linear-algebra

[wintry steppe]

about the red axis

[quartz compass]

well no matrix can do that with this as your origin

you'll have to translate it first

[dusky epoch]

you can do an affine map i guess

[quartz compass]

so you have two tasks: determine the distance from the origin to the red line, and determine the angle between the xy plane and the upper point at that angle there, you can use a dot product

you can do a translation with a matrix if you're using homogeneous coordinates if you really want to be fancy, sure

but probably not a good idea to try to do it that way if you're asking this question in the first place lol

@wintry steppe so do that, find a vector from the origin to the red line that's the shortest length

this should be easy cause it's nice and symmetrical

hint: it's the average of two vectors

[dusky epoch]

(1/2, 1/2, 0)

i guess then that the sequence of transformations would be
translate by (-1/2, -1/2, 0)
rotate by some angle around (1, -1, 0)
translate by (1/2, 1/2, 0)

[wintry steppe]

how do I rotate about (1,-1,0)?

I understand how to do it along the axis

[lavish jewel]

you can do that in steps as mentioned above, or try the rodrigues rotation formula

[wintry steppe]

I mean after I translate by (-1/2,-1/2,0)

oh, do you mean like rotate so I can rotate about an axis and then rotate back?

[dusky epoch]

yeah i think thats what it boils down to

[glad hamlet]

is it impossible to simplify fractions if the numerator or denominator is an odd number?

[dusky epoch]

this isn't linear algebra, just fyi.

sometimes it's impossible, sometimes not.

[glad hamlet]

what is linear algebra

ohh wait nvm

[dusky epoch]

see pins

[glad hamlet]

yea i did

[quartz compass]

you can get the angle from the dot product

[zealous junco]

In finite fields, do we need to be careful doing row reduction?

I found this problem pretty interesting

Suppose F was Z/2Z (It cant happen here because it is a subfield of C) then this is no longer true

like my question is we can still do row reduction right for finite field except just doing mod

[dusky epoch]

yes

[wintry steppe]

Sup

[wary lily]

is it correct to say that a transform that maps from a higher order dimension to a lower one, has always infinite many solutions, if any?

[native rampart]

You mean Ker is non trivial?

Yes

[wary lily]

I don't know Kernel. Just R^n -> R^m mapping

where n > m

[dire thunder]

ker T = nul T = {v: Tv=0}

[wary lily]

and the system is consistent

[dire thunder]

if n > m then nul T includes not only zero vector

yo az

i suggest you to prove that set of vectors v s.t Tv = 0 is subspace of V where T : V->W and V,W are both vector spaces

[wary lily]

I'm still not that far

will be there in a few months IG

[wintry steppe]

just translate the statement into subspaces of euclidean space and linear transformations between euclidean spaces if you're uncomfortable with abstract vector spaces

there's no loss of generality after all

[dire thunder]

all functions are analytic?

[wintry steppe]

[dire thunder]

privyet tterra

kak dela

[wintry steppe]

good

first time getting 8 hours in two weeks

[dire thunder]

why good

[dire thunder]

8 hours of what

[wintry steppe]

sleep

[dire thunder]

nice

you have to follow sleep regime, kid

[wintry steppe]

kid

[dire thunder]

how old are you

[quartz compass]

he's like 9 or 10

[dire thunder]

how do you know mero

i know that you are 5

[wintry steppe]

at mods

,ban mero

[stoic python]

This may only be done by a moderator!

Mar 27, 2021, 2:50 PM

[quartz compass]

😮

[wary lily]

A is just [v_1 v_2], right?

[tame mural]

Does anyone have any recommendations for introduction to finite groups?

[wary lily]

maybe ask in #book-recommendations

[lavish jewel]

your matrix A looks ok az

[coarse marsh]

is there a simple way to get the determinant of this matrix?

[wary lily]

starting from the left subtract columns from their counter part on the right

this will leave you with 2 diagonals and the rest 0

the main diagonal will be all 1000

the other diagonal will be all -1000

then add those corresponding rows and you will have a matrix full of zeros

@coarse marsh

[coarse marsh]

i think i get it

but wouldnt that just leave me with a matrix where the det is 0?

[wary lily]

yes

you even don't need to do that all

just subtract first column from last column

then add first row to the last row

you find the last column is 0

a matrix with a column/row of zeros has determinant zero

done

[coarse marsh]

ooooh

i got it now

thanks

[wary lily]

you welcome

[dusky epoch]

honestly i would probably try and do something with characteristic polynomials

but that's highly specialized fuckery

[coarse marsh]

lol

i dont think i've learned that yet

[dusky epoch]

lemme try and see if it works out.

[coarse marsh]

oki

[dusky epoch]

looks like it did work out

and the answer is -972 * 10^21 exactly

[coarse marsh]

wait

damn

[dusky epoch]

let me doublecheck with matlab

but i'm pretty sure i'm right

[coarse marsh]

my first reaction was putting it on symbolab

then it gave me 1028 * 10^24 as an answer

[dusky epoch]

1028 * 10^24 hm

i might've made a sign error somewhere.

oh yeah.

i did

oops!

plugged in 1000 where i shouldve had -1000

[coarse marsh]

oh

lol

it happens

[dusky epoch]

wait. 10^24? not 10^21?

or was it 1**.**028 * 10^24

[coarse marsh]

10^21

sry

you are right

[dusky epoch]

you won't get the last col as zero, az

there will still be a -1000 in the lower right corner after doing what you did

if you want i can explain what i did.

[coarse marsh]

it would be nice

[dusky epoch]

let $u = \bmqty{0\1\2\3\4\5\6\7}$ and $v = \bmqty{1\1\1\1\1\1\1\1}$, then we are asked for the determinant of the matrix $A = uv^T + 1000I$

[stoic python]

Ann

[dusky epoch]

is that much clear?

[coarse marsh]

i think so

thank you so much

[dusky epoch]

uh

i'm not done yet lmfao

[coarse marsh]

im sry

im doing other exercises at the same time

[dusky epoch]

now if we consider the eigenvalues of $uv^T$, we get the following:
\begin{itemize}
\item 0, with multiplicity 7 (eigenspace consists of everything orthogonal to $v$)
\item $v^Tu = 28$, with multiplicity 1 (eigenvector $u$)
\end{itemize}

[stoic python]

Ann

[dusky epoch]

so we have that $\det(uv^T - \lambda I) = -\lambda^7(28 - \lambda) = \lambda^7 (\lambda - 28)$

[stoic python]

Ann

[dusky epoch]

to find $\det(A)$, plug in $\lambda = -1000$

[stoic python]

Ann

[coarse marsh]

oh, i forgot

thank you once again for your patience

[wary lily]

I don't see why this doesn't work

if you subtract first col from last col you are left with [-1000 0 0 0 0 0 0 1000]

now add first row to last row and the last column becomes all zeros

@dusky epoch

[dusky epoch]

you're left with -1000 in the top right corner then

[wary lily]

damn

[dusky epoch]

your second operation affects only the last row

[wary lily]

I'm row reducing like Thanos

you're right

then my first method works tho

[wary lily]

this

it won't become all zeros like I said

it will become a diagonal

which we can easily multiply

[oblique rune]

differential n-forms are a type of tensor right?

What about their transformation properties?

how would something like $dx\wedge dy$ transform when changing coordinate systems?

[stoic python]

Exynouz

[wintry steppe]

well you'd look at how dx and dy transform

and then simplify a bit using some wedge product properties

[oblique rune]

Oh, so like this?

$dx \wedge dy = (\sin\theta dr+r\cos\theta d\theta) \wedge (\cos\theta dr-r \sin\theta d\theta) = -r dr \wedge d\theta$

Is there a way to represent this using Jacobians?

[stoic python]

Exynouz

[oblique rune]

Using Jacobians is the way I usually see coordinate transformations being done

[wintry steppe]

you have the right idea, yeah

although i think it should be x = r cos theta and y = r sin theta (i think you did it the other way around)

[uncut bison]

Could someone help me prove that this polynomial is LI?

[faint lintel]

there's no way to sum x^3 and x^2 to get x^4

I mean really what you want to show is that

ap(x) + bq(x) + cr(x) = 0 implies a = b = c = 0

so like just work with that

set up the matrix, row reduce and die cause row reduction sucks most of the time and then you'll see everything is 0

[ancient perch]

how was this last step simplified?

[faint lintel]

row reduction it seems like

[ancient perch]

i plugged it into a calculator to see but i got this

[blissful vault]

can someone help me with this?

except C and F i dont' know which ones are linear

[limber sierra]

that's also a valid row reduction, yes.

your first image provides a row reduction but not a full row reduction

whereas your second image is a full row reduction

i don't know why they didnt fully row reduce, hard to tell without context.

[stoic python]

slimvesus

slimvesus

mirzathecutiepie

slimvesus

slimvesus

slimvesus

slimvesus

slimvesus

slimvesus

[fickle skiff]

elementary matrices are basically row reduction matrices

ie you can represent a row operation with an elementary matrix

[nocturne jewel]

I mean you can use the 1st line to find the inverse of H

$H=E_1E_2...E_nI\implies (E^{-1}nE^{-1}{n-1}...E^{-1}_1)H=I$

[lavish jewel]

they're the same thing. doing row operations is the same as multiplying by elementary matrices

you can use matrices that swap rows, add them up, scale them, etc

[stoic python]

moshill1

[nocturne jewel]

so $H^{-1} = E^{-1}nE^{-1}{n-1}...E^{-1}_1$

[stoic python]

moshill1

[nocturne jewel]

which is consistent with the 1st line, invertible H^-1 is a product of elementary matrices

[distant gate]

Is matrix calculus considered on topic for this channel?

[lavish jewel]

that's a better fit for multivariable calculus

[distant gate]

If only it was covered in my multi-variable calculus class.

I don't even know if stewart calculus covers it.

Maybe parts of it.

[lavish jewel]

what's the q

[distant gate]

I don't understand how the differentiated (8). Did they skip a lot of steps?

[lavish jewel]

mhm

[distant gate]

$d\theta = d(e'e) =?= e'd(e)$?

[stoic python]

Steve3

[distant gate]

Not sure I see why that's true.

Is e' here treated as a constant?

And if so why?

[lavish jewel]

you can try and see for yourself by expanding it as a sum

is this over the reals or the complex numbers?

[distant gate]

This is the reals, is e' treated differently if it's a row vector?

[lavish jewel]

with complex numbers yes, with reals, no

just expand it as a sum

[distant gate]

over complex it's just conjugate transpose.

[lavish jewel]

e'e is the 2-norm squared of the vector

so $e^Te = \sum_n e_n^2$

[stoic python]

Edd

[distant gate]

Yeah I know this.

[lavish jewel]

this is literally the only thing you need though

now take the gradient w.r.t. e, which is the vector of partial derivatives w.r.t. e_n

[distant gate]

The book has $d(x'y) = d(x)'y + x'd(y)$.

[stoic python]

Steve3

[distant gate]

So I'm guessing they skipped some steps.

[lavish jewel]

you can also do it like that if you want and it'll be equivalent

[distant gate]

Yeah but I'm trying to understand matrix calculus b/c literally everything in modern statistics/ml uses this language.

I guess it's a good idea to be able to translate back and forth.

[lavish jewel]

it doesn't matter which one you use

i have to say that this is inconsistent regarding the vector dimensions though

[distant gate]

What do you mean?

[lavish jewel]

let's do it another way. look at what you wrote up there in terms of differentials

if you divide both sides by d(x)', it tells you the derivative of x'y w.r.t. x' is y + x' dy/dx

y usually does not depend on x, and so d(x'y)/dx' = y

you treat the whole of x' as if it were a single variable in your usual 1-variable calc

if y = x, however, the term x'd(y) is no longer 0, but rather x'd(x)

now, x' and d(x) are vectors of the same size, and that expression is a dot product that yields a scalar. over the reals, you can transpose a scalar and get the same scalar

so x'd(x) = d(x)'x

so now you have that d(x'x) = d(x)'x + d(x)'x = 2d(x)'x

[sturdy portal]

According to the entry-wise definition of matrix multiplication,
$c_{ij} =[A_{m \times n}B_{n \times p}]{ij} = \sum\limits{k=1}^{n}A_{ik} \cdot B_{kj}$. Do I understand correctly that for $c_{ij} \in \mathbb{F}$, where $\mathbb{F}$ is a field, $c_{ij} \neq \langle a_{i \bullet}, b_{\bullet j} \rangle$?

For example, for complex inner product space V, $\langle \cdot, \cdot \rangle = w_1 \overline{z_1} + \cdots + w_n \overline{z_n}$.

In Einstein summation notation $c_{ij} = a_{ik}b_{kj}$. What's the name of that operation? Dot product, according to many sources, is defined either for real or complex vectors.

[distant gate]

That clears a lot up.

Have you seen something like this before?

[stoic python]

JohnDark

[lavish jewel]

about the matrix calc?

[distant gate]

Yeah I feel dumb not being able to figure this out.

[lavish jewel]

i get paid to do it for a living

[distant gate]

I even studied most of this in college just not all together.

[lavish jewel]

(even though i'm terrible at it)

[distant gate]

Honestly, if you have any resources on it it would be much appreciated.

I was reading from the book Matrix Differential Calculuswith Applications in Statisticsand Econometrics

I wish undergrad would go over this stuff as it seems important to literally all modern research.

But I've talked about this in the past and they say oh, it's just notation....

Everything you've said makes sense but I would have never thought it about on my own lol

[lavish jewel]

they are inner products, you just define the rows of A appropriately so that the product takes that form. over the reals, you directly get cij = <ai, bj>. over the complex numbers, you let the rows be conjugate transposes of something.

[lavish jewel]

i would agree there. all you have to do is write the stuff as a sum and differentiate one variable at a time

i don't have any good books on it off the top of my head, though wikipedia has a bunch of nice properties listed with no justification

the people in multivariable calc will probably have better suggestions, so give it a shot there

#multivariable-calculus

[sturdy portal]

According to ``Linear Algebra Done Right" by Sheldon Saxler, inner product requires conjugate symmetry: $\langle u,v \rangle = \overline{\langle v,u \rangle}$ for all $u,v \in V$

[stoic python]

JohnDark

[lavish jewel]

mhm

[sturdy portal]

And I'm not sure that conjugates are defined for fields.

[lavish jewel]

in some fields they do nothing

[distant gate]

Can I ask how you use this for work?

[lavish jewel]

what you showed up there was the basic "least squares" or "pseudo inverse"

i tend to work a lot with so-called multi-channel image processing

stuff like x-rays, ultrasound, radar

[nocturne jewel]

if V is complex, then conjugate-symmetry is needed

[lavish jewel]

in which you end up solving regularized versions of that same problem

[nocturne jewel]

Is minimizing $||Ax-b||$ the same as finding least-squares solution(s)?

[stoic python]

moshill1

[lavish jewel]

yes

[nocturne jewel]

[sturdy portal]

@nocturne jewel I have to think if $a_{1k}b_{k1}$ is really conjugate-symmetric.

[stoic python]

JohnDark

[sturdy portal]

Conjugates are not defined on abstract fields, so conjugate symmetry cannot be defined as well.

[lavish jewel]

maybe nami or someone else can comment. i don't know enough to answer your question

[sturdy portal]

@lavish jewel Thank you anyway ❤️

According to the entry-wise definition of matrix multiplication,
$c{ij} =[A{m \times n}B{n \times p}]{ij} = \sum\limits{k=1}^{n}A{ik} \cdot B{kj}$. Do I understand correctly that for $c{ij} \in \mathbb{F}$, where $\mathbb{F}$ is a field, $c{ij} \neq \langle a{i \bullet}, b_{\bullet j} \rangle$?

For example, for complex inner product space V, $\langle \cdot, \cdot \rangle = w_1 \overline{z_1} + \cdots + w_n \overline{zn}$.

In Einstein summation notation $c{ij} = a{ik}b{kj}$. What's the name of that operation? Dot product, according to many sources, is defined either for real or complex vectors.

[stoic python]

JohnDark

[sturdy portal]

Inner product requires the definition of a conjugate, which is not defined on an abstract field.

Thank you, I was trying to figure out what this sum was for quite a while

And before I stumbled upon Einstein notation, I couldn't even conveniently write it down

[lavish jewel]

you forgot to transpose l in l' T

mhm

[stoic python]

mirzathecutiepie

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

[sturdy portal]

I guess the only thing known about $a_{ik}b_{kj}$ for $A_{m \times n}, B_{n \times p}$ that $a_{ik}b_{kj}: \mathbb{F}^{m \times n} \times \mathbb{F}^{n \times p} \to \mathbb{F}$, if we abstract away from i and j

[stoic python]

JohnDark

[zealous junco]

for finite vector space do c1f(a1)+...+cnf(an) = f(c1a1+...+cnan) = 0 then by bijective this is only true when c1=...=cn=0

first equality assuming ur isomorphism satisfy cf(a) = f(ca) but idk

so pick a basis a1,...,an

f(c1a1+...+cnan) = f(0) = 0

and if f(a) = f(0) then a = 0 since f is bijective, but that means c1=...=cn=0 since a1,..,an lin indep

now u pull the c out

ye so f(a1) through f(an) r indep

and ur done

they span too

like pick any other a in V then f(a) can be written as linear combo of f(a1) through f(an)

ye

i dont understand this pf

they say if B' = {a1',...,an'} is ordered basis then it is clear that... this part i understand, but i dont know how the next part works

doesn't this equality use the assumption that B' is an ordered basis, which is what the pf is all about

like first of all how do you know there is a B' such that i) is valid

[stoic python]

mirzathecutiepie

mirzathecutiepie

[zealous junco]

ok i think i got itnvm

[stoic python]

mirzathecutiepie

[novel hamlet]

how should i go about showing this

Matrix D diagonal entries are matrix A eigenvalues and P colums are matrix A eigenvectors

(sorry for epic ms paint skills)

[dusky epoch]

well

let's say P = [v_1, v_2, ..., v_n] where v_i are column vectors

and let's write λ_1, λ_2, ..., λ_n for the entries along the diagonal of D

your goal is now to show that $Av_i = \lambda_i v_i$ for $i=1,2,\dots,n$

[stoic python]

Ann

[novel hamlet]

somehow like this?

and then lambda1 correspond to vector V1?

[dusky epoch]

yes

[novel hamlet]

Im still confused where does the P^-1 come to play

[dusky epoch]

$P^{-1}v_i = e_i$

[stoic python]

Ann

[novel hamlet]

that V_i is colum of the P? A= obviously matrix A and lambda is row of matrix D?

[dusky epoch]

...

v_i is the ith column of P yes

λ_i is a number..

[novel hamlet]

been trying to google this on math stacckexchange but with few success

https://math.stackexchange.com/questions/3235534/eigendecomposition-proof

basically this?

[lavish jewel]

you are pretty much being asked to justify the first step that person took in the answer

[novel hamlet]

just came to mind, probably wont affect this, but i found proof of this for real values, i think the same works for complex numbers too?

[lavish jewel]

shouldn't make much of a difference here

[novel hamlet]

im still a bit confused if this is called eigendecomposition or spectral decomposition

or they are synonyms

[lavish jewel]

for a matrix, the spectral decomp is the eigen decomp

spectral decomp is more general

[novel hamlet]

yeah, i found this nice proof while i was googling how to do this

should work?

[lavish jewel]

sure. it's the same thing you were given above 😛

well, also more general since they don't assume Q is invertible, though it kinda happens as a consequence

[novel hamlet]

tbh im starting to think this "higher math" is just take some random D insert to these conditions and poof D matches them and the desired result is achieved

while im still thinking like in high school wtf is D and how i know i can put it here

[novel hamlet]

how do i calculate Schur lemma?

i dont get where the numbers come from

[limber sierra]

numbers?

oh youre referring to the schur form

sorry, thought you meant the result from rep theory for a second

anyway, there's a variety of sources on the internet covering schur decomposition; it's a bit of a large algorithm to try and summarize over discord messages

a google search brings up this: http://web.math.ucsb.edu/~padraic/ucsb_2013_14/math108b_w2014/math108b_w2014_lecture5.pdf see section 3 for a practical algorithm

a more "naive" algorithm is just repeatedly gram-schmidt'ing

to generate an orthonormal basis

and that gives you the columns of your matrix eventually

(this is detailed in the big 7-part proof in my link)

[pseudo thicket]

is orthogonal matrix the same as orthonormal matrix?

Quite confused with the terms

Orthogonal matrix has orthonormal row/columns

then there is no such thing as orthonormal matrix?

[limber sierra]

some people say "orthonormal matrix" to mean "orthogonal matrix"

some sources even use the terms differently

one requiring square-ness and meaning A^-1 = A^T, which is equivalent to orthonormality of rows/columns

and another that does NOT require square-ness (hence A^-1 doesn't make sense without further work) [https://en.wikipedia.org/wiki/Orthogonal_matrix#Rectangular_matrices]

just stick to whatever definition your source/course/text is using

[pseudo thicket]

thank you

[zealous junco]

can someone explain why "it is now clear that R is uniquely determined by W?

proof of this, im losing the big picture

(the last paragraph)

like i understand everything before that but how that ties together to say R is unique row reduced echelon i have no idea

[native rampart]

You understood how k_1 ,k_2,...k_r are uniquely defined by W?

Now,just use 2 rref forms are row equivalent,to show rref is unique

@zealous junco

[native rampart]

Suppose there are 2 distinct rrefs call them A and B,atleast one row should be distinct. That row vector in B can be obtained from row vector in A,via some row operation

[zealous junco]

yea, but actually before that

im confused the whole purpose of this paragraph

[native rampart]

That's to show pivots of rref of the matrix will be in columns k_1,k_2...

For an example,Take
p_1=(1,2,3,4) p_2=(0,0,1,2)

k_1=1 and k_2=3

[zealous junco]

ok so since for all B in W, we can write B = (0,...,0,0,bks,...,bn) for some s = 1,...,r and bks≠0

is that what its writing?

[native rampart]

Yes

[zealous junco]

i thought that was obvious by definition

of the rhos

wait i must be missing something, if rho_1,...,rho_r are defined as the row vectors of R so that their leading 1 is in column k_1,...,k_r, then why do you still need to show this?

[native rampart]

If you just put those vectors as row vectors,You end up with a row reduced form

Not rref

[zealous junco]

i mean once you order them they are rref

[native rampart]

Not really sure why he did that

[zealous junco]

oh anyways i dont understand this also

wait actually.. i think i do

so is it because of the fact about B he proved

if B≠0 in W then its first nonzero entries has to be in one of k_1,...,k_r, so if there was another R'~A with another rho that didnt have leading 1 in one of those columns then its contradiction

[native rampart]

Do you understand why it's contradiction?

[zealous junco]

yea since the leading 1 would be not be one of k_1,...k_r so specifically 1*rho would be in W

right

[native rampart]

Yes

[zealous junco]

ok nice then i think i got the proof

thanks

[wary lily]

this is false, a uniqueness questions is "Is T one to one"?

Meaning there is at most one solution for every c in R^m

someone check, please

[dusky epoch]

what's a uniqueness question

[wary lily]

whether the solution to Ax = c is unique

[dusky epoch]

okay

then surely is c in the range of T? is an existence question

[zealous junco]

for this question i found that there is 2 solution for alpha and beta

but im confused if that is even allowed or i messed something

alpha can be (cosø,sinø) but that means beta = (sinø,-cosø) or (-sinø,cosø)

[dire thunder]

[zealous junco]

but they give 2 different coordinate

[dire thunder]

why you want to transfer to trig here

[zealous junco]

cuz i thought its easier

to represent solution set as trig

[dire thunder]

why won't you just show that these vectors are lin independent

which automatically make them basis

[zealous junco]

no the question im confused is the coordinate part

the basis is ok

[dire thunder]

well anyway

why you think they should give the same coords

[zealous junco]

oh ur right

my bad i was confused

thanks

[wary lily]

True. Let T(x) = Ax be a mapping for R^m |-> R^n. If for every c in the codomain of T there exists a solution x in the domain, then T is onto R^n.

[zealous junco]

Suppose i have B = {a1,...,an} and B' = {b1,...,bn}.

B is a ordered basis but B' starts off as any set

but I found P invertible so that $b_j = \sum_{i=1}^nP_{ij}\alpha_{i}$ for all $j = 1,...,n$, how do you show that B' spans V? To conclude that B' is the unique ordered basis by theorem 8 works on it

[stoic python]

Anticipation

[zealous junco]

wait i think i figured

i show if c1b1+...+cnbn = 0 then ci's = 0 by invertibility of P

[blissful vault]

i need help for this

the half magic number is 1+1

if we remove 2 from every entry in the basis it should give -1 everywhere.//.

[zealous junco]

i think ur B is not the simplest but still possible

so for the first one [1 1;1 1] well what u get is [-1 -1 ; -1 -1] and that is -1a1+ 0a2

and for second one u get [0 -1; -1 0] which is -1a1+a2

so ur [L]BB should be just taking the coefficient of a1 and a2 for first line as first column, and second line as second column

[pseudo thicket]

is there a way to check if a matrix contains an orthgoonal sets?

to quickly check*

[dusky epoch]

what would that even mean

[quartz compass]

if the columns of a matrix A are all mutually pairwise orthogonal then A^T A = diagonal matrix since you're looking at every combination of dot products

[blissful vault]

THANKS ALOT ANTICIPATION!

[wary lily]

I'm a little confused when they talk of a "line" in R^n.

[dire thunder]

why they should not

[wary lily]

yeah, makes sense

this way of thinking will also extend to having 2 parameters and being a plane

this way of thinking helps to take the plane to dimensions above 3

[dire thunder]

i mean az, if you want you can always visualise stuff by R^2 and R^3 since then you can generalize

[wary lily]

yeah, I didn't think before about them as parameters

I knew it but it wasn't that bold for me

now it's easier to generalize from 2d/3d

[native rampart]

Maybe non square matrices?

For square matrices,both are same

EEEE will be left inverse

You know linear transforms?

You know a linear transform from a space to itself is injective implies it's surjective?

That's why right inverses and left inverses are same(and both exist,if one exists)

You know you can represent a transform by a matrix?

and vice versa

If a matrix has a left inverse,it means it's corresponding linear transform has a left inverse

and same for right inverses

and vice versa

Yes

And there's one more very important theorem to finish this:
A function is injective iff it has a left inverse

and A function is surjective iff it has a right inverse

Take the matrix, let's say it has a right inverse

Take the corresponding linear operator,it will be surjective by that

So, It's also injective,which means it has a left inverse

which means the original matrix also has a left inverse

if a function has both left and right inverses,both will always be same

Yes

[waxen flume]

https://i.imgur.com/pNpQGzz.png

For A I got always, B I got sometimes
C always
D, sometimes

can someone check or counter prove any of them

[wintry steppe]

seems okay

well

c might have to be sometimes, because i know there's sometimes some fuss over whether or not 0 is called an eigenvector, and 0 is certainly a scaled eigenvector

@waxen flume

[stoic python]

az

[wary lily]

can someone have a look, please

[faint lintel]

How do I test if a stochastic matrix is regular?

I know that A is regular if there is some power m such that A^m contains only positive values

(note A^m+1 could contain non-positive values, just there has to be at least one such m)

Is there a good test for this?

[tiny palm]

anyone happen to know what i'm doing wrong here? this is the question (i tried plugging in y=1 and z=1 after this attempt as well but it didn’t work)

and here is my work

when i plugged in my original solution from my work i was also marked wrong

[waxen flume]

Thanks @wintry steppe

[sudden narwhal]

@tiny palm how did you do the gauss pivot ?

Line 2 <--- Line 2 - 5 x Line 1

1 - 5(-4) = 21

You did a mistake

[tiny palm]

ohhh my gosh thank you, dumb mistake

[sudden narwhal]

no problem

[wintry steppe]

so here

in this equation

Why do you only simplify 1/7 and 5

but not with (x - 1)

pretty sure u just can do

1/7 * (x - 1) = 5
x - 7 = 35

like thi

s

why not 7 * -1

i messed something up

lol

uhm

weir

d

lmfao

[lavish jewel]

try equation* to get rid of the number

[stoic python]

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

[wintry steppe]

who coded the bot

hm

okay

$\begin{equation}
\frac{1}{7} (x-1) = 5
(x-1) = 35
x = 36
\end{equation}$

[stoic python]

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

Shaky

[wintry steppe]

lol

[lavish jewel]

#bots

i saw

it... sure looks that way

i also have no good leads about the volumes, other than saying "yeah it works in 2 and 3D"

must be some fancy trick to build parallelotopes

idk

or integration as you say

understanding it, on the other hand...

here's what you want

https://math.stackexchange.com/questions/3813429/find-the-volume-formula-of-a-simplex-proving-that-int-01-int-01-x-n-int

https://math.stackexchange.com/questions/1718021/intuition-for-volume-of-a-simplex-being-1-n

i didn't fully read nor understand it, but there it's done with integrals

[west orchid]

why did you put equation* environ in $...$

[lavish jewel]

they mean to ask why you used begin{eq} at all

i only realized right now xD

guess it makes sense if you wanted to get something that wasn't in-line, but there was no accompanying text anyway

[west orchid]

you can use align* for that

you're going to use gather like doggy?

[lavish jewel]

throw in some page breaks

[prime drum]

sooo, i may be in a worse position then i thought for this class lmao

[reef sleet]

Umm, can I ever use Cramer's Rule to test for linear independence?

Because if my system is homogenous then won't Cramer's rule just give me a determinant of zero for each solution?

[wintry steppe]

why is this wrong? I factored -2 out of the denominator and then cancelled the terms in the numerator and denominator

and the reason its negative is because you get this: $$sqrt((1/-2)^2) = 1/-2$$

its the modulus right?

[wintry steppe]

yes

[stoic python]

request new nickname

[wintry steppe]

|a/b| = |a|/|b|

[wintry steppe]

is this correct

can you just cancel the square and sqrt

or does it produce 1/2

its 1/2

why

oh, i see. thx!

[reef sleet]

How do I do this???

If I let any arbitrary vector in R^3 = <u1, u2, u3>, then I get

<u1, u2, u3> = a<1, 0, 0> + b<0, 1, 0> + c<0, 0, 1> + d<-1, 2, -3>

Therefore,

a-d = u1
b+2d = u2
c-3d = u3

Letting d = t, we get

a = u1 + t
b = u2 - 2t
c = u3 + 3t

What am I supposed to get from that?

I mean, I guess we can tell that coefficients exist for any real parameter t?

Or that there is some parameter t for every vector in R^3 such that that vector is a linear combination of aS1, bS2, cS3, and dS4?

[nocturne jewel]

I mean, it clearly spans R^3 since you have the canonical basis of i, j, k vectors

and it's clearly dependent since the 4th vector is in the span of the canonical basis

[reef sleet]

So it’s not a basis for R^3

[nocturne jewel]

nope

[reef sleet]

But it does span R^3

[nocturne jewel]

yes, the 1st 3 vectors span R^3

so adding another R3 vector doesnt do anything

[reef sleet]

Can you explain what the equations I got mean? I understand your explanation but I’d like to know if there’s any value in what I got

Or if it’s just nonsense

[reef sleet]

This sounds right to me but I’m not sure

[nocturne jewel]

I mean what you wrote was to see if the vectors span R^3, namely every R^3 vector can be written as a lin comb of the vectors

[reef sleet]

And they can, because there is a parameter you can find to get each coefficient?

Sorry if I’m not making sense

[nocturne jewel]

They can span R^3, cause a subset of that set is a basis

namely the 1st 3 vectors span R^3 (clearly)

[reef sleet]

Alright, I'll just use that then. Makes more sense anyways. Thank you Moshill!

[nocturne jewel]

For example, you shouldn't have to do much work for 32

given it's the exact same, the canonical basis of R3 is a subset

[reef sleet]

Okay 😁

[wintry steppe]

what am i doing wrong?

[stoic python]

request new nickname

[wintry steppe]

are you sure sqrt3 is an input recognized by the system

also ur off by signs

[wintry steppe]

yes it is a correct inpuit

[wintry steppe]

how so

,w 2e^(-2i pi/3)

[stoic python]

Results provided by WolframAlpha

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

Requested by TTerra#5291

[wintry steppe]

ohhh its in the third quadrant... i see lol

ty

[nocturne jewel]

yes, since -2pi/3 is between -pi and -pi/2, it's Q3

[wintry steppe]

for this one I rewrote it as $$e^{2+4i\pi}e^{6+i\pi/2} = e^{8+i\pi/2}$$ but then when I solve $arctan(\frac{b}{a}) = \frac{\pi}{2}$ it has no solution?

[stoic python]

request new nickname

[nocturne jewel]

that means it's a purely imaginary number

[wintry steppe]

oh yeah i got it

a = 0, b = e^8

[nocturne jewel]

$Arg(z)=\pm\frac{\pi}{2} \implies z = bi, b\in\mathbb{R} b\neq 0$

[wintry steppe]

oh yeah i see. thx!

[stoic python]

moshill1

[zealous junco]

im not sure if this is right way to solve

so x is my coordinate of (a,b,c)

and P1,..P3 are columns of P, and ~ is row equivalent

cuz i found solution online that is different answer

idk why online solution put the alphas in rows rather than columns

[dusky epoch]

doesn't matter

a matrix is invertible iff its transpose is

[zealous junco]

but the coordinate obtained is different

[zealous junco]

the coordinate matters though and idk if im wrong or solution wrong

[dusky epoch]

oh

then it looks like the solution is wrong

[zealous junco]

ok nice lol i thought i did it wrong, yea cuz i got confused why solution put it in rows

for this example if R has rows of 0 then does that mean b4 in the boxed part can be anything? and the answer will be correct no matter the choice of b4 right

wait nvm

yea do you just place a placeholder for b4 to be anything? cuz u need the matrix multiplication to work out so the 3rd element in row vector cant be gone

[mortal juniper]

Hi can i ask if a nxn matrix have more than n eigenvalues and how do i possibly proof it?

[dusky epoch]

???

it is impossible for an n by n matrix to have more than n eigenvalues.

[lavish jewel]

you can use the fundamental theorem of algebra and the characteristic polynomial to prove what ann said

[mortal juniper]

oh right

so its not possible for a 1 dimension eigenspace be digonalizable if the det(λI−A)=(λ−2)squared * (λ−1)?

[dusky epoch]

?

[mortal juniper]

becos im not sure whether dimension eigenspace has to do with eigenvalues

[dusky epoch]

...

det(λI - A) = (λ-2)^2 (λ-1) does not by itself prevent A from being diagonalizable.

[mortal juniper]

but if i constrain it such that the eigen space is only 1 dimension will it still be diagonalizable

thats what im confused

[dusky epoch]

if you know that det(λI - A) = (λ-2)^2 (λ-1) and that all eigenspaces of A have dimension 1 then A will not be diagonalizable.

[mortal juniper]

ok i c thanks!

[dusk sage]

hyperplanes are n-1 dimentional affine spaces for an n-dimentinoal vector space right?

then doesn't that imply that if we rk(A) = 1 for that highlighted part?

i don't understand why rk(A) is necessarily 1 there

[tulip glacier]

guys for eigenvectors, there would never be a standalone eigenvectors which is unique right? cos we are essentially solving for non trivial solutions of the homogenous eqn

[nocturne jewel]

eigenspaces have a basis, yes

[tulip glacier]

can i say that eigenvectors is always a subspace?

[nocturne jewel]

eigenspaces are subspaces

[tulip glacier]

but eigenvectors may not be?

[nocturne jewel]

eigenvectors form the basis

$\lambda-$eigenspace is $ker(A-\lambda I)$

[stoic python]

moshill1

[tulip glacier]

are eigenspaces always square matrix?

[nocturne jewel]

Not always

[tulip glacier]

this is what i saw on my teachers notes

[nocturne jewel]

yes that's equivalent to finding the kernel

[tulip glacier]

and it seems to suggest that eigenvectors is always a subspace?

kernel is a subspace i guess

[nocturne jewel]

E_o is called an eigenspace

the eigenvectors are any vector in the eigenspace associated with that eigenvalue

[tulip glacier]

but wouldnt a whole set of the vectors become a subspace on its own for eg. (0,1,1),(0,2,2)...

one smaller than the eigenspace

[tulip glacier]

is there a difference between kernel and subspace, my school hasnt mentioned this term b4

[nocturne jewel]

I mean hard to go one smaller than an infinite set

kernel means what gets mapped to 0, subspace is a subspace

$ker(A-\lambda I) = [A-\lambda I|0]$

[stoic python]

moshill1

[tulip glacier]

$ker(\lambda I-A) = [\lambda I-A|0]$ does it work the other way round?

[stoic python]

amphibian sergeant

[nocturne jewel]

yes

[tulip glacier]

so is there a relation between kernel and subspace or are they completely different concepts?

[nocturne jewel]

completely different

[tulip glacier]

ok thks i guess its wrong for me to see eigenvector as a subspace

[tulip glacier]

my teacher told me that anything with span(anything) is a subspace lol

[nocturne jewel]

yes span(something) is a subspace, but span(eigenvector) is the eigenspace not the eigenvector

[tulip glacier]

does that mean that there are 3 eigenspace in this case?

or are they just grouped as one

[nocturne jewel]

there are 3

the 0-eigenspace, -1-eigenspace, 2-eigenspace

[tulip glacier]

oh ok i see so the eigenvalue is like a name for the eigenspace

[mortal juniper]

hi can i ask if is possible for all symmetric matrices to form a basis?

[wintry steppe]

all symmetric matrices
no, this set is linearly dependent

after all, it contains the zero matrix

[wary lily]

we can show that f(x) + f(y) = mx + b + my + b = m(x+y)+2b = f(x+y) + 2b which contradicts the linear property of f(x+y) = f(x) + f(y)

The answer key of my book tho, gives another explanation that I don't understand.

can someone explain this and if possible make a connection with my solution so that I understand it?

[wintry steppe]

linear functions take 0 to 0.

if T is linear, T(0) = T(0 + 0) = T(0) + T(0). subtract T(0) from both sides to get T(0) = 0

[wary lily]

O that makes sense

linear functions don't translate by adding or subtracting, only through multiplication

the zero vector must always give zero for a linear function

[dire thunder]

for a linear transformation*

[wintry steppe]

linear function, linear transformation

all the same thing

😌

[dire thunder]

privyet tterra

kak dela

[wintry steppe]

tired

[dire thunder]

why

[wary lily]

I'm using wrong terms maybe

[wintry steppe]

attending first analyssi lecture in like a month

it is boring

[dire thunder]

are you missing classes, kid?

[wintry steppe]

yes

[dire thunder]

bad kid

go back to your room

[wintry steppe]

he just reads right out of the book

it's so boring

[wary lily]

no good prof

[dire thunder]

pugh

[wary lily]

if just reads from book

I could do for free

[dire thunder]

ok

@wary lily read category theory for me for free

[wary lily]

O, just learned the correct terms

linear transformation vs. affine transformation

b = zero, b \neq zero

[wintry steppe]

https://media.discordapp.net/attachments/268882317391429632/785727935050809354/771287728305471528.gif

[dire thunder]

just speak about functions you know

[wintry steppe]

i am falling asl;eep

this is a section of the book i could read in 10 minutes

[dire thunder]

especially upper arrows?

and proofs by pictures?

[wintry steppe]

oh my god

don't even get me started on the "obvious from figure abc" proofs

[dire thunder]

well but this is obvious from figure abc

[wintry steppe]

,ban vimes

[stoic python]

This may only be done by a moderator!

Mar 29, 2021, 2:43 PM

[dusky epoch]

[wintry steppe]

"Proof: obvious from figure 145"
figure 145: contains the proof in the caption

[dire thunder]

@wintry steppe look at this amazing proof of bolzano-weierstrass

isn't this obvious?

[wintry steppe]

you forgot the entire proof in the caption

[dire thunder]

sorry this is left as an exercise to the reader

[wintry steppe]

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

[dire thunder]

well many profs tho do proofs by "obvious"

well but i used such books so mainly it was indeed obvious

[wintry steppe]

"the proof is obvious if you read the proof"

[dire thunder]

"the proof is obvious if you write the proof"

[faint lintel]

can someone look over a proof for me?

[native rampart]

Go ahead

[faint lintel]

That's my proof that any such transformation T is diagonalizable

[dire thunder]

T(x,y,z)=(z,y,x)

[faint lintel]

T^2 =/= T right?

[dire thunder]

wot

[faint lintel]

cause T^2 = identity

[dire thunder]

i gave an example where T^2=T and T is not an identity

nvm

i am dumb

[faint lintel]

F

I have an example that works I don't need help with that

I just wanted to prove that such a T is always diagonalizable

I used this fact which I am allowed to assume

(I mean the proof of said fact is in the textbook which is why I'm allowed to assume it)

[dire thunder]

i mean ok

[faint lintel]

this is the only part that worries me that might be shaky

[native rampart]

I don't think that's necessary

[faint lintel]

Isn't it tho?

[native rampart]

Nvm,it is

[faint lintel]

cause then I get v = u-w rather than u + w

[native rampart]

Yea, That's clearly true

[faint lintel]

and I guess I could work with that but direct sums are better worked with as sums :P

[native rampart]

Since you are multiplying (T-I)v=0 with -1 on both sides

[faint lintel]

yea

so proof checks out?

[native rampart]

Write u is in E_1 and w is in E_0

Other that that, It's correct

[faint lintel]

dope

[native rampart]

In general,intersection of distinct eigenspaces is always zero subspace

[north steeple]

if im given two vectors

2x1 + 2x2 + x3 - x4 = 0

how do i setup the matrix to find the basis

i forget it its straight across or vertical

[faint lintel]

same variables are aligned vertically

@north steeple

[north steeple]

ok so ```1 1 1 1
2 2 1 -1

thanks

[zealous junco]

can someone help w/ my question in the #multivariable-calculus its also related to linalg

[sharp idol]

Correct me if I am wrong, is square matrix multiplication commutative when one of the matrices is a diagonal matrix?

[dire thunder]

nah it fails

[sharp idol]

Example?

[stoic python]

Commander Vimes

Commander Vimes

[lavish jewel]

multiplying by a diag matrix from the right scales the columns. from the left, it scales the rows

[sharp idol]

Despite the seizure you seem to be having I see the issue

[wary lily]

only matrices that always commute are multiples of identity matrix, AFAIK.

[zealous junco]

yea quick way to see the scale column/row is to remember 2 ways of matrix mult

AB, then first way is linear combo of rows in B by reading off coefficients of rows in A

[dire thunder]

just read linear mappings no need in matrices yknow

[zealous junco]

and then theres lin combo of columns in A by coeff of columns in B

[sharp idol]

Hmm, this definitely presents some problems with what I am doing :/

[zealous junco]

hm thats nice way to see it actually

[sharp idol]

Actually, there aren't any issues here. I can make this work. Thanks!

[crude falcon]

I have this operation: m^2 + n^2 where m and n belong in Z and I want to check if this operation have an identity element

[sharp idol]

Is this a linear algebra question 🤔

[crude falcon]

am I looking for this? an element such that: g^2 + e^2 = g

this is from my lineal algebra course, sorry if this is not the place

[sharp idol]

Also, if m and n are the identities, then 0 is indeed the additive identity here

Ohh, I see the dilemma. Yea, not sure how well this fares for you

[dire thunder]

0

e = 0

[sharp idol]

@crude falcon Since you are in the integers, this is the same as finding the roots of
$g^2-g+1=0$

[stoic python]

dackid

[dire thunder]

wait

[crude falcon]

but g^2 != g?

[dire thunder]

this cannot have idenity

[crude falcon]

yeah, thats what Im guessing

[sharp idol]

And the roots of this equation is not in the set of integers

[dire thunder]

another way is

assume e exists

then f(e,e)=2e^2=e

[sharp idol]

In any case, you will have exactly two elements that satisfy this equation

[dire thunder]

this only possible in integers only for 0

[sharp idol]

But it will not be true for all the integers

[dire thunder]

thus e should be 0

but m^2 != m in general

[sharp idol]

Yea. Here is my way to go about it: assume the additive identity exists in Z. $Let k\in\Z$ be the additive identity.
Then $g^2-g+k=0$. This equation has exactly two solutions and is therefore not true for all the integers, regardless of the value of k.

[stoic python]

dackid

[crude falcon]

i understand this reasoning

[crude falcon]

but I'm not following you there

why do you use g^2-g+k=0? since the operation is just m^2+n^2?

[sharp idol]

In that case the quadratic would be $g^2-g+k^2=0$. But k is constant so the argument is exactly the same

[stoic python]

dackid

[crude falcon]

ooh I see now

[sharp idol]

You are trying to show an identity element e.
So $g^2+e^2=g$. I am saying if we assume e exists, let e=k for some integer Z.

[stoic python]

dackid

[crude falcon]

alrigh thank you!

btw isnt this really lineal algebra? its about algebraic structures

[sharp idol]

This is definitely abstract algebra

[vast iron]

To find inverse of a matrix, in one book they used the method where they would write square matrix and attach an identity matrix to it, and then use row reduction to turn the original matrix into identity, then the identity matrix would become the inverse of the original. Is this generally true or does it work for 3x3 / some special matrices?

[dire thunder]

it is generally true if i understand you correctly

you mean they take A
then make an augmented matrix [A I] and do rref on it?

if so, then yes this method is applicable

[vast iron]

Yes exactly

Thanks

[sharp idol]

It just gets extremely tedious very quickly

[vast iron]

It's better than adjoint/determinant tbh

[sharp idol]

Adjoint is messy, but pretty straightforward. But there are certainly better options yes

[vast iron]

Like matlab?

[sharp idol]

I like the product of elementary matrices approach. Probably not the most efficient of all the options, but I do like the option

[dire thunder]

[vast iron]

I'll Google that, thanks wholesome guy that draws but who's name I'll never be able to recall

[nocturne oracle]

pablo picasso i think

[vast iron]

Thanks, I can't remember names ever.

[modern otter]

Guys I am having trouble

I know I have to find out the length of the base of the triangle

How do I do that

Basically how do I find the length of the bottom side of the triangle given the height and slope if it

[wintry steppe]

this ain't linear algebra

[wary lily]

[nocturne jewel]

There are lines on the picture, close enough \s

[crude falcon]

I have this set A = {-1,1} with the multiplication operation, and I want to check if the set is asociative

How can I do it? I would need 3 elements right?

or can i just do (-1 x 1) x -1 = -1 x (1 x -1)

[wintry steppe]

operation is associative

u just check all possible multiplications lol

[wary lily]

are you trying to prove matrix multiplication associativity?

Like A(BC) = (AB)C

[crude falcon]

no, I'm checking if that set is an algebraic group or not

and for that I want to see if its asociative

[wary lily]

OK, I'm not that far

[crude falcon]

np

[crude falcon]

rly? isnt any analytical/abstract way to do it?

im lazy

like the way I put it shows its associative for that given example

[stable kindle]

if you're really lazy you could say something something inherits associativity from integers? idk

[crude falcon]

mm I dont think thats allowed for this exercise

[stable kindle]

then brute force it

associativity is always boring

can't think of much

set theory? elementary algebra?

[crude falcon]

matrices/systems of ecuations

[wintry steppe]

those are typically things you learn in linear algebra, no?

[limber sierra]

systems of ecuadors

[forest knoll]

Ik my school was supposed to teach matrices in Alg II but they cut it out cause of remote learning

[limber sierra]

honestly the main thing you need is comfort doing algebra

[stable kindle]

also basic set theory i think, idk

[limber sierra]

i'd be surprised if that wasnt taught during the LA class.

[calm yoke]

Are the identity matrix and a number conmutative?

Like I * 5 = 5 * I

Its an i, not a 1

[nocturne jewel]

yes those are both 5I

[calm yoke]

So a matrix with just a diagonal of 5s and the rest is 0?

[nocturne jewel]

yes

[calm yoke]

Ok, thanks

[wintry steppe]

Anyone could help me explain the Null space and column space basis?

I still dont get the Null space

[desert rapids]

The null space is the set of all vectors $\vec{v}$ such that $\textbf{A}\vec{v} = \vec{0}$

[stoic python]

LifeSource

[wintry steppe]

That I know

how do I find them for a not so simple rref matrix such as this?

[desert rapids]

Quick question: Are you Polish?

[wintry steppe]

No

[desert rapids]

Oh

[desert rapids]

That is Polish

at the bottom

[wintry steppe]

https://matrixcalc.org/slu.html

That is what I am using

[desert rapids]

Oh

[wintry steppe]

idk why is Polish there

[desert rapids]

XD

[wintry steppe]

Anywho, what the row 0 0 0 1 means?

no solution?

[desert rapids]

Yes

[wintry steppe]

Does that translates to the null space?

That's whats not clear to me

[desert rapids]

Well, going back to your example, you want to first get rref

The augmented matrix makes things confusing

[desert rapids]

The augment is unessisary

[wintry steppe]

and?

[desert rapids]

Then you get it of the form $\begin{pmatrix} 1 & 3 & -9 & -1 \ 0 & 1 & -3 & \frac{-3}{5} \ 0 & 0 & 0 & 1 \ 0 & -10 & 30 & 4 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \ v_3 \ v_4 \end{pmatrix} = \vec{0}$

[stoic python]

LifeSource

[wintry steppe]

I know that, what's next?

[desert rapids]

You write it in equation format as you did

[wintry steppe]

next

[desert rapids]

Well, the one that was done by the calculator is not quite right

It accounts for the augment

the augment here should just be all 0s

[wintry steppe]

that's irrelevant, I know how to write the general solution

what I need to know is how to write the null space basis

Would you have time for a voice chat?

[desert rapids]

You get $\begin{cases} v_1 + 3v_2 - 9v_3 - v_4 = 0 \ v_2 - 3v_3 - \frac{3}{5} v_4 = 0 \ v_4 = 0 \ -10v_2 + 30 v_3 + 4v_4 = 0 \end{cases}$

I can't rn

[wintry steppe]

oh, alright

Well thanks you for bearing with me

[stoic python]

LifeSource

[wintry steppe]

Yeah, that's what I wrote down

[desert rapids]

Good

Then, you know how to write the general solution? Right?

Get the pivot variables in terms of the non-pivot variables

[wintry steppe]

Yes, and what do I do with the x4 = 0

is that part of the Null space?

Yeah, this wail require more Khan Academy

[desert rapids]

What you do is you want to set each individual pivot variable equal to 1, and all others equal to 0

Then you solve for you non-pivot variables

[wintry steppe]

is x4 = 0 a non-pivot?

[desert rapids]

Sorry, I mean the other way around

[wintry steppe]

Yeah, not a single youtube video address that instance

[desert rapids]

It might help to watch a video

[wintry steppe]

wft do I do with a row that has a single 1

[desert rapids]

It is a very tedious process

It is fixed

Nothing

Obviously

[wintry steppe]

Is that part of the null space, yes or no?

[desert rapids]

Yes! It is always the fourth component of the vector!

[wintry steppe]

Thank you

so I basically have 3 rows to work with

[desert rapids]

Yes

[old dirge]

When it says position (2,1) is it (column,row)?

or (row,column)

[wintry steppe]

row column

always

[old dirge]

thanks

[old dirge]

According to the picture I posted above

This is what i'm writing and its create a matrix [1,0,0;0,1,0;0,0,1]

But i'm getting it incorrect

[wary lily]

what language is this?

[old dirge]

matlab

[stoic python]

az

[wary lily]

is that correct?

[dire thunder]

yes

[wary lily]

thanks, vimes

[dire thunder]

np, az

[pallid rampart]

wrong

you forgot the condition that it needs to be linear

[glossy parcel]

maybe a stupid question, but ive got [cost sint], and i reparametrize it with t = 2 sin^-1 (s), that limits t to -pi/2 to pi/2. how do i get past this domain restriction?

[chrome current]

What would make it possible to express a vector u as a linear combinations of v1 and v2 in more than one way?

[lavish jewel]

v1 being a scaled version of v2

[pseudo thicket]

Ax=0
When A is LI, x is 0.
Why is x=0?

Prof asked this in class today, but I can’t figure out why

All I think can think of it’s because columns of A are LI, the matrix is full rank

[wary lily]

another way to see it that when vectors are linearly independent, their corresponding matrix of coeffs has no free variables.

Or you can see it this way: for a non trivial solution to the equation Ax = 0 to exist, there must be scalars c_1, c_2,... c_{n-1}, not all zero, that multiplied with n-1 columns of A, will produce the nth column (we assume that the nth is the dependent column)

This will allow us to have a sum of all columns equal zero, where the scalars are not all zeros.

But if we can't produce such scalars, not all zeros, we end up with the trivial solution.

[pseudo thicket]

My classmate said similar lines but apparently that’s not what he wants

He’s asking like “why”

And I have got no more ideas

[wary lily]

is he looking for a proof based on the linearity of A?

[dire thunder]

columns are images of basis vectors of domain

and Ax is just sum(x_ic_i)

where x_i is i-th coordinate of x and c_i is ith column

thus linear independence of columns forces x = 0

@pseudo thicket