linear-algebra

[limber sierra]

another subspace containing (1, 1, 1) would be, for example, a plane containing this line - which is certainly larger

[thorny hemlock]

ye

i got it

thank you

[arctic karma]

hi, am i allowed to ask for help here? i posted the problem earlier in the questions channel but it got taken over and they're all kinda occupied

[tawny tulip]

@arctic karma if it is related to linear algebra

[arctic karma]

yes

[fickle citrus]

You can ask for help

But it's up to chance if anyone helps

Honestly I think it just depends on who's passing and whether the audience that can help you is available

[arctic karma]

ah sorry forgot to say but i was able to put it in a questions channel again, thanks

[thorny gate]

does anyone know how I would go about even starting to solve this problem

im kind of completely lost

[wintry steppe]

hi somebody helps me diagonalize this matrix?

[tawny tulip]

@wintry steppe Hey, can you show us what you tried? Where do need help

[wintry steppe]

c from ex 5.2

[round coral]

if you just want the solution, just diagonalize it in symbolab, why wasting time here

[wintry steppe]

i have the solution

i just want to kwo how to reach it

[tawny tulip]

@wintry steppe Did you try to find the eigenvalues?

[round coral]

https://yutsumura.com/how-to-diagonalize-a-matrix-step-by-step-explanation/ looks alright

[wintry steppe]

i used adjugates

in order to simplify the determinant and avoindit a third grade equation

[tawny tulip]

@wintry steppe try to expand $R_2$ in the determinant, you won't get a third degree equation this way.

[stoic python]

Nyrre

[wintry steppe]

i did that

the thing is that one of the steps generates 2 degree equation with complex numbers that I failed twice to solve

thats why i need some help

[tawny tulip]

Generally, send here a picture of what you tried and people will help you from where you got stuck, don't send the whole question.

Now, can you send what you tried to do? i'll see if I can help

[arctic canopy]

Can someone give me examples of either scalar or vector random sequences?

[round coral]

@wintry steppe give the equation you can't solve

[acoustic zodiac]

what methods are there to find a basis for a vector space? proving that a set of vectors is one is easy, but what about finding one yourself?

[native rampart]

Depends on the vector space

Write down a general term and identity the free variables

[round coral]

there are 2 methods, either use construction, or collapsing , construction you do by taking a set of linearly independent vectors in vector space then building it up, until it can span the whole vector space, collapse I call, when you take a list of vectors that span the vector space, and kill the redundant vectors, if you have studied linear independence lemma, killing the useless guys, until we get finally a linearly independent list

there are many other ways too

but they all follow from these two

[thorny hemlock]

why not just let j = m ?

[round coral]

if a_m !=0, then j=m

[thorny hemlock]

yeah

same thing as letting j be the largest from that set tho

[ocean sequoia]

thats what we are doing

we are choosing j as the largest element

all they did was move the right side over

and divivde by aj

[thorny hemlock]

ik

but

[ocean sequoia]

we are showing vj is a linear combo

[thorny hemlock]

just introducing a new symbol for what

[ocean sequoia]

because its dependent

thats why its not m

[hollow finch]

its possible am=0

[thorny hemlock]

am != 0

[hollow finch]

i don't see that stated anywhere in what youre given for the proof

[thorny hemlock]

...

nvm

[ocean sequoia]

are you asking why we arent removing m

[thorny hemlock]

no

i was saying

there was no need to introduce j at all

[ocean sequoia]

yes there is

are you asking or stating

those are very different

[thorny hemlock]

asking

[ocean sequoia]

because we can remove the jth vector and still span the space

i guess we could remove m here to if you wanted

but like nix said

it doesnt say anywhere that am isnt 0

[thorny hemlock]

ugh

[hollow finch]

i believe they could have said WLOG let a_m be nonzero

[ocean sequoia]

here lets look at it this way

what is this lemma telling us

if you understand that

who cares

[thorny hemlock]

yeah

lol

[ocean sequoia]

ok

then cool

[thorny hemlock]

ok so i dont think i understand this second part of the lemma

after taking something from the span of the list

$u = c_1v_1 + \dots + c_j(-\frac{a_1}{a_j}v_1 - \dots - \frac{a_{j-1} }{a_j}v_{j-1}) + \dots + c_mv_m$

[stoic python]

Yes

[thorny hemlock]

idont see how that shows us that u is in the span of list that has the jth term removed

[wintry sphinx]

because you can see that u can be written as a linear combination of v1, ..., vm, except vj

you'll note that the equation contains no mention of vj

[thorny hemlock]

o

yh

i see

amazing

[half forge]

can someone me with this one im stuck

how do i determine if its linear dependent or linear independent?

[thorny hemlock]

c1v1 + c2v2 + c3v3 = 0 would be linear independant if the only way to achieve zero if making all scalar multipliers zero

@half forge

[half forge]

oh but how would i do that

hmm can we just take the determinant and determine from there if its = to 0

[wintry sphinx]

no don't do that

[ocean sequoia]

No

[half forge]

nvm lol

ahh im stuck but am i on the right track

[thorny hemlock]

c1(2-5x) + c2(3+7x) + c3(4-x)

you basically want to see if you can make any real polynomial of degree 1

if you can, it spans V

[stoic python]

SUNSHINE

[half forge]

ohh okay, so then

[thorny hemlock]

and if you can make c1(2-5x) + c2(3+7x) + c3(4-x) = 0

if c1 = c2 = c3 = 0

then linear independant

if not, linear dependant

[half forge]

hmm, what if you have something like mine c1 = -31/29 c3

[little citrus]

can someone help me with spectral decomposition

problem

[thorny hemlock]

idk i dont think youve done it correctly

[little citrus]

i got v1 to be 1/srt2and v2 to be 1/srt2

How do i find spectral decomposition

[ocean sequoia]

I’m assuming the spectral theorem just wants it in A=Q lambda Q^-1

I could honestly be wrong I know that’s the eigendecomposition

[half forge]

can someone help me im stuck 😦

[wintry steppe]

[acoustic path]

[hollow finch]

[half forge]

i did above lol

[wintry steppe]

[tame mural]

Would it be conventional to assume that F^n only refers to finite-dimensional vector spaces?

[wintry steppe]

🤨

F^n refers to the n-dimensional vector space over F of n-tuples of elements of F

(assuming F refers to a field)

[steady fiber]

yes, it would be conventional to assume that

[wise flare]

oes the inverse of an inverse matrix give the original matrix

[wintry steppe]

yes

[tame mural]

I see, thanks!

[wise flare]

@wintry steppe sorry is that yes for me ?

[wintry steppe]

it is.

[wise flare]

ok thank you

[near nova]

how would i start this?

[nocturne jewel]

Firstly det is multilinear, so $det(cA) = c^ndet(A)$ for scalar c and n x n matrix A

[stoic python]

moshill1

[little citrus]

can someone help me with this

I got v1 value to be 1/sqrt 2 and v2 value to be 1/sqrt 2

but how do I find the spectral decomposition

I am lost

[hollow finch]

what is the form of spectral decomposition?

[little citrus]

@hollow finch Spectral decomposition is the factorization of a matrix into a canonical form, whereby the matrix is represented in terms of its eigenvalues and eigenvectors.

got it from google

but idk how to proceed with this since I got same eigenvalues

[hollow finch]

spectral decomposition has a really specific form. do you have it in your notes or textbook?

[near nova]

if i transpose a matrix by turning it counterclockwise, does the determinant change if i was to simply swap the rows/columns?

i think u change the sign for 3x3 but does the sign even out for 4x4? or is it still swapped once

[red prawn]

transpose is reflection across the diagonal

[near nova]

but i saw some tutorials do it by rotating counterclockwise

[red prawn]

@little citrus You want to write $A = Q\Lambda Q^{-1}$. $A$ is your given matrix, $Q$ is the change-of-basis matrix, and $\Lambda$ is diagonal with eigenvalues

[stoic python]

Apopheniac

[tawny tulip]

You can take 2 and 3 out of the ()

[near nova]

so the answer is just 2*3*5*4?

[tawny tulip]

Yes

[near nova]

cool

[red prawn]

Wouldn't the 5 get cubed?

[near nova]

but since its c^n do i do

5^3

[red prawn]

yes

2 * 3 * 4 * (5^3)

[near nova]

im lost on the basics of proving

how would u know which matrices to Let be something

[limber sierra]

i'm not exactly sure what you mean

you're asked to prove that the set consisting of diagonal and symmetric matrices is a subspace

if you recall, the typical way to prove something is a subspace is to show it:

• is nonempty (obvious here)
• is closed under vector (in this case matrix) addition
• is closed under scalar multiplication

[near nova]

ah okay

[fallen karma]

So double dual space of a vector space V is the space of the linear functionals whose inputs are the linear functionals on V?

[wintry steppe]

ya

[fallen karma]

Ok

[wintry steppe]

<@&286206848099549185>

what is an example of z

idk what the squared on the set meeans

[real stirrup]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[wintry steppe]

like 75 and 57?

probably just the cartesian product of {5, 7} with itself

so tuples alright

also for future reference this belongs in a different channel, take a peek at the pinned message in here

[pallid rampart]

{5,7}^2 can be turned into a vector space if we define 5+5=5, 5+7=7, 7+5=7, 7+7=5, 5*5=5, 5*7=5, 7*5=5, 7*7=7

[wintry steppe]

i too denote the elements of Z/2Z by {5, 7}

[shrewd mortar]

@pallid rampart @wintry steppe it is time

[wintry steppe]

?

[shrewd mortar]

time to initiate the LoL road to gold grind

[steady fiber]

imagine playing LoL

I was an average LoL fan

but now I am an average abstract algebra enjoyer

much more fun

[near nova]

do i plug in p(t) for t in S

[gray dust]

no. find a,b where p(t)=a(t+1)+b(t-1) for all t, then (a,b) is the coords we want

[royal tangle]

hello

I need help with matrices

[wintry steppe]

[royal tangle]

how do i do part b?

can someone help? ^^

[hexed mural]

Question when they talk about a Matrix A being, A=CR (Column) (Row) do they Mean 1 Single matrix or Some other Matrix Times Another Matrix

and they treat the second matrix as a sorta Row Operation

[hollow finch]

probably a product

[hexed mural]

ok

[hollow finch]

i know of a way to write A as a product CR where C has the pivot columns of A and R has the nonzero rows of the rref form of A but that may not be what they're talking about.

[hexed mural]

so its not just take a singluar matrix and break into rows and columns

you do that after a product

of two matricies

[hollow finch]

i think its a decomposition of A in terms of its columns and rows

[hexed mural]

So from reviewing my good ole linear algebra

[hollow finch]

if $A$ is $m\times n$ and has rank $r$, then $C$ will be $m\times r$ and $R$ will be $r\times n$

[stoic python]

nix

[hexed mural]

I find stuff I obviously did not learn

I see here this Column Space Row Space stuff

and they use it to determine Indepence and Rank

Indepence or dependency in order to i guess see how a transformation affects another Matrix right?

if its a subspace of some two vectors

or if its a vector space

[hollow finch]

i wasnt taught it either tbf i just found it
but rank is defined to be the common dimension of the row and column space. the rank will also be the dimension of the range of the matrix transformation. because the rank is the dimension of the column space, it will also be the number of independent columns (and rows).

[royal tangle]

woah

are u guys talking about my problem?

pls explain

[hexed mural]

I saw that Max Independence is

R^N - 1

so in 3d vector space

you would have max 2 vectors that are independent

to make a 2d space to describe all the combinations

what is N defined as when they use it to determin solutions for Ax=0

[hollow finch]

i was answering static's question. but i guess its tangentially related to yours

[hexed mural]

n-r

r = rank obviously

this counting theorem

is interesting

if you have the sum of two column spaces

[hollow finch]

yeah N is nullity, and its the dimension of the null space (or solution space of the homogeneous system Ax=0)

[hexed mural]

and you add them and their sum equals the third

then the third is dependant on the other two column spaces

Ah

[hollow finch]

im not sure where youre getting this, though.

[hexed mural]

So suppose you have

3x3

right

[hollow finch]

im with you

[hexed mural]

Column 1 [1,3,2] Column 2 [4,2,1] Column 3 [5,5,3]

Rank is 2

because Column 1, Column 2 independent

well the column space

So if you set that matrix

to 0

you need to find a matrix multiple X that gives you the null

right

so C1 + C2 = C3

now if you do this

C1 - C3, C2 - C3

and solve for X as a multple

you get

(1,1,-1)

and that is your X

[hollow finch]

yeah that is absolutely a basis for the null space of that matrix

exactly right

[hexed mural]

when you say basis

the word basis and independent are exchanged as the same thing?

[hollow finch]

basis implies independence
independence does not necessarily imply basis

[hexed mural]

the rank

then

see im getting super confused with the vocabulary

check this out

https://youtu.be/azzrfdysfI0?list=PLUl4u3cNGP61iQEFiWLE21EJCxwmWvvek&t=643

how is he pulling out the x = (1,1,-1) using that counting method I described?

[hollow finch]

So you noticed that c1+c2=c3 right?

[hexed mural]

yeap

[hollow finch]

That's also c1+c2-c3=0 of course

[hexed mural]

ah

ok carry on

[hollow finch]

That means if I take 1 of the first column, 1 of the second column, and -1 of the third column I will get the zero vector

[hexed mural]

x(c1+c2-c3) = 0

right?

O shit

wtf

really

how do you know this

like where did this trick come from

[wintry steppe]

Y’all niggas smart?

[hollow finch]

I'd call it the column perspective of matrix multiplication. Some professors teach it, some don't

[hexed mural]

"column perspective"?

[wintry steppe]

[hexed mural]

interesting

[hollow finch]

Yeah a way to think of matrix multiplication in terms of columns

[wintry steppe]

[thorny hemlock]

occupied ?

[wintry steppe]

No but Fr tho can someone help me with geometry

[thorny hemlock]

n word tho

[hexed mural]

Im thinkin here

[hollow finch]

I believe you can use that to define matrix multiplication entirely

[thorny hemlock]

im i okay to post a Q here, are you guys r busy?

[hexed mural]

with just columns?

[hollow finch]

Yeah in terms of columns. It's one of the most useful ways to look at it imo

[thorny hemlock]

i know how to answer this, a(i + i) + a(1- i) = 0 , for the first the a gotta be real and the second a gotta be complex. But what does C as a vector space over R mean visually ?

[hexed mural]

what is R?

the dimension?

[thorny hemlock]

real number

[hexed mural]

and C is the vector space

[hollow finch]

There are other perspectives of course like in terms of rows or entries and sometimes they're useful. but column tends to be the most useful most of the time, especially conceptually.

[thorny hemlock]

i dont get this

[hexed mural]

I will google into this and keep it in mind

[thorny hemlock]

i havnt learnt isomorphism in linear algebra yet

how

i see

[hexed mural]

Ok hate to ask

[thorny hemlock]

so

[hexed mural]

but regarding x=(1,1,-1)

which columns are you refereing too

in that video

I dont seem to see which colums

[thorny hemlock]

we have the list (1 + i, 1-i)

if we think C as a vector space over R

we need to re write the list

if R^2

[hollow finch]

@hexed mural
https://ghenshaw-work.medium.com/3-ways-to-understand-matrix-multiplication-fe8a007d7b26

[thorny hemlock]

(1,1) ,(1,-1)

[hollow finch]

I'll watch it and see

[thorny hemlock]

but then it wouldnt be linear independant ? @ornate valve

a) says that it IS linearly independant if we think C as a vector on R

right

for part b)

C as a vector space over C is just a0(1+i) + a1(1-i) = 0

where a is a complex number right ?

what does it mean visually tho

Like having C as a vector space over R

So a vector space that contains all real points

If we have a infinity large 2 dimensional square

irl

that would be a R^2 vector space over R ?

uh

why not

um

as infinite as Real numbers are?

ok

imagine we could

would R^2 is a vector space over R

[tame mural]

I think it's okay to base linear algebra on real-world physics intuitions

and generalize from there

you are free to imagine a coordinate system imposed on our universe, one which is infinite

even if our universe isn't technically infinite

[thorny hemlock]

not the magnitude and direction part

[tame mural]

that geometric intuition is how a lot of people teach linear algebra though

Strang does that for example

in his video lectures

so does Axler

Axler says that vectors are arrows before he discusses inner product spaces

but the arrow perspective is clearly euclidian-motivated

[hollow finch]

@hexed mural at 3:19 he goes over matrix multiplication in the column perspective
I was referring to matrix A1 which he talks about around 10 minutes in
But if we have a relationship like C1+C2-C3=0 it doesn't actually matter what the columns are. We know what combination of the columns will give us 0.
That's 1 of the first (1,*,*)
Then 1 of the second (*,1,*)
And -1 of the third (*,*,-1)
In total that gives us (1,1,-1)
If we knew that 1C1+7C2+19C3=0 then (1,7,19) would a the vector in our null space

[hexed mural]

Wait your saying that

the x(1,1,-1) is the combination

itself

encoded

in that vector

[thorny hemlock]

The vector space over R is infinite tho right ?

or is it more complicated

ok so R as a vector space over R

[tame mural]

R^n is is finite-dimensional if n is finite

[thorny hemlock]

R^2 over R then lol, would that be infinite?

[tame mural]

R^2 over R would be finite-dimensional

But it has aspects of infinity inside it, yes

it's finite-dimensional because R^2 -> 2

[hexed mural]

so x=(C1,C2,C3)

"the combination"

[thorny hemlock]

what does the arrow mean

[hollow finch]

Yeah when you multiply a matrix on the right by a column vector (1,1,-1) you get 1 of the first column+1 of the second column-1 of the third column. That's the a way to think of matrix multiplication

[tame mural]

I mean the fact that R^2 is to a finite n is the important part

There are also vector spaces like over {0, 1}

and there are vector spaces that represent function spaces

[thorny hemlock]

so a vector space is called finite dimensional if the span of all vectors in it span the entire vectorspace ?

[tame mural]

F^n over F is called finite-dimensional if n is finite. Otherwise it is infinite.

[thorny hemlock]

i see

so we cant measure infinite dimensions but we can measure finite dimensions?

[tame mural]

No they're just two different areas of study

[thorny hemlock]

oh ok nvm

[wintry steppe]

a vector space is called finite dimensional if the size of a basis for it is finite; equivalently, the size of any linearly independent subset of it is at most finite; equivalently, if it has a finite spanning set of vectors

[wintry steppe]

we can measure infinite dimensions, at least to the extent of countable dimension and uncountable dimension (that i know of)

[hexed mural]

So you're saying that if I take any matrix and multply it by that A matrix I will get that relationship Column+1 Column-1 of the third

[thorny hemlock]

so kinda similar concept of countable and uncountable sets

[wintry steppe]

something like function space would be an example of a vector space with uncountable dimension, whereas something like the space of all sequences, all but finitely many of whose terms are zero, would be a space of countable dimension (why?)

[thorny hemlock]

idk this is beyond me lmao

[tame mural]

Most first classes are finite-dimensional linear algebra

that's why it's fair to consider them "separate studies" from the student POV

[wintry steppe]

i see

understandable

[thorny hemlock]

a1v1 + ... + amVm = 0 where all a's are zero

we re arrange the new list as 5c1V1 + (c2 - 4c1)V2 + c3v3 ... + cmvm

5c1 and c2-4c1 are just another scalar

so we can write it as bv1 + ... + bVm

thus this is also linearly independant

this proof works right

just making sure

[tame mural]

another way I would argue is that row operations don't change independence

[thorny hemlock]

i noticed that too

[pallid rampart]

or multiplying by an invertible matrices

[gray dust]

@thorny hemlock it's all over the place with some unnecessary statements and statements that are necessary but missing

definition, {v_1,..,v_m} is LI iff c_1v_1+...+c_mv_m=0 implies each c_i=0

so to prove {5v_1-4v_2,v_2,...,v_m} is LI we assume {v_1,...,v_m} is LI, let c_1(5v_1-4v_2)+c_2v_2+...+c_mv_m=0, then show each c_i=0

you are correct in the algebra you did is part of a proof, but you must explicitly state {v_1,..,v_m} is LI somewhere along the way to show each c_i=0

[native rampart]

Or for a more general approach, write the vectors as rows of a matrix and show that row operations preserve span of row vectors

So,if {v_1,v_2..v_n} is a basis, {5v_1-4v_2,v_2 ...v_n} will also be a basis

[thorny hemlock]

this is just rough proof but yeah

[gray dust]

there's no real need to make a matrix of the v_i's. proving a set of vectors is LI is a standard exercise at the start of an LA book where all you need to know are how scaling/adding works and defns of linear (in)dependence. reframing as row operations on a matrix is overkill

[round coral]

I rarely make a matrix, unless it is difficult to do just plainly in these linear independence questions, the way Drake used, is good to use in this case. I would have done the same way

just that I won't describe it as row operations, just directly show that it spans V and then I am done

[little frigate]

so like I was trying to find the determinant of a matrix by doing row echelon and then just taking the diagonal because the matrix would be upper triangle but for some reason

when I perform row elementary operation the determinant changes for a reason that I ignore

let me just take a pic of

[limber sierra]

performing elementary row operations changes the determinant, yes.

but it does so in predictable ways.

[little frigate]

yes but in this case i just did addition

so i think that determinant should remain the same

bcuz it just changes - negates when i swap rows or when i multiply row by a scalar i thonk

the good one is "-2"

wait why did i put a 5 am i high smh sorry wait

[limber sierra]

sure seems like youre multiplying by a scalar to me

when you take L_3 -> 3L_3

that's multiplying by a scalar

even if you add another row after it

[little frigate]

oh shit im fucking dumb

[limber sierra]

you still multiplied a row by a constant multiple

[little frigate]

sorry for wasting your time and thanks

[limber sierra]

no worries, it can be easy to miss this stuff

[little frigate]

If I have a matrix where rows represent equations and columns represent dimensions

if I'd like to see if those dimensions are independent then I could simply swap rows and columns and put it in echelon no?

[wintry steppe]

yes

best way to check linear independence is to take a determinant.

if computable

[little frigate]

Hm

so I'm being asked to find if some equations constitute a base

so first I wanted to see if they were generating the space and they could since the determinant was different than 0

but apparently they still weren't a base because they were not linearly dependant

hm, I just got super confused now, if I want to check if they are linearly dependent, should I look at the equations themselves (row vectors) or the column vectors?

nvm

[half forge]

#groups-rings-fields

[round coral]

@little frigate so that doesn't look to me wrong anywhere, you showed that v_1, v_2, v_3 are linearly dependent.

now you just want a basis for R^4 ? what do you really want to do?

if you just want a basis, take the standard basis of R^4

[little frigate]

?bruh

😂

and i got it now

basically I just wasted a goddamn hour on a damn exercise because I didn't correctly write it down

[native rampart]

Is it true(in context of finite dimensional spaces) that a vector space V_1 over Field F_1 is isomorphic to a vector space V_2 over Field F_2 iff F_1=F_2 and dim(V_1)=dim(V_2)?

[round coral]

for the condition of isomorphism, we require that both V_1 and V_2 must be over the same field F , by definition

The mention of this specific field F is often omitted once the field is fixed, but it is still implicitly always part of all statements

[native rampart]

Ok

[round coral]

if two v sp are over two different fields F_1 and F_2 , then the fact that they are isomorphic over F_1 does not imply that they are isomorphic over F_2

[wise flare]

is span an map the same thing ?

for example a 4x3 matrix cannot map R4 because it can't have a pivor position in every row

[near nova]

hi

does this make sense?

the first one. ignore c)

only question 2 to 2b

[near nova]

@ me please

[wise flare]

does that last column count as a free varible ?

why or why not

[steady cargo]

It is a leading one therefore not a free variable

Also does anyone know how to use Sage Math?

[native rampart]

I think #computing-software might be more helpful

[steady cargo]

Thank you

[acoustic path]

0x1+0x2+0x3=3

seems plausible

[thorny hemlock]

can someone help me understand this proof

I get that putting another vector into the list will make it linearly dependant

"step j"

so we have the list u1 , w1, ... , w_n . Which is of length n becuz we removed a w

do we keep adding u's and removing w's

until j-1

$u_1 , \dots , u_{j-1}, $

in step j

shouldnt that include some amount of w's

[torpid wedge]

when are 2 planes perpendicular? i mean if i have the equation of 2 planes

1 has a parameter, how do I check if they are perpendiculat?

[thorny hemlock]

theres a formula

[torpid wedge]

3x-y+b*z=3 and -2x+2x-3z=5

i tried searching for it, didnt find anything

[thorny hemlock]

cross product thing iirc

[torpid wedge]

i know how to use that for lines but not sure about planes

[thorny hemlock]

hint plez

ive managed to show that w is in span(v1 ... vj)

<@&286206848099549185>

[hoary osprey]

what's j

[thorny hemlock]

largest index where a_j != 0

same j as here

I took the fact that theres a certain vector such that isnt in the span of the previous

set up an equation

and rearragned and showed that w is a linear combination of v1 ... vj

stuck since then

-_-

is it just that

$w \in span(v_1, \dots ,v_j) \in span(v_1,\dots,v_m)$

[stoic python]

Yes

[thorny hemlock]

-_-

@hoary osprey

[hoary osprey]

just write it as a1v1 + ... anvn = -(a1 +...an)w and work from there

[thorny hemlock]

i did that

w = a linear combination of v1 -> vj

i think ive got it

[hoary osprey]

yea then ur done

divide both sides by -(a1+...an)

[thorny hemlock]

yeahhh

[hoary osprey]

why cant this be 0?

[thorny hemlock]

oof

i didnt think about that

[hoary osprey]

well good thing is that it cant

think about why

[thorny hemlock]

well

all the a1 -> an arnt all zero at once cuz linear dependance

@hoary osprey

[hoary osprey]

yes but why cant their sum equal 0

[thorny hemlock]

arnt we actually dividing both sides by

(1-a1) + ... + (1 - an)

but idk

why the sum cant be zero

[hoary osprey]

cuz if it was 0 youd have

a1v1+....+anvn=0

and this is a problem

cuz of what we know about v1...vn

[thorny hemlock]

but

vj + w = a1(v1 + w) + ... + aj-1(vj-1 + w)

leads to dividing both sides by (1-a1) + ... + (1-aj-1)

(j-1) - (a1 + ... + aj-1)

if their sum was j-1

j-1 - j-1 =0

@hoary osprey

[manic drum]

does anyone know how to write the slope-intercept form of a vertical line through (-1,3)? This is on my homework and I'm confused by it as a vertical line has a slope of undefined

[thorny hemlock]

not really sure how to prove this

like

should i assume that its finite dimensional and then contradict it?

idk how to write it mathematically tho

why take a linear independant list

@wintry steppe

[near nova]

can someone skim over my answers on a lab? ill send pdf

[limpid fiber]

Why is it true that in elementary row operations, replacing one row by the sum of itself a a multiple of another row preserves the solution set

I guess it makes sense because we're always adding equivalent quantitites to both sides, but I feel weird about cancelling out variables

[royal ore]

https://cdn.discordapp.com/attachments/504653099793645568/788810396521857084/image0.jpg

[wintry steppe]

just go through the vector space axioms and see which one fails

[thorny hemlock]

finite dimensional if the span of some list of vectors in it spans the entire space

no

thats the next chapter

what about this one

i can take the list 1,x^1,x^2 .... forever

[near nova]

does this make sense?

[thorny hemlock]

how do i show that span(1,x^1 ,x^2 .... ) doesnt span every value in the interval [0,1]

do you know how ? @wintry steppe

(ping me if anyone helps)

all this book says is, an infinite dimensional vector is one that isnt finite-dimensional

[near nova]

@wintry steppe does my screenshot of my answer make sense?

is it correct?

[wintry steppe]

why are you pinging me

[near nova]

i thought you were a helper

[native rampart]

Ah yes,The yellows are helpers

[wintry steppe]

i'm not, and even then, you don't ping individual helpers

[near nova]

ah, sorry

[wintry steppe]

@everyone

[native rampart]

Suppose there was a person named everyone

Would that ping the person or everyone?

[thorny hemlock]

ill read the next chapter and learn about "basis"

and come back to the Q

ah ok

ye P(F)

span(1,x^1 ,.... ) would be the span of all continuous real functions

ok

ok

sure

for the polynomial Q

ive found a solution online

tho i dont understand it

if m was 1

a0 + a1x = 0

how is that linear independent

if x is just a real number

ok

i see

so we treat is like a vector

ok

so after realising that its always linearly independant

how do they conclude its infinite dimensional

hm

because it carries on forever?

and finite dimensional would stop

[magic light]

hey slimvesus

[thorny hemlock]

like F^3 is finite dimensional cuz (1,0,0) (0,1,0) (0,0,1) is a finite list size

witht the polynomials, youll have an infinite list size? and thats how they conclude its infinite dimensional?

or am i completely wrong lmao

[stoic python]

slimvesus

[thorny hemlock]

Ok

We

Err

I do

When u say consider sequences DJ

Ej*

ej

I don’t get what you did exactly

Ok

You take elements of F infinity such that they are a linear independent

[acoustic path]

then you need an infinite amount of column vectors

[thorny hemlock]

Yes you are

Theorem states linear independent lists are always less than the spanning list

Yes

Ok so basically

For all n in f^n you see that the linear independent list has a greater length

F^n

n is natural

Why not

Yes

Ok

ok right

Contradicting that the length is smaller than the length of the spanning list?

Ok

Right

That makes sense, my confusion is when finding a linear independent

Are we taking a linear independent list from F infinity?

Lol

The ej list isn’t linearly independent in F^n tho?

True

[stoic python]

slimvesus

[thorny hemlock]

Ok

If we look at when n = 3 as example

F^3

as example

It’s not making sense to me that you can take a linear independent list outside of Fn

You’re argument is that The linear independent list is largest than spanning list

When n does equal 3

100 010 001 is smallest spanning list

oh

Ok I think I get why I’m wrong

Yeah

But

Yes

But

I get it I think

A linear independent list of length infinity is possible

but spanning list is only as large as n

What’s the difference

Okok I see

Lol

A linear independent list of any length is possible but spanning list is only n ?

Is that not how you would say it

Right

I was imagining it wrong in my head and was confusing myself

[robust pond]

@royal ore the second one allows you to solve for 2 of the values

then you can use the first to solve for the other two

probably an easier way using bases or something

id just treat it like a system

id be curious what the way with bases is

[round coral]

you can choose your basis here as ((1,1) , (0,2))

and then do it the simple way

as for why you can choose that a basis for R^2 , I will leave it you to think about it

@thorny hemlock

[limpid fiber]

Is there a good explanation for why substitution works to solve systems or is that going to come in cramers rule?

[robust pond]

i dont understand the basis method

bummer bc i just took this stupid class

[wintry steppe]

can somebody help me with exerc.5.4?

[robust pond]

🤔

[wintry sphinx]

basically you treat [1, 1] and [0, 2] as your new basis

so to compute the value of Av for some vector v

you write v in terms of [1, 1] and [0, 2]

then multiply by the matrix with columns [-1, 2] and [3, -1] (call this M)

so you can write it as A = M <change of basis matrix> v

[split badge]

Y’all know this?

I need help

[wintry steppe]

this a test?

ya left in how much points each one's worth and what not

not even linear algebra smh, read the pinned message in thischannel

[limber sierra]

why does this test have airbnb advertisement

what is this product placement

[shrewd mortar]

professor making bank

[scarlet yarrow]

Guys, Is there a channel for vectors?

[jolly rivet]

left side is in W and right side is in the orthogonal complement

[scarlet yarrow]

Im guessing this is the right channel

[jolly rivet]

i think i understand

so

the the left and right side are in the intersection

which makes both the zero vector

and the only solution is a1...ak b1....bm = 0

thx

[thorny hemlock]

taking list a1p0 + a2p1 + ... + am-1p_m

it can be zero when all polynomials have input of 2

thus not linear independent

does this work?

:(

whats wrong with it

wait

linear independant is if the only way to make the linear combination zero is when all scalars are zero, agreed?

but here, if 2 is the input it can be zero without scalars needing to be zero?

ok

yes

ok

supposed

we have polynomials

$4 + (3 + x) + (4 + x + 3x^2 ) + (5 + x + x^2 + x^3) $

when m is 3

each polynomial is zero when x is 2 ?

pj(2) = 0

what does that mean then

errr

my bad

bad example

but

what

yeah

ik

err

i mena

supposed we have polynomails that do vanish at 2

we are only left with what ever p0 is right

lemme give a proper examplee

$4 + (-2 + x) + (-6 + x + x^2)$ when m is 2

[stoic python]

Yes

[thorny hemlock]

p1 and p2 vanish at 2 but the 4 remains yes?

yes

4 would be polynomail of degree 0

bruh

a(4) + b(-2 + x) + c(-6 + x + x^2) = 0

at 2

:/

ok

0 + (-2 + x) + (-6 + x + x^2)

a(0) + b(-2 + x) + c(-6 + x + x^2) = 0

at 2

a,b,c all being zero at once isnt the only way to make zero right

so not linearly independent ?

ehhh

what the correct solution then :/

...

yeah my bad

sorry

is considering the list p0 + p1 + p2 + p3 + .... + pm = 0

the right starting approach

[surreal thistle]

can anyone help me approach this problem

[wintry steppe]

what's your definition of a unitary matrix

also, the standard inner product on C^n is given by

[stoic python]

TTerra

[thorny hemlock]

spanning list of Pm is of length m but p0 -> pm list is of length m+1 so cant be L.I ?

[wintry steppe]

P_m(F) has dimension m + 1

not m

[thorny hemlock]

wdym

[wintry steppe]

okay, sure, you can just post the answer

[thorny hemlock]

all polynomials are subspaces of R^2 ?

[tawny tulip]

$(Ux,Uy) = (x,U^{*}(Uy)) = (x,(U^{ *}U)y) = (x,Iy) = (x,y)$

[stoic python]

Nyrre

[wintry steppe]

the space P_m(F) has dimension m + 1, so the set p_0, ..., p_m being of size m + 1 does not automatically guarantee that it is linearly dependent

it'd have to be greater than m + 1 to automatically guarantee linear dependence

so your reasoning fails

[thorny hemlock]

what does dimension mean

[wintry steppe]

page 31 of axler

review the definition there, and the theorems there and on the following pages, and you should be able to solve this problem

er

this is axler right?

[tame mural]

it is impressive people recognize the book so readily

[thorny hemlock]

er

[wintry steppe]

hold on i might be looking at a different version

anyways

[thorny hemlock]

page 31 has no mention of dimension of polynomials

[wintry steppe]

.

axler chapter 2 should have a section on dimension

[thorny hemlock]

yeah it does

but

this Q is before then

[tame mural]

The dimension of a vector space is the size of the basis.

[wintry steppe]

okay strange i guess my version of the book is vastly different than yours

because my copy of axler has a full section on dimension in chapter 2, before this problem

[wintry steppe]

[thorny hemlock]

ah

[wintry steppe]

which version? lemme get the right version, so i dont have an outdated one

[thorny hemlock]

i have third edition

[tame mural]

Every finite vector space can be "generated" from a subset of its elements using linear combination

The smallest generating set is the basis

And the size of that basis is the dimension

[thorny hemlock]

ok

you say that the basis of Pm(F) has a basis of length m+1

[wintry steppe]

1, x, ..., x^m

[thorny hemlock]

i se

[thorny hemlock]

oh

how does that fact help

i thought u were talking about the other persons Q lmao

one less

m-1

yep

m

length of the list?

yesss

m+1 i mean

yeah contradicts that certain lemma

i gotchu

well it cant be LI cuz its bigger than span

you mean that right

not LI

the scalars dont all need to be zero to equal zero

lol

[stoic python]

slimvesus

[thorny hemlock]

yes

i can try ..

So this new list, isnt LI in Pm-1(f)

which is what youve shown ther

how do we know this

ok

im not really sure how to do finish this tbh'

adding an element from Pm(f) will make list m+2 in length

if we take for fact that Pm(f) is spanned by list m+1

thats my guess

length will be m+1 ?

i mean it will be

but are you saying

since its LD in a Pm-1 it will also be LD in Pm

[stoic python]

slimvesus

[thorny hemlock]

eek

thats in the space of Pm-1(F)

ah yes true i see

the underlying field is R^2 ?

ok

ohk i see

[near nova]

question

why do matrix transformations shape the grid behind it not just the vectors itself

https://m.youtube.com/watch?v=uQhTuRlWMxw

[limber sierra]

"the grid behind" is really a representation for a basis

[nocturne jewel]

Why do I feel like this is a- YEP

[near nova]

example: happens here

[nocturne jewel]

3b1b

[limber sierra]

assigning a coordinate system to a space is just writing vectors in terms of basis vectors

and linear functions are determined entirely by how they act on a basis

[near nova]

ill read that slowly

[limber sierra]

hence why it transforms the underlying coordinate system.

[nocturne jewel]

every point is represented as a vector, so every point moves w/ the transformation

grid lines = points = movement from the transformation

[near nova]

ohhhhh

tysm u two this is my favorite discord channel 😍

server 🤔

[wise flare]

i have a question

i did a and b

i dont see how c is possible without the physical matrix...

its the dimension of the column space

least (n,m)

(n,m) being the size pof the matrix

[acoustic path]

4

[wise flare]

im not following you

like what are you trying to say ?

[acoustic path]

Isnt rank also the number of leading pivots in each row/column

[wise flare]

dim col(a) = dim row(a) ?

im not sure i agree with that. is it the same for nul a ?

[acoustic path]

I love the rank nullity theorem

[wise flare]

rank + dim nul = colum

nul a is 0

dim bul a

nul*

because rank + dim nul = #of colums

rank of a (c in the problem)

is 4

because its the lowest number of (n,m)

that is the maximum a 6x4 matrix can be

it is 4

the problem is purely hypothetical tho

and that answer is 4

i know what row a is

[acoustic path]

9

[wise flare]

row a is 6

[acoustic path]

9 upside down

[wise flare]

what do you think im asking ?

right

yes

my prof said it is # of columns tho

[acoustic path]

Thats just a condition for c tho not a

[wise flare]

his note always say tha the rank of a is the dim of col a

also*

ok now you're confusing me

basis of dim

or*

the basis of row a

or dim row a

which are you talking abouit

matrix transposed right ?

the row space

the column space of a transposed ?

its rank

yes youre riught

oh shit

its the pivot positoins

its the column of the pivot positions basically

[fallen karma]

So the column space is the span of the columns taken individually as vectors, and is equal to the range of the linear transformation represented by the matrix

I'm gonna drop a quick question

34 I'm confused by the notation

It's this function Lambda:V->V''

So input must be vectors in v, but it's also got that fancy p (I forget what that letter is called )

[high wyvern]

Hi. I'm having trouble understanding this.

[fallen karma]

Fancy p is a functional in the dual space

Phi got it. Anyway it's Lambda: V->V'' and not Lambda: V×V'->V''?

[wise flare]

oh shit

its the number of free variables

reeee

[fallen karma]

Ok so why didn't they say Lambda: V'->V'' is that a typo

Because it looks like the input is a functional in V'

Ok yes

[wise flare]

1 0 0 1
0 1 0 1
0 0 1 1
0 0 0 0
Since that last Row can’t ever have a pivot position by default. Does it still count as a free variable ?

Esp since the system is inconsistent

[fallen karma]

Got it

[stoic python]

slimvesus

[fallen karma]

Yeah

Is the matrix of a linear transformation T the same as that of T''?

That's what I thought. Well that's good

Ok thanks for your help!

[fallen karma]

@high wyvern I haven't really studied this too deeply but i think they explicitly tell you two eigenvalues, and third one is implied in the problem. 3 is the most you can have if dim(V)=3. Trace is the sum of eigenvalues multiplied by their multiplicities.

[round coral]

@gilded solstice frankly speaking, this isn't linear algebra question

put it in one of the question channels that are free

[thorny hemlock]

any others except the trivial one?

[hearty gyro]

how would i rotate vector a to turn it into vector b
like what matrix would i use
in a general sense
like lets say vector a is (1,0,0) and b is (x,y,z)
what rotation matrix should I apply to a, to make it b?

[wise flare]

can someone pl check my work for this

F T F F F F T F T F F T

[thorny hemlock]

false for the first one?

[wise flare]

yes

oops

sorry for caps

[thorny hemlock]

a+t^2 + b + t^2 still in P2 ?

and so is c(a+t^2) ?

im not completely sure tbh

but thats my initial thoughts

[tame mural]

How does one type the long vertical bar in a set for Latex?

{ x | x ∈ N }

[thorny hemlock]

\vert

[tame mural]

hm, but that has the same effect as just doing |

and it won't scale right

[thorny hemlock]

does \bvert work

[tame mural]

hmmm nope

oh well ~_~ how sad

latex is frustrating

[thorny hemlock]

wdym scale

i recall seeing something that takes care of it

[gray dust]

$\brc{x\mid P(x)}$

[stoic python]

RokabeJintarou

[gray dust]

$f(x)\eval_a^b$

[stoic python]

RokabeJintarou

[tame mural]

mmm mid I see

thx

[gray dust]

sure

[exotic tusk]

when you read a matrix aloud do you read it row by row or column by column

[native rampart]

Depends on the situation,If the matrix is used for solving a system of linear equations, row by row. If the matrix is used for describing a linear transformation,Column by column

[wintry sphinx]

always left to right, up to down

if you do not do it that way, you are a monster

[fallen karma]

@thorny hemlock did you think of any other vector spaces with exactly one basis

[wintry sphinx]

there can't be lol

[native rampart]

Why not

[fallen karma]

what if your field is Zmod2

[native rampart]

Well,it's also called F_2

[fallen karma]

ooo

[wintry sphinx]

I guess that also works lol

[near nova]

can someone translate this to words?

i want to understand it

so A is similar to B because:

matrix A is = P inverse * B * P -1 and det of P isnt 0

why is 2 substituted for 1?

then below “substitute 2 in 1“ loses me

[limber sierra]

you take equation 1

A = P^{-1}BP

and substitute in equation 2, B = Q^{-1}CQ

A = PBP^{-1} = P^{-1}(Q^{-1}CQ)P

substituting in equation 2 for B

and this suffices to prove similarity, since thanks to associativity and the fact that (ab)^-1 = b^-1 * a^-1

we can rewrite this as (QP)^{-1} C (QP)

so A = somematrixinverse * C * somematrix

[and for "somematrix" they wrote D]

[near nova]

hm

[scarlet yarrow]

$ \lambda{1} a + \lambda{2} b + \lambda_{3} c = 0 $

why doesnt it work>>

[limber sierra]

a bunch of reasons

$\lambda_1 a + \lambda_2 b + \lambda_3 c = 0$