linear-algebra

[wintry steppe]

it's true but you should give a detailed argument as to why

[delicate zealot]

Dang

Big rip

[wintry steppe]

unless that's a fact you were given beforehand

[delicate zealot]

But hold up, and I'm gonna keep messing with this, but how can there be a set of linearly independent vectors that don't span their respective dimensions?

[wintry steppe]

what does "respective dimensions" mean

[delicate zealot]

Like how do two n dimensional vectors that are linearly independent not span all of n dimensional space

^^^^^

Sorry I said what I said before super poorly

[wintry steppe]

take (0,0,1) and (0,1,0) in R^3. they're linearly independent but they don't span R^3 (i e. here n = 3)

if n = 2, then two linearly independent vectors will span the entire space

[delicate zealot]

OH WAIT

[wintry steppe]

and if n = 1, you cannot have two linearly independent vectors

[delicate zealot]

Because I know that they are in a square matrix then the number of vectors will always equal the number of dimensions

[wintry steppe]

uh if it's telling you that you have n distinct vectors in R^n...

[delicate zealot]

So that means that each vector will necessarily have to span all of R^n

[wintry steppe]

unless im horribly misunderstanding what you're being asked to prove

[delicate zealot]

Here I'll type the problem again

[wintry steppe]

a single vector only spans the line it's on, be more careful with your wording

[delicate zealot]

If A is an n x n matrix, why are "The columns of A are linearly independent" and "The columns of A span R^n" equivalent"?

[wintry steppe]

phrased like that, this is a lot easier

rank nullity perhaps?

[delicate zealot]

We haven't gotten to that term in class yet

[wintry steppe]

rank A + nullity A = n

[delicate zealot]

Right but we haven't gotten to what those mean yet

[wintry steppe]

have you seen anything like that at least?

rank is dimension of the space spanned by the columns, nullity is the dimension of the null space

[delicate zealot]

We haven't talked about null space yet

Definition of rank I would understand tho

[wintry steppe]

hmm well i guess rank nullity isn't helpful here cause it just boils down to the same issue as before with needing the replacement theorem

and if you haven't seen that then i don't know what to say :(

[delicate zealot]

Oh well

Thanks for trying :)

[wintry steppe]

maybe you can use a thing or two you learned about row reduction? i recall you asking about that some time ago

[delicate zealot]

Maybe that could help I'll look into that

Would the identity matrix be of help?

[wintry steppe]

it'd probably show up

well, i have a strong feeling it will

anyways i feel like that's the solution they want you to go for, so ill let you work on it now

[delicate zealot]

So if A reduces to the identity matrix then the columns would be linearly independent and therefore have to span R^n

OK sorry Ill shut up my b lol

[wintry steppe]

soooomething along those lines ;)

[delicate zealot]

Youre help is appreciated lol

[wintry steppe]

see now that i remembered you were asking about row reduction stuff i know how to point you in the right direction

||i haven't performed a serious row reduction since i took linear algebra so i always forget you can do cool things with it||

[delicate zealot]

I love how my professor says "you don't have to prove each one separately" but when going to submit the points are split up so you have to prove each one separately lol

I'm still fine just kinda funny

[clear sparrow]

Are power series members of the polynomial function vector space?

[brittle juniper]

nope

not all of them

[clear sparrow]

Why so / why not?

I bring this up because of

Not sure what I'm missing

[brittle juniper]

for example, would you think that the exponential function
$$\exp:x\mapsto\sum_{k=0}^{+\infty}\frac 1{k!}x^k$$
is a polynomial function?

[stoic python]

Tuong:

[clear sparrow]

It is not
But I don't see how, in its series form, it doesn't fit in the space

It is a linear combination of a set of scalars and the basis {f(x) = x^n: n = 0, 1, 2, ...}

[brittle juniper]

exp cannot be written as a (finite) linear combination of elements of that basis

[clear sparrow]

Right
So the book raises the point "infinite basis =/= infinite linear combinations"
But why not? What issues arise when one allows infinite linear combinations?

Assuming convergence

[brittle juniper]

infinite sums belong to topology and analysis, and linear algebra is algebra

there are definitions

[clear sparrow]

Alright that's fair
Thanks!

[hollow finch]

Would the power series of exp(x) exist in the infinite dimensional polynomial space P_inf?

[limber sierra]

we would generally consider that a space of analytic functions

not really a "polynomial space"

because again the functions involved arent polynomials

they're analytic functions (i.e. the limit of a series of polynomial terms)

and yes, exp(x) is analytic so it would be a member of the space of real analytic functions

[opal osprey]

hello cool people!

I'm doing a linear transformation from R3 to M_2x2

when it comes to finding out the matrix form of the transformation, I am stuck

should I expect to apply the general method (ie, images of the canonical basis as columns of the transformation matrix) and find a 2x6 matrix in this case?

oh I see!!!

[gray dust]

the size of the matrix of the map, wrt whatever bases, is determined by the dimensions of the domain & codomain

[opal osprey]

I'm missing the writing of the coordinates of the basis!

[gray dust]

sure that's part of it. looking at the dimensions, the matrix you seek is 4 by 3

[opal osprey]

alright

I see how the dimensions play a role too!

thank you buddy! 🙂

[gray dust]

no prob

[jaunty compass]

how would you find the determinant of a 5 x 5 matrix

[opal osprey]

you can do row operations

and predict how that will affect the determinant

I'd say the most versatile method is laplace

[jaunty compass]

if I do row operations i just multiply all the pivot columns then right

[latent ledge]

To show $\dim(U_1+U_2+\cdots+U_m)\leq\dim(U_1)+\dim(U_m)$ where $U_i$ are subspaces of $V$

[stoic python]

Konoha:

[latent ledge]

I will use induction, $dim(U_1+U_2)\leq\dim(U_1)+\dim(U_2)$, then if $dim(U_1+U_2+\cdots+U_m)\leq\dim(U_1)+\dim(U_2)+\cdots+\dim(U_m)$, we should have $$\dim(U_1+U_2+\cdots+U_m+U_{m+1})=\dim(U_1+U_2+\cdots+U_m)+dim(U_{m+1})-\dim(U_{m+1}\cap(U_1+U_2+\cdots+U_m))\leq\dim(U_1)+\dim(U_2)+\cdots+\dim(U_m)+\dim(U_{m+1})$$

[stoic python]

Konoha:

[wintry steppe]

anyone have fun/interesting/difficult linear algebra exams? preferably with an answer key. please ping or DM if so!

[wintry steppe]

if you're at a uni/college/institution-of-sorts then they may let you see past exams

mine does, thats where i go for practice

yeah i’m not lolol

[limber sierra]

what sorts of material was covered in your course

was it proof-based? computation-based? did it introduce vector spaces over arbitrary fields, spaces of polynomial functions, quotient spaces, modules? did it discuss isomorphism theorems or the spectral theorem? what normal forms, if any, were covered? etc

theres a lot of variance between different LA courses

[latent ledge]

Let $T$ be a linear operator on a finite dimensional real inner product space, and let $m(x)\in\mathbb{R}[x]$ be the minimal polynomial of $T$. Prove that $T$ has a matrix representation that is upper-triangular if and only if $m(x)$ splits into linear factors in $\mathbb{R}[x]$.

[stoic python]

Konoha:

[latent ledge]

stuck at the reverse direction

[wintry steppe]

mine covered like intro to linear algebra, abstract vector spaces, isomorphisms, spectral theorem, singular value decomposition, but no modules

[latent ledge]

I am working on the berkerly math problems book

it has solution

[cursive narwhal]

@wintry steppe Why not try a problem book instead? The Linear Algebra Problem book by Ikramov is nice (but it's a bit old and may not contain enough problems to satiate you)

There's also another one by Fuzhen Zhang that's quite nice so it's worth a look at

[wintry steppe]

yeah i found a pdf thanks @cursive narwhal !

[cursive narwhal]

You're welcome

[wintry steppe]

@latent ledge if T has an upper triangular basis representation then its characteristic (and hence minimal) polynomial split into linear factors.

for the converse, maybe you can use the fact that any linear operator with a splitting characteristic polynomial has an upper triangular form?

(if you haven't seen that before, it's an induction proof iirc)

[latent ledge]

@wintry steppe thanks, I will look this, linear operator with a splitting characteristic polynomial has an upper triangular form, up

[wintry steppe]

i recommend proving it yourself, it's a good exercise

the induction, if done right, also gives you a way to compute that upper triangular form, if you're interested

[latent ledge]

I will try that later

[wintry steppe]

@dusky epoch i have the most trivial question ever

in the history of questions

[dusky epoch]

okay what is it

[wintry steppe]

if you want to find the matrix which projects vectors onto (a;b), then all you need to do is solve the system (a,b;c,d)(-b;a) = (0;0) and (a,b;c,d)(a;b) = (a;b)

right

[dusky epoch]

no

why are you reusing your variables

[wintry steppe]

ugh

sorry

tired

let me rewrite it

$\begin{bmatrix} a&b\c&d \end{bmatrix}\begin{bmatrix} -v_2 \ v_1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$

[stoic python]

polynomial:

[dusky epoch]

i mean sure if you wanna look at it this way i guess that's fine like
yeah you're specifying a linear map on R^2 by what it does to two LI vectors

[wintry steppe]

where we want to project onto vector v

but you sound like there's a better way

or something

[dusky epoch]

there is, but it'll sound needlessly overcomplicated if you limit yourself to R^2

[wintry steppe]

R^3?

wait

does your way involve eigen anythings

[dusky epoch]

in a way i guess?

[wintry steppe]

i haven't learned that yet

[dusky epoch]

i would construct the matrix via its diagonalization basically

[wintry steppe]

would anyone be able to explain to me in why this is true?

[dusky epoch]

which part

[wintry steppe]

the first part

Ax = b is consistent if and only if b is in col(A)

[dusky epoch]

do you know what it means for a system to be consistent

[wintry steppe]

yes

[dusky epoch]

what does it mean for a system to be consistent then

[wintry steppe]

unique or many solutions

[dusky epoch]

it means it has a solution

ie there exists an x such that Ax = b

now, do you know what col(A) is

[wintry steppe]

column vector of A?

[dusky epoch]

no.

col(A) is the column space of A.

[wintry steppe]

so what does it mean when they say b is in col(A)?

isnt b already in the augmented matrix of Ax = b?

[dusky epoch]

[A|b] is not the same matrix as A

when they say b is in col(A) they mean b is in col(A). it means b is in the span of the columns of A

since that's what col(A) is

[wintry steppe]

is there an example?

[dusky epoch]

of what

[wintry steppe]

i feel like im getting what you are saying but i feel like i need a visual of why this is true

[dusky epoch]

why what is true

the iff statement that started this?

[wintry steppe]

yes

[dusky epoch]

ok now we can get to that

so

[wintry steppe]

the span is the space generated by the basis vectors being the column vectors @wintry steppe

[dusky epoch]

col(A) by defn consists of all linear combinations of the cols of A

but every such combination can be written as the product of A with a vector holding the coefficients of the combination

[wintry steppe]

that i dont quite comprehend

[dusky epoch]

okay so

are you ok with matlab syntax for matrices or would you rather i write them out in latex

[wintry steppe]

latex is fine

[dusky epoch]

ok

[wintry steppe]

so what im understanding is that the col(A) spans R^n

[dusky epoch]

so let $A = \bmqty{3 & 1 & 4 \ 1 & -5 & 9 \ -2 & 6 & -5}$ and $b = \bmqty{1 \ -18 \ -9}$

@wintry steppe no

no, that need not be the case

[wintry steppe]

the linear combination* of the col(A)

[dusky epoch]

also col(A) is a subspace and so taking its span will just give you col(A) again

col(A) doesn't mean "the columns of A"

[wintry steppe]

ill get that off my head

[stoic python]

Ann:

[dusky epoch]

notice that $b = \bmqty{3\1\-2} + 2\bmqty{1\-5\6} - \bmqty{4\9\-5}$

[stoic python]

Ann:

[dusky epoch]

at least that's what i intend it to be in case i screwed up the arithmetic

which uhh

i think i did in the third component oops

b = [1; -18; 15] sorry

[wintry steppe]

i get the point, all good

[dusky epoch]

ok well

b is in col(A) clearly

since i expressed it as a linear combo of the cols of A

that make sense?

[wintry steppe]

so column space is the span of the linear combinations of column vectors of A?

[dusky epoch]

column space is the span of the columns of A

[wintry steppe]

okok

[dusky epoch]

or the set of all linear combinations of the columns of A

[wintry steppe]

👍

[dusky epoch]

anyway ok so like

yeah b is in col A

[wintry steppe]

i understand now

[dusky epoch]

but on the other hand

i could also write it as A * [1; 2; -1]

[wintry steppe]

b is in the linear combinations of columns of A

so Ax = b is consistent if b is a linear combination of the columns of A

does that sound right?

[dusky epoch]

close enough

[wintry steppe]

thank you!

very much appreciated @dusky epoch

[hazy gull]

for part (a) would it be the absolute value of each term of <f,g> using the difference between the two vectors?

like |-3^2|+ |1^2|+|4^2|

[wintry steppe]

@hazy gull what are S and E

[hazy gull]

s denotes the sum norm

[wintry steppe]

E is euclidian?

[hazy gull]

E denotes the euclidian norm

[wintry steppe]

i see

so then yes

but the difference in sums

so $|u-v|{S} = |u{x}-v_{x}| + |u_{y}-v_{y}| +..$

[stoic python]

robo™:

[wintry steppe]

or another way you can do it is

[hazy gull]

so you dont use the inner product here?

[wintry steppe]

$||u-v||_{S} = <u-v, u-v>$

[stoic python]

robo™:

[wintry steppe]

by definition of the inner product

then use the definition of the inner product space that they gave you

(im pretty sure the two methods should match up)

[hazy gull]

so it would be |f(0)g(0)|+|f(1)g(1)|+|f(2)g(2)|

[wintry steppe]

$<u-v, u-v> = ((u-v)(0))^2 + ((u-v)(1))^2 + ..$

[stoic python]

robo™:

[wintry steppe]

= $(u(0)-v(0))^2 + (u(1)-v(1))^2 + ..$

[stoic python]

robo™:

[wintry steppe]

makes sense?

[hazy gull]

ya

itd be 26

[wintry steppe]

yessir

[hazy gull]

noice

feels good having a simple problem every once and awhile

[wintry steppe]

actually i think it should be $\sqrt{26}$

[stoic python]

robo™:

[wintry steppe]

but yea other than that its easy

i think in general $||u|| = \sqrt{<u,u>}$

[stoic python]

robo™:

[dusky epoch]

okay can you please not use <> for inner product

[wintry steppe]

right?

why

[dusky epoch]

$\left< u, u \right>$

[stoic python]

Ann:

[wintry steppe]

ohh

thx

[hazy gull]

are you sure it's not just |-3|+|1|+|4|?

[hazy gull]

anyone else care to weigh in?

[dusky epoch]

can you post the problem and your question again

[mild loom]

if you span 2 2x1 vectors that arent parallel, is a plane spanned?

[quartz compass]

yep

[mild loom]

holy shit i actually get something in this god forsaken course

[wintry steppe]

Is this a clockwise or counterclockwise rotation ?

I used this formula, is there one for clockwise rotation or do I have to do a reflection after ?

[dire thunder]

rotation conventionally assumed to be counterclockwise. but your book may define it differently

is it strang?

[wintry steppe]

David C. Lay

Linear Algebra and it's applications 5th edition

[dire thunder]

so

you want to do clockwise rotation or what?

[wintry steppe]

Yeah I checked my prof's notes and he defines that matrix as "rotation of .. angle about the origin"

So I suppose it's counterclockwise

[dire thunder]

ye

but anyway, if you want clockwise just plug in negative angle

[wintry steppe]

Looks good ?

And ok gotcha, I was wondering that

[hazy gull]

@dusky epoch

part a

the sum norm is the sum of its constituents so eg |u_1|+|u_2|+|u_3|

[dusky epoch]

uh

hol up

[hazy gull]

but do i need to use the inner product

[dusky epoch]

hold up.

what are the defns of $\nrm{\cdot}_S$ and $\nrm{\cdot}_E$?

[stoic python]

Ann:

[hazy gull]

s would be the sum norm E would be euclidian norm

[dusky epoch]

no

give me the actual definitions

oh

wait

hold on

u and v are in R^3

my bad

i mean, parts a and b do not involve any inner products ¯_(ツ)_/¯

[hazy gull]

well thats what i had assumed, but in a different problem I had the euclidian norm use an inner product

[hazy gull]

I think you only need to use the inner product for the euclidian norm because you're multiplying it

so its $ \sqrt{\left<u,u\right>} $

[stoic python]

gramcracker:

[hazy gull]

but since the sum norm only involves summation , I dont need to.

[stoic python]

datguy:

[jaunty compass]

What would be a good place to start in calculating the determinant of this matrix?

[hazy gull]

2nd row

you only need the second column

really easy

have you learned about laplace expansion yet

if not, look it up because thats the idea here

[jaunty compass]

ok gotcha

[opal osprey]

and then you can do some row operations afterI think

lots of 8's there

[wintry steppe]

Lots of 0’s and 8’s

[wintry steppe]

Row 1 and 3 are a linear combination away from eachother

[clear sparrow]

Why are the statement and the proof of this theorem structured like they are?
That is, why is it necessary to introduce subspace W, or to mention subset S?

I was thinking more along the lines of:
Theorem: given some finite-dimensional space V, any independent subset S0 of V is a subset of a basis of V
Proof: if S0 is a basis, then we're done
Else, find b1 in V not spanned by S0, then let S1 = S0 union {b1} be another independent subset
Rinse and repeat up to a maximum of the size of V, and then a basis is guaranteed (but in more formal language)

Since this is true for any space, it must hold for any subspace
What does this shorter version lack compared to the one the book provides?

[pallid rampart]

Since this is true for any space, it must hold for any subspace
I think the big take away from this theorem is that subspaces of finite dimensional vector spaces are finite dimensional

That isn't a given fact

[clear sparrow]

I see - that explains the purpose of W
What about the set S though?

[pallid rampart]

It's a name for a general linearly independent subset of V

[clear sparrow]

What is its purpose in the theorem?
Its existence isn't used to justify any fact, as far as I can see

[pallid rampart]

About that question, I'm not sure either. I would just remember that every linearly independent subset of a finite dimensional vector space can be extended to a basis, and linear subspaces of a finite dimensional vector space is finite dimensional

It also seems a bit redundant to me too

[clear sparrow]

Alright, thanks!

[stiff frost]

@clear sparrow I disagree with Whoever and think the bit about S serves a purpose, though the proof isn't written well

the parenthetical "in not more than dim V steps" part requires justification along the lines of "If S is a linearly independent subset of W...then S contains no more than dim V elements"

[clear sparrow]

🤔 that makes sense
I shouldn't have been taking the <= dim V for granted

[stiff frost]

I mean, another author would not have bothered to write down justification for it. And this one put the justification far away. So it's completely understandable.

[wintry steppe]

can someone tell me what i did wrong

Let \begin{align*}
&\mathbf{A} \text{ rotates vectors by $\frac{\pi}{4}$ counterclockwise},\
&\mathbf{B} \text{ is the matrix } \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}.
\end{align*}

[stoic python]

polynomial:

[wintry steppe]

i need to find A^102 B^102 - B^102 A^102

by induction, i found that B^102 = (1,n;0,1)

and A^102 = the equivalent of rotating by 3pi/2

so it should be (102,0;0,-102)

but that's not it

why?

@dusky epoch ?

[wintry steppe]

<@&286206848099549185>

[dire thunder]

@wintry steppe for the beginning write A explicitly

[wintry steppe]

ok

[quartz compass]

looks right to me

[wintry steppe]

$\begin{bmatrix} \sqrt{2}/2 & -\sqrt{2}/2 \ \sqrt{2}/2 & \sqrt{2}/2\end{bmatrix}^{102}$

[stoic python]

polynomial:

[wintry steppe]

@quartz compass how did you check it?

[quartz compass]

102 mod 8 is 6

[wintry steppe]

yup

that's what i did

[quartz compass]

6*pi/4 = 3pi/2 which is 90 degrees clockwise

[wintry steppe]

yup

that's what i did as well

[quartz compass]

just multiplied with same closed form for B^n as well, since that's correct

[wintry steppe]

but look here

,w {{cos pi/4, -sin pi/4}, {sin pi/4, cos pi/4}}^102 . {{1,1},{0,1}}^102 - {{1,1},{0,1}}^102 . {{cos pi/4, -sin pi/4}, {sin pi/4, cos pi/4}}^102

[stoic python]

Results provided by WolframAlpha

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

Requested by polynomial#4757

[wintry steppe]

so wtf is wrong?

[quartz compass]

cos(pi)/4

is not cos(pi/4)

[wintry steppe]

oh my fucking god

[dire thunder]

you divided cos

[wintry steppe]

HAHAHAHA

[quartz compass]

lmfao

[wintry steppe]

NO

[dire thunder]

lmao

[wintry steppe]

FUCKMING WAY

[dire thunder]

happens

[wintry steppe]

IVE BEEN STCUK FOR AN HOUR

BECAUSE OF THIS SHIT

THINKING I WAS WRONG

i went through my work

5 times

jesus chrsit

[dire thunder]

anyway, i guess this matrix is idempotent with A^4=A

and 102 is the same as A^2

[quartz compass]

lol bad way @dire thunder

[dire thunder]

why

[quartz compass]

I just explained earlier

102 mod 8

[dire thunder]

oh lmao

i am dumb

he needed to find A^102B^102

not just A^102

[quartz compass]

his problem was in plugging into wolfram alpha hahaha

[wintry steppe]

i know

😦

[dire thunder]

but is my reasoning with idempotency correct in case of just A^120, mero?

[wintry steppe]

i mean i would laugh as well if it didn't waste so much time and stress

sigh

[quartz compass]

well A^8 = I so A^4 = -I not A^4=A

[dire thunder]

lmao

[quartz compass]

that's basically what mod 8 means, A^8 = A^0

[dire thunder]

need to learn LA

[quartz compass]

since rotating an 8th of the way around a circle 8 times is the same as doing nothing

[wintry steppe]

actually i don't know why wolframalpha thinks i want to divide everything by 4

i mean i guess i should've included parens but i expected more from WA

"computational intelligence™"

[quartz compass]

it's wolfram alpha's fault

blame the computer lol

[dire thunder]

even man not aways able t otrack operations btw lol

1+++2++

[wintry steppe]

@quartz compass have you read LA books in english

[quartz compass]

I haven't

[wintry steppe]

ah ok

[quartz compass]

read any LA books

[wintry steppe]

what

how'd you learn it then

just lecture notes?

[quartz compass]

just psychic

[wintry steppe]

..

[quartz compass]

I don't even read

[wintry steppe]

ok

merosity (and most other mathematicians) learn things by buying the books physically and consuming them whole at the start of the semester

that's why textbooks are so expensive

if you think you're funny, you're not

just so you are aware

im well aware

perhaps you shouldn't type then

[dire thunder]

merosity (and most other mathematicians) learn things by buying the books physically and consuming them whole at the start of the semester
@wintry steppe ye

librarians hate me

[wintry steppe]

📚 😋

[quartz compass]

I'm not a mathematician

[wintry steppe]

and

you are one in my heart 😳

[spice storm]

Need help reducing the second row. The goal is to find the eigenspace

[gray dust]

merosity (and most other mathematicians) learn things by buying the books physically and consuming them whole at the start of the semester
no we learn by putting the books under our pillow and absorb the content in our sleep

[pallid rampart]

I eat them

[delicate zealot]

If I have a2x3 Mary's then it has 2 rows and 3 columns yes? Or is it the other way around?

[gray dust]

2 by 3 matrix, yes @delicate zealot

[old swift]

I don't understand how a list of numbers can be thought of as a function between {1, 2, ..., n} and a sequence of numbers in F

[dusky epoch]

there's a missing comma there.

[old swift]

yea I still don't understand how the first clause of that sentence is true

[dusky epoch]

list of n numbers in F <-> function from {1,2,...,n} to F

[old swift]

yea

[dusky epoch]

a function s: {1,2,...,n} -> F can be thought of as the list [s(1), s(2), ..., s(n)]

this correspondence goes both ways

[old swift]

isn't that a function to F^n then

oh is it just applied n times?

[dusky epoch]

ok no like

im trying to describe a correspondence between F^{1,2,...n} (a set whose elements are themselves functions from {1,2,...n} to F) and F^n (a set whose elements are F-lists of length n)

like

under this identification, the list 55, -14, 17 would be identified with the function from {1,2,3} to R defined by the rule "1 ↦ 55, 2 ↦ -14, 3 ↦ 17"

[old swift]

hmm

lemme think about this

ok what messed me up was thinking that the function would take like the entire list as a parameter

wait how does this even make sense

because 4, 10, 8 would be defined by "1 ↦ 4, 2 ↦ 10, 3 ↦ 8"

so oh

nvm

makes sense now

[weary isle]

Could anyone explain the learning objective equations (1) and (2) in https://hal.inria.fr/hal-00815374/document#page=4 ?

[plain hamlet]

So I had a applied linear algebra project on Fecundity that we weren't given a little bit of information on and we had to do a lot. I was wondering if someone is willing to give me feedback on my report. I've attached project instructions as well. I am also willing to DM video links professors had sent to us.

[plain hamlet]

Feedback on the content

If i'm on the right track of what is asked of me, suggestions of what to do to change,etc.

Since fecundity , all we know is what's in the project instruction and the two videos we were provided(That I can share in DM since it shows personal information of instructors and university)

[hazy gull]

so the rank of a transformation is the dimension of the image, and the range is a set of vectors that the transofmation maps to, so is the rank the dimension of the range?

[dusky epoch]

yes exactly

image is the same as range

[hazy gull]

but if rank is not equal to the range then a tranformation is not onto?

[dusky epoch]

...

rank is a number

also range is not the same as codomain

[hazy gull]

Ya it didnt make sense to me either. My teacher wrote it down, but I dont think that's what she meant. How would you actually determine if a transformation is onto?

[cursive narwhal]

You would just show that the range is equal to the codomain

"onto" is something that applies to functions more generally. So, when you're asked to determine if a given linear map is onto, do exactly what you would do if you weren't dealing with functions between vector spaces

[wintry steppe]

for a function or for a linear map, you can just bash through the definition of onto. just be aware that there are a few special things you can do for linear maps between finite dimensional spaces:

if it's a matrix or if you have a matrix representing it, gaussian elimination and counting
if the dimension of the domain and codomain are the same, show that it's injective (this is usually not that hard; rank nullity gives you surjectivity/onto-ness)

those are the two main ways for linear maps between finite dimensional spaces

if you're working in infinite dimensions, cry do what abhi said i guess

[clear sparrow]

Is there a nice intuition for this formula?

[pallid rampart]

If you write it as $\text{dim}(W_1 + W_2) = \text{dim}W_1 + \text{dim}W_2 - \text{dim}(W_1\cap W_2)$, you get that the dimension of $W_1+W_2$ is the sum of the dimensions of $W_1,W_2$ minus the dimension of the part that overlap

[stoic python]

Whoever:

[gray dust]

ah sniped

[clear sparrow]

Ahh makes sense
Thanks!

[gray dust]

does the book attach a proof of this?

[clear sparrow]

Yea I have the proof

It's all algebraic though

[pallid rampart]

The word algebra is in linear algebra for a reason

[clear sparrow]

that is a true statement, yes it is

Oh yeah you can \dim btw

[gray dust]

it's ok whoever likes extra typing

[pallid rampart]

$\dim$

[stoic python]

Whoever:

[pallid rampart]

Frick

I didn't realize it's in my preamble

[clear sparrow]

Should for some reason \dim isn't available, you can \DeclareMathOperator{\dim}{dim} in preamble

[pallid rampart]

Takes too long

The mods always take forever to add stuff

[gray dust]

whoever do you have a cmd for span

[pallid rampart]

I don't think so

[clear sparrow]

\DeclareMathOperator{\span}{span}

[gray dust]

$\Span\brc{\xi_1,...,\xi_n}$

[pallid rampart]

nvm I do

[stoic python]

RokettoJanpu:

[pallid rampart]

$\Span\brc{v_1,\dots,v_n}$

[gray dust]

smh forgetting brc

[stoic python]

Whoever:

[pallid rampart]

Oh I actually have a lot of commands for linear algebra

[gray dust]

wait we had the same idea of capitalizing S bc texit already has \span for who knows what reason

[pallid rampart]

Yeah

Actually

[clear sparrow]

Hm you can override

[pallid rampart]

I might have copied my preamble from lightning

[gray dust]

i def copied his brc,br,sbr like everyone else did

[clear sparrow]

\let\span\relax

And then \DeclareMathOperator...

[pallid rampart]

lmao

[clear sparrow]

Isn't really "best practice" though, but whatevs

[gray dust]

i'd be too scared to disable any already existing cmds

[clear sparrow]

What even is \span

$\span$ test \span

[stoic python]

Hello World:

[clear sparrow]

If you can't see it, it doesn't matter

[gray dust]

stackexchange says span does smth in multicolumn but still doesn't clarify for me https://tex.stackexchange.com/questions/33264/span-as-a-math-operator

[clear sparrow]

Ah \span is primitive
Forget everything I just said, don't touch it

[shy atlas]

too late

mmm im touching it real gud

[wintry steppe]

What method do you use when finding the invert of a matrix ?

I usually do an augmented matrix such that [A | I] and make A into I by row operations

and then whatever is on the right side after that is my inverted matrix

As in [A | I] -> [ I | A^-1]

[storm python]

that's how i'd do it

[wintry steppe]

Would you also use that to check whether a matrix is invertible before trying to invert it ?

do you know what the determinant of a matrix is?

I know that if det != 0 it has one solution, and if it is = 0 there are no solutions or an infinite number of solutions

Well, for the coefficient matrix of a linear system of equations

so you want to find out whether a matrix is invertible or not?

yeah ?

then you can check a couple of different things

you can check if its columns are linearly independent, which is equivalent to solving [A|0]

you can check if its determinant is different from zero

you can reduce the matrix and check whether its rank is equal to the number of columns

I see

I don't see the "check if it's determinant is different from zero" in the invertible matrix theorem

Is it worded differently ?

For some reason it's not in my book, but when I look it up online that rule you mentioned is there.

[unique quarry]

Determinant is different from zero is equivalent to all those listed above

do you know how to calculate inverse matrix with determinant?

[wintry steppe]

No

The formula in yellow ?

Oh I see

if the determinant is 0 then it's not invertible since a division by 0 occurs

if you know about cofactor matrices and laplace expansions there's a way to generalize to square matrices of arbitrary size

Not there yet unfortunately 🙂

Well, haven't seen that in class yet, idk if we will

[zealous widget]

hey

i have a question

wai

*wait

how do I do latex here?

[wintry steppe]

surround with $ $

[stoic python]

eac:

[zealous widget]

lemme try

$ e $

[stoic python]

catfood:

[zealous widget]

ah ok

thanks!
so my question was

my linear algebra textbook by strang says that the projection matrix for projecting a vector b onto C(A) is $ P=A(A^T A)^{-1} A^T $

huh

why can't I do ^-1

[unique quarry]

put between {}

[zealous widget]

ah OK

thanks!

[stoic python]

catfood:

[zealous widget]

can't that just be simplified to I

because $ (A^T A)^{-1} $ is just A^{-1}(A^T)^{-1} $

[stoic python]

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

[zealous widget]

and slotting that into our original formula $ P=A(A^T A)^{-1} A^T $ gives $ P=AA^{-1}(A^T)^{-1} A^T $

[stoic python]

catfood:

[zealous widget]

which then becomes $I$

[stoic python]

catfood:

[unique quarry]

is $A$ a square matrix?

[stoic python]

DAILI:

[zealous widget]

hmm

is that of any relevance?

tell me what happens if A isn't square

and also what happens if A is

[unique quarry]

Tbh i'm not that sure myself, but if $A$ isn't square then you can't use $(A^TA)^{-1} = (A^{-1})(A^T)^{-1}$

[stoic python]

DAILI:

[zealous widget]

wait, i think i get it - the expression only simplifies to I if A is invertible - if A isn't invertible, then $ A^{-1} $ and $ (A^{-1})^T = (A^T)^{-1} $ can't exist, while because A in this context always has full column rank, then $ (A^T A)^{-1} $ always exists - hence why the expression always works

[stoic python]

catfood:

[zealous widget]

@unique quarry thanks for the help anyway, it was appreciated! <3

[warm harbor]

hey guys i have a question about 1to1 matrix transformation. Why is saying "Ax=0 has only trivial solution" is equivalent to saying it's a 1-1 transformation?

i dont see why an equation equation that only has trivial solution mean there is an unique output for each input

[limber sierra]

this can be proved using properties of linear transformations

[gray dust]

there's a proof of a linear map T being injective iff ker(T)={0}

[limber sierra]

^

i can prove both directions real quick

first assume T(x) is one-to-one, i.e. T(y) = T(x) implies y = x

[gray dust]

let $T$ be a linear map. suppose $T(x)=0\implies x=0$
$$T(x_1)=T(x_2)\implies T(x_1)-T(x_2)=T(x_1-x_2)=0$$
we now use $T(x)=0\implies x=0$ to say that
$$x_1-x_2=0\implies x_1=x_2$$
therefore, $T$ is injective $\qed$

[stoic python]

RokettoJanpu:

[limber sierra]

thats the other direction

i'll continue my directiokn

T(y) = T(x) implies y = x, and T(y) - T(x) = 0

so by linearity T(y-x) = 0

but y-x = 0

since y = x

so T(0) = 0

and if you don't have y = x, then you can't have T(y) = T(x)

so the only case where T(z) = 0 is when z = 0

i.e. T(x) = Ax = 0 has only the trivial solution, x = 0

therefore 1-to-1 implies only trivial solution

roketto gave the other direciton of the proof

only trivial solution implies one-to-one

(which is basically the same thing but backwards)

[warm harbor]

wow that's really clear, I got it now thank you both!! @limber sierra @gray dust

[gray dust]

cool

[limber sierra]

this is a very powerful statement about linear transformations

it makes it real easy to check injectivity

although note that it, of course, only works if the function is linear

taking the (not linear) function f(x) = x^2, f(x) = 0 has only the trivial solution, but obviously this function isnt one-to-one

similarly the function f(x) = x + 1 is one-to-one, but f(x) = 0 doesn't have the trivial solution (the only solution is x = -1)

so the fact only works for linear functions

but it's a very powerful fact indeed, and is one of the reasons why linear functions are so well-behaved

[and relatively easy to work with]

[warm harbor]

gotcha, thanks @limber sierra !

[wintry steppe]

In gaussian elimination

Can Line 2 become Line 1 - Line 2?

Or does it have to be Line 2 becomes Line 2 - Line 1 ?

[spice storm]

So i did the calculation to find the eigenvalue by finding the characteristic polynomial which I got $ -(x+2)(x^2-4x+5)$ and the only real eigenvalue was x= 2. Does this mean this matrix is not diagonalize because the other eigenvalues are complex numbers?

[stoic python]

KillerWhale2498:

[wintry steppe]

... I just posted man

[limber sierra]

Can Line 2 become Line 1 - Line 2?
Or does it have to be Line 2 becomes Line 2 - Line 1 ?

no, both are fine

if you prefer, note that

you can make L_2 into L_2 - L_1

then multiply by -1

to make it into -L_2 + L_1

i.e. L_1 - L_2

this is why it's fine to turn line 2 into line 1 - line 2

[wintry steppe]

Cool, thanks 🙂

[limber sierra]

@spice storm it means that, if M = SJS^-1, then J will have complex entries (and so will S and S^-1)

[spice storm]

But we haven’t learned complex eigenvalues.

[limber sierra]

ah, hm

[spice storm]

Yea

So what should I do here 😥

after completing the characteristic polynomial the only real eigenvalue is x=2. Therefore, this matrix is not diagonalize? @limber sierra

Because we cannot not have a basis of eigenvalues of R^3 by theorem

[wintry steppe]

@spice storm are you working with real valued matrices / endomorphisms on real vector spaces?

[spice storm]

@wintry steppe it says find the real eigenvalues but the only real is x=2 the other two eigenvalues are complex numbers

[wintry steppe]

ok

you're only working with martices of real numbers right?

[spice storm]

it's not diagonalizable in R, but it's diagonalizeable in C. So should I just mention that?

Yes only real

[wintry steppe]

well a necessary condition for it to be diagonalizable is that its eigenvalues are all in the field

so the answer depends on what field you're on

you can definitely say that it's not diagonalizable in R

[spice storm]

Alright, sounds good. Thank you Gio

[wintry steppe]

If I use gaussian elimination to solve a system with x_1 to x_6 as variables

When I get my final solution in terms of r, k, and w.

When I get specific numbers after plugging r = 1, k = 2, and w = 3 - should those solve my original system ? or only the system in rref

if you used gaussian elimination all of the systems you got should be equivalent to the original one

yeah that makes sense

[zealous widget]

hey guys

i have a question

what result would you get if you attempted to project a vector b that is in space $R^3$ onto the space $R^3$?

[stoic python]

catfood:

[zealous widget]

help anyone?

[gray dust]

just b

[zealous widget]

hmm

though it seems so strange

with equations it would just be b yes

hang on lemme think about this for a sec

[limber sierra]

what do you think a projection onto a space is?

[zealous widget]

taking the vector closest to the vector to be projected

this vector of course also has to be on the subspace @limber sierra

[limber sierra]

ok sure

so if you project it into a space it's already in

the "closest" vector is surely the vector itself

[zealous widget]

good point

now it all makes sense

i was trying to understand this through a geometrical standpoint

and you've just done it for me

thanks!

much appreciated

[wintry steppe]

If I'm asked to rotate a point (0, 2) by (11*pi)/12

Should I put that angle in degrees first before using the rotation matrix ?

[zealous widget]

your choice

radians is usually preferred

[wintry steppe]

So it doesn't matter ?

This one accepts both radians and degrees ?

im terrible at trig

[limber sierra]

cos(45 degrees) is the same thing as cos(pi/4 radians)

and similar for other angles and other trig functions

whether the angle itself is written in degrees or radians is irrelevant

as long as the correct version of the trig function is used

(i.e. if using a calculator, it's set to degrees or radians appropriately)

[wintry steppe]

I see, thanks

[wintry steppe]

Does this look like AB has been rotated by 60 degrees, then reflected through the origin and then applied a shear of factor 2 ?

[wintry steppe]

@wintry steppe compare the points of A and B to A' and B' (' meaning after being transformed) and see if that is what it does

And be aware that order matters as matrix multiplication as AB = BA

[wintry steppe]

does anyone here ever use cramer's rule

i mean

does it actually have any practical applications

i.e. does anyone remember what it actually is without looking it up

cool

[gray dust]

cramer gives an explicit formula for computing a solution to a linear system with an invertible coefficient matrix

[wintry steppe]

i know

but is it useful in practice?

[cursive narwhal]

@wintry steppe Take that to #point-set-topology

[gray dust]

if you’re only focused on computation, it’s quite fast for 2 by 2 & alright for 3 by 3 systems, the bigger the system the more tedious dets you must compute

[wintry steppe]

so you would use it to compute a 2x2 linear system?

well ok..

[quartz compass]

I'd only recommend it if you are trying to solve for one variable in a system of equations and not all of them

[gray dust]

key, only works if its coefficient matrix is invertible

[ivory basin]

is it ok if someone checks my work i feel like i messed up somewhere. And does this mean there are infinte solutions

[quartz compass]

yeah that's right

there are infinitely many solutions, you can pick any value for z you want, and that uniquely determines what x and y have to be for it to be a solution

[fickle citrus]

Better than saying there are infinite solutions, can you characterise the solutions?

[ivory basin]

@quartz compass thank you!

[tribal lodge]

hi can someone help me with this exercise

[dusky epoch]

what's giving you trouble here?

this is a definition-pushing exercise at most

[tribal lodge]

o uh im sorta just learning the definition
to prove, do i need to show the following?

1. W orthogonal is non-empty
2. W orthogonal is closed under scalar multiplication
3. W orthogonal is closed under vector addition?

[tribal lodge]

oh ok so if i said:
Let u,v ∈ W^T, and k be scalar
<ku, w> = k <u, w> = k(0) = 0
so (ku) ⊥ w
and ku ∈ W^T
-> which means W^T is scalar?
@wintry steppe

[tribal lodge]

o um for this one im confused, is this asking about a addition or determinant property?

what properties?

oh

[zealous widget]

do orthogonal projection matrices commute?

or orthogonal matrices in general for that matter

[tribal lodge]

oh

um is it where (A-ki) = (λ - k )
@wintry steppe

(A-ki) = (λ - k)i

ah ok oop

okay so you then get (A-Ki) - (λ - k)i = 0

uhm

i am not quite sure

take the derminant?

uh do i det(A-ki) - det((λ-k)i) = det (0) ?

oop so uh det(A-Ki) - (λ - k)i )= 0

ok so you have
det( (A-Ki) - (λ - k)i ) = 0
det( (A-Ki) - (λi - ki) ) = 0
det( A-ki - λi + ki ) = 0
det( A - λi ) = 0
😮 @wintry steppe

uh no, why?

ohh

i see

i noted it. thank you!

[torpid horizon]

anyone here know diff equations

anyone here

[dire thunder]

@torpid horizon depends on which level

just ask

[torpid horizon]

ok

ik how to do part a but idk how to do part b

thats the answer key

[dire thunder]

it is analgous to linear approximation

i mean it is linear approximation

like look

0.1 stands for change in x

because 1.1-1 = .1

and by the similar reasoning you have -.1 in y

@torpid horizon do u know linearization of function of several var?

[torpid horizon]

nop

how did u know it is 1.1 - 1

[dusky epoch]

$F(x) \approx F(x_0) + F'(x_0) (x - x_0)$

[stoic python]

Ann:

[dire thunder]

that for one var case

[stoic python]

Commander Vimes:

[dire thunder]

that's for two var case

@torpid horizon

[torpid horizon]

ohhh

ok

let me see

[wintry steppe]

i agree with your approximation @dusky epoch

we should approximate more things

[dusky epoch]

are you trying to piss me off

[dire thunder]

me?

[torpid horizon]

what is f (x0)

[dusky epoch]

x_0 here is [1; 1]

the point at which you found the derivative of your function

[torpid horizon]

ahuh

[dusky epoch]

F(x_0) is

well

the value of F

at x_0

[torpid horizon]

im following the formula

so is this problem wrong

this is the problem

and someone in my class solved it

so instead of (-0.1 -0.1) shouldnt it be (-5.1 -0.1)

?

[dire thunder]

wy 6 3

derivative of xy with respect to y is x

and x = 1

[torpid horizon]

ahuh...

[dire thunder]

and delta looks fine

[torpid horizon]

how did they get it then

[dire thunder]

-.1 in both lines is good

[torpid horizon]

i thought the first line should be 0.9-6?

[dire thunder]

oh lmao i am dumb

6 3 is nice

because it is not derivative it is H

[torpid horizon]

hmmm

im still a little confused >><

[dire thunder]

well this part stands for value of H in point (1, 3)

[torpid horizon]

yes

but idk how he got the 0.1

[dire thunder]

(1, 3) is given point

[torpid horizon]

yes

[dire thunder]

and (0.9, 2.9) is point to approximated

in each coordinate -.1 change

[torpid horizon]

yes'

so then when calculating that coordinate i subtract approx value from H' value

i think i see where i was confused i thought i was supposed to evaluate it then subtract it

[dire thunder]

is this everyithing clear to u

[torpid horizon]

yes it is now lol

[dire thunder]

nice

[torpid horizon]

for 5 and 6 idk how to do those

thats the answer key but im confuseed on how to set up number 5

[zealous widget]

hey guys

why does the product of two n x n projection matrices being 0 imply that these two matrices commute

i.e, $ P_{1}P_{2}=0 $ implies $ P_{2}P_{1}=0$

[stoic python]

catfood:

[zealous widget]

given that $ P_{1} $ and $ P_{2} $ are projection matrices of the same size

[stoic python]

catfood:

[torpid horizon]

ahuh

im not sure how to set it up

[wintry steppe]

you could try with simultaneous diagonalization but that seems farfetched

[torpid horizon]

ahuh

ok ill check

[tribal lodge]

hi quick question
for b, do i just find the eigen values of P^-1AP and square it?
So: the eigen values are: 3,2,4 for A.
So A^2 = 5,4,16?

[dusky epoch]

A^2 is a matrix and not a list of numbers

[tribal lodge]

sorry the eigen values of A^2

[dusky epoch]

and also, do you think 3*3 = 5?

[tribal lodge]

whoops typo i meant 9

[dusky epoch]

do you think 3*3 = 6?

[tribal lodge]

reee

[dusky epoch]

9, 4 and 16 are the eigenvalues of A^2 yes

[tribal lodge]

haha sorri tyty

oh also what about d)?

[wintry steppe]

you have three different eigenvalues so the respective eigenvectors are linearly independent

[tribal lodge]

is this asking that the eigen vectors found for these values need to be different?

ah yes

ok

ty

[wintry steppe]

@zealous widget

Its because the basis vectors of their column spaces are orthogonal to eachother

Meaning ker(P1) = img(P2)

[tribal lodge]

uhh can someone help me through this proof

[pallid rampart]

My hint is that if you can find a nontrivial solution to $A{\bf x}={\bf0}$ then it automatically follows that $(A^TA){\bf x}={\bf 0}$

[stoic python]

Whoever:

[tribal lodge]

hm. are you suppose to use A^T Ax = A^Tb

[pallid rampart]

?

Ok if Ax=b is inconsistent when b=(1,...,1), what does that tell you about A?

Think about the adjectives you can use to describe a matrix

[tribal lodge]

Ax=b is inconsistent then A would have no solutions?

[pallid rampart]

Does the statement "A has no solution" make sense?

[tribal lodge]

uh, yes?

[pallid rampart]

No it doesn't

A is a matrix

A is not an equation

You can say Ax=b has no solution in x and it's a perfectly good statement

But what the word you're trying to look for is invertible

If A is invertible then Ax=b has a (unique) solution for every b

But in this case it doesn't have a solution for a b

So what does that mean about A?

[median forum]

or has a nontrivial kernel

[tribal lodge]

ok wait so:

1. A being inconsistent for the given b -> A has no solutions
2. If i is invertible, then Ax=b has an unique solution for every b
   3)however, we just said that it doesnt have a solution for b
3. ---> then this means A is not invertible?

[pallid rampart]

Good

Yes

[tribal lodge]

😄

[pallid rampart]

Now

A is invertible if and only if Ax=0 only has the trivial solution x=0

Since A is not invertible in this case, what does that mean about the equation Ax=0

[median forum]

you could think of invertible as having a 1-1 correspondence between the domain and codomain of the transformation(and also spanning the whole codomain)

[tribal lodge]

so if A is only invertible IFF Ax=0 has a trivial solution where x=0
then because we said that A is not invertible, this would mean that Ax=0 never has a trivial solution where x=0?

[pallid rampart]

Lmao

No

[tribal lodge]

:/

[median forum]

has only the trivial solution

[pallid rampart]

Yes

How do you negate that?

[half storm]

The reason why what you suggested is not the correct answer - "Ax = 0 never has a trivial solution where x = 0"? - is because A(0) = 0 is always true for any matrix A.

Just to elucidate why that is not the correct answer.

Use your knowledge of the fact that A is invertible if and only if Ax = 0 has only the trivial solution.

[tribal lodge]

uhh im confused D:
okay so A is invertible If and Only If -> Ax=0 has a trivial solution x = 0
I thought we said that A is Not invertible

[half storm]

Yes.

[median forum]

you could think of A as a map if that helps
if it is invertible, then only zero gets brought to zero
otherwise, a set of nontrivial vectors is also brought to zero

[pallid rampart]

uhh im confused D:
okay so A is invertible If and Only If -> Ax=0 has a trivial solution x = 0
I thought we said that A is Not invertible
@tribal lodge no not quite, it is not "Ax=0 has a trivial solution x = 0", but "Ax=0 only has a trivial solution x = 0"

x=0 is always a solution to Ax=0 like JohntheDon said

[half storm]

In order to answer this question, use what you know about conditional statements and contrapositives.

[pallid rampart]

Lmao we shouldn't throw in other concepts like maps and contrapositives when he's already pretty confused

[half storm]

Sorry. I'm presupposing some knowledge.

[tribal lodge]

oH wai t
-> A is invertible IFF Ax=0 only has a trivial solution x=0
-> however, we said that A is not invertible
-> So that means theres other stuff(vectors) that are not 0 will also make it 0?

[pallid rampart]

Yes

So A is not invertible means that Ax=0 has a nontrivial solution. In other words, there exists an x, not equal to 0, such that Ax=0

Now, let x' be the vector we just found. What can you say about (A^T A)x'?

[tribal lodge]

(A^T A) x' = 0
?

[pallid rampart]

Yes

Now

x' is not 0

So this means (A^T A)x=0 has a nontrivial solution, namely x'

Guess what

[median forum]

Id just factor out

At(Ax)

but thats not obligatory

[tribal lodge]

😮

i see

that was kinda hard ngl

but thanks 🙂 also its a she* haha

[pallid rampart]

Oh sorry

[median forum]

you only have a few steps if you use the correct properties
A has an inconsitent solution for a b=> A is not invertible => there is an x' s.t. Ax'= 0 => (AtA)x' = At(Ax')=At0=0

(x' nontrivial)

[tribal lodge]

i know now, but i am smooth brain :0

[median forum]

not saying it shoild be easy

just take it carefully and you get used to it

it helps to write it all out

shouldve done in latex

[tribal lodge]

yeeee its just hard because im trying to self learn it before i actually take linear next semester

and not having someone to talk it out with is sad

[median forum]

$A$ has an inconsistent solution for $b =(1,..,)^T \implies A$ is not invertible $\implies$ there is an $x'≠0$ s.t. $Ax'=0 \implies (A^T A)x' = A^T(Ax') = A^T 0 = 0$

[stoic python]

Fractal:

[median forum]

jeez
thats p cool of you

couldnt even dream of linear before having it

[tribal lodge]

thanks 🙂 theres not much else to do cause of corona

so math it is c:

haha

[median forum]

I have tons to do
but good initiative

[pallid rampart]

Imagine being productive

[median forum]

dont even tell me

my sleep schedule is being the weakest link

and answering questions in this server

[soft tundra]

Is the error vector the vector that is closest to b in part a?

[quartz compass]

no, it's the difference between b and the vector closest to b in the span of V

a way to think of it is, $\vec b = \vec s + \vec e$

[stoic python]

Merosity:

[quartz compass]

s being the closest to b in the span of V

[gray dust]

the vector in a is commonly called the projection of b onto V while the vector in b is what i'd call the rejection of b from V or the normal of b wrt V

[soft tundra]

So its Ax?

[gray dust]

idk what A is

[half storm]

A must be the vector he constructed from that set

V

[soft tundra]

A is the matrix made from the span of v

I think

yeah what john said

[half storm]

the matrix i mean

[soft tundra]

^

Is the correct answer for a. the vector (4,2,2)?

[half storm]

You probably could do some multivariable calculus for this

It's a multivariable optimization problem if you decide to resort to that. I'm not sure what other methods you would use in LA.

I'm just now relearning LA

[gray dust]

Is the correct answer for a. the vector (4,2,2)?
seems off. show work? @soft tundra

[median forum]

just do the normal equation

[half storm]

The equation for the plane?

[median forum]

the least square problem is solved by using the proejction matrix

[half storm]

I'm pretty sure the answer is is the vector given by (4/3, 2/3, 2/3)

with the minium distance being 1/sqrt(3)

So this is a least squares problem? I don't know least squares regression analysis 😦

[gray dust]

try not to straight up give an answer, even if right or wrong

[half storm]

whoops

my bad.

[ocean sequoia]

does anyone have a hint on how to set up a linear transformation of a given dimension

im not sure if im over thinking this

give an example of T \in L(V,W) such that dim of null T is 3 and dim of range is 2

so that means W has a dim of 2, and V has a dim of 5 right?

$ T(v_1,v_2,v_3,v_4,v_5) = [v_1 - v_2,v_3 \cdot v_4 - v_5 ]$

[cursive narwhal]

\cdot

[stoic python]

brzig:

[pallid rampart]

$T(v_1,v_2,v_3,v_4,v_5)=(v_1,v_2)$

[stoic python]

Whoever:

[ocean sequoia]

ok i was actually going to ask that next if thats ok lmao

i felt like that was cheating a bit tbh

because since the range is 2 and the domain is 5 that means by definition the nullspace must be 3 i

[pallid rampart]

Not by definition

But by rank and nullity theorem

[ocean sequoia]

oh i thought thats what the fundamental theorm of linear alegbra said

ahhh

ok so poor choice of words again

[pallid rampart]

Well still

It’s a theorem

Not a definition xD

[ocean sequoia]

ok

so i should say by the rank-nullity theorem

[pallid rampart]

Yeah

[ocean sequoia]

and i actually really appreciate it when people on here are picky with the words it helps alot lmao

[pallid rampart]

Lmao

The whole point is to be rigorous

[ocean sequoia]

agreed

[pallid rampart]

So rigorously speaking, it’s wrong to say “by definition” here

[ocean sequoia]

100 percent thats why i like it

is $F^m$ an infinite dimensional vector space?

[stoic python]

brzig:

[half storm]

No

It has dimension m

Fuck gave answer again.

[ocean sequoia]

nah you didnt