a prof using that term to describe the solutions to nonhomogeneous linear odes

[half ice]

Oh like + a vector

Yeah that's a non-homo DE solution right there

[paper egret]

wat

[half ice]

Know any differential equations? That's a great example of using vectors that aren't usual vectors

Or, that aren't arrows, I should say

[paper egret]

nop dont know any DE, dont ever plan on doing that ever

[half ice]

We say two vectors u,v are collinear if, for some scalar k,
u = kv

[paper egret]

uh huh

[half ice]

Actually, you just want this in your own words? Imagine a plane. Now imagine translating it. These planes are parallel, and there's an affine transformation between them

[paper egret]

that's it

?

[half ice]

Well, each plane is actually a vector space

But yeah

[paper egret]

Let V and W be vector spaces, and Let T:V->W be a linear transformation

Why is the ker(T) a suspace of V, and Im(T) a subspace of W

do i just refer back to the definition of a subspace

1. show 0 vector exists in ker(T)/Im(T)
2. show vector addition holds true
3. show scalar multiplication holds true

[half ice]

Yeah, if addition and multiplication are regular

You'll need the axioms of a linear transformation

[paper egret]

gotcha

another question

Let U be a subspace of V

Show that the following are equivalent
(1) 0<=dim(U)<=dim(V)
(2) dim(U) = 0 <=> {0} = U
(3) If dim(V) is finite, then dim(U) = dim(V) <=> U = V

how the hell do you tkae the dimention of a vector SPACE

like

waT

[half ice]

So all vector spaces have a dimension

[paper egret]

how tf

[half ice]

Dimension = number of elements in the basis

[paper egret]

ohhhh

but couldnt the number of elements in a basis vary?

oh wait it can't

because if the number of elements in a basis vary, then the property of linear independence won't hold

bro this makes so much sense

there is like a sweet spot of number of elements in a basis of a vector space

[half ice]

That mineswell be the "fundamental theorem of linear algebra"

[paper egret]

wait

that's the fundamental theorem of lin alg?

[slow scroll]

then the property of linear independence won't hold
what if you are short vectors in your basis?

[half ice]

Any two bases of a space have the same number of elements. It's an invariant to the space

[paper egret]

if you are short vectors, then the span won't hold

[half ice]

And no it's not the fund thm, but it should be lol

[paper egret]

it's kinda like

you got n dimensions < = >each vector needs to have n entries

nxn

[half ice]

Remember some vector spaces don't have "entries"

[paper egret]

oh

hol p

[half ice]

Actually, no forget I said that. Even a vector space with 3 "spots to put numbers" can be of dimension 2

[paper egret]

wat

[half ice]

Like a plane in 3 space

[paper egret]

but then

it wouldnt span

[half ice]

Is a 2D vector space, but you can't picture it without a 3D picture

I'm using two different kinds of "D"s here oop

[paper egret]

nice D

damn doing mathematics is almost like english holy shit

the definitions are so precise

[half ice]

They are lol. Sometimes it's thicc

Layers of layers of definitions. Pure math does a lot of this

[paper egret]

i secretly like english

actually no fk this

this is torture

like

but the thing i

it's actually making a ton of sense

ah i foudn the right word

maximal linear independent set and minimal spanning set, usually means an n*n

wait another question

what does isomorphism mean in the context of linear algebra

An invertible linear transformation T:V->W is an isomorphism if V nd W are vector spaces and there is a linear transformation T:V->W that is an isomorphism, then V and W are isomorphic

[half ice]

@paper egret
A linear transformation that has an inverse that is also a linear transformation

"Isomorphism" is an arrow that has another arrow that undoes the first

To be general about it

[paper egret]

[steady fiber]

bruh

[paper egret]

nani

[half ice]

Isomorphism = linear transformation that has an inverse

[paper egret]

how can a linear transformation have an inverse

[half ice]

Also a matrix that has an inverse

[paper egret]

i know matrices can have an inverse

you got any numerical examples?

[half ice]

This could also be the fundamental theorem of linear algebra lol:
Every linear transformation can be represented with a matrix

[paper egret]

ok yes

i get that

[feral mountain]

finite

[half ice]

Finite, yes

Shit breaks when stuff is infinite dimensional

[paper egret]

i understand thate

every lin trans can be represented with a matrix

[half ice]

So if you can choose to work with a matrix, or a linear transformation any time you want

An invertible linear transformation = a matrix with an inverse

[paper egret]

wait is that it?

so what you call

matrices invertible, you call transformations isomorphic

[half ice]

We say that two vector spaces are "isomorphic" if there's an isomorphism between them

[paper egret]

wait

welp back to square one

[half ice]

This essentially means that, in terms of linear algebra, these are the same space.

[paper egret]

same space as in

same dimensions?

i dont understand what they mean by the same space

what defines two spaces to be the sae?

same*

[half ice]

If there's an isomorphism between them

Oop, I'm getting circular

Basically, if an element in the first space can be matched to an element in the second space, such that algebra works the same way on each element

[paper egret]

got an example?

or is it too long

[half ice]

You're going into one probably soon, change of basis is an example of this

Take two bases. Show there's an invertible matrix that goes from one basis to the other. The space these bases span must then both be the same space

[paper egret]

hmmm

yeaaaaaaaaaa i dont get it

😦

sry

[half ice]

Sry, this might be getting too technical. In easier terms, if one space maps onto another with an invertible matrix, they're actually the same space

[paper egret]

i dont get the idea of mapping onto another spae via an invertible matrix

like how tf is that supposed to work?

something like this?

v -> A -> w
w -> A^-1 -> v

?

[half ice]

Take a vector, use a linear transformation to map it.

OR
Take a vector, multiply it by a matrix

Both the same thing

[paper egret]

ok yiea i understand that

that makes sense

i dont get how the invertible plays a part in ths

[half ice]

If that matrix/transformation is invertible, then the two spaces you're mapping between are actually the same space

[paper egret]

OH

IT FINALLY CLICKED

I THINK

wait but why is that

[half ice]

We say these two spaces are "isomorphic" to make this rigourous

[paper egret]

is there a proof for this?

[half ice]

Sadly it's a definition. We're defining "same space" this way

[paper egret]

oh

a definition

no proof

gotcha

so we define invertible to be isomorphism for vector spaces, gotchu

see, i was trying to like picture the math in my head

and was like

nani

[half ice]

Because if you change what the elements "look like", this doesn't change what the space "is"

And an isomorphism captures this idea

You'll see much more of this if you go onto abstract algebra

[paper egret]

hmm

is there like a more mathy way fo saying this

using equations

n shit

[subtle burrow]

There is hardly anything more mathy than abstract algebra

[paper egret]

hmm

[strong bison]

say U and V are isomorphic vector spaces --- Although U and V may be different sets, they may be regarded as indistinguishable
(or equivalent) algebraically. AKA U and V both represent just one underlying vector space but perhaps with different “labels” for the elements.

an excerpt from a tbook

[paper egret]

damn that isnt confusing at all .-.

[half ice]

Here's an example. Let's say I'm working in R² but rather than using the vector (1, 1) I just start calling it "bob"

bob + (2, 3) = (3, 4)

[paper egret]

uhhh okay?

[half ice]

I think you'd agree this didn't change anything. It's just a name

[paper egret]

bob and (1,1) are isomorphic

lol

[strong bison]

bob <-> [1,1]

[paper egret]

wait wtf

[half ice]

This is still R² I'm working in

[paper egret]

yes

but im not understanding what this really means in terms of vector spaces

[half ice]

So this "new" space I made up is isomorphic to R²

[paper egret]

like i want to say i get it, but i dont think i get it enugh to explain it to someone

uh huh

[half ice]

We call this the "bob space".

The bob space is isomorphic to R²

Because there's an isomorphism between them

[paper egret]

uh huh...

[wintry steppe]

bob space is very useful

[paper egret]

at this point i'm just gonna thnk of isomorphism as same

[half ice]

OH how didn't i think of this

R² and C are isomorphic

In terms of linear algebra

[wintry steppe]

we should use bob_2 to denote the 2-dimensional bob space

[half ice]

Like, 2 + 3i is the same as (2,3)

If you're ignoring that complex multiplication can happen

[paper egret]

ignoring complex multiplication can happen?

[half ice]

You can't multiply two vectors in R² together

You don't need vector multiplication for a vector space anyway

[paper egret]

ima just die

[half ice]

(1, 1) + (2, 3) = (3, 4)

(1 + i) + (2 + 3i) = (3 + 4i)

[paper egret]

yes

[half ice]

In that sense, you can choose to work in R² or C. They're isomorphic

[strong bison]

It's different "labeling", algebraically without labels they're identical

[paper egret]

oh

i get it

but idk how that paplies to vector spaces still LOl

[half ice]

Even though they represent very different things, they're both the same vector space

[paper egret]

like i understnad what an isomorphism is in other contextx, in graph theory for example

but vector spaces wtf

[half ice]

Same idea. A way to say two graphs are the same

Without considering the unimportant stuff

[paper egret]

wait random question, lets say you have

Ax =b
does that mean

A^-1b = x

wait no duh

im retarded

so invertible means map to each other

in both ways

[gloomy arrow]

Does linear independence also imply othorgonality

[half ice]

No it does mean that, goon

One might say that we use this to solve a system of equations

[steady fiber]

(1,0) is linearly independent to (1,1)

but very obviously not orthogonal to it

[paper egret]

oh

[half ice]

Ax = b, you know A and b, find x

x = A'b

A system has a unique solution if A is invertible

[paper egret]

is it right to say inverbiel is like inverse function

[half ice]

Every matrix can be thought of as a function, yeah

Specifically a linear transformation

Which is a special function!

[paper egret]

a function has an inverse if it has a bijection

bijection inverse

i think im starting to get it

[half ice]

Very nice! It's true

Also, isomorphisms are bijective linear transformations

By that same logic

[paper egret]

yes...

ok now i understand the definition

now comes the intuition]

kk ima go hang myself

[half ice]

Grats, this will make it much easier when you get to abstract, when isomorphisms mean everything

[strong bison]

it's basically just saying we can just swap back and forth what kind of "labels" we put for an isomorphic vector space. a "transform"

[paper egret]

hmmm

[half ice]

It's a weak condition on vector spaces, because there's very few examples where two regular spaces aren't "obviously" isomorphic. For more exotic spaces this notion is useful

[paper egret]

do you have any more numerical examples for isomorphic vector spaces?

the one you gave R^2 to C was helpful

[steady fiber]

the span of any k linearly independent vectors in R^n and R^k, where k<n

are isomorphic

[half ice]

Oh yeah duh

Good one

[paper egret]

wat

[half ice]

Any two planes in 3 space

[steady fiber]

are isomorphic with just the xy plane

for example

it's a pretty simple thing if you think about it

[vast torrent]

Row vectors to column vectors

[half ice]

You can map one plane onto another

[paper egret]

ooOh

OH

[half ice]

Or map backwards

[paper egret]

OH

wait thatm akes so much sense

nani

the

fuck

how can something be so confusing, yet not confusing

[strong bison]

[half ice]

You can choose which plane to do math in, they'll both give the same result

[steady fiber]

being able to find isomorphisms to R^n

is incredibly useful

[paper egret]

hmmm gotcha

[steady fiber]

and you can find an isomorphism of any n-dimensional vector space with R^n

[half ice]

And an invertible matrix takes you from one to the other

[paper egret]

welp, this was a freakin awesome study session for me lol

thanks to everyone who helped out

[half ice]

Lol np. Got swept up into it. Feel free to ask if you have anything else

[paper egret]

will do will do 😄

thanks a bunch again, really appreciate it

[strong bison]

not relevant but I just noticed you're an emoji mod Kaynex

my favorite emojis are all on this server

[paper egret]

+1

[half ice]

I've not done like any of them oop

I probably still shouldn't have that role

Mnnip and Woog and Flimflam are responsible for a lot of them

[steady fiber]

this is one of the major reasons to do isomorphisms

you can convert a linear function between two abstract vector spaces

to a matrix

[half ice]

Actually pretty cool to understand why you can matrix, rather than transformation

[paper egret]

hmmmm

[]_a and []_b we just learned that

i think

dude lin alg is such a mind fuck

[steady fiber]

lin alg is everywhere

[paper egret]

it feels like a mind fuck

because it's like challenging common sense

[steady fiber]

it shows up a lot more in calculus than I thought it would

like all of it after a certain point

[paper egret]

actualy not so much challenging common sense

but you get the point

[half ice]

It's all very natural. The theorems are usually something you could guess, but the terms are the hard part for many

I like lin alg btw didn't know if you knew that

[steady fiber]

lin alg is great

if something can't be solved with lin alg, just approximate it somewhat so it can

:^)

[unkempt robin]

The question asks when this matrix is invertible, but I don't think it can be inverted. How can it? A is a (n + 1) x (n + 1) matrix. a is an unknown n-vector $A = \begin{bmatrix} I & a\a^T & 0 \end{bmatrix}$

[stoic python]

skippi:

[steady fiber]

well immediately we can see that it's not invertible if a is all 0s

so we can try to see if that's the only case or if there's more cases

[half ice]

What's I?

[steady fiber]

Identity

i assume

[unkempt robin]

The block identity matrix of n x n

[steady fiber]

the n x n identity

[half ice]

Identity n-vector?

[unkempt robin]

The way I've been trying to approach this problem is by finding a solution for $ A^T A = I $, but I have to be approaching it wrong 🤔

[stoic python]

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

[quartz compass]

look at its determinant

[steady fiber]

if the kth value of a is non-zero, then the last row can only be linearly dependent on the previous rows if the kth value of a^T is 0

which cannot happen

obvious contradiction

[quartz compass]

expand along the last row or column, det = 0

[half ice]

Indeed. You just need to be sure the determinant is not the 0 matrix

[steady fiber]

and so I think it's invertible as long as a isn't all 0s

[half ice]

aa' = 0

[unkempt robin]

I see. I can see the intuition for it being non-invertible when a is the zero-vector, but struggling with the rest.

I read the determinant answers as well, but I'm trying to not prove like that because this class hasn't introduced it at all.

[half ice]

Well, you could use guassian elimination too

But apparently we have both vectors and matrices in this matrix so idunno

[steady fiber]

when the kth value of a is non-zero, that means the final row cannot depend on the kth row of the matrix. This is because the (n+1), (n+1) value is 0, so it cannot depend on all 0s and a non-zero element. The kth row has a 1 in the kth entry (because of the identity matrix), which is the only non-zero element in that column. That means that the only way the final row is not dependent on that row is if the kth element of the last row is 0. Our premise was that the kth value was non-zero, so requiring the kth element to be 0 is a contradiction

you get the same conclusion with aa' = 0

since aa' >= 0 always

with equality only when all the values are 0

[quartz compass]

found a funny thing just playing around, I pick a constant C for the last spot:

$\begin{pmatrix} I & a \ a^T & 0 \end{pmatrix} \begin{pmatrix} a \ C \end{pmatrix} = (C+1) \begin{pmatrix} a \ C \end{pmatrix} $

[unkempt robin]

Thanks @steady fiber . Need some time to consume that 😰

[stoic python]

Merosity:

[quartz compass]

you can force this equation to be true by setting C to be one of the solutions to the quadratic:

$2a^Ta = C(C+1)$

[stoic python]

Merosity:

[quartz compass]

when C = -1 then we have a non invertible matrix

I dunno, probably not super helpful in hindsight, but it seems natural to want to multiply it by a vector containing a with a free constant at the end and poking around

(obviously what I just said is not that useful because aside from the 0 vector no vector is perpendicular to itself)

[north sierra]

How come for this one a is true but d isn’t

[half ice]

What's the solution to the equation Ax = b when b = (0,0,1)?

[north sierra]

Inconsistent

[half ice]

There isn't one, so it's not true for a

As such, it's false for all 4

[north sierra]

I didn’t ask for when b = 001 though lol

I asked for when b =320

[half ice]

It has to be for any b you can think of

"for each b in R^m"

[north sierra]

Oh so for every possible vector b?

[half ice]

Yus

[north sierra]

Ahhh makes so much sense now

Thx bro

[half ice]

Np, feel free to ask if you need anything else

[north sierra]

Kk 😀

[half ice]

They're hiding a term from you. We call this transformation "surjective" or "onto" because everything in R^m is a possible output

[unkempt robin]

Is an invertible matrix necessarily orthogonal?

[steady fiber]

no

[unkempt robin]

omfg wait, now I understand your explanation

Trying to prove this via orthogonality is dumb because it's nonsense... so that means the sensible way to solve it would be to prove that the set of rows and columns are linearly independent.

Or actually, I only need to solve one of those because the other is implied

[unkempt robin]

If a matrix A is nilpotent for $A^k = 0$, is it implied that $A^n = 0$ for all $n < k$?

[stoic python]

skippi:

[unkempt robin]

I'd assume yes because matrix product is associative, but I'm not sure why the definition of nilpotent is defined specifically in this fashion.

Why exactly k 🤔

[slow scroll]

hm what? Do you mean n>k?

[unkempt robin]

No, I mean n < k.

Well actually, it goes both ways 🤷

[dusky epoch]

uhh

[slow scroll]

for all n<k implies A is the zero matrix

[dusky epoch]

n=1 tho

[slow scroll]

^ yea

[unkempt robin]

🤦

So I guess it stands true for all k != 1?

Ah but that doesn't make sense 🤔

Cause k = 0...

[dusky epoch]

uhh

what

[slow scroll]

im not sure why you would think that you can say anything about n<k

[dusky epoch]

no, nilpotence is defined as the existence of a number k s.t. A^k = 0; it then follows that A^n = 0 for all n >= k

but for n < k it is impossible to say anything

[unkempt robin]

Wow, that confuses me in more ways than I want

[dusky epoch]

if k is the nilpotence index of A, i.e. the smallest exponent that annihilates it, then of course A^n will not be zero when n < k.

it's not that complicated a concept though is it

[unkempt robin]

Just reading that, I want to figure out a counter example where A is nilpotent for, idk, k = 5 and say $A^4 \neq 0$

[stoic python]

skippi:

[unkempt robin]

💢

[dusky epoch]

"nilpotent for k=5"?

so like

you want a matrix A with A^4 != 0 but A^5 = 0?

[unkempt robin]

Yeah ig

I want a counterexample to understand why nilpotence is defined as you wrote it

[slow scroll]

try $A = \begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}$

[stoic python]

kxrider:

[dusky epoch]

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

[stoic python]

Ann:

[dusky epoch]

voilà, this matrix has a nilpotence index of 5

i.e. raising it to the 4th power won't annihilate it but raising it to the 5th power will

@unkempt robin

[unkempt robin]

Ah, okay I'm going to tinker with that for a bit

Do I assume the blank entries are 0?

[dusky epoch]

blanks stand for 0 yes

[unkempt robin]

Ok just checking, a lot of this syntax is new to me 🤣

[dusky epoch]

the 0s on the diagonal are written explicitly to make clear the location of the 1s

[unkempt robin]

Okay, now I see that nilpotence definition 100% makes sense

Hm. That should mean $(I - A)^{-1}$ converges for any nilpotent matrix 🤔

[stoic python]

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

[unkempt robin]

Wait... okay yeah, I think that's right. It should make sense from what I learned just now. Nilpotent matrices will always converge to 0 once they reach a certain power.

[dusky epoch]

you mean (I-A)^-1 exists.

and that's a good observation.

[unkempt robin]

Is it wrong to say that it converges?

Hm, I guess that particular expression doesn't really converge, but it can at least exist

[half ice]

Convergence is something that a sequence does

A matrix has an inverse, or an inverse exists

[buoyant chasm]

I can get d but not sure how to get R

[latent hare]

,rotate

[latent hare]

,rotate

[simple vector]

How can I solve this?

I thought (1/2)^3 * det(A) * det(A) = 1/det(A)
det(A) = 8, but the answer sheet says it is 2

[dusky epoch]

how did you get det(A) = 8

[simple vector]

(1/2)^3 * det(A) = 1

[dusky epoch]

and how did you get that

[simple vector]

because (1/det(A))/det(A)

=1?

or wait no lol

[dusky epoch]

(1/x)/x = 1? are you sure of that?

[simple vector]

nope, it will give 1/x^2

[dusky epoch]

yeah so

[simple vector]

now im more confused

[dusky epoch]

the exact same thing will happen if you divide both sides by det(A) in your equation

you won't get 1/8 det(A) = 1; you'll get 1/8 det(A) = 1/det(A)^2.

[simple vector]

yes I see

but how did they get det(A) = 2

[dusky epoch]

are you able to solve the equation $\frac18x^2 = \frac1x$ for $x$?

[stoic python]

Ann:

[simple vector]

x^2 = 8/x ?

[dusky epoch]

are you able to solve the equation $\frac18x^2 = \frac1x$ for $x$?

[stoic python]

Ann:

[simple vector]

$x = \sqrt(\frac{8}{x})$

[stoic python]

mehmet:

[dusky epoch]

do you know what it means to solve an equation for x?

[simple vector]

to get a value for x?

[dusky epoch]

no, it means to rewrite the equation in such a way that it has x (and only x) on one side, and the right hand side does not have any x's

[simple vector]

oooh I get it lol 😄

then I get 1/8x^3 = 1

and x^3 = 8
x = 2

oh wow this is elementary school math, why was I doing such weird things lol

[buoyant chasm]

bruh

[paper egret]

Kms Kaynex isomorphism was on my exam

[latent hare]

<@&286206848099549185> I asked long time ago please help when u r free 🙏

[dusky epoch]

care to repost your question?

[proper crescent]

@honest swift

[clear spoke]

$\begin{pmatrix}1&-2&-1&1-2a&1\ :2&-3&a&5&3\ :-1&1&a+1&a-1&2a\ :0&2&-2a&-2&2\end{pmatrix}$

[stoic python]

<3 Hoodie <3:

[clear spoke]

What is the simplest way to solve this matrice?

I always manage to go the long way and increase my chances of making a mistake

[dusky epoch]

what do you mean by "solve this matrix"

is this a matrix representing a 4 by 4 system of equations

[clear spoke]

Yes

Sorry, I meant transform it into a triangle matrice using gauss eliminations

[north sierra]

So I got the answer for this

[dusky epoch]

ok

[north sierra]

But I was wondering why I can’t do AA^(-1)(x1,x2) instead

[dusky epoch]

wdym

AA^-1 is just the identity matrix

by definition

so like

what exactly were you going for these

[north sierra]

Sorry I took long to draw this lol

So why can I do the first but not the second

[dusky epoch]

uh wdym

[north sierra]

They somehow equal two different things

[dusky epoch]

okay so like

first off

$\begin{bmatrix} -9\11 \end{bmatrix} \begin{bmatrix} 1&1\-\tfrac75 &-\tfrac85 \end{bmatrix}$ isn't even defined

[stoic python]

Ann:

[north sierra]

Oh ya true

[dusky epoch]

and also

just what the fuck did you even do

$ABC \neq (AC)B$

[stoic python]

Ann:

[dusky epoch]

especially not when A and C aren't of compatible dimensions!

[north sierra]

LOL

yeah true didn’t think about that

[north sierra]

@dusky epoch what did a b c correspond to

[dusky epoch]

??

what

[cobalt tartan]

Hrm given an orthogonal matrix P = [p_1... p_n], if I square it , what happens?

i.e, if I do P * P

Then that's [p_1... p_n] * [p_1... p_n], does that equal to [p_1^2 ... p_n^2]?

[dusky epoch]

what is p_1^2

[cobalt tartan]

Oh

Hrm

Ok I see

Thats' not defined

[dusky epoch]

more like doesn't make sense

[cobalt tartan]

Yea

Err that yea

[dusky epoch]

i mean the orthogonal matrices form a group so the square of an orthogonal matrix is again orthogonal

[cobalt tartan]

yes

[dusky epoch]

and is also guaranteed to have determinant +1

but other than that nothing of interest can be said in general

[cobalt tartan]

well

This is what the question says

So my idea is that uh because A is symmetric, we have that there must exist some P orthogonal, D diagonal such that $A = PDP^T = PPD$

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

And I thought that maybe there'd be something to do with the properties of PP?

[north sierra]

Is matrix multiplication just row reducing?

Because I just realized I can do this by row reducing

[sonic osprey]

No

It tells you how to do it

[north sierra]

Ya but I also found the answer my row reducing

By

I did it both ways

[sonic osprey]

That's fine

[north sierra]

So what’s the deal with this?

What’s the pattern between this and row reducing and inverse

I’m curious

[sonic osprey]

You're right that they're kind of the same

At least, row reducing can be thought of as multiplying by a particular matrix

[north sierra]

Oh

[half ice]

@north sierra
If you row reduce a matrix, and do the opposite steps onto an identity matrix, that gives the inverse

With a block matrix, you are computing A'b without computing A' first

[wintry steppe]

@everyone i need help

[dusky epoch]

@wintry steppe what the actual fuck do you think you're doing, pinging everyone in a server with literal THOUSANDS OF PEOPLE ONLINE AT ANY GIVEN MOMENT

it's disabled for a reason

[wintry steppe]

i need help

[dusky epoch]

yeah, go ahead and keep saying that instead of... idk, being more clear about what you need help with

[wintry steppe]

can you help me

[dusky epoch]

since you haven't posted any problems, no i can't, because i don't even know what you're having trouble with.

[north sierra]

@wintry steppe lol don’t ask for help man just send the question

[dusky epoch]

i'm surprised my message was apparently not something they could take as a cue to do so already

or maybe they chose not to

[cobalt tartan]

Wait if I have an orthogonal matrix multiplied by an orthogonal matrix, the result is orthogonal, which means that it has nullspace = {0} right?

[dusky epoch]

yes all orthogonal matrices are invertible so of course the nullspace is trivial

[cobalt tartan]

If I have $A = PDP^T$, then I also have $A^k = PD^kP^T$ for some orthogonal P and diagonal D

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

Can I go and do anything to the $PD^kP^T$

[stoic python]

Liria ^(;,;)^:

[dusky epoch]

wdym

[cobalt tartan]

Like I want to show that

So I'm thinking taht theres' something to do with PD^kP^T

But uh P^T is orthogonal so it's nullspace is 0 and so it's a linear mapping right?

[dusky epoch]

.....

you really are not thinking in the right direction here i'm afraid

[cobalt tartan]

Hrm

[dreamy depot]

Can anyone help me with this step on the proof of the Vandermode Determinant

http://mathb.in/37524

[cobalt tartan]

Do you have any hints or what direction I should be looking at instead, Ann?

[dusky epoch]

i am afraid i can't say much w/o spoiling the problem but consider the minimal polynomial of A

[cobalt tartan]

Uh I don't know what a minimal polynomial is

Is that like the characteristic polynomial...?

[wintry steppe]

well no one can help me.

just ignoring it

[dusky epoch]

yeah bc you post an incomplete problem statement. and in the wrong channel, too.

[cobalt tartan]

If A is symmetric, then A is diagonalizable, and the diagonal matrix that it's similar to will have values equal to the eigenvalues, so it will have a characteristic polynomial of the form
$(\lambda - \lambda_1)(\lambda - \lamda _2)... (\lambda - \lambda_n)$, where there is some $\lambda_i = \lambda_j$ for some $i \neq j$ right?

[stoic python]

Liria ^(;,;)^:
Compile Error! Click the reaction for details. (You may edit your message)

[lunar sable]

Given Matrix A in the question I posted

Why is it -2 -2 and -3

and not 2 -2 and 3

Isn't it + - + for the first row when using cofactor

[blissful vault]

linear isn't even that hard but i can't seem to get over 80

wtf

[lunar sable]

Depends on the teacher

Some make simple concepts seem way harder than they are supposed to be and then you youtube the information and it clicks

[tardy wadi]

lol i'm terrible at it. i got a 57 on my last midterm (the median was 52 but still) 🤕

[north sierra]

Why does j and k mean? I’m confused

[gray dust]

recall that if a matrix A is invertible, then

$AA^{-1}=A^{-1}A=I$

[stoic python]

RokettoJanpu:

[quartz compass]

j and k are just saying there's a left inverse and a right inverse

incidentally the left and right inverses are equal and it's about a 1 line proof

[north sierra]

Ohhh

Would this be no because it would be linearly dependent?

Or would it be no for another reason

[dusky epoch]

what's "it"

[north sierra]

The matrix sorry

[gray dust]

one speaks of LI sets of vectors, not single matrices

[dusky epoch]

linear (in)dependence is not something that makes sense for a matrix

[north sierra]

Oh

I see

Ok so

Would a reason be no because the set of vectors are dependent.

[dusky epoch]

what set of vectors

[north sierra]

Nvm

[gray dust]

@north sierra just be more specific when referring to a set of vectors or a matrix whose columns consist of those vectors

[dusky epoch]

there are many ways to explain why a 5 by 5 matrix whose cols don't span R^5 cannot be invertible

[north sierra]

Yeah I got the answer that’s why I said nvm

[dusky epoch]

sounds like a typo to me

should be A(BW)

[north sierra]

True

Ty

[grave halo]

Hey! I'm trying to do a ellipse regression on a curve in 3D space. I suppose I should apply some kind of least square's method, but I'm sure how to do this in 3D-space. Should I first apply a change of basis to the plane's coordinate system that the ellipse lies in, and then do the least square in that plane? Any help is appreciated! =)

[eternal quest]

hello

can anyone give me a clue on how to solve this

?

[cobalt tartan]

Does this make sense so far? I just want to double check, I think it makes sense but I'm not sure

[clever cedar]

graphically speaking, is the vector -u just the vector +u except flipped?

[clever cedar]

how does one express $\mathbb{R}^4$ verbally

[stoic python]

mike0x81:

[clever cedar]

R is a set of numbers in real space with 4 numbers?

[quartz compass]

I say "R four"

[clever cedar]

okay ty

[quartz compass]

I think most people do too, in my mind I can hear other people saying "R two" and "R three" much more often though

[clever cedar]

oh okay its just my first time seeing it so i didnt know if theres a specific way

thanks for ur help

[gray glen]

R tesseracted

[subtle burrow]

It is just R 4

[ionic steppe]

Im working with the Leontief Equation and could someone please explain how we calculated (I-C)^-1

[clever cedar]

its

1 / det(Matrix) * adj(Matrix)

its divided 1 by the determinant of the orignal matrix

then multiplying that by the adjucate matrix

thats how u find inverse

with determinant

[ionic steppe]

Thanks so much

@clever cedar

[clever cedar]

anytime

[north sierra]

If we have matrix A multiplex by matrix B (mxm matrix) and it equals to the identity matrix can we say that A and B are both linearly independent

Like for this equation how do we know that ax=b has a solution for any b

[pliant thistle]

Is there a quick way of checking if i^(n), where n is a even number , how do I know if its +1 or -1

[near torrent]

I can't figure this out

if the transpose of a is an n x m matrix and b is in R^m then transpose b is an n x m matrix right, but that implies it equals y which is an element of R^n

[north sierra]

How could I show this.

[subtle burrow]

if E*F = I then F = E^-1 and E = F^-1

so E * F = E * E^-1 = I

and F * E = F * F^-1 = I

they are each others inverses

[paper egret]

i dont want the answer, just to clarify, to change bases

here's what I got

P_(E <- B) = [[v1]_E | [v2]_E | [v3]_E]

how do i evaluate each of those individual vectors

nvm

im dumb

are stirling numbers an actual thing 👀

[clever cedar]

in order to find vector equation of a line why does the formula involve the vector of two points rather than just two points

q = R + t(P-R)

[subtle burrow]

I like this formula

L = (1-k)p + kq

[clever cedar]

is k a constant

[half ice]

That's the line between two points. Between the two lines is 0 ≤ k ≤ 1

I don't know what you're asking though, Mike

[clever cedar]

me neither kinda, i was just introduced to the topic of "Vector Equation of a Line"

and it seems more complex than necessary

the formula given is q = P + t(R-P), where q, P and R are vectors

with space R^3

[half ice]

It is more complex than necessary, yeah. The common one is
L = P + tR

That's a line that passes through a point P, and has direction R

[clever cedar]

how can u tell direction from a matrix

i mean vector

[half ice]

I can't say I really know how to answer that without saying something circular

"It goes that way"

[clever cedar]

fair enough

appreciate ur help nontheless

[half ice]

Like, in 2 space, (1,1) goes diagonally up and to the right

[clever cedar]

do we assume the tail is at origin

[half ice]

L = (0,1) + t(1,1)
Is a line that goes diagonally up-right (or down-left) and passes through (0,1)

Yes, vectors start at the origin here

You can find any point on it by choosing a value for t and simplifying

[clever cedar]

oh i see that makes more sense

Hm okay i will think upon it some more

[half ice]

OH mb I see what your form is now

Yours is the line that passes through two vectors P and R

[clever cedar]

Yes i shouldve been more clear

apology

[half ice]

Mine is the line that passes through a vector, and has the direction of another vector

[clever cedar]

oh okay, my text "found" the direction

by taking P - R

[half ice]

In yours, if you let t = 0, you get q = R
If you let t = 1, you get q = P

That's a good interpretation, yeah. Our two equations are very similar

[clever cedar]

Ah I see I guess i should plug in some values for t and play around and see what I get

to learn general behaviour

awesome, thank you

[half ice]

Np, feel free to let me know if there's anything else

[clever cedar]

🙂

[native ore]

Idk if my brain is not working or what

But how did they get from 2nd last eqn to last one

[half ice]

Let's say you want to multiply rref(A) by [x, y, z], and get [0,0]. What equations do x, y, and z need to satisfy?

[native ore]

See im confused becausw now theres lesd than 3 rows

[half ice]

Are you having trouble with the multiplication?

[native ore]

No I just dont understand whats going on here to reach that end point

[half ice]

So you're having trouble getting to the second last part as well?

[native ore]

Yea maybe im not understanding whats going on here in the second part either

Wait no I do

Its really is just the third part Idk the steps they took to get there

[half ice]

Do you see how they got rref(A)?

[native ore]

Yes, they just row reduced the matrix which is the column picture of the system

[half ice]

Okay, so we're back to finding the nullspace

Let's say you want to multiply rref(A) by [x, y, z], and get [0,0]. What equations do x, y, and z need to satisfy?

Basically, after doing the multiplication, you get this:
x + z = 0
y = 0
See that on rref(A)?

[native ore]

Yes I see that

Oh

OHHH

that makes sense