linear-algebra

[thorny hemlock]

ok

well

i dont see anything particularly useful :/

[stoic python]

slimvesus

[thorny hemlock]

right

i dont see where youve shown this

yeah i agree with everything

[stoic python]

slimvesus

[thorny hemlock]

7yes

where has the second inequality been proved

[stoic python]

slimvesus

[thorny hemlock]

ok

[stoic python]

slimvesus

[thorny hemlock]

oh ok i got it

thanks so much

[thorn robin]

What a funny looking inequality

Does it have any interesting interpretation?

[thorny hemlock]

wdym

[thorn robin]

What does it mean, geometrically?

[thorny hemlock]

idk havnt seen anything

uses an orthogonal decomposition in the proof if that means anything idk

[thorn robin]

For example if n=1 and |u|^2 + |v|^2 = 1 then the tighter estimate you proved sqrt(1-|u|^2) * sqrt(1-|v|^2) \leq 1 - |u||v| implies that the largest area of a rectangle inscribed in the first quadrant of the unit circle has area 1/2

[thorny hemlock]

@wintry steppe

you replaced 1- <u,v> with 1-|u||v|

a^2 <= b^2 ?

how you replaced the RHS

because

$1 - |u||v|$ is less than $1 - \langle u,v \rangle$

[stoic python]

Yes

[thorny hemlock]

but $(1 - |u|^2)(1-|v|^2) \leq (1- \langle u,v \rangle )^2$

[stoic python]

Yes

[thorny hemlock]

how are you sure that 1 - |u|v| is larger or equal to the LHS

ugh this is tough

Let me re explain myself

[thorn robin]

It may help you to rewrite the into a single chain of inequalities

rewrite the entire proof*

[thorny hemlock]

we want to prove this

but

you said that it is enough to show $(1 - |u|^2)(1-|v|^2) \leq (1- |u||v|)^2$

[stoic python]

Yes

[thorny hemlock]

when

$(1 - |u||v|)^2 \leq (1- |\langle u,v \rangle|)^2$

[stoic python]

Yes

[thorny hemlock]

You are replacing the RHS with somethign that could be less than what was already there, so the new inequality could be not true?

ok

oh ok

otherwise it makes sense to me

ok

ty

[split wedge]

{A,B, C } over modulo 2

Is not a subspace

But any example or proof to show how?

Where {A,B, C} are subsets of a set

[tropic turret]

Hello, is that statement is correct?
det(AB)=det(A)det(B)=det(B)det(A)

[limber sierra]

Indeed.

this means that det(AB) = det(BA)

even though AB might not equal BA

this is a very handy fact

[tropic turret]

That what i thought because det(A) or det(B) is just a scalar so it is commutative

Thank you all

[wintry turret]

To prove c=8 $\implies$ linearly dependent, would it be correct to just show that the vector with c can be written as a linear combination of the other two vectors, or would I have to use the formal definition?

[stoic python]

beeswax

[limber sierra]

that suffices because:

if $\lambda_1 v_1 + \lambda_2 v_2 = \mu v_3$, then we can rearrange this to [\lambda_1 v_1 + \lambda_2 v_2 - \mu v_3 = \mathbf{0}]

[stoic python]

Namington

[limber sierra]

and setting lambda_3 = -mu gives the definition of linear dependence

so yes, that works.

(although you may want to quickly justify it)

[tropic turret]

Just if -mu is not zero

[limber sierra]

well if mu is 0 then v_1, v_2 are linearly dependent anyway

since we've shown that lambda_1 v_1 + lambda_2 v_2 = 0

but in this case, the first two vectors are not linearly dependent

so thats not a concern

[wintry turret]

Oh, I see. So I can start off with writing the vector with c as a linear combination of the first two, then manipulate it to look like the definition of lin. dep.?

[tropic turret]

How do you know that lambda_1 and lambda_2 is not zero as well?

[limber sierra]

by assumption that v_3 is a linear combination of them

and the fact that v_3 is nonzero

@wintry turret correct

and that "'manipulation" is just subtracting it from both sides

[wintry turret]

Wait, why is it valid to assume that the 3rd vector is a linear combination of the first two? Would I first just have to show the computation to figure out what the scalars are?

[limber sierra]

well yeah you cant assume that the third vector is a lin comb

you have to show it

for c = 8

my "by assumption" line was just answering bardak's question

[wintry turret]

Gotcha

[misty storm]

what does P ⊆ R mean?

[limber sierra]

P is a subset of R

that is to say, every element of P is also in R.

[misty storm]

I'm kinda confused:
the practice exercise is giving me 2 vectors (a and b) and a line l=a+b*t | t belongs to R

and it's asking me to find all points in R³ in which the distance to the line is 1

so basically it's asking me to find all points that meet the criteria that there are?

doesn't the line extend forever?

[limber sierra]

the line does extend forever

visually youll be creating an "infinitely long cylinder" around the line

[misty storm]

that much I got

but it's asking me to "find all points"

I assume it's not asking me to right down an infinite series of numbers

at least not in the way that I thought

[limber sierra]

i'd assume it just wants you to use set builder notation

[nocturne jewel]

write out every point, might take you a bit

[misty storm]

how would I go about accomplishing such a task?

[limber sierra]

well, it's the set of all points p where ||(a+b*t) - p|| = 1

for some real number t

im not sure if this is the format theyre looking for

maybe try looking at examples from class/textbooks to see if they have any similar problems?

but it does describe all points

[misty storm]

I do have the values of a and b

if that helps

[misty storm]

already did, no help

[misty storm]

In retrospect, it very likely does

[limber sierra]

i mean thats what im saying

do you know how to write

"the set of all points p in ℝ³ such that ||(a+b)*t - p|| = 1 for some real number t" in set builder notation

[misty storm]

I'm not even sure I know what set builder notation is really

I know how to use some symbols

[limber sierra]

well you already gave an example:

the line L is the set {a+b*t | t belongs to R}

[misty storm]

yeah, that particular fashion is how most of the exercises are laid out

which is why I know

[limber sierra]

yes, and that's an example of set builder notation

[misty storm]

but to say I understand the underlying rules would be quite a stretch

(of this notation)

[limber sierra]

{things take this form | but they have to satisfy these rules}

[misty storm]

oh that's cool

[limber sierra]

in this case you'd get simply

{p in ℝ³ | ||(a+b)*t - p|| = 1 for some t ∈ ℝ}

[misty storm]

so the exercize is just asking me to write it out?

[limber sierra]

i think? im not sure

i dont know what else they'd be looking for

unless you want to describe the shape more... geoemtrically?

[misty storm]

nah, I doubt that's it

[limber sierra]

"the cylinder with radius 1 along the line..."

[misty storm]

we haven't had classes on anything related

well, thanks a lot

[limber sierra]

i mean maybe they want you to expand ||(a+b)*t - p|| using the definition of magnitude/norm

into sqrt(entry1^2 + entry2^2 + ... + entryn^2) or whatever

[misty storm]

The weird thing

is that they go through the trouble of giving me numbers

for the vectors

a and b

this is the first time that they give me numbers that aren't relevant

[pure tangle]

Could someone please help me with this question. I was able to row reduce with no issue but I'm having trouble with the questions. Would it be 4 equations and 4 unknowns since it's a 4x4 matrix? And the part I'm completely stuck on is knowing how to tell if there is a non-trivial solution. And for the last part I got t*[-1,-3,1,0]

[limber sierra]

Would it be 4 equations and 4 unknowns since it's a 4x4 matrix?
yes; each row corresponds to an equation, each column to an unknown
And the part I'm completely stuck on is knowing how to tell if there is a non-trivial solution
does the row echelon form of the matrix have a 0 row? what does that tell you about nontrivial solutions?
And for the last part I got t*[-1,-3,1,0]
seems like you do think it has a nontrivial solution.

(i think "what you got" is correct btw)

(but i just did the row reduction in my head, mightve made a mistake)

[pure tangle]

Thanks so much for the response. Does non trivial mean there is a Bx = 0 where x isn't 0?

[limber sierra]

right

x isnt the zero vector

[pure tangle]

great. Thanks so much

[limber sierra]

homogenous systems Bx = 0 will ALWAYS have the solution where x is the zero vector

so we use the term "nontrivial" to refer to the other solutions

(which don't always exist - in fact, they exist precisely when the Row Echelon form of B has a zero row)

[pure tangle]

thanks so much. Just to double check my understanding, the zero row is needed so that we can get a free variable that will provide other solutions (non trivial) and the zero vector solution?

[limber sierra]

right

i'm assuming the matrix is square here btw

[pure tangle]

exactly

thank you so much for the help and great explanations

[limber sierra]

if you dont have a zero row, then the RREF form of B is just the identity matrix, so Bx = 0 will have the same solutions as x = 0; that is to say, it will ONLY have the zero vector as a solution

the other direction is a bit harder to prove but, as you suggested, comes from the fact that it'll have a free variable

[pure tangle]

that's awesome. linear algebra yields a lot of cool stuff

[pure tangle]

if i wanted to see if there was an inconsistant vector b, is the best way to create an augmented matrix with the solutions being b1,b2,b3,b4, then row reducing it to see if i get a 0= some equation of b_n(s) that isn't true?

[fiery obsidian]

hi, not sure if this is the correct place to ask, but, in the k-means clustering algorithm, the maximization process readjusts the the centroid of the cluster by taking the average of assigned data points. But how do you take the average of coordinates (i.e [2,3], [5,3]) to get a new centroid?

Is it average of X's and average of Y's?

[tropic turret]

How to prove that if $m>n$ for some $A\in M_{m\times n}$ so there is $b\in \mathbb{F}^m$ s.t for Ax=b there is no solution?

[stoic python]

Bardak

[wintry steppe]

yeah.

[wintry steppe]

Use the rank argument. Find the rank of A, this is if it's allowed tho.

[dire thunder]

@tropic turret hint: dim V = dim null T + dim range T

and to each matrix there is associated linear map

i mean consider dimension of codomain

and maximal dimension of range T

[tropic turret]

Ok cool

How do I find the inverse of a linear transformation?

What are the general steps?

[tawny tulip]

@tropic turret Can you find a matrix representing your linear transformation with some basis?

[nocturne jewel]

double checking something: Let's say I have a vector space V with dim(V) = 5, if I have a set of 6 vectors in V, are the vectors automatically dependent?

since $|{v_0,v_1,...,v_5}| > dim{V}$

[native rampart]

Yes,The set is linearly dependent

[stoic python]

moshill1

[nocturne jewel]

ok thanks

and the reason is that the set can never be a basis by equicardinality property?

[native rampart]

Yes(Assuming you mean All basis of a space have the same number of elements)

[nocturne jewel]

yeah

we called that equicardinality of bases

[magic light]

Q: If I want to find the base of KerT and its dimension do I just pluck the equations to be equal 0? 3x - 2y = 0 etc

If I want to find base and dim of ImT do I need to put them into a matrix like (3 -2 0) as the first row in that matrix and eliminated?
https://i.imgur.com/V4d713l.png

also, what does it being from R^3 to Matrix tell me if anything?

[edgy wing]

help me

[toxic imp]

I tried multiplying it out and subtracting to get the matrix = 0 but I didn't get the identity matrix as a solution

so I assume I did something wrong

[pallid rampart]

Well first you explicitly write out what the product of the matrices are, then the left hand side and the right hand should should both be some 2x2 matrices where each entry is some expression involving A,B,C,D,x1,x2,x3,x4. Now two matrices are equal if and only if each entries are equal, so you'll get 4 equations, then solve these equations for x1,x2,x3,x4

[pseudo thicket]

Could a set of 3 vectors in R4 span all of R4? The answer is no because not every row has a paivot
What if every row has a pivot, but not every column?

Also if there is free variable, it means infinitely many solutions, does it mean the vector set can span all of R4?

[pallid rampart]

Well think about it

The columns in which the pivots are on have to be different

So that means that there are 4 pivot columns

And there are 4 columns

[hollow finch]

What if every row has a pivot, but not every column?

That requires having more than 4 columns in your matrix, which just shows that 5+ vectors can span R4.

Having a free variables just means the columns/rows are linearly dependent. If the columns don't span R4, a free variable doesn't change that.

[stoic python]

meow

Commander Vimes

[tame mural]

vector spaces made from GF(2)

[onyx cypress]

For matricies A,B has anyone seen this product tau before? Does it have any known properties?

[native rampart]

That looks like trace(AB)

<A,B>=tr(AB) is a bilinear form

[onyx cypress]

Yeah, damn you are right, didnt even notice that. Thank you so much

[wintry steppe]

for Q6, could we just show that it doesn't contain the zero vector?

by doing p(0) = a + 0 = a

[dire thunder]

@wintry steppe suppose a+t^2 = 0 for all t in R

this is impossible

since polynomials of second degree have exactly two roots counting multiplicity

[magic light]

I have this question and I'm not sure my proof is correct, can someone check my logic please?
T:V->V linear transformation
for every v in V, there exists w in KerT, u in ImT s.t u + w = v

prove T(v) = 0 if and only if T(T(v)) = 0
one side is simple
If T(v) = 0
then T(T(v)) = T(0) = 0

If T(T(v)) = 0, then T(v) in kerT, and if T(v) in kerT then T(w) + T(u) in KerT, but u is already in kerT therefore:
T(w) + T(u) = T(w), thus T(w) in kerT

In the end, T(v) = T(w + u) = T(w) + T(u) = 0 + 0 = 0

I didn't really use the fact that w is in the image

[thorny hemlock]

"but u is already in kerT"

i think you mean w

looks good otherwise

[nocturne jewel]

Is Im(T) analogous to the range of a function?

[native rampart]

It is the range of a function

The function being the Linear Transformation T

[gray dust]

image & range are synonymous in almost every use

[dire thunder]

[gray dust]

yes, use im not Im

too much work, capitalizing I is sufficient

[stoic python]

Commander Vimes

[dire thunder]

do you wanna go hard?

[nocturne jewel]

just drop the I to avoid any confusion

m(T)

[dire thunder]

physics package ig

$\im{x}$

[stoic python]

Commander Vimes
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gray dust]

$\im,\Im$

[stoic python]

RokabeJintarou

Commander Vimes

[thorny hemlock]

[nocturne jewel]

$(S[f])(x) = \int_0^x f(t) \dd{t}$

[stoic python]

moshill1

[nocturne jewel]

so would Im(S) be all functions..?

[thorny hemlock]

ig

?

[dire thunder]

not all functions are integrals of some function

[thorny hemlock]

oof

[nocturne jewel]

yeah it was a multiple choice participation question but idk why e^x - 1 is in Im(S)

options were e^x - 1, e^x + 1, e^x - x

[dire thunder]

i prolly do not get defn of S properly

[dire thunder]

wait why tf you have e^x-1 and e^x+1

[nocturne jewel]

options of the question were $e^x - 1 \ e^x + 1 \ e^x - x$

[stoic python]

moshill1

[nocturne jewel]

it's the white box at the bottom of the ss

[stoic python]

Commander Vimes

Commander Vimes

[dire thunder]

actually

it equals

[stoic python]

Commander Vimes

[dire thunder]

@nocturne jewel

[nocturne jewel]

Oh i see it now

so is f(t) = e^t the only input of S?

OH I didnt sub in bounds properly that's why

I put in f=e^t - 1 and forgot that e^0 = 1

S(e^x-1) -> e^x - x -1

[jaunty sage]

can anybody explain to me as to what i am supposed to do in this question

im very confused

just @ me please

[dire thunder]

@jaunty sage suppose a is 1x3 matrix and x is 3x1 matrix (i.e a is row vector, and x is column vector)

then find ax by usual matrix product

[jaunty sage]

ye

sure

that i did

and got a

3x3 matrix

@dire thunder what am i to do after that

[stoic python]

Cyoger

[dire thunder]

how after multiplication of 1x3 matrix by 3x1 matrix you got 3x3 matrix

[jaunty sage]

oh wait

nvm

fuck

1x1

i fucked up

[dire thunder]

i mean recall what is dot product

[jaunty sage]

alright

is this not what im supposed to do?

[dire thunder]

sorry channel is busy

@jaunty sage yes but you use wrong order

[cunning pier]

alright then

[jaunty sage]

oh how so do u mind correcting my mistake

[dire thunder]

you multiply row vector a by column vector x from the right

[gray dust]

@cunning pier eigenvalues are exactly the roots of the charpoly

[dire thunder]

ax, not xa

(generalized eigenvalues, rokabe, no?)

[jaunty sage]

OHHHHH

so u get an equation

[gray dust]

nah

[stoic python]

Commander Vimes

[cunning pier]

okay, is there some requirement (looking at the calculation of charpoly) or does this hold always true?

[jaunty sage]

ye i understand but also what about the second part @dire thunder

[stoic python]

Cyoger

[dire thunder]

recalls what it means for dot product to be zero

[jaunty sage]

perpendicular

innit?

also can the values of x1 x2 x3 be anything that satisfy the equation

[gray dust]

@cunning pier always

[jaunty sage]

or is there another method to use?

[cunning pier]

alright thanks

[jaunty sage]

@dire thunder

[dire thunder]

i mean just think

you have constant vector a

and let x vary

solutions of ax=0 are then?

[jaunty sage]

infinite?

[dire thunder]

[jaunty sage]

since x varies?

am i wrong?

im confused

[dire thunder]

what it means for dot product of a and x to be zero?

[jaunty sage]

that a and x are perpendicular

[dire thunder]

or can you say that x is perpendicular to a?

[jaunty sage]

we can

[dire thunder]

so?

[jaunty sage]

the solutions are perpendicular to x?

is that what you mean?

[dire thunder]

how tf x which is soultion can be perpendicular to itself

[jaunty sage]

im not sure i am following

[dire thunder]

x is a solution

x is perpendicular to a

connect these two facts

[quaint heath]

how is x perpendicular to a

[dire thunder]

[quaint heath]

dudeeeeeeeeeeee

[jaunty sage]

understood

got it thanks

[somber lintel]

Quick question, does i hat always need to be equal to [1, 0] and j hat always need to be equal to [0, 1], or can i hat and j hat been any two linearly independent vectors which span the entire plane?

[dire thunder]

i and j are usually defined to be (1,0) and (0,1) but you can have any two orthogonal unit vectors as basis for plane

[somber lintel]

do they have to be orthogonagal?

[dire thunder]

well in fact you can have any basis

you just want it to span vector space and be linearly independent

[somber lintel]

ok, and for it to span vector space, that means if your were to multiply i, j, and k by different scalars, the vector created by their sum can reach any place in space

[gray dust]

yes

[magic light]

What are the rules when trying to gaussian elimination inside a determinant?

I know that if one of the rows gets canceled then the determinant is 0

but you can't just willy-nilly eliminate like in a normal matrix right

[gray dust]

the only row op that preserves det in general is adding a multiple of a row to another

[magic light]

Can't multiply by yourself as well?

like R_2: 2R2 - 3R3

[gray dust]

idk wym

[magic light]

R2 = 2 * R2 - 3 * R3
this might cause R2 to be 0 in a normal matrix

like just in general

as an example

lol

can I multiply a row by itself is what I'm asking

like if the row was (-2, -4, -6) I can't just call it (1, 2, 3) because that changes the determinant

[gray dust]

R2=2R2 is NOT multiplying R2 by R2. it's scaling R2 by 2

also by row op i mean elementary row op. 2R2-3R3 isn't an elementary row op. it's scaling R2 by 2 then adding -3R3 to R2

[magic light]

ah I see

[gray dust]

adding -3R3 to R2 doesn't change det. scaling R2 by 2 also scales the det by 2

[magic light]

if I'm trying to prove the determinant is 0 then this doesn't matter right?

So I'm basically asked if 8 is an eigenvalue of this 4x4 matrix
I just put in the determinant and scale matrices and if the determinant is 0 the scale didn't matter, I can just say that as a footnote?
is that okay to do?

if 8 is indeed an eigenvalue then Det(A-8I) = 0
so I don't need to worry about scaling ops?

[gray dust]

after scaling R2 by 2, the det is scaled by 2. after adding -3R3 to R2, R2 becomes a 0 row, so det=0. the det of the original matrix is 2*0=0

[magic light]

right right

so I think that's what I said

is this an iff situation?
Det(A-xI) = 0 <=> x is eigenvalue?

[gray dust]

the other row op, row swap, negates the det. so scaling & row swap in general changes det but as long as you get a matrix whose det is 0 then all the extra factors you picked up don't matter, the det of the original matrix is 0

[magic light]

but what if I row swap R2 and R4

why would that negate

[gray dust]

it follows from det being multilinear

det(A-xI) is A's characteristic polynomial. eigenvalues are DEFINED as exactly the charpoly's roots so that's an iff

[magic light]

thanks

Another question, I've asked here before but actually I misunderstood

Why do skew-symetrical matrices only have non-zero eigenvalues in C?
so I've been trying to do this operation to show it
Av = xv => (Av)^T = (xv)^T
v^TA^T = xv^T
v^T (-A) = xv^T
and I got kind of stuck

basically trying to show -x = x, and thus the only answer in R is x=0

[quartz compass]

work the same way but start with v^T Av

[magic light]

$v^TAv = v^Txv -> (v^TAv)^T = (v^Txv)^T$

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

uhm

like so?

I'm not sure how to continue here eh

[quartz compass]

keep going

how do you transpose a product of matrices

[magic light]

$A^Tvv^T$?

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

wait no..

[quartz compass]

yeah doesn't make sense dimensionally

so not a chance

I guess the real issue is how do we take this and use it for 3 matrices? $(XY)^T = Y^T X^T$

[stoic python]

Merosity

[magic light]

$(v^TAv)^T = (Av)^Tv$

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

is that not right?

[quartz compass]

aha good

[magic light]

okay so

$(v^TAv)^T = (Av)^Tv = v^TA^Tv$

[stoic python]

blackmamba[Beatrix Kiddo]

[quartz compass]

yes good now you're getting it

[magic light]

thats the left side, let me try the right

$(v^Txv)^T = (xv)^Tv = v^Txv $

since x^T = x

$v^TA^Tv = v^Txv -> v^T(-A)v = v^Txv ->
-v^TAv=v^Txv -> -v^Txv = v^Txv$

[tame mural]

Are there any well known introductory texts for $F^n_2$?

[stoic python]

meow

blackmamba[Beatrix Kiddo]

[magic light]

can I just change scalars position here?

like

[quartz compass]

yeah always

scalar multiplication commutes with matrix multiplication

[magic light]

$-xv^Tv = xv^Tv$

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

$-> -x = x$

[stoic python]

blackmamba[Beatrix Kiddo]

[quartz compass]

yeah, and specifically why is it that we can say this?

[magic light]

v is non-zero I think

because it's an eigenvector?

can eigenvectors be 0?

[quartz compass]

yeah exactly

nope, they can't

[magic light]

yeah that doesn't really make sense because Av = xv holds true for all As and all x's for v=0 so yeah.

[quartz compass]

yup

[magic light]

nice

thanks, got it

[quartz compass]

you're welcome

there's a similar kind of proof to show symmetric matrices have all real eigenvalues

and another similar one that shows that different eigenvalues of a symmetric matrix have orthogonal eigenvectors

[magic light]

$A^T = A$ is symmetrical?

[stoic python]

blackmamba[Beatrix Kiddo]

[quartz compass]

yeah

although instead of using the transpose you'll have to use the complex conjugate and transpose

I wouldn't worry about it for now, it'll come up eventually, just wanted to mention it so you don't forget the trick as a one time thing is all

[magic light]

ah thanks

My test is tomorrow actually lol.

[quartz compass]

fun lol good luck

[magic light]

I've been hard screwed over by it

My professor likes proofs and stuff like that, he didn't do a ton of super technical stuff taking ages, sadly because corona and stuff like that the test was changed like 2 days beforehand

into being very technical

which is what I'm weakest at lol

[quartz compass]

that sucks yeah

[pure tangle]

if i have two variables that become free variables can they be equal to the same parameter t, or do they need to be independent. ie could i have x3 = t, x4 = t or would I need to do x3 = t, x4 = s

[pseudo cobalt]

are the dot product and cross product commutative?

[limber sierra]

what do you think? have you tried out some vectors?

[pseudo cobalt]

yes

and so far

the dot product seems commutative

just tried it

and the cross product doesnt seem commutative

strangely enough

[marble lance]

If you think about what the dot product is, it should make sense why it is.

[pseudo cobalt]

$\vec{a} \cross \vec{b}$ seems to be equal to $-(\vec{b} \cross \vec{a})$

[stoic python]

The Rock Obama

[marble lance]

Exactly

[pseudo cobalt]

that would make sense though, if were using the right hand rule

[pseudo cobalt]

I tried to get an intuitive answer, but I can't see why

so if I'm not wrong:

the dot product projects a vector onto another one, and then multiplies the magnitude of the projected vector and the other vector

[limber sierra]

well if you think of the dot product formulaically

you're just multiplying vector elements and adding them, right?

and we know that elements in a vector commute

[pseudo cobalt]

yeah

[limber sierra]

under + and *

[pseudo cobalt]

oh

[limber sierra]

so certainly the dot product commutes as well

[pseudo cobalt]

that makes a lot of sense

yeah

thank you

[limber sierra]

but, as you correctly obesrved, the cross product does not

in fact, we have a term for a x b = -(b x a)

"anticommutative"

[pseudo cobalt]

haha that's cool

[limber sierra]

standard subtraction is another example of an anticommutative operation.

[pseudo cobalt]

oh yeah

$a - b = -(b - a)$

[stoic python]

The Rock Obama

[limber sierra]

right

[thorny hemlock]

i dont understand this proof

specifically how the part under 1 < k < j simplifys

<@&286206848099549185>

when j = 2 and k = 1

im getting $$\langle e_2 , e_1 \rangle = \frac{\langle v_2,e_1 \rangle - \langle v_2,e_1 \rangle \langle e_1 , e_1 \rangle - \langle v_2,e_2 \rangle \langle e_2 , e_1 \rangle}{x} $$ where $x$ is the denominator. too long to write

-_-

[magic light]

gram schmidt process is from hell

[stoic python]

Yes

[thorny hemlock]

somehow <v2,e2><e2,e1> is zero but i dont see how :/

[rocky hill]

Is there an operation that computes the product of two sums? I have two vectors which I'm multiplying after summing each individually, but would like to use numpy to do it more efficiently. It's essentially an outer product, but then I want everything summed up too.

[neat dragon]

I think all the cancelling comes from the assumption that (e1, ... e_j-1) are orthonormal

[thorny hemlock]

er

[neat dragon]

So all the terms in the numerator where you have an inner product <something*e_i, e_k> (where something is a scalar from the <v_j, e_1>'s inside of the big < > and i =1,...,j-1 ) are zero

[thorny hemlock]

why

[neat dragon]

Cause taking the inner product of two orthogonal vectors gives you 0

And the something doesn't matter, since it's a scalar

Does that make sense?

[thorny hemlock]

it does

[neat dragon]

Okey 👍

[magic light]

I have this matrix
4 1 1
1 4 1
1 1 4

What does this tell me about the determinant?

I know there's something about a matrix being symmetrical

I just don't know what it is

or I think I do mm

[thorny hemlock]

thank you

[neat dragon]

Sure! 🙂

[thorny hemlock]

ok slight issue

in the example

i have shown above

we dont know that e2 and e1 are orthogonal

@neat dragon

so when j =2 and k=1

$\langle e_2,e_1 \rangle = \langle \frac{v_2 - \langle v_2,e_1 \rangle - \langle v_2,e_2 \rangle e_2}{x} ,e_1\rangle$

[stoic python]

Yes

[thorny hemlock]

we cant cancel the <e2, e1> ??

since we are actually tryna prove that they are orthogonal

[neat dragon]

Do you know about linearity of <>?

[thorny hemlock]

yes

[neat dragon]

The e's are orthogonal up to j-1

That's part of the induction hypothesis

Try to split up the big <> into separate terms by using linearity

Something like <v_j, ek> - <e_1, ek> - ... - <e_j-1, ek>

(ignoring the scalars and the denominator)

[thorny hemlock]

ah okkkk

I see now

[neat dragon]

Great 🙂

[cunning pier]

.

[thorny hemlock]

😅

[cunning pier]

f*

[thorny hemlock]

,rotate

[cunning pier]

thanks

I've calculated the kern for this matrix
I want to know if it's surjective now since it's not injective (Kern(f) =/= 0)

[neat dragon]

I think you mixed up + and - in the second row ?

[cunning pier]

it's the finite field F4

[neat dragon]

Ah

[cunning pier]

so it should be correct

(little bit unsure about the 1 and alpha)

but alpha +1 is guaranteed correct

[neat dragon]

Yep

The image is at most 2 dimensional

And the kernel is already 2-dmsl

[cunning pier]

what does that tell me?

[neat dragon]

Have you had a theorem about the dimension of the kernel / image ?

I forget what it's called

dim(im)=n-dim(ker)

Where n is the dimension of the space that's being mapped into

So what is n in your case

[cunning pier]

dunno, can't anything in my notes

[neat dragon]

It's the number of rows of the matrix

So 2

(the theorem is rank-nullity theorem)

[cunning pier]

found it the prof's notes

but there isn't anything with surjectivity/injectivity

[neat dragon]

What is the definition of surjectivity?

[cunning pier]

For matrizes I don't know

[neat dragon]

You can apply the general definition here

[cunning pier]

each y gets assigned atleast 1 x

how'd I do that?

[neat dragon]

Yeah

That all points in the space that you're mapping into are reached

So if everything is going into the kernel

There's "nothing left" for the image

[pure tangle]

if i have two variables that become free variables can they be equal to the same parameter t, or do they need to be independent. ie could i have x3 = t, x4 = t or would I need to do x3 = t, x4 = s

[cunning pier]

I'd do s,t

[pure tangle]

great, thanks so much

[nocturne jewel]

yeah if they're both t then that implies 1 free variables instead of 2

[pure tangle]

that makes perfect sense. Thank you

[magic light]

Hello - I had another difficult(I think) question and was hoping you can tell me if I got it correct
Q:R^2 -> R^2 linear transformation
dimKerT = 1
sqrt(2) is an eigenvalue of Q
w !=0 in ImT, u != 0 in KerT
prove that w and u are linearly independent.

First of all, since sqrt(2) is an eigenvalue, then T(v) = sqrt(2)v for some v, this means that sqrt(2) is in ImT and thus v is in ImT
Because dimKerT = 1, we can conclude dimImT = 1 as well, thus w and v are linearly dependent. In other words, w = cv for some c

Secondly, if w and u were linearly dependent, then w = tu, and thus cv = tu or v = (t/c)u
T(v) = T( (t/c) u) = (t/c) * T(u) = 0
In other words, T(v) = 0
but this is impossible, because T(v) = sqrt(2)v, and v cannot be the 0 vector as v, w ,are non-zero vectors - we conclude that w, u must be independent

[still elbow]

are integers modulo 8 over integers modulo 2 a vector space?

[north sierra]

Can anyone explain how the zero matrix is in here?

[red prawn]

If a, b, d equal 0

[north sierra]

oh ok

thank you

[red prawn]

no prob

[pure tangle]

for gauss jordan reduction, can you have two 1's in a row where the first one has zeros above and below it, and the second one doesn't?

[hollow finch]

we only care about the first 1 (the leading 1)

[pure tangle]

is it possible to have a second 1 that becomes a free variable or does that make the system inconsistant?

[hollow finch]

why wouldnt it be possible?

the only thing you really care about when doing rref is the leading ones and their pivot columns as diagrams such as this try to stress
https://puu.sh/HaRtW/6e4ee798e3.png

[pure tangle]

@hollow finch gotcha. thank you so much for taking the time to explain and find those charts to show me. I really appreciate it

[night kite]

can someone help me understand how I'm not properly inversing this matrix?
https://images.irenebot.com/bonusbaby_everglow_n2r5td.png
https://images.irenebot.com/exo_after_school_y1gyoo.png

correct answer:
https://images.irenebot.com/3ye_dreamcatcher_tv34ik.png

looking back I know I could have started off with R2 - R1, but shouldn't I still get the same outcome?

[limber sierra]

recheck what you did in this step

-1 - (-1) is not +2

might be more mistakes but thats the first one i saw

[night kite]

ah ok, I didn't see that

ty

[lofty yacht]

Anybody interested in studying University level Linear Algebra together?

[lofty yacht]

Long term (8-10 weeks)

[hollow finch]

all that work is proof that the adjugate is the best way to get the inverse of a 3x3 if it's not possible by inspection

[digital bough]

How would we do it?

I am week 2 in out of 7. Currently at the chapter discussing vector spaces and linear spans. I am only constraining myself to R^n, matrix spaces and the polynomial vector spaces.

[lofty yacht]

I am currently following Linear Algebra done Right.

I just want a study partner so that we can motivate each other and finish the course in a designated time frame. I was thinking on doing 1 chapter per week. We can log out doubts, do assignments and sync up in 2-3 days or 1 week depending on your availability

[astral epoch]

idea ain't bad

[dark valley]

ayo guys, i have a question, doing the matrix adjugate of a 3x3 matrix the plus minus will be like this:
`+ - +

• • +`
    following the 1st plus and alternates, but doing a 2x2 isnt the same? im getting a wrong answer doing an exercise

[fading plaza]

I think it's just

+ -
- +

[grave bison]

det(2x2 matrix) =ad-bc

[nocturne jewel]

#probability-statistics is more akin to correlation / pearson's co-efficient

[quasi vale]

@wintry steppe Which scatter plot do you think has the worst "line of best fit".

[humble oak]

hello, what does it mean for a set to be independent in a vector space V?

[wintry steppe]

if any (finite) linear combination of any of its vectors is zero, then the scalars of the linear combination are zero

[humble oak]

when proving a set is a vector space is it necessary to prove all axioms or does it depend on the question?

like for this axiom for example

if it's really obvious that it passes, generally do i still need to write it

[ebon veldt]

need to do all of them if u want a proof

but yeah a lot of them are very short

[humble oak]

oki doki

tyvm, and ty @wintry steppe

[acoustic path]

isnt it just additivity, scalar multiplication and 0 vector

theres others but thats the 3 we use to prove a vector space

[marble lance]

You can show a subset of a vector space contains the zero vector, and is closed under addition and scalar multiplication to show it is a subspace. But you need to show all the axioms, if you don't know your set and operations comes from a known vector space.

[nocturne jewel]

that sounds like the Naive test for subspaces

[acoustic path]

yea

but its just extra noise

[marble lance]

[acoustic path]

like we aint writing full on proofs in our LA assignments

so we use the naive test whatev u wanna call it

unless if the assignment is askin to check for all axioms

ur gonna do those 3

[nocturne jewel]

If you're asked "Is this a vector space" and you just check the operations are closed and 0 vector is in it, then you cant conclude it's a vector space

cause, for example, it might not have scalar distribution

[acoustic path]

u can check its a subspace tho

[marble lance]

We aren't talking just about subspaces.

[acoustic path]

oh

subspaces>

i thought we was talkin about subspaces then my b

[humble oak]

another quick question, about additive inverse. it says something about given a vector (v for example) its inverse (-v) their addition should equate to 0. if my scalar multiplication is defined as something other than the regular operation i need to apply the newly defined multiplication to -v right?

or can i just make a vector that satisfies the conditions so that (v + something = 0)

[marble lance]

@humble oak -v is (-1)*v but that's after you know it's a vector space. -v is just how we write the vector w that satisfies v+w=w+v=0. So you need to show such a vector w exists using the addition defined for that set.

It has nothing to do with the scalar multiplication

[humble oak]

thanks!

[marble lance]

Np

[reef sleet]

When performing the elementary row operation of interchanging two rows, do I interchange the variables?

That was really badly worded

Ohhh wait it doesn't matter

Never mind lol I'm dumb

[reef sleet]

I'm so confused where to start with this :/ the leading entry of each row is even but then there's always some odd entry later and I feel like if I start working with fractions I overcomplicate this

I don't know much yet, just elementary row operations

Would prefer to solve using Gaussian Elimination

[quartz compass]

yeah good thinking, avoid fractions until the end

[reef sleet]

Due to their evenness I can't subtract anything or any multiples to get a leading 1 in R1 right??

[quartz compass]

first thing I'd do is get rid of everything below the first entry of the first column

I'd subtract 2 of the second eqn from the third eqn

then subtract 3 of the first eqn from the second eqn

then just leave a 2 there

[reef sleet]

Okay, so the resulting augmented matrices would be

1. -2R2 + R3 --> R3
   [ 2 3 3 | 3
   6 6 12 | 13
   0 -3 -25 | -24 ]

2. -3R1 + R2 --> R2
   [ 2 3 3 | 3
   0 -3 3 | 4
   0 -3 -25 | -24 ]

Hmm

Still don't see any way to turn that 2 into a 1 :(

Oh wait

Hmm

[quartz compass]

that's why I say just leave it

[reef sleet]

Would I be turning it into a 1 later? The problem says solve using Gaussian or Gauss-Jordan

[quartz compass]

yeah in the end you will

next take the middle row and use it to get rid of everything above and below it in that column

[reef sleet]

Gotchya, I could add R2 to R1 into R1, then -R2 + R3 --> R3

[quartz compass]

perfect yup

[reef sleet]

Do you think it'd be a problem if I typed the matrices in chat like before?

[quartz compass]

do whatever you like

[reef sleet]

3. R2 + R1 --> R1
   [ 2 0 6 | 7
   0 -3 3 | 4
   0 -3 -25 | -24 ]

4. -R2 + R3 --> R3
   [ 2 0 6 | 7
   0 -3 3 | 4
   0 0 -28 | -28 ]

I can multiply R3 by -1/28? 😅

Hmm

[quartz compass]

yeah definitely

[reef sleet]

Just out of curiosity, why work through it like this? Everything I've seen recommends trying to work it row by row, trying to get R1 perfect and then moving on to R2

Or at least that's what I understood

[quartz compass]

it is working it out perfect except we're not making the pivot 1

this is just to avoid dealing with fractions

[reef sleet]

What's a pivot?

What the leading entry should be?

[quartz compass]

just mean the entries on the diagonal

[reef sleet]

Ohh

[quartz compass]

the first column and second column has all 0s in it

now you're in a good place cause the last row will be really simple just 0 0 1 1

[reef sleet]

Will they eventually turn into 1? Don't they need to be 1 for it to be in REF or RREF?

[quartz compass]

so cleaning out the 3rd column is dead easy

yes

[reef sleet]

Yeah, -1/28R3 --> R3

[quartz compass]

we're nearly done so you'll see in a second

[reef sleet]

5. (-1/28)R3 --> R3
   [ 2 0 6 | 7
   0 -3 3 | 4
   0 0 1 | 1 ]

I could add R2 and R3 to R1

Makes all of R1 even

For a perfect division by 2

[quartz compass]

hmm?

just subtract the last row from the first and second rows to make it all 0s above it in the 3rd column

[ 2 0 0 | 1
0 -3 0 | 1
0 0 1 | 1 ]

now divide

[ 1 0 0 | 1/2
0 1 0 | 1/-3
0 0 1 | 1 ]

and that's it

by leaving these diagonal entries with numbers instead of 1 on them, you save the division to the very last step

you can still divide if it's not a fraction though ofc

[reef sleet]

You did -3R3 + R2 --> R2 and -6R3 + R1 --> R1?

This is so fun lol, it's like a puzzle 😝

[quartz compass]

yeah

lol can be yep

you'll probably enjoy evaluating determinants cause there are some fun tricks to do kind of similar to this too

[reef sleet]

Thanks for all the help man 😁 ah yeah I hear that name come up every single time linear algebra is mentioned LOL, I've been watching some of Grant Sanderson's Linear Algebra videos for fun and they're interesting

But not as intuitive as the calculus series

[quartz compass]

I never watched any of those, but if they're working for you, good

[native rampart]

cu+V

You define c•(u+V) to be (cu+V)

You know quotient groups?

The idea of a quotient space is to make a vector space with quotient groups as the abelian group

cosets

For a ring you would need to define a multplication Between 2 elements of the quotient group

Multiplication here is just scalar multiplication

[pure tangle]

how do i find a vector that would make a system inconsistant? for Ax = some vector b

[zealous junco]

When does there always exist a linear operator T that maps the basis from V to basis of W?

[half moss]

When the basis have the same number of dimensions?

that's what I immediately think of at least

[zealous junco]

Make it so that A is m > n, or such that m = n and det(A) = 0 I think, where b ≠ 0

oh

[zealous junco]

Is it also true if dimV > dimW right

[half moss]

Hmm well if dim V > dimW then you could establish a linear map, but it loses it's symmetry. You can't establish a linear map both ways. In linear algebra terms this means that the linear transformation matrix created would be rectangular instead of square, and it wouldn't be invertible.

So, I'm not sure if this part would apply to the question. You can definitely establish a mapping, but it would lose a lot of qualities like injectivity and whatnot.

I get the feeling that this question is trying to talk about how all vector spaces of the same finite dimension are isomorphic. If that seems about right then specifying dim V = dim W is what you're looking for.

[zealous junco]

Yea I see, thanks

I just forgot most of what I learned in lin alg, which I suddenly need again..

[pure tangle]

@zealous junco thanks for the response. So the matrix I have is a set 4x4. What do I do from there?

[zealous junco]

you want to find a vector b given the matrix A?

[pure tangle]

I want to find a vector b that makes Ax = b inconsistant

[zealous junco]

I guess you first do gaussian elimination on A

[pure tangle]

so ive done that

[zealous junco]

lemme think..

Oh

[pure tangle]

could I make b equal to one of the rows and change one entry?

*columns

[zealous junco]

you get all the row vectors and find a vector that the row vectors doesn't span

that would be your b

[pure tangle]

perfect

thank you so much

[zealous junco]

wait i could be wrong, and sorry i dont have more time to think about this...

yea i actually forgot how to do this..

[pure tangle]

i'm looking at a similar solution and the have the definition as some weird P(A|b) not equal weird P(b)

and they set b equal to one of the columns with one different entry

and then they put that as the solution to an augmented matrix with A

and then row reduce

and get an inconsistant answer

but i'm not sure why it works

or what the weird P means

<@&286206848099549185>

[zealous junco]

oh ok yea i got it

so when you row reduce you should have gotten another b1 right

like b should become something else

then as long as the thing b becomes doesn't have a zero in 4th entry, i believe it's inconsistent

[pure tangle]

so could if i have a column like [1,2,3] could I make an entry of b a non factor of A like b[1,2,4]?

[reef sleet]

How do you solve a matrix (find the solution to the system of equations) with more variables than equations using Gaussian or Gauss-Jordan Elimination?

THis is the problem in question

I performed -R2+1 --> R1, 3R1 + R2 --> R2, -R1 --> R1, and (1/2)R2 --> R2 in that order to get the system (matrix)

1 3 -12 | 20
0 1 5/4 | (-6/4)

How would I solve from here? Express one of the variables in R2 in terms of another variable and then back-sub into R1 and simplify?

[native rampart]

That means you have free variables

if you choose a value of x_3,x _1 and x_2 will be uniquely determined by that value of x_3

[reef sleet]

Oh gosh, I wasn't paying attention to the parametric solution bit 😅 people have tried explaining it to me before (I don't think in this server) but I don't get it

Do I set y = t and then express t in terms of z?

Because in my final matrix R2 only has y and z

[native rampart]

Set z=t

And write x and y in terms of t

[wintry steppe]

Why am i gettin this wrong?

can someone try this question and see whether yall are gettin teh same answer as mine?

[native rampart]

Let's say v,T(v)... Spans V. U(v) will be in V,i.e.,
U(v)=g(T)v for some polynomial g

U(Tv)=T(Uv)=T (g(T)v) = g(T)(Tv)

Similary you get U(f(T)v)=g(T) (f(T) v) for any polynomial F

i.e. U(w)=g(T)w for all w in V

@pastel saffron

i.e.,U=g(T)

U(v) will be some w(where v is a generator of the T cyclic space)

And w will be g(T)v

Since V is T cyclic

That's your g

[native rampart]

You mean this step?

[native rampart]

This?

You assume UT=TU

Because V is a T cyclic subspace, any element in V will be g(T)v for some polynomial g

And U(v) will be an element in V

Yes

Constant term is a_0

p is just a_0+a_1t+a_2t^2...

p(T)=a_0 I + a_1 T + a_2 T^2...

Here I is identity

And P(T)v=a_0 v + a_1 Tv+ a_2 T^2v...

P(T) is a linear operator

Yes

Yes

[keen flame]

What's the difference between base and kernel?

[wintry sphinx]

The kernel is a subspace

a basis is a set of vectors

[keen flame]

yes. however set of vectors can be also describe a subspace.

A basis won't be a definition if it's the same as a set of vectors like kernel won't be a loose term for subspace

[half ice]

What?

[digital bough]

i havent reached kernels yet, my book says that a bases are Unique linear combinations of vectors

[tame mural]

@_@

this might be better fixed with 3B1B imo

[half ice]

Sacc makes a very good point, that a basis needs not be a subspace. It's the least amount of info necessary to express a vector space

On the other hand a Kernel is just a specific type of subspace

They're not really even similar objects haha

[tame mural]

yeah the question is confused enough

that answering it directly seems only half adequate

that's why I think 3B1B might be better

[keen flame]

so V={(1,0),(0,1)} would generate the entire subspace. Does that make V a kernel or basis?

[tame mural]

-_-a

A kernel and basis are really different!

Saying that something is a basis is like saying that something is an ingredient

ingredient for what?

[half ice]

{(1,0),(0,1)} is a basis for R², or the xy plane.

Not the only basis, but a natural choice

[digital bough]

"A subspace to R^2 can be one of the following: the kernel, a line through the origin or the whole vector space"
So the whole vector space, R^2, is just the plane, and the whole space R^3 is just the 3D-room?

"{(1,0),(0,1)} is a basis for R², or the xy plane."
What does {} mean in this context? The linear span? The set of all linear combinations of the vectors (1,0)(0,1)? Just asking because i wanna familiarize mysef with english notation

[half ice]

Where's that quote from? I didn't say it lol

[digital bough]

first quote isnt you, second is

[half ice]

All subspaces of R² are one of the following:

• A line
• R² itself (all vector spaces are subspaces of themselves)
• The 0 space

This is just because R² has dimension 2, so the subspaces all have dimension 0, or 1, or 2 which is pretty limiting

.
{ } denotes a set, which is roughly a collection of things. I used it to say "this is the collection of vectors that is a basis"

[digital bough]

thanks

so it is wrong to say R^2 is a plane because it is a set of vectors and those set of vectors may conatin linearly dependent vectors that doesnt span a plane?

[keen flame]

It comes from discreet math for group / collection. We study that and so the lecture uses it to mean a general group.

[ember matrix]

R^2 is a plane. Who said otherwise?

[tame mural]

What is a plane?

[ember matrix]

R^2

[keen flame]

2D shape within in a 3D space. x(1,0)+y(0,1) is a plane within R^3

[tame mural]

hmm

[digital bough]

a plane contains a set of points that is spanned by two non-parallel vectors, or that is how i understand it but i may be wrong i started with this a week ago

[tame mural]

I see

hmm

[ember matrix]

so you add inner product to it as well.
Nice

[tame mural]

to me, a plane is from geometry

but I intuitively know what you mean

I would also ask, do you insist that a plane must be inside a 3D space?

What about a plane in 4d space, or just a plane by itself?

[keen flame]

@tame mural Good point

My bad

[half ice]

I might say "plane" if I'm being lazy or just want someone to picture the problem. No linear algebra course will put a proper geometric definition of a plane forward

"2-dimensional vector space" is safer haha

[fallow jolt]

Where does the A come from in the second step?

I am confused

Oh would it be

R2 += AR1?

[dark valley]

its the only information about the exercise?

[fallow jolt]

wdym

[dark valley]

thats the whole exercise?

[fallow jolt]

This is just a proof I'm reading

I'm still not completely clear tbh

On each step's operation

[keen flame]

what does this mean?

[half ice]

I'm thinking NULL(B) means the nullspace of B

And sp means span

So any vector you can make with a linear combination of those two is the span

[uncut wraith]

Idk im only 5th grader

[half ice]

And to put a picture in your head, will make a plane in 5D space

[keen flame]

I'm listening

[fallow jolt]

<@&286206848099549185> sos

[wintry steppe]

Can a vector space contain a finite number of vectors?

[native rampart]

Yes

Take some vector space over say F_2

Now, pick a vector. The vector space generated by that vector will be finite

[wintry steppe]

@native rampart what is f_2?

[native rampart]

Field of 2 elements

[wintry steppe]

so, can a vector space be for example a square?

[native rampart]

What are your operations

[digital bough]

Square as in the set {x^2 : x in R} ?

[wintry steppe]

let's say that we have a vector space that contains the vectors i and j, that space over the field of F_1

is just three lines right?

[limber sierra]

F_1 does not exist

also when we talk about "vector spaces over a field" we need our vectors to somehow be compatible with the multiplication from that field

just talking about vectors "i" and "j" doesnt make sense unless you define what the multiplication from the field is

and how they interact with it

[native rampart]

What about the trivial field

[limber sierra]

is not a field.

anyway, i have no clue how to address "is just three lines"

i feel like you fundamentally misunderstand what a vector space is

vector spaces dont necessarily have to have a direct analogue to euclidean vectors

so you cant necessarily visualize them as "lines"

(in fact, "line" in the context of general linear algebra is often used simply to refer to the span of a single vector - since the geometric notion doesnt really make sense for most spaces)

[digital bough]

If I span his set of two vectors then I get a vector space right?

[limber sierra]

spans of vectors are vector spaces, yes

they form a subspace of the space the vectors are from.

it's a good (though easy) exercise to prove this fact

[digital bough]

I have some excercises to define two operations on a set and then prove it is a vector space, but I am skipping it unfortunately. The pace this course is going at is too fast so i ought to settle with just knowing how to prove that a set is subspace of a certain vector space.

[native rampart]

That's a pretty important exercise

Do it

[digital bough]

I am not a math major, I am already behind, I should know least square and gram schmidt by now.

[limber sierra]

i mean regardless, the fact i mentioned is of the form prove-this-is-a-subspace

so if those are the exercises youre doing, that qualifies

its not really... much of a proof though

proof by "duh"

[digital bough]

There are like 8 axioms for vector spaces that need to be satisfied or whatever you call it, i remember that we used something that kinda was circular reasoning to prove that a*0=0 (a is a real number) using a set of axioms and we used tables to prove it. Not sure if that counts as proof, I reckon you can do something similar to vector spaces?

[limber sierra]

uh i sure hope it wasnt circular reasoning

since thats not valid

but yes, if you use a table and you exhaust all cases (or at least enough cases that you can WLOG the rest away), that suffices as a proof

it's usually a very inefficient proof method for not-small things though

[wintry steppe]

can a vector field be defined over the field F{0}?

[limber sierra]

there is no field F_0.

im not sure what F{0} means

could you describe the field F{0}?

[wintry steppe]

F(x,y,z) = (0,0,0)

[limber sierra]

...

so F is... a function

and not a field

you cant define a vector space over a function, no

that doesnt make senes

[wintry steppe]

wait wait

my bad

let's say that this field is consisted of only the 0 vector

[limber sierra]

fields dont have vectors, vector spaces have vectors

in a linear algebra context, we often call field elements scalars

[wintry steppe]

pfff, I'm messed up

[limber sierra]

anyway

a field that consists only of the 0 element doesnt exist

we generally require fields contain at least 2 elements

the typical way we do this is by saying 0 must not equal 1

there are some good mathematical reasons for this, mostly that if we allow F_1 to be a field, we have to throw out basically all our theorems about finite fields

or add a special case for F_1

[wintry steppe]

oh, so that they can have the necessary properties

for example associativity

you can't do that if you have less than 2 vectors

[limber sierra]

eh you could

the group of 1 element for example makes sense

as then the theorem just becomes

e * (e * e) = (e * e) * e

if e is your only element

[wintry steppe]

yeah...

then why do we need 2 elements?

why is it necessary?

[limber sierra]

well as mentioned

the ring with 1 element doesnt really behave like a field

whatsoever

the notion of "characteristic" doesnt make sense, for one

geometry over it is nonsense*

• (we can KINDA do geometry over F_1, but it wont be the same as geometry over the trivial ring)

[limber sierra]

it also screws up the theorem

"all finite fields have order p^k for a prime p"

i guess you could just allow k = 0 and that still holds but

it kind of goes against the "spirit" of the result

which is really a deep statement about the characteristic of fields

(see https://math.stackexchange.com/questions/3756806/why-isnt-the-zero-ring-the-field-with-one-element)

[wintry steppe]

thanks a lot for your help

I will try to further investigate what you just said

[limber sierra]

the honest reason is that theres very deep mathematical reasons to not consider the "field" (ring) of 1 element to be a field

but these arent really obvious in a first course

typically speaking

[wintry steppe]

how long have you been doing this kind of math?

why do you know all these stuff?

(and I don't)

[limber sierra]

uh, almost a decade now but i knew this after my first abstract algebra course

[wintry steppe]

I finished linear algebra but I don't know this stuff as well as you do

[limber sierra]

linear algebra kinda "speedruns" the definition of a ring/field since it only really uses them to talk about vector spaces

which are the object linear algebra is really studying

abstract algebra, specifically ring theory, studies fields and rings in their own right

in much greater detail

finite fields in particular are very well-behaved, which is a nice way to say "not very interesting"

[wintry steppe]

pfff, I have this problem that I can't move on if I don't fully understand something to it's roots... It costs me a lot, wasting my time trying to figure out how and why everything works from skratch

that's why I will never succeed

(in math and physics)

anyway, again thanks a lot for your time and effort

[wintry steppe]

by the way @limber sierra for a transformation to be invertible, does it also have to be an isomorphism?

[gray dust]

@wintry steppe ok you crossposted to q0. in the context of linalg (along with other algebra topics), isomorphisms ARE invertible maps

[humble oak]

when dealing with the vector space P_n does it only have one element?

i keep seeing that it's suppose to have element(s) but from this

isn't it just one element

[gray dust]

no

a_i in R is short for 'each coefficient a_0,a_1,...,a_n can individually vary over real values'

so P_n is very much an infinite set

[humble oak]

so to name an example of an element, a_0 is one element, a_1x is another?

[gray dust]

no. each element of P_n is a polynomial of degree at most n

[humble oak]

so i could have polynomial degree 0, polynomial degree 1, ... , polynomial degree n. those are all the elements?

[gray dust]

say n=2. P_2 is the set of polynomial of deg at most 2. examples of elements in P_2 are the constant polynomial 6, a degree 1 poly 3x+8, and degree 2 poly 4x^2

[humble oak]

ah okay

[nocturne jewel]

dk why my messaged are being deleted but ok

[humble oak]

ty all i understand now 😄

[gray dust]

you're welcome

[marble lance]

[nocturne jewel]

someone's trigger happy on the delete button oop

[hoary osprey]

happened to me before, quite curious

[sweet vine]

which book offers a decent amount of exercise on similarity transformation?

[remote gorge]

Anyone knows a good intro linear algebra book that also goes over bilinear forms and quadratic forms?

[wintry steppe]

@remote gorge section 6.8

[remote gorge]

@wintry steppe thanks a lot!

[wintry steppe]

there's a whole part in there on the second derivative test in MVC lmao

[tame mural]

Mmm I've noticed that many linear algebra texts introduce vectors as straight arrows almost right away

that Friedberg book does that

so does Axler

[wintry steppe]

gotta ease the high school kids into the book somehow

[native rampart]

Try Hoffman Kunze

[wintry steppe]

[humble oak]

what does it mean for a set to span a vector space?

[gray dust]

every vector in the space is a linear combo of the vectors in the given set

[humble oak]

so, for instance if i had the vector space R^2

if i had a set that contained 2D vectors that set would span R^2?

[gray dust]

gimme a set

[humble oak]

uhhh