linear-algebra

[wintry steppe]

what does overlapping mean?

intersecting only at 0?

cause that's not true, e.g. (0,1) and (1,1)

[tame mural]

but the dot product of those two vectors isn't 0

or am i seeing things

[wintry steppe]

uh

you asked if "nonoverlapping subspaces" implies orthogonal

i gave you an example of non orthogonal vectors whose spanned subspaces are the closest thing to "not overlapping" i can think of

[tame mural]

I see

mm I'm just phrasing it poorly

does it make sense to say

"they don't contribute in other subspaces"

[wintry steppe]

i still don't know what that means

[quartz compass]

sounds like you're asking about having 0 projection

but since projection is related to dot product, that answers the question

[tame mural]

Mmm I see, I was trying to see if you could define orthogonality without inner products

but it was silly

[quartz compass]

I think it's a good question since defining an inner product is kind of dependent on the vector space or field, it just doesn't really have some clean general form that I know about unfortunately

[supple seal]

hey I'm learning transformations and my prof is terrible. can anyone help me with this question?

[wintry steppe]

what part do you need help with? what have you tried?

[supple seal]

I just dont know how to answer it and my prof is suupper russian

said this will be on the test

[wintry steppe]

🤨

[supple seal]

how do I find domain and codomain of a transformation?

[wintry steppe]

well it's a function, so you have to look at what goes into it and what comes out

here, T takes in vectors with 3 entries and spits out vectors with 3 entries

[supple seal]

Would it just be R3 and R2 then?

[wintry steppe]

why do you say R2?

[supple seal]

I'm not confident on why, I've just seen most solutions in the textbook saying that

[wintry steppe]

well

it spits out vectors with 3 entries

which implies that the codomain should be...?

[supple seal]

also 3?

[wintry steppe]

can you be a bit more precise?

[supple seal]

like I said, the topic is new and my prof hasn't been explaining things very well

[wintry steppe]

you gave the correct domain but the incorrect codomain

what was your reasoning for the domain being R3?

[supple seal]

because it takes in 3 values right?

[wintry steppe]

sure

now if you apply the same reasoning to find the codomain

what do you get?

[supple seal]

so domain is how many it takes in and codomain is how many it spits out?

but what does that have to do with R?

[wintry steppe]

sure

presumably the x_1, x_2, x_3 are real numbers

i mean they could be complex numbers as well, but the question doesn't say

let's pretend they're real 😌

[supple seal]

oh I'm thinking of 3d space then

thanks

[brittle orchid]

Hey, could someone eli5 this to me, I understand what this is saying but I have no idea why?

nvm I totally missed the fact that there's more below

[humble pumice]

Hey guys, i was wondering if you could help me check my answer on this practice problem and lmk which options are correct thanks

[rustic panther]

I got to this point

GCD is 3 so somehow I need to manipulate the rows/ columns so I have 3 where 1158 is at right now

[red prawn]

There is a theorem saying "positive-definite". Multiply stuff by -1 somewhere in there.

[round coral]

this is a baby version of the theorem, in fact you can factorise even non-invertible matrices as A= BC, just a comment

[red prawn]

Gilbert Strang has a pretty cool course wherein he does decompositions very nicely

[hollow finch]

on youtube?

[red prawn]

yeah MITopencoursewares

There's a short version, and he has a longer lecture series as well. I've only recently started watching the short ones

[round coral]

@jolly rivet this is a general case, Let A be a m×n matrix of rank r.
Prove that there exist matrices M and E, of size m×r and r×n respectively, such that M has linearly independent columns and E has linearly independent rows, and
A=ME.

[red prawn]

This is the short version.
https://youtu.be/YrHlHbtiSM0

[round coral]

@red prawn thanks for sharing that

[red prawn]

He has another Linear Algebra course on the same channel in which he spends more time on each topic. But the short ones are impressive, how fast he demonstrates everything

[round coral]

Strang is ok, his problem is just that he never goes into the theory, at least his course and book which I have roughly seen, it is more like just result based and computational

for application

[red prawn]

yeah that's his focus, for applied stuff. No jordan decomposition, or else he'd have to do chinese remainder theorem?

[round coral]

well I would be interested to see a proof of jordan decomposition through chinese remainder theorem, never seen it. Looks quite interesting if its true

[red prawn]

See Humphreys' Lie Algebras

[round coral]

maybe I will, I need to study some more till I go on to Lie Algebras

[spare crystal]

I've heard ppl say the inner product or dot product measures the amount of similarity between two vectors

I feel like a much better term is "overlap"

What do you guys think

[spice storm]

A dot product is a special case of an inner product. Consequently, every dot product is an inner product, but not vice versa. @spare crystal

[native rampart]

Every inner product is a dot product with the right choice of basis

[tame mural]

Right choice of basis?

[native rampart]

I meant that a inner product is a dot product wrt an orthonormal basis

[tame mural]

I see, that's interesting to know

not obvious to me at all

[hollow finch]

Overlap is nice. I like to think of the inner product as a definition of "distance" in a sense.

[native rampart]

Have you heard of Gram–Schmidt?

[tame mural]

Yeah I know the gram schmidt technique

it's to generate orthogonal basis

[hollow finch]

Would you mind showing me how that's true with an integral inner product?

[native rampart]

Yea,So that point kinda follows from gram schmidt

[hollow finch]

I can't picture it atm

[near nova]

i got this matrix but not sure how to continue

[stoic python]

nix

[tame mural]

Is this just orthogonality of two functions?

[native rampart]

Pn is polynomials of degree less than or equal to n?

[hollow finch]

Ye

I'm trying to imagine how exactly a dot product like thing would come out of it but I'm just about to pass out lol

[native rampart]

Take P2 as an example

[hollow finch]

Try turning the third row of your augmented matrix back into an equation (like you started with) and see what it tells you (it's telling you something very important)

[hollow finch]

Sure

[near nova]

3a-b-c= 0 so does that mean i can plug in 0 into the rest of the rows

[fallow coral]

guys

when we say

ker(L)

do we mean the basis for the kernel

of all the vectors in the kernel

[tame mural]

all the vectors in the kernel

not merely the basis

[fallow coral]

okok

ty

[native rampart]

Consider the basis {1, (x-1/2) , (x^2-x+1/6)}

(ok, not exactly orthonormal,but just divide by their norms)

[fallow coral]

to show V and W are not ismophoric is it enough to show a linear transformation from V-> W is not one to one

[hollow finch]

Yes

[fallow coral]

but could there not exist another linear transformation from V->W that is an isomorphism?

[hollow finch]

Oh wait I thought you meant if a transformation is not an isomorphism

[fallow coral]

ohh nah i meant the 2 subspaces being ismorphic or not

like i have a linear transformation

i proved that its not an isomorphism

does this mean the 2 subspaces arnt isomoprhic

or not nessessairly

[hollow finch]

No definitely not

There are transformations V->V which aren't isomorphisms

Two vector spaces are isomorphic iff they have the same dimension I believe

[fallow coral]

ah

okok

one more thing

what does it mean when they say

"finite dimensional vector space"

like whats the significane of the word finite

[hollow finch]

${1,\sqrt{3}(2x-1),\sqrt{5}(6x^2-6x+1)}$

[stoic python]

nix

[native rampart]

Yeah,that works

[hollow finch]

I think that's the orthonormal basis

[fallow coral]

is R3 finite dimensional

[hollow finch]

I definitely need to sleep now but I'll check back in tomorrow morning
In general tho thank you Drake for taking the time and effort to help a tired fool lol

[hollow finch]

Yes

[fallow coral]

okok

i get it now

thanks

[near nova]

why is this asking why these 4 things are the same

b doesnt necessarily have to be 0 unlike a though

[limber sierra]

correct

"only the trivial solution" means "only the solution where all the variables are 0"

whereas the "exactly one solution" referred to by part (b) will not be all 0

[but there's a way to make it homogenous so that (a) applies; think about the system it represents]

[nocturne jewel]

It's just asking about FTIM

[rustic panther]

I need to find matrices P and Q so that D = PBQ where D is the Smith normal form.

My smith normal form, D is

1 0 0 0
0 3 0 0
0 0 6 0

I know that P is the transformation matrix

and I've done like 20+ steps to get from B to my SNF

I did rows and column operations to go from B to D

[round coral]

well, is that really the right way?

[rustic panther]

uhm

my D is the right answer so..

[round coral]

but how will you be able to figure out what P is ?

[rustic panther]

that's my problem lol

find the basis of B?

I can't do a Jordan normal form on a non-square matrix. Or can I?

[gray dust]

@rustic panther eigen/jordan decomp is for a square matrix which represents a linear map from a vector space to itself. nonsquares don't represent these and so eigen/jordan decomp doesn't make sense for em

but for nonsquares an analog exists called singular value decomp

[rustic panther]

@gray dust thought so

what's an analog decomop/ singular value decomp

Can someone kindly help me with the transitional matrix to get from D to B?

[wintry sphinx]

you can literally look up singular value decomposition on Google

[wintry steppe]

I have the set A = $B(0_2, 1) \ {0_2} ⊆ R^2 $ where B is a closed ball
and I need to find the int, fr of A
and find if A is a closed or open set
is it correct to say that int(A) = A\fr(a)?
and that fr(A) = {(x_1, x_2 ∈ R^2 | (x_1)^2 + (x_2)^2 = 1 }
and A is a closed set?

[thorny hemlock]

anyone know if a solution manual exists for sheldon axlers Linear algebra book?

[marble lance]

Yeah

In your brain

You just need to transcribe it

[thorny hemlock]

lol

im not that good

[marble lance]

You are

It might just take a while

[thorny hemlock]

i feel like i can go through questions and get them all wrong without knowing

[marble lance]

Seeing solutions won't do anything for you

If you aren't sure about a solution. First: Go through every claim you make and ask yourself how you can justify it. Second: Ask here for us to take a look.

[thorny hemlock]

self learning without solution manual is kinda tough no?

[wintry steppe]

but how can you know if you solved something correctly?

[marble lance]

No, it's how you should learn

[wintry steppe]

for example, I had this problem and I'm not sure if it's correct

although I "solved" it

[marble lance]

Ask

[round coral]

discuss with peers

[thorny hemlock]

(0.5,6,-7/2,0.5)

correct?

[marble lance]

Why aren't you sure about that one?

[thorny hemlock]

idk

i feel like its wrong until i know its right

also, im new to linear algebra

[marble lance]

You can just substitute it in, and see if LHS = RHS, and yes it does, so it's right

I can understand more if you are unsure of proofs, but this you just plug in and see if it's right

[thorny hemlock]

well i wasnt sure if i was actually supposed to do what i did

2(x1,x2,x3,x4)

[marble lance]

What is fr?

[wintry steppe]

frontier

[marble lance]

I honestly don't know what that is, lol

And this isn't a linear algebra question

[wintry steppe]

https://math.stackexchange.com/questions/3005724/find-the-closure-frontier-and-boundary-of-the-following-set?rq=1

then in which channel I should ask?

hm, in topology probably

[marble lance]

I guess that's more appropriate, or analysis. Seems like it's a real analysis question

[marble lance]

You did the right thing, and your answer is fine

[gray dust]

@marble lance eh just closure-interior

[marble lance]

Oh, thanks

[tame mural]

Does an inner product necessarily mean a norm?

And vice versa?

[gray dust]

inner product naturally defines a norm by $\norm{x}=\sqrt{\ip x}$

[stoic python]

RokabeJintarou

[gray dust]

as for the other way, a norm must satisfy the parallelogram law to naturally define an inner product

[tame mural]

Hmmm... thx!

[hallow lodge]

So I've got this problem with its solution. But i don't understand what the ratio between the two vector's components represents:

[limber sierra]

note that "parallel" in the context of linear algebra means that one is a scalar multiple of the other

so $u = kv$ for some scalar $k$

[stoic python]

Chillin' with Quillen

[hallow lodge]

ok

thanks

got it lol

[limber sierra]

and yeah, this computation "behind-the-scenes" is jsut determining what that scalar is

note that vectors "don't care about" position (in a linear algebra context)

so you can imagine all vectors as coming from the origin (0 0 0), hence parallel vectors are "overlapping"

[hallow lodge]

so say this was in R^2 and the components were (4, -3) and (x, -18). would 4/x = -3/-18?

[limber sierra]

right

the point is that the ratios will be the same, since one vector is just a "stretched" version of the other

$\begin{pmatrix}4\-3\end{pmatrix} = k\begin{pmatrix}x\-18\end{pmatrix} = \begin{pmatrix}kx\-18k\end{pmatrix}$, so $4 = kx$ and $-3 = -18k$, hence $4 / x = k = -3 / -18$

[stoic python]

Chillin' with Quillen

[limber sierra]

(the last equation comes from solving both prior equations for k)

so $\frac{4}{x} = \frac{-3}{-18}$

[stoic python]

Chillin' with Quillen

[limber sierra]

and you can solve for x from there since there's only one unknown.

[hallow lodge]

haha i haven't cross multiplied in about 17 years

that brought me back

[wintry sphinx]

you could start by assuming that ku + lv =0 for some nonzero k, l, (u and v are linearly dependent)

Then try applying A to it

[thorny hemlock]

whats this Question asking?

is it just that infinity isnt even part of the Real set?

[hexed mural]

Question, when some sets for a matrix is said to be Det = +1

does this mean the transformation is preserved?

[hollow finch]

wdym by preserved

[hexed mural]

for a square matrix

for example SO(3) for the Orthogonal set

3x3 matrix it says with Determinate = +1

[hollow finch]

right if the determinant is involved its definitely square i just dont know what you mean by 'preserved'

[hexed mural]

So if a determinate is > 1 then the transformation is scaled linearally?

or positive

maybe I am mixing up vocab here

[hollow finch]

the determinant cant be interpreted geometrically as the factor by which things are scaled, with a negative value implying a change in orientation. so det(A)=1 means that area is preserved, but that doesnt necessarily mean its an isometry which might be what youre referring to

[hexed mural]

so what do values >1 mean

if i were to plot the determinate 1-D

[hollow finch]

things get stretched basically

[hexed mural]

Indeed

so more positive more stretch

negative reverse stretch

0 no stretch at all

[hollow finch]

more magnitude means more stretch. if its negative then i guess you could say somethings getting flipped. zero means things are squashed into a smaller dimension.

for 3x3s a determinant of zero means things are squished onto a plane, a line, or just a point at the origin if you want to think geometrically

[hexed mural]

that makes sense

[hollow finch]

if 0<|det(A)|<1 then theres a bit of shrinkage

[hexed mural]

K just wanted to make sure

because I always seem to come back to Det's and looking at them wrt to rotation matricies

side tangnet but nix are you familiar with Quaternions?

[hollow finch]

yea its good to have a good geometric understanding of things in linear, it can help a lot. rotation/reflection matrices are definitely one way to have |det(A)|=1 but theres also shears.

and no not really

[hexed mural]

are you familiar with Rotation matricies?

[hollow finch]

yeah i guess

[hexed mural]

Well I am a bit familar with Quats currently I am trying to map the operations from a Rot Matrix and do the analogous in Quaterion land

So regarding a Rotation Matrix, If I construct a Rotation Matrix from Two Axis Vectors, V1 (Remaining Fixed) V2, being some other axis lets say Z-axis thus Y-Axis being the new computed (Cross product between) X-Y Where Y = X cross Z where would axis of rotation be?

for this to make more sense say I have some vectors A B C in 3d space

A=(1,0,0) Aligned with the X-axis of some 3D coordinate system, and I want to Align that reference frame with some New Vector Call it B, Z=(1,0,0)

A's Z axis is aligned with (0,0,1)

but I want to align with a new Z so as shown (1,0,0)

then my Vector A gets rotated 90 degrees about the Y axis

if I make a Rotation Matrix from a vector along the x axis and rotated to to align to a new Z reference then the cross product would the be the rotation between those two which would be my Y-axis

?

[hollow finch]

that is a very interesting problem. i cant say im familiar enough to be of much help. that said, for 3 space you can obtain the axis of rotation by finding the eigenvector with eigenvalue 1 (so if you have the matrix for that transformation A basically just find the null space of A-I). in four space, however, im fairly confident 1 will not always be an eigenvalue, so im not sure how an axis would be determined. a four dimensional rotation is quite a difficult thing to imagine.

[hexed mural]

So I wrote it in C++, I take two vectors one X-Axis vector in some 3d space and the Z-Axis vector I want to align too

and it works

so for example

say in a video game you have gravity aligned

Z=(0,0,-1)

suppose you want to rotate your character to be upright on another 2D Plane

but you need to rotate gravity itself

to align with the new reference frame

you can use the Rotaion matrix to do this

the other option besides a rotation matrix is to use a quaterion to perform rotations

but to do this you need to construct a quaternion that describes the same axis angle rotation that the Rot Matrix would output

then take your system as a quat

and right multiply it by the quaterion you just constructed that is rotated by that quat to align with the new Z

We can cheat Nix

and create 1:1 mapping to 3d

for the axis of rotation

where we set a quaterions real component to 0

and the vector part to be some vector

hence q = w + vxi+vyj+vzk, pure quat w = 0 q = 0 + 0i + vyj + 0k means a axis of rotation about the j axis which is pitch

anywho thanks tho nix

I will dig deeper

[hollow finch]

sorry i couldnt be of more help. i wish you the best of luck though

[hexed mural]

thanks happy holidays

[hollow finch]

you too 🙂

[acoustic path]

merry christmas

[thorny hemlock]

how to formally prove the forward the statement?

ik its obvious to see it holds

[gray dust]

call the set U. the forward direction is 'if U is a subspace of F^4 then b=0'. we show it by assuming U is a subspace of F^4 then showing b=0

[thorny hemlock]

yeh

and how would you show that b = 0?

[gray dust]

from the definition of U & the assumption U is a subspace

[thorny hemlock]

-_-

[last gazelle]

what does rank tell me about linear independance

[gray dust]

for a matrix A, rank(A) gives the maximum number of linearly independent cols A has

[hallow lodge]

Can someone explain to me how this works?

[limber sierra]

the idea is that we know what T does to (1, 0) and we know what it does to (0, 1)

we dont know what it does to (-1, 5), but we can use linearity to figure it out

[acoustic path]

the basis vectors

[limber sierra]

since T(-1, 5) = T(-1, 0) + T(0, 5) = -T(1, 0) + 5T(0, 1)

again this is just linearity; "linear" means that:

• T(a+b) = T(a) + T(b), and
• T(sa) = s*T(a) [where s is a scalar]

so we just apply these rules to "break up" T(-1, 5) into -T(1, 0) + 5T(0, 1)

[you might've heard something like "linear transformations are determined by how they act on the basis" to describe this idea]

and since we know what T(1, 0) is

and what T(0, 1) is

we can just substitute those in

then simplifying gives us our answer.

[hallow lodge]

thanks, might take me a few mins to grasp it

is there an equivalent kind of procedure done on normal functions I can abstract this from

like, i get it, but i feel like i've never done it before, but suspect i have with functions

[limber sierra]

this doesn't work for general functions since it relies on linearity

[hallow lodge]

ahh

[limber sierra]

in fact, this ability to study linear functions based entirely off what they do to a small defining set of vectors (a "basis") is why we care about linear functions/linear algebra at all

though analogues do come up in other areas, e.g. group homomorphisms are determined by what they do to the generators of that group

[in fact, linear functions can be regarded as homomorphisms of vector spaces - but thats a bit beyond this conversation]

[hallow lodge]

seems like it

[brisk sparrow]

hi is it just me or is this wrong? https://i.imgur.com/561ebyO.png

that linear combination doesnt work right?

[steady fiber]

I don't think it's right either

lol, pretty sure all three additions are wrong

[brisk sparrow]

ya ok im going crazy no one pointed this glaring mistake and its too late to tell my TA

[hollow finch]

if all three are wrong then maybe they were originally different vectors for which it was true, and then the writer changed them but forgot to change that part?

[true goblet]

this example is shockingly wrong 😮

[steady fiber]

they could've changed the vectors but not the coefficients

maybe

[true goblet]

it's actually linearly independent as given

lol

[steady fiber]

ya

if the 2 was swapped to -2, and -1 in (0, 1, -1) was swapped to 1

then it would be linearly dependent

two signs off

[last gazelle]

wait is basis of range the same as basis of column space

[hollow finch]

pretty much. the column space can be thought of as the range of the matrix transformation

[last gazelle]

is basis of the subspace just the linearly independent vectors?

[wintry sphinx]

a basis of a subspace is a list of linearly independent vectors with length equal to the dimension of the subspace

(though tbh that's how "dimension" is defined, so it's fairly useless)

it's a list of vectors that spans the subspace and is linearly independent

[last gazelle]

so i put the vectors into a matrix, do RREF, and then any column with a leading 1 corresponds to a vector in my subspace

[wintry sphinx]

which subspace

[last gazelle]

because any column with a leading 1 is linearly independant

[wintry sphinx]

the row space or the column space?

RREF will give you the columns that are linearly independent, but you have to remember that you have to go back to the original matrix to figure out the actual basis vectors

[last gazelle]

okim confused

theres basis of row space, basis of column space and basis of subspace right?

[wintry sphinx]

the row space and column spaces are subspaces of vector spaces

say you have an mxn matrix; this represents a linear transformation from R^n -> R^m

the row space is a subspace of R^n

and the column space is a subspace of R^m

[last gazelle]

so basis of subspace is something that is found through finding the basis of columnspace and rowspace

or am i just making that up

[round coral]

I think you need to first understand what is the basis of vector space?

could you tell what it is?

[last gazelle]

i mean i know it involves RREF and linearly independat vectors

[wintry sphinx]

no

[last gazelle]

oop

[wintry sphinx]

it involves the concept of linear independence

but "involves" is basically a useless word

and taking an rref might be used to calculate something

but almost all definitions of "basis" do not mention the rref of a matrix

[round coral]

a basis for a vector space, say V is a list of linearly independent vectors that span V

[last gazelle]

well cant i used rref to find which are learly independant

or determinate

[round coral]

in your case in order to find basis of row space and column space, rref is ok , using that you can easily tell which are linearly independent rows and columns respectively

[rocky wolf]

basically if you take the span of the basis then you'll get the whole vector space

[round coral]

but you don't need rref always. It is just one way of doing it.

[gray dust]

all these mentions of span without someone defining it

[last gazelle]

i understand span

because of 3blue1brown

my hero

[round coral]

follow a good book with 3blue1brown as well , his videos are not enough

3blue1brown also uses determinants way much when it isn't needed

[last gazelle]

I just watch his videos for context really

im done with lin algebra after this final exam as far as i can tell

ya but i understand how to do it

just not the concept

[round coral]

well

[last gazelle]

ik im not setting myself up for success

[hollow finch]

True or False

If $A^2+3A+4I_3=0$ for a $3\times 3$ matrix $A$, then $A$ must be invertible.

[stoic python]

nix

[hollow finch]

I believe this is true but I want to prove it either way.

[native rampart]

True

Explicitly construct an inverse

[hollow finch]

aha ill try that

[native rampart]

A^-1 will be ||(A+3I)/(-4)||

[hollow finch]

thank you! ill be able to check my answer when im done 🙂

[hollow finch]

that is indeed what i got. very interesting

would the fact that A is 3x3 imply that the eigenvalues of A are the roots to the quadratic x^2+3x+4, and one of them has an algebraic multiplicity of 2?

[native rampart]

The eigenvalue could just be one of the roots of x^2+3x+4

That is the minimal polynomial is a factor of x^2+3x+4

[hollow finch]

i see. that makes sense

so then any matrix which satisfies a quadratic like that where the coefficient on I is nonzero must be invertible

i said the coefficient on I is nonzero, so that wouldnt apply

[native rampart]

Yea,you could construct an Inverse by "dividing with A" on both sides

However,If The coefficient of I is zero,The matrix could still have an inverse

Take A^2-A=0 Identity satisfies this and has an inverse

Minimal polynomial of T has a root 0 iff T is not invertible

[little frigate]

So apparently in order to find if a matrix is invertible one could just find out if there's a pivot in every column

now may someone explain me why

[hollow finch]

well what matrix do you have when youve row reduced to have a pivot in every column?

[little frigate]

a matrix where rows may be linearly independant and 0 0 0 at the bottom?

uhhh

where you basically have pivots also

[hollow finch]

well if theres a pivot in every column there wont be any rows of zeros since we're talking about square matrices i assume

[little frigate]

yes square

yes you are right

oh

okay I see now thanks

i just realized that another way of thinking about it would be to think about the determinant with echelon form

because like if you are in R^3 and there is a row of zero you're going to get a degenerated parallelogram hm

i think that this formula can be used for finding the determinant and the inverse,

but if im not mistaken and if i recall correctly

there was a scenario where we don't consider the b_ij

i dont know if it's exactly the b_ij but does anyone know what scenario it is

[hollow finch]

i guess it could be the cofactor expansion of a matrix full of 1s

[little frigate]

hmmmmm

i should just check my book rq

[hollow finch]

yeah thats the adjoint

it works for all sizes of square matrices

[little frigate]

nvm im stupid

i just didnt know my formulas oof, 2 days before the final

https://cdn.discordapp.com/attachments/306153640023031820/787923720739356682/unknown.png
https://cdn.discordapp.com/attachments/689948583562444858/787922751486558268/unknown.png
so it's in the calculation of the determinant that you must consider the thing at c_ij
not required for the calculation of an inverse- you just take the inverse of that determinant

gotcha

[hollow finch]

the adjoint is the matrix of cofactors transposed

so for getting the inverse of a 3x3 using the adjoint, you do nine 2x2 determinants and of course the one 3x3 determinant of your original matrix

[little frigate]

hey

if you had let's say a 4x4 matrix and you wanted to find the determinant

instead of doing this cofactor expansion, would it simply be possible to find the "volume" of the shape that would be generated in 4d by multiplying the column vectors magnitude

sounds kind of weird because it's in 4d doe so idk

considering you put it in row reduced matrix obv

or maybe it should actually be row reduced

ive tried it and it doesn't seem to wrong, not sure if i've done something wrong or if my reasoning is just ape

oh no that wouldn't make sense if it would be row reduced

because the "volume" wouldn't be the same because of row operatoins even doe the system itself would be equivalent

[hollow finch]

for 3x3s and up i usually row reduce

you just have to account for the operations

so if you switch two rows then add a minus sign out front. if you divide a row by 3 then put a 3 on the outside

luckily if you just add a multiple of one row to another you dont need to change anything

[little frigate]

oh ok phew

because i was like how he is going to do that with multiple added to rows

makes sense then, thanks

will probably do that for 3x3 and higher

haha i like linear algebras because you can just figure random stuff like that out of nowhere

we never really calculated the determinant in class with row reduce, always cofactors so yep

[hollow finch]

there are an uncountable number of shortcuts and tricks they dont teach

and you can indeed come up with them yourself

[fair vector]

Can anyone help me with this question?

for a part we have to show it is linear so when we put x+y in the left hand side how do we show Qf(x+y)=Qf(x)+Qf(y)

[fallow coral]

can anyone answer a question i have about orthonomal bases

[round coral]

@fair vector what have you done till now?

[fair vector]

For b part i just set the right side =0 and found the kernel to be the space of even functions

[round coral]

that's right

[fair vector]

For a part m honestly stuck at the beginning

[round coral]

do you know the definition of linear map?

[fair vector]

yep

[round coral]

then it must satisfy linearity and homogenity

a simple way I use to satisfy both Q( f + a g) = Q(f) + a Q(g)

f, g belong to V

and a belongs to the field over which V is

[fair vector]

cant it also be Q(bf+ag)

[round coral]

in this one equation you can satisfy both the conditions , it can be anything, just prove it linear using definition and you are done, you wrote the same thing as me anyway

[native rampart]

Go ahead

[fallow coral]

okay

so

I have an orthogonal basis for a subspace of M2x2(R)

the inner product is <A,B>=B^J times A

^J just means switch out the a11 and the a22 values with each other

anyway

so i found an orthogonal basis for this

and 2 of the matrices in the basis

have norm of 0

so i was unsure on how to normalize them

to make the basis into a orthonormal basis

[native rampart]

Are you sure this is an inner product

[fallow coral]

yes

[native rampart]

Then,You can't have a nonzero element having zero norm

[fallow coral]

wait soryr

the inner product is

waitttt

omg

i mis read the question

omlllll

LOL

it was B^t not B^j

smh

anyway ty

[crimson snow]

Whats the difference between isomorphism and homomorphism in simple terms?

[native rampart]

Isomorphism is a bijective homomorphism

[crimson snow]

ok

[thorny hemlock]

i dont understand this example

if we are just summing

what am i missing ?

[stoic python]

Yes

[dusky epoch]

no

U+W consists of all possible sums of every vector in U with every vector in W

[thorny hemlock]

ye

[dusky epoch]

so really what you would've done instead is $$U+W = { (x_1, x_1, y_1, y_1) + (x_2, x_2, x_2, y_2) \mid x_1, x_2, y_1, y_2 \in F }$$

ugh

[thorny hemlock]

$U+W = { (x_1, x_1, y_1, y_1) + (x_2, x_2, x_2, y_2) \mid x_1, x_2, y_1, y_2 \in F }$

[stoic python]

Yes

[dusky epoch]

\{ and \} but whatever

[thorny hemlock]

i see

[stoic python]

Ann

[thorny hemlock]

ok

so in

the

example

he just let

x1 +x2 be x

y1 + x2 be y

?

[dusky epoch]

yes basically

[thorny hemlock]

i see

woulda been easier to see if he subscripted

[dusky epoch]

i mean

this is rough wor

work*

not the final result

[thorny hemlock]

wdym

[thorny hemlock]

whats rough work

[dusky epoch]

scratch work

notes you make for yourself while doing a problem

yknow, shit like that

stuff you don't submit for grading

[thorny hemlock]

yeh

are you saying that his example is scratch work?

im confused on what is better

[dusky epoch]

no

god

ijghj\s

dgghuhufhg

ugh

no

im saying

the x1 and x2 thing that i gave you

is rough work

in an attempt to explain to you why the answer isnt what you wrote

[thorny hemlock]

if its correct why is it rough work

confusion

[dusky epoch]

did i ever say rough work == incorrect work

[thorny hemlock]

well

you did say u dont submit rough work

u usually only submit correct work no?

[dusky epoch]

rough work and incorrect work are both things not submitted but the two are not the same omg

[thorny hemlock]

ok

[thorny hemlock]

what does U intersect W even look like?

if (x,y,0) \in U and (0,0,z) \in W. why is the intersect 0 ?

are we doing the intersection element wise?

why is the result only 1 element and not the same amount of elements as U or W

[dusky epoch]

if (x,y,0) \in U and (0,0,z) \in W. why is the intersect 0 ?
the only vector which can be represented in the form (x,y,0) AND in the form (0,0,z) is the zero vector

are we doing the intersection element wise?
what does that even mean...

intersection is intersection

[thorny hemlock]

i dont know ...

but i understand what you said

i thought about it another way tho

If all vectors start at zero

thats the only point they have in common

[round coral]

0 is there because every subspace, or rather vector space has 0 as a vector

[thorny hemlock]

yeah

[round coral]

so kind of an analogy for two subspaces to have intersection as 0 , would be that two sets are disjoint

though this is not a nice way of explaining that

[thorny hemlock]

yep

for the forward

U + W only has one representation

i dont get the 0 = v + (-v) part

[native rampart]

If a non zero v exists in U int W,0 will have atleast 2 representations

0+0 and v+(-v)

[thorny hemlock]

-v could also be in U tho?

[native rampart]

-v can be seen as an element of W

[wintry steppe]

hi

so I had this question:

Find a∈R so that the list v=[v1,v2,v3]t is a base of R3, where
v1=(a,1,1), v2=(1,a,1), v3=(1,1,a),

I've found a = R \ {-2, 1}

now how can I prove that the vectors span the space?

[fickle citrus]

Did you try proving by definition

Meaning, for anything in R3, there is a linear combination of your list that is equal to that?

And IIRC, the linear combination is unique, there should be one and only one solution for any vector in R3

[round coral]

if you show the list v is linearly independent, you are done

[fickle citrus]

^that's a little indirect

I mean it's true yes but you wouldn't be so sure that's what the marker would accept

And if the marker doesn't accept a proof by definition then nothing would work LOL

[round coral]

I am writing to explain him not writing a solution for my assignment

I won't write like that obviously

[fickle citrus]

I mean, I assume it's a question from homework tbh

[wintry steppe]

I want to say that <v> = {a1v1 + a2v2+ a3v3 | a1, a2, a3 $\in$ R}

[fickle citrus]

If it were a normal question yeah I'd just get the rank

[stoic python]

hivo

[round coral]

to write the correct way is his job, we are just to guide him

[wintry steppe]

it is, yes:))

[wintry steppe]

because this would be the definition, no?

[fickle citrus]

Ya

The 'span the space part' is 'there exists a1,a2,a3'

For any vector $v$

[wintry steppe]

so I need to show that <v> is included in a1v1 + a2v2+ a3v3 | a1, a2, a3 $\in$ R

[stoic python]

ShatteredSunlight

hivo

[wintry steppe]

and a1v1 + a2v2+ a3v3 | a1, a2, a3 $\in$ R is included in <v>

[stoic python]

hivo

[mint dagger]

The longest side of a triangle is 11 meters longer than the shortest side. The medium side is 15 meters long. The perimeter of the triangle is 46 meters. Find the length of the shortest side of the triangle.

[mint dagger]

Can anybody help me with that?

[round coral]

sorry, this is the wrong channel for this question

@mint dagger

[mint dagger]

What channel?

[native rampart]

#geometry-and-trigonometry

[mint dagger]

Oki thx

[wintry steppe]

actually to show that v is a basis of R^3, I should say that <v> spans R^3.

then <v> = {a1v1 + a2v2 + a3v3 | a1, a2, a3 ∈ R}

and that <v> ⊆ R^3, which is obvious

and that R^3 ⊆ <v>

[fickle citrus]

_> nothing is obvious

[wintry steppe]

😦

can you give me a hint how to prove that <v> ⊆ R^3?

[fickle citrus]

I mean, you're kinda saying the definition - the 'proof' is somewhat not there

Do you know what's a determininant

[wintry steppe]

yes

[fickle citrus]

So the 'there exists a1 a2 a3, for any v' forms a system of linear equations

Then the statement becomes is there a solution for any unknown v

Then the statement becomes, is the determinant of the coefficient matrix of the system of linear equations 0

then you show it's not 0, then all the iff statements connect and you get your 'vectors span space' proven

^ if anyone has a better proof please do share

[wintry steppe]

don't we have an infinity of linear equations?

[fickle citrus]

No you have 3 equations

For any vector in R^3

I am saying, for any $v\in\mathbb{R}^{3}$ with components $v_{1},v_{2},v_{3}$, that
$$\begin{pmatrix}v_{1}\v_{2}\v_{3}\end{pmatrix}=\alpha_{1}\begin{pmatrix}a\1\1\end{pmatrix}+\alpha_{2}\begin{pmatrix}1\a\1\end{pmatrix}+\alpha_{3}\begin{pmatrix}1\1\a\end{pmatrix}$$
for some $a$ you determine, where $\alpha_{1}$, $\alpha_{2}$ and $\alpha_{3}$ depend on $v$ itself.

[stoic python]

ShatteredSunlight

[fickle citrus]

Wait actually

Those vectors shouldn't span R^3

[wintry steppe]

isn't this similar to what I did here?

[fickle citrus]

Oh wait nvm, it is spanning for those a you found, yup

[fickle citrus]

Yup seems simiilar

[wintry steppe]

I wanted to prove that there isn't any other a outside {-2, 1} for which the system is linear

[fickle citrus]

It's the same

It shows the same thing

[wintry steppe]

okay, thank you:D

So what I wrote above proves that <v> spans R^3?

A "basis" for a vector space of dimension n is a set of vectors that has three properties:

The vectors span the space.
The vectors are independent.
There are n vectors in the set.

And they are equivaent

So I guess yes

:-?

[fickle citrus]

I'd say it's the necessary working

I mean 😦 it kinda implies that it is not the case there is not a solution rather than there is a solution, which is kind of different you know?

[fickle citrus]

Actually honestly yeah I think you should write further "since the determinant ..." "hence ..." it's a few more lines of English and you already have the working so why not

[wintry steppe]

yeah, I know. I'm not sure how to prove that there is a solution

[wintry steppe]

yup, I will!

[fickle citrus]

Existence of solution is tied to the determinant

It ties in to invertibility

I mean, it's why invertibility of matrices has a huge "The following are equivalent"

https://en.wikipedia.org/wiki/Invertible_matrix#The_invertible_matrix_theorem

[wintry steppe]

thank you. I didn't know about this one

[wintry steppe]

hmm, then there should be a solution for any b in R {-2, 1} because the determinant is not equal with 0

[fickle citrus]

Yup

I guess you mean for any a not equal to -2, or 1 - and then for any vector in R^3

[wintry steppe]

couldn't I take a vector y in R^3 (with y1, y2, y3) and try to write it like this

[wintry steppe]

yup

[wintry steppe]

to see if for any a in b in R \ {-2, 1} v generates R^3

[wintry steppe]

Hi:)) I have one more question. I'm solving this problem

First I proved that b is a basis by calculating the determinant (which is always equal with -18, thus it is always linearly independent)

and now I want to find the coordinates of x in relation to the basis

should I take each vector one by one?

I said that [x]B = [x,y,z,t]

and x*[1,2,-1,2] + y[1,2,1,4] + z[2,3,0,-1] + t[1,3,-1,0] = [2,3,2,10]

I wrote it as a system

then solved it using the inverse of a matrix

so the coordinates are [1,2,0,-1]

is this right?

[surreal thistle]

yo if i have two linear transformations that map R3 to R3 and they both have rank 2

what are the possible values of the ranks be if i compose them

[frosty orbit]

Have no idea what to do

[dusky epoch]

do you know the defn of linear independence

[frosty orbit]

I don't know how to write it

When the vectors are in this form

Just a1f1 + a2f2 + a3*f3= 0?

[dusky epoch]

remember what the zero vector of C^2[0,pi] is, & what it means for two functions to be equal

but yes the form isnt any different

[tall thunder]

Is anyone able to explain this like I’m 5?

[stoic python]

Nyrre

[tall thunder]

Ok so I do that formula

Like lambda + 11 and then all the way across diagonally

Right ?

Then just get the det of that

But what about the invertivle matrix part

[round coral]

you need to diagonalize the matrix A, if possible. So first find the eigenvalues, you can either use gaussian elimination to find the characteristic polynomial or just use the determinant

[tall thunder]

Ok so I got thus far

Now what haha

Determinant ?

[tawny tulip]

@tall thunder

[round coral]

look it up online how to do diagonalization, we can't help you like baby. First learn how to find eigenvalues and eigenvectors for an operator

[stoic python]

Nyrre

[round coral]

first study then come here

[tall thunder]

@tawny tulip is your negatives mixed up ? For the example you showed

Or was that like a thing your supposed to do

Nvm I got it 😇☠️ thanks

[hollow finch]

maybe not for 5 year olds but if youre fairly comfortable with the column perspective of matrix multiplication:
multiplying the eigenvectors of A by A is just scaling them. so if you put the eigenvectors as the columns of a matrix P then AP is just P with its columns scaled by each eigenvectors' eigenvalues. but that can also be achieved by multiplying P on the right by a diagonal matrix with the eigenvalues D. so AP=PD. from there you can solve for A or D by multiplying P^-1.

[tall thunder]

Hey nix, thanks for explaining I really appreciate it

[surreal thistle]

yo can someone help me out with this

i dont really understand how you can just multiply a vector b that number

[wintry steppe]

any vector takes on the form (x, y), and you identify that with the complex number x + iy

now multiply complex numbers

when you get the result in the form a + ib, what vector does it correspond to?

[surreal thistle]

0

+2i?

idk bruh

[wintry steppe]

okay, let me write it out in a bit more detail

When it says "Consider $\bC$ as a two-dimensional real vector space with basis ${1,i}$," it means that our vectors are literally complex numbers, of the form $a + ib$ for real numbers $a, b \in \bR$. If you're used to writing vectors as tuples, then you can write $(a, b)$, and think of this as the complex number $a + ib$. Now, we want to figure out what multiplication by $1+i$ means. Given any vector $(a, b)$, where $a$ and $b$ are real numbers, the multiplication is given by $$ (1 + i)(a + ib). $$ Now write this out.

[stoic python]

TTerra

[wintry steppe]

you're literally just doing complex number multiplication lol

just the "consider C as a two dimensional real vector space with basis {1, i}" part may be a bit confusing

[split heart]

Hi, if I have a f:R3 -> R2, with canonical basis in domain and domain, and a matrix A, what process should I follow? Sorry by mistakes

It's about linear transformations.

[wintry steppe]

what are you trying to do

[split heart]

find the transformation rule

or law it's called I think.

[surreal thistle]

ok i understood that part but I kinda just have trouble creating a matrix from that information

[gray dust]

we want to find f(x,y,z)=...

[split heart]

Yes

[surreal thistle]

this the right idea?

i put the left vector on the wrong side by mistake i think

[kindred locust]

^I'm also confused about this problem 😖

[safe root]

Does slope count as algebra?

[wintry steppe]

Not this algebra

[limber sierra]

see #prealg-and-algebra

[thorny hemlock]

How do i show that its closed under addition?

[quartz compass]

take two arbitrary elements and show that their sum obeys the identity

[thorny hemlock]

ok

what would two arbitrary elements from this Q be?

[quartz compass]

call it h(x) = f(x)+g(x)

f and g satisfy that identity

[thorny hemlock]

ok

[quartz compass]

just try something, play around a bit and see if you can get it

[thorny hemlock]

when it says R^(-4,4)

thats the openinterval?

R^0.5 exists?

[quartz compass]

R^(-4,4) is the set of real valued functions on (-4,4), don't think this is like R^n it's a separate notation

[thorny hemlock]

ohhhh

ive got it, thanks

[tame mural]

Axler's talks about a product of vector spaces, but it seems hard to search up

Is that just his own thing?

[gray dust]

@tame mural it's just cartesian products

[tame mural]

I see

[thorny hemlock]

for the forward, under the assumption that its closed under addition. Taking another real valued continuous function on interval [0,1] g $\int^1_0f + g = 2b$

[stoic python]

Yes

[thorny hemlock]

b = 2b and thus b = 0

does this work?

[hoary osprey]

yes

[thorny hemlock]

just another subspace of V ?

[limber sierra]

another?

recall that $U + U$ is the subspace composed of all $u_1 + u_2$ for $u_1, u_2 \in U$

[stoic python]

Chillin' with Quillen

[limber sierra]

but also recall that $U$ is closed under addition, since it's a subspace

[stoic python]

Chillin' with Quillen

[limber sierra]

what does this tell you?

[thorny hemlock]

oh

the same subspace

[limber sierra]

right, $U+U = U$

[stoic python]

Chillin' with Quillen

[tame mural]

What is the difference between isomorphism and equailty?

[limber sierra]

equality says two things are literally the same, isomorphism says that they behave the same/can be considered the same

so theyre very related notions

[tame mural]

I see, so I'm not going to go wrong by thinking of them as the same

[limber sierra]

but with a slight semantic difference

[tame mural]

I see, so linear isomorphism = equal as far as linear algebra is considered?

[limber sierra]

again its not QUITE the same thing

but thats how you can think of it

[tame mural]

I think I see, because R^2 has slightly different properties than C

thx!

[thorny hemlock]

when it asks which subspaces have additive inverses

isnt it none

[tame mural]

I have a feeling this is for a test, but

Shouldn't all subspaces always have inverses and identity?

[thorny hemlock]

no, its from the axler book

lol

yay

yep

exactly my thoughts, just needed to confirm

yeah

i see

[thorny hemlock]

How is something like this done?

Working backwards for multiple linear equations?

[hoary osprey]

you want to find a sub space that intersects trivially with U and behaves in a way such that V=U+W

so try out a couple of options

[thorny hemlock]

my initial thought its (0,0,-y,y,0) but that obviously doesnt work

[acoustic path]

W=0 vector 🕶️

[thorny hemlock]

W = {0} doesnt work

[acoustic path]

cap

[thorny hemlock]

something like{1,1,1,1,0.6} isnt in F^5 if W = {0}

...

[acoustic path]

if W = {0} we just get U which is in F^5

[thorny hemlock]

dont we need the union to equal F^5 ... not subspace it

@acoustic path

[acoustic path]

yea im not sure lol

[thorny hemlock]

:/

<@&286206848099549185>

[split heart]

How can I develop this binomial using newton’s binomial theorem?

I'm confused about -1/2

[hoary osprey]

do you have any guesses @thorny hemlock

not the right channel @split heart

[split heart]

Isn't linear algebra?

[hoary osprey]

no

try #precalculus

or any of the question channels

[split heart]

Right, ty

[thorny hemlock]

nope

like

i dont see it

[hoary osprey]

try (t,2t,w,2w)

[thorny hemlock]

ok

that works

thanks

ill try some more

[slender tangle]

Ok I need a quick verification because this is driving me nuts. I have a 6x4 matrix which represents a homogeneous linear system

When is it possible for this system to have non-trivial solutions?

I thought I had it didn't work

[hollow finch]

there are nontrivial solutions to the homogeneous system when the nullity of a matrix is nonzero. since nullity=# of columns - rank, the matrix would need a rank less than 4

[slender tangle]

Hmm ok so far so good, is there a way of finding the rank in general as a function of the elements in my matrix?

I vaguely remember looking at the determinants of 4x4 submatrices

[hollow finch]

hm that definitely wont always work

[limber sierra]

"as a function of the elements"

does "perform gaussian elimination" count?

since thats the general algorithm

[hollow finch]

the normal way to compute rank is to get it into ref or rref and count the pivots

[slender tangle]

I was hoping to get a polynomial out of this

[hollow finch]

is there something different about this matrix that would lead you to think something like that is possible?

[slender tangle]

About half the elements are 0

[hollow finch]

i see

[slender tangle]

I'll make a quick picture

[hollow finch]

yea was just about to ask

[slender tangle]

There are 4 complex variables arranged in this matrix

[hollow finch]

right i see what you mean now

[slender tangle]

I guess I can do a Gauss elimination but I'd hoped there was an "cleaner" way to solve it

The annoying part is, I know that it has non trivial as soons as a single variable is 0

So the only remaining problem is all 4 being non zero

[hollow finch]

elimination is very tedious with variable entries yeah

well the rank is also the number of linearly independent columns, right?

[slender tangle]

I do think so

My linear algebra is quite rusty

[hollow finch]

yeah so i think id try examining the conditions for one of the columns to be linearly dependent. then it would have nullity of at least 1.

since there are so many zeros this should be pretty feasible

[slender tangle]

I could of course cheese it by putting it in a calculator but that would be the easy way out

Them again my deadline is in 18 hours and I still have much to write

Hmm wait this is interesting

"the rank of the matrix is strictly smaller than n-d it and only if all n-d square minors vanish"

With n being the number of columns

[hollow finch]

what is d?

[slender tangle]

In this case the dimension of a variety but it's arbitrary in the general case

I think

But I fear this will just lead me into a circle for my specific problem

[prisma bough]

hey uh sorry to interrupt

but i got a dumb question

is this statement true?

[slender tangle]

Project matrix is a diagonal matrix of full rank? As in no 0's on the diagonal?

[prisma bough]

hm.. we defined a projection matrix as "if it corresponds to a projection onto a subspace"

or if a linear transformation P satisfies P^2 = P

[slender tangle]

What you've written down is basically "Any square matrix can be Gauss Eliminated"

[prisma bough]

for context, this is a question that I am given a possibly correct/incorrect statement, and I have to explain if it's true or false

[slender tangle]

Are you working over the real numbers?

[prisma bough]

yeah

[slender tangle]

I'm not sure about that P^2 = P part

Because indeed, any matrix can be gauss eliminated (those elementary matrices) into a matrix with only elements on the diagonal

But I don't think that it's possible to have that P^2 = P property for reals

[prisma bough]

oh

[slender tangle]

Though you might wanna wait for a second opinion on that

[prisma bough]

yeah, but thanks anyway

[slender tangle]

*I don't thinks it's always possible

[hallow lodge]

anyone want these? Professor gave these at the start of the semester and said good luck on your exams.

[acoustic path]

yes please

can i have the exams

[wintry steppe]

are all unitary operators hermititian. or is it the other way round?

[native rampart]

I don't think it's either

[raw magnet]

so to show this i have to make sure that

1. f is one to one and onto
2. it preserves the property of linear transformation
   but how do i do it if i dont have an explicit definition of the transformation?

[hollow finch]

you can just make one. thats what its asking you to do

i wouldnt think too hard about it. keep it simple

[raw magnet]

so let say, i make v be p2(c) and w be C_3, then the transformation is just T(a_0 + a_1 X +a_2 X^2 ) = (a_1,a_2,a_3) ?

[raw magnet]

then from that i show these

[hollow finch]

your index is off but i like the idea

[native rampart]

Well, Check the numbering,but yea it works

[raw magnet]

Oh thanks both, notice that 🙂

[raw magnet]

P_2(C) over C is isomorphic to C_3, would P_2(C) over R be?

is it no, because p_2(C) over R has dim 6 while C_3 only has dim 3?

[native rampart]

Yea,It won't be

[raw magnet]

Is this a good reasoning?

[native rampart]

Yes

2 vector spaces of different dimensions cannot be isomorphic

[raw magnet]

Thanks

[native rampart]

Also,2 vector spaces of same dimension need not be isomorphic(when the corresponding fields are different)

For example V over F_2 is not isomorphic to V over F_3

[stoic python]

xD_Lopez

[arctic karma]

when getting the bases of row and column space of a matrix, do i use rows and columns of the original matrix?

the row space should be R1 and R2 since they're non-zero, and column space should be C1 and C4 since they have the leading entries

[native rampart]

You can use the final thing as well

Span{original rows}=span{rows obtained in rref}

[stoic python]

xD_Lopez

[native rampart]

Yes

[arctic karma]

thank you!

[native rampart]

The row space wouldn't be span{(1,-3...) ,(3,-9...)} though. you also need to include the other 2 rows

[arctic karma]

but if it's asking for the basis then it would only need the first two rows right

[native rampart]

Span of those 2 is not row space

So they don't form a basis

[thorny hemlock]

Not sure i understand that last part

I get that the span contains each individual vector

I dont get how he deduced that its the smallest subspace of V

[thorny hemlock]

<@&286206848099549185>

[limber sierra]

do you see why, if a vector space contains each v_j, then it must contain span(v_1, ... v_m)?

[thorny hemlock]

no

[limber sierra]

well span(v_1, ... v_m) is the set of all linear combinations of vectors from {v_1, ... v_m}

[thorny hemlock]

yes

[limber sierra]

and if your space contains {v_1, ... v_m}

as well as maybe some other vectors

we can certainly take linear combinations of {v_1, ... v_m}

even if we entirely ignore the other vectors in that space

hence we have all linear combinations, i.e. we have span(v_1, ... v_m)

so ${v_1, \dots, v_m} \in V$ implies $\mathrm{span}(v_1, \dots, v_m) \subseteq V$

[stoic python]

Chillin' with Quillen

[limber sierra]

that is to say, any vector space containing {v_1, ... v_m} must contain the span of these vectors, and hence be "at least as large"

so the smallest is the one that contains only the span

(i.e. when that subset sign is an = sign)

[thorny hemlock]

i understood most of what you said

[limber sierra]

basically:

vector space contains {v_1, ... v_m}

vector spaces are closed under + and scalar *

which means we can take any linear combinations of this set

[thorny hemlock]

ye

[limber sierra]

and they'll still be in the space

[thorny hemlock]

ye

[limber sierra]

hence span(v_1, ... v_m) is in the space

[thorny hemlock]

ye

[limber sierra]

so if V is our vector space, span(v_1, ... v_m) is a subset of V

[thorny hemlock]

ye

still thinking about the smallest part tho

[limber sierra]

since span(v_1, ... v_m) will always be a subset, the smallest possible vector space is span(v_1, ... v_m) itself

uh okay forget the linear algebra, maybe we just need to think about sets then

suppose i have the set {1, 5, 6}

[thorny hemlock]

ok

[limber sierra]

what is the smallest set with {1, 5, 6} as a subset?

[thorny hemlock]

{1,5,6}

[limber sierra]

right

replace {1, 5, 6} with span(v_1, ... v_m)

and the conclusion follows*

*well, we also need to make sure that span(v_1, ... v_m) is actually a vector space [that is, a subspace], but that's what the first half of the proof did

[thorny hemlock]

ok

so

i think i understand

if we have the vector (1,1,1)

the theorem states that a(1,1,1) is the smallest subspace of V that contains (1,1,1)

[limber sierra]

not 100% sure what the notation a(1, 1, 1) means

[thorny hemlock]

a is a scalar

[limber sierra]

oh

the set {a(1, 1, 1) | a is a scalar} is the smallest subspace containing (1, 1, 1) yes

[thorny hemlock]

that visually doesnt make sense to me

{a(1,1,1) | a is odd scalar} is a smaller subspace containing (1,1,1) ?

[limber sierra]

that's not a subspace

[thorny hemlock]

ohhhh

[limber sierra]

subspaces should be closed under scalar multiplication

[thorny hemlock]

i see

yes

[limber sierra]

for three-dimensional real vectors, you can visualize subspaces as lines (if one-dimensional), planes (if two-dimensional), or the entire space (if three-dimensional)

{a(1, 1, 1) | a is a scalar} is then one long line that you get if you "stretch out" the vector (1, 1, 1) [centered at the origin] infinitely in both directions

[thorny hemlock]

i understand

[limber sierra]

and this line is certainly the smallest subspace with (1, 1, 1) inside it