linear-algebra

[subtle walrus]

(in coordinates of B)

the rest is showing this is really a basis

that P^{-1} acts that way is just computed

(or just say its obvious)

but tbh i think this result can be represented a lot nicer

[elfin ingot]

okay i think i got it

[subtle walrus]

i wonder how axler does this

[elfin ingot]

this is like WAAAAY worse than df imo

this is just chap 2 XD XD

[subtle walrus]

oh right, he doesn't

[elfin ingot]

okay so

the theorem jsut says

that you can do this..

that you can change coorddinates by changing the basis

and there is a matrix that does that for you

[subtle walrus]

yeah

[elfin ingot]

okayy

are all theorems this hard?

[subtle walrus]

i mean the book is written more "mathematically"

in the sense it presents important theorems

that are needed to actually do stuff

[elfin ingot]

i can do exercises tho 😄

[subtle walrus]

like, you need this theorem to justify that you can actually choose a basis

[elfin ingot]

yea

so its basically back end s tuff

[subtle walrus]

and all your computations are also valid, if your friend uses another basis

[elfin ingot]

thats his proving

[subtle walrus]

ye

[elfin ingot]

bad question

is this book considered hard

the exercsises arent hard ik that

but the writing itself the proofs especially

are all just weird

[subtle walrus]

its considered complete

its also rather old

[elfin ingot]

yea thats why ig

[subtle walrus]

like

i would not assign this a student to read

i think a lot of stuff can presented more

pedagogically

but tbf i never read the whole book lol

[elfin ingot]

oh ur a teacher?

cool okay

[subtle walrus]

nah, speaking more hypothetically

i know i couldn't have learned from that book

i mean i do TA work, but i rarely make decisions on what material is covered and never on how

[elfin ingot]

okay can u help me with this problem?

let a= (x_1,x_2) and b = (y_1,y_2) such that x_1y_1 +x_2y_2 =0 and (x_1)^2 +(x_2)^2 = (y_1)^2+(y_2)^2 = 1

show that B = {a,b} is a basis for R^2

so i am trying to show that any vector in R^2 can be representred as a linear combination of a and b

and a and b are linearly independant

so in my like draft stuff outside the proof

i supposed a(x_1,x_2) +b(y_1,y_2) = (t,d) for some vector (t,d) in R^2

and now i am trying to find a and b

right approach?>

for a and b in R

[subtle walrus]

if you show linear independence, you would be already done

[elfin ingot]

why?>

shouldnt i show it spans

the space

also

[subtle walrus]

yes, but 2 linearily independent vectors in R^2 will always span R^2

[elfin ingot]

why lol

never knew that

[subtle walrus]

you can extend every linearily independent set to a basis

so if they did not already span R^2, you could extend them to a basis, that would have more than 2 elements, which is absurd

[elfin ingot]

you can extend every linearily independent set to a basis
@subtle walrus tht was a theorem wait

i took that but just didnt check the proof lol

[subtle walrus]

its a really important one

basis extension theorem or something

[elfin ingot]

wait no its not

its not written

for me

i thikn the only ones i know are like if S is a set that spans V then any linearly independant subset doesnt have more elements than S

[subtle walrus]

i mean if you don't know that, your approach would be the right one

[elfin ingot]

yea

but now how i can solve this system?

or wait

i will try

[subtle walrus]

i temporarily forgot the name, but it's a theorem due to steinitz

[elfin ingot]

yea i doint htink its written for me

wait?

steintz

not the chess guy right

nah nvm its written forme

if W is a subspace of a finite-dimensional vector space V then every linearly indpendant subset of W is finite and a part of the basis of V

?

[subtle walrus]

its not that, but it might work anyway

you would need that every linearily independent set spans a vectorspace

(which is obvious)

[elfin ingot]

yea

[subtle walrus]

and then that the only subspace of a vectorspace that has the same dimension, is the vectorspace itself

for finite vectorspaces

[elfin ingot]

yea i got it now

thats better

[subtle walrus]

actually, its corollary 2 of theorem 5

[elfin ingot]

uh oh

okay the exercises for the coordinatese section arent hard

i thikn i get it

but there are too manyu theorems

and i get confused

[subtle walrus]

🤷

[elfin ingot]

tysm for the help

i htink i got it the section now

[mellow oar]

hey, so umm, I dont really understand subspaces to well

could someone help me out with this

[wintry steppe]

define subspace

just write out the definition and check it

[mellow oar]

ok but like

idk what the first question means

"W is a subset of what cartesian space?"

[limber sierra]

yeah, that's what i assume it's supposed to mean

it's not worded well

[mellow oar]

yeah lol

[mellow oar]

oof

[wintry steppe]

The question is: For what values of b the vectors (-6, b, 2) and (b, b^2, b) are orthogonal?

I solved it with my calculator doing solve(dotp([-6, b, 2], [b, b^2, b]) = 0, b)

Is there a way to do this manually or ?

[cursive narwhal]

Use the definition of orthogonality. The inner product of the two vectors must be 0.

[subtle walrus]

just plug it into your scalarproduct

sniped i guess

[cursive narwhal]

So, -6b+b^3+2b = 0. Can you solve that analytically?

Lul

[wintry steppe]

Oh interesting... thanks for showing !

[cursive narwhal]

I don't know what's so interesting about it, it is literally the definition of orthogonality.

[subtle walrus]

depending on what scalar product you use, maybe its interesting

[wintry steppe]

Idk, I started this stuff Monday lol

[cursive narwhal]

Fair

[subtle walrus]

im more surprised at calculators with dotp built in

[wintry steppe]

ti nspire cx cas

[subtle walrus]

engineers have nice tools

[cursive narwhal]

Lol i remember a time when calculators were relevant

[subtle walrus]

they never were

you have been tricked

[cursive narwhal]

Rekt

IB pranked me for 2 years

[elfin ingot]

can i say that

(1,0,0) (1,1,0) (1,1,1) must be a basis since they are indpenedant

cuz ift hey are not

then its missing atleast one lemeent to span

and if it does then the dimension would be greater ?

than the original space which is absurd ?

@subtle walrus

[cursive narwhal]

Sure.

[elfin ingot]

tysm

[cursive narwhal]

Also, linearly independent

Not independent

I saw that

Independent vector who don't need no man - Lochverstarker 2k20

[elfin ingot]

hahaah

[wintry steppe]

If i = (1, 0 , 0), j = (0, 1, 0) and k = (0, 0, 1)
I need to find a unit vector that's orthogonal to both i + k and i + j

I found i + j = (1, 1, 0) and i + k = (1, 0, 1)

But since the zero is in a different in both vectors, I'm not sure how I can get a 0 vector

[gray dust]

@cursive narwhal accidentally correct grammar. it's don't need no man

[cursive narwhal]

Oh shittt copied it wrongly

[gray dust]

you disappoint me

[cursive narwhal]

I'm sorry buddy. I'll let your mom punish me tonight

Prank

Relax

Memes

Chill

[wintry steppe]

lolll

[cursive narwhal]

❤️

Oooh lala

If you start talking like that, then what will the kids say? Gotta keep the kids away from adult content, you know what I mean?

[gray dust]

Gotta keep the kids away from adult content
no use in trying, you're already corrupted

how dare you thonk me

@wintry steppe you're lucky these things live in R^3. the cross product is useful here since by definition it produces a vector orthogonal to the two input vectors

[elfin ingot]

help

with b

[wintry steppe]

@wintry steppe you're lucky these things live in R^3. the cross product is useful here since by definition it produces a vector orthogonal to the two input vectors
@gray dust Hmm, sorry, I'm not sure how to make use of that info 😂

Oh so the cross product of the two vectors will give a orthogonal vector

Hm, if I multiply the 2 vectors I get [1, -1, -1] which has a length of sqrt(3)

[subtle walrus]

you will have to use the fact that $\mathrm{e}^{\mathrm{i}x} = \cos(x) + \mathrm{i}\sin(x)$ @elfin ingot

[stoic python]

Lochverstärker:

[elfin ingot]

FUCK okay

ty i will try

[gray dust]

@wintry steppe now make it a unit vector

[river jasper]

can someone tell me how to identify an abnormal matrix?

[subtle walrus]

what's that?

[wintry steppe]

is that the same thing as an irregular matrix?

[river jasper]

a matrix that is not normal

[slow scroll]

well most matrices are not normal, so ig that makes most matrices abnormal

[wintry steppe]

be more specific @river jasper

[river jasper]

oh nevermind

Is the spectrum of a matrix the highest absolute power eigenvalue?

[wintry steppe]

In mathematics, the spectrum of a matrix is the set of its eigenvalues. More generally, if is a linear operator over any finite-dimensional vector space, its spectrum is the set of scalars such that. is not invertible. The determinant of the matrix equals the product of its eigenvalues.

[pallid rampart]

...

Anyone can google it

Don't just send wiki definition

[wintry steppe]

The spectral radius of a square matrix is the largest absolute value of its eigenvalues

@river jasper is this what u meant

[dreamy iron]

So say I have a linear combination $\sum_{i=1}^{n}\left(\lambda \vec{x}_i\right)$, but all the lambdas are one. Is there a special name for a linear combo with scalars all being one?

[stoic python]

ninnymonger:

[limber sierra]

i'm not aware of a "special name" besides just "the sum of all x_i"

its not really a super meaningful notion if you're talking about, say, a basis

since bases "behave the same" up to multiples of their constituent vectors

but this of course affects the sum of the vectors

[dreamy iron]

i was thinking more like "oh, these are your everyday normal expenses, assuming nothing extra, but if there are added losses at this particular value or that particular value, than the scalar for those variables will go up by an amount greater than 1"

so like $\lambda_i > 1 $ means there's an "inefficiency" related to this particular value, the steps around $x_i$

[stoic python]

ninnymonger:

[alpine estuary]

Can somebody help me with this

[alpine estuary]

Anyone ._.

[wintry steppe]

this is not linear algebra lol

#precalculus

[sick dragon]

Is this true?

[wintry steppe]

i don't think so but i'm not an expert

[sick dragon]

Seems true when I do it geometrically

🤷

[latent marten]

how to workout the general solution

i have reduced it to rref

[dreamy iron]

i have reduced it to rref
@latent marten https://docs.google.com/spreadsheets/d/1nDJHiQGB7mB-IYoDcfPfiwg7Y7CqWjkp2Ywlaei0RE4/edit

[wintry steppe]

Put in matrix form

@latent marten

[latent marten]

i have done taht

[wintry steppe]

Ok

[latent marten]

but i get 0 on bottom 2 row

and x-z-2w = -1 for first

y + 2z +3w = 3

and idk if that is a general solution of the system

is that correct for my q

[half ice]

Yes

[elfin ingot]

why do these vectors form a basis

id get why on their own ig by showing this and that

but he used 'thus'

what ebfore that implied that its a basis

[half ice]

Because the matrix is invertible

[elfin ingot]

?

[half ice]

The column space has to be surjective. That is, it spans R³ or whatever the output space feels like it wants to be

[elfin ingot]

what does invertible mean

[half ice]

A matrix M is invertible if it has an inverse M^(-1)

[elfin ingot]

yea okay why does this imply basis

[half ice]

Let's use the "linear transformation" interpretation of a matrix haha. So P represents a linear transformation that has an inverse which means it's bijective, which means it's surjective

[elfin ingot]

thats next section

and the book says that the rest of the book uses this more

so im done with this boring ass shit

so i just wanna get this please :p

[half ice]

You know what surjective means! The column space covers all of R³ or whatever the output is

[elfin ingot]

yea

i get that btu thats cheating

what does the author want me to know

[half ice]

What's the book again?

[elfin ingot]

hoffman kunze

im sorry if im being annoying with u helping me but

tbh this shit is abit hard than expected

[half ice]

Imma download

[elfin ingot]

damn im such a nuisance

u dont have to bother im jusut going to try it on my own bro

[sick dragon]

Which does not belong and why?

[half ice]

@elfin ingot
Waht page?

[elfin ingot]

... coordinates section

last page

63

[half ice]

Lol breakfastbattle is asking the exact same question

[elfin ingot]

i think it should be a and d

cuz

A is invertible ---> basis ---> spans ---> lin indep

( dk why )

idk what the fuck those mean for a and d

but just elimination ig

@half ice am i right XD?

[half ice]

I think you mean e when you say d

[elfin ingot]

yea

lmao

xD

the systems ones

[half ice]

Note that x = 0 is always a solution to Ax = 0

[sick dragon]

That's right

So the other four are "equivalent"?

Is this saying that all four of these must be true for this particular n by n matrix

[half ice]

For any function ever:
Invertible
⇔ Bijective
→ Surjective
So the column space spans the codomain

[elfin ingot]

no breakfastbatle i think it means more

that every thing implies the other

[half ice]

Then, the dimension of the output is the same as the input, which means the output has to be a basis. Therefore linearly independent

[elfin ingot]

yea cool

okay so i want to recap abit stuff

[half ice]

Wait mo I looked I have a good solution for you

[elfin ingot]

a matrix has columns and rows

okay sur

[half ice]

At this point in your book, "invertible" means "is a product of elementary matricies"

[elfin ingot]

what are elemenatry matrices?

1s and 0s in entries?

[half ice]

They represent the identity matrix with a single "row reduction" applied to it.

You hate this part though haha

[elfin ingot]

okay so

okay now i know what invertrible means

what now

[half ice]

For example
[1 1]
[0 1]
Is elementary because I took the identity and added the rows together

[elfin ingot]

yeaa okay so

elementary just means am atrix that can be reached

from I and row operations

okay cool

[half ice]

Oh fk wait I'm not positive where I'm going with this

[elfin ingot]

oh my god i just wanna get fucking done with this shitty chap

THATS IT

now i get to linear transformatiosn which are easy right?

[sick dragon]

literally just rows and col swapped

[half ice]

@elfin ingot
Let u,v,w be a basis on the domain. Then Au, Av, Aw is linearly independent in the codomain.

Let's say they weren't. Then you could make a dependency out of them, and just take A^(-1) to show that u,v,w are not linearly independent, a contradiction.

[elfin ingot]

yea i get the proof

okaay

first linear algebra proof i actually get

and understand lol

[half ice]

I genuinely can't think of a way to not say "if Ax is an invertible mapping, then it must be a surjective mapping"

They must use that idea, but hide it haha

[elfin ingot]

yea i am going to use that ig

okay

tysm i get it now

[half ice]

Cool cool, feel free to ask if you have anything else!

[elfin ingot]

tysm

[slow scroll]

you need injectivity to conclude that the columns form a basis too

[elfin ingot]

why is everything on the columns

what about the rwos

[slow scroll]

injectivity gives you linear independence and surjectivity gives you spanning

[elfin ingot]

rows*

same?

[half ice]

I do kinda assert it's use in my proof, but I call it invertibility instead haha

[slow scroll]

its the definition of matrix multiplication, when you multiply a matrix by a vector, it corresponds to some linear combination of the columns

[half ice]

You can retry my proof and find it works with just injectivity. That invertibility is extraneous, mo

[elfin ingot]

okay

1 more thing b4 i finally stop bothering u

what ddo i need to know b4 jumping on to linear transformations of vec spaces?

so far im supposedly done the rest is just the author doing examples

the matrices have like

columns and rows that span vector spaces?

that are basis(s) for vec spaces?

yea tysm anyways @half ice

and @slow scroll

[half ice]

Honestly you'll be much more "used" to this idea if you understood homomorphisms

[slow scroll]

linear transformations are pretty much just a more abstract take on what you are doing now. It might(?) help things make more sense anyway.

[half ice]

I think you'll be like "oh this is why people like lin alg"

But that's just me

If you get the idea of dimension and basis, that's enough to tackle field extensions and galois theory

[west spade]

Is plugging linear transformations into polynomials something that happens? Never taken LA so maybe the question is dumb but it sounds cool

[subtle walrus]

yes

but you have to make sure the linear transformation is into itself

otherwise powers are not defined

[ripe hemlock]

Quick question

are all transpose conjugates of matrices unique

[subtle walrus]

this is one of the reasons why the theory of linear functions of a vector space into itself is richer than the more general theory

[ripe hemlock]

like are there two 3x3 matrices which have the same adjoint

[subtle walrus]

taking transpose conjugate is its own inverse

[ripe hemlock]

what do you mean

so transpose conjugate of A times A is the identity matrix?

[subtle walrus]

if you do it twice, you get the original matrix back

that, no

(A*)* = A

[ripe hemlock]

no i meant are all transpose conjugates unique

[subtle walrus]

this answers the question?

[ripe hemlock]

like is it possible for the transpose conjugate of A and the transpose conjugate of B to be the same

[subtle walrus]

if A* = B*, then A = (A*)* = (B*)* = B

[ripe hemlock]

is there a proof for that?

[subtle walrus]

of what

[ripe hemlock]

wait i dont understand

how does that answer the question?

youre just saying that the adjoint of the adjoint is the matrix itself

[subtle walrus]

if A and B have the same transpose conjugate, then A = B

[ripe hemlock]

ohhhh

thank you

[subtle walrus]

i.e. * is a bijection from the set of all matrices to itself

so in particular injective

[ripe hemlock]

so my question is worng then

my teacher must have mixed up adjoint with adjugate

[open pivot]

I have an equation involving 3, 2x2 matrices: (A+B)^2=I, where I is the identity and B is known. My approach is to let A = {{a,b},{c,d}} and solve, will this work/ is this the best way? I ended up with 4 equations in 4 variables but some of them are non linear terms so I'm not sure how to solve this

[subtle walrus]

will probably work

you could also use the fact, that A + B has to have determinant +-1 and be it's own inverse

but not sure this makes it easier

[open pivot]

I'm midway through first year LA and haven't looked at determinants yet

I'm just stuck with the equations I ended up with

[subtle walrus]

determinant just gives you another equation

[pallid rampart]

A=I-B

[open pivot]

Big think

[pallid rampart]

I mean that is a solution

[open pivot]

There are more solutions?

[pallid rampart]

A=-I-B

This question is the same as finding all matrices A such that A^2=I

Which I think is the set of matrices P^{-1}DP with D having +-1 on the diagonals

[open pivot]

How do you get A=I-B? Can you also explain why (A+B)^2=I is the same as A^2=I?

[subtle walrus]

as i said, you just have to determine (A+B) such that it is its own inverse

as those are 2x2 matrices it's easy

[open pivot]

So, (A+B)= +-I

[pallid rampart]

Not necessarily

$\begin{bmatrix}0&1\1&0\end{bmatrix}$

[stoic python]

Whoever:

[pallid rampart]

The square of this matrix is I

[subtle walrus]

just set up the equation, its literally 4 linear equations in the entries of A+B

[pallid rampart]

How is it linear tho

[open pivot]

Here is the value for B:

If it helps..

[stoic python]

Zophike1:

[dreamy depot]

I don't get how you obtain $U+W$ I thought directly applying the definition would be enough like the previous example can someone explain ?

[stoic python]

Zophike1:

[subtle walrus]

so, first of all, you copied the example wrong

my version of the book says U ={(x,x,y,y)} and W as you stated

[dreamy depot]

Oh my bad

[subtle walrus]

then you get U+W = {(2x, 2x, x+y, y)}

and the exercise is verify that this is equal to {(x,x,y,z)}

[dreamy depot]

yeah

[subtle walrus]

where x,y,z are arbitrary field elements

so this boils down to veryify that for every $x_2 \in \mathbb{F}$ there is a $x_1 \in \mathbb{F}$, such that $2x_1 = x_2$

[stoic python]

Lochverstärker:

[subtle walrus]

and similarily that every field element can be written as the sum of 2 field elements

for the second component

and i just noticed that my notation in "U+W = {(2x, 2x, x+y, y)}" is bad

actually, wrong even

[dreamy depot]

How come if I may ask ?

[subtle walrus]

(1, 1, 0, 0) is in U, (2, 2, 2, 3) is in W, their sum is (3, 3, 2, 3)

[dreamy depot]

all right

[subtle walrus]

like,

if you add vectors from U and W

use different variables for them

then do the addition

and check that the simplification in the book can be made

[dreamy depot]

ahh yeah that makes sense then why dosen't axler do that ?

[subtle walrus]

(which boils down to using field axioms)

because he does not present how to calculate the sum

he just writes down the spaces and re-using variables is common then

like the x in the definition of U is different from the x in the definition of W is different in the "representation" of U+W

[dreamy depot]

Because when you add them you would get this,

$\vec{e}=(a, a, b, b)+(c, c, c, d)=(a+c, a+c, b+c, b+d) \in U+W$

[stoic python]

Zophike1:

[subtle walrus]

ye

[dreamy depot]

but then what do you do from there

[subtle walrus]

well

the first 2 entries are clearly identical

and they are again field elements

so we may just set a+c = x

and for every field element x, we may find a and c, such that this is true

[dreamy depot]

ahhhh that makes sense so you can just write them as linear equations ?

[subtle walrus]

so every vector of the form (x, x, b+c, b+d) for arbtirary x, c, b, d in F is in U+W

well, you can just set c=0, a = x

then a+c=x

at this point you just have to verify that every vector of the form (x, x, y, z) can be written as (a+c, a+c, b+c, b+d)

[dreamy depot]

Ok when you do that you get (a, a, b, b) ? by doing some algebra

and I think that's it

[subtle walrus]

it's just algebra yeah

solving a system of linear equations

[dreamy depot]

@subtle walrus I got it thanks what was the eariler comment you were making about Axler's notion I think I missed it it in the dicussion

[subtle walrus]

huh?

[dreamy depot]

you mentoineded that he isn't clear at times ?

[subtle walrus]

no, he is

[dreamy depot]

@subtle walrus you said that he just writes down the spaces and re-using variables is common then

[subtle walrus]

yes, my point is

let me think, it's hard to put into words

[dreamy depot]

because if he just reuses varibles it can make things unclear

[subtle walrus]

like, he uses x and y both in the definition of U and W

but when you deal with elements of U and W in the same equation

e.g. you add them

that doesn't mean that x and y is the same

but that is normal notation

[dreamy depot]

oh I understand now

[subtle walrus]

the x, y in the definition of U are arbitrary

and the x, y in the definition of W are arbitrary

[dreamy depot]

That makes sense I got confused on thought they were the same for a second

because in the solution I looked over the MSE poster mentioned that you should use different variables to be clear

[subtle walrus]

yeah

[dreamy depot]

But thx @subtle walrus I understand it now

[subtle walrus]

yw

[wintry steppe]

Why did he draw vector b as a straight line?

Also where is his z axis?

[limber sierra]

presumably that drawing is just there for visual intuition

[dusky epoch]

he's not drawing any axes at all

[limber sierra]

i dont think its meant to be accurate or anything

it's drawn so that the vector b is horizontal

which means the axes would be "tilted"

[dusky epoch]

the plane of the board need not be any of the three coordinate planes

[wintry steppe]

Well he defined be as having -2 on the y-axis

so wouldn'it have made more sense to represent it that way ?

but yeah like you said it's drawn so that vector b is horizontal

I feel like it makes it more confusing for me tho, idk

I understood much better with this video: https://www.youtube.com/watch?v=fqPiDICPkj8

[wintry steppe]

Did I draw these vectors correctly ?

[quartz compass]

eh

it can be helpful to label a little tick mark on the axis to help make it clearer

like a 5 on the z axis and a -1 on the x axis

your picture looks fine, idk why you're doubting yourself so I can't give you any meaningful feedback

[wintry steppe]

I graphed the (1, 3, -2) vector with this tool and it's pointing in the opposite direction than mine is

That's why I doubted

Or I might just be seeing it wrong

[bitter glade]

the only technicality is that your (-1,0,5) vector isn't tall enough but it doesn't matter

you're seeing it wrong

that vector matches yours

[wintry steppe]

Okay cool

thx guys

[grand ingot]

Hey guys, could someone help me get some ideas for this problem: Give an example of a linear map R4 -> R4 such that the rank is equal to the nullity

[odd kite]

@grand ingot do you know what the rank and nullity must be then?

this is a fairly simple problem if you understand what those terms mean

[grand ingot]

I kinda understand what they mean, but the idea of a linear map is confusing to me.

[limber sierra]

have you learned matrices?

each linear map corresponds to a matrix, and vice versa

[grand ingot]

Yeah

I have.

[odd kite]

So what (number) is rank and nullity in this case?

[grand ingot]

Would be 4 right

[odd kite]

rank and nullity are two separate things

neither are 4 here

[quartz compass]

maybe try to explain to us what rank and nullity are for some simple maps to or from R^3 or R^2 @grand ingot

[odd kite]

yes, like, are you able to find the rank and nullity for these

$\begin{pmatrix}
1 &0\
0& 1
\end{pmatrix},;;;;
\begin{pmatrix}
1 &0\
0& 0
\end{pmatrix},;;;;\begin{pmatrix}
0 &1\
1& 0
\end{pmatrix}$

[stoic python]

Timon:

[elder robin]

Can I not switch rows when finding rref looking for an inverse matrix?

I can’t figure out what is wrong (if anything) with my row reduction, but it doesn’t match wolfram alphas

Oh I inputted it wrong into wolfram lmao

Now it’s the same

[elfin ingot]

[wintry steppe]

yes u can switch rows

[elfin ingot]

does T_a mean T(a)

T is a linear transformation a is a vector

so im given definiton of a null space all vectors such that T_a = 0

id guess T_a = T(a) and that sjust notation

and null space is analogous to kernel

right?

[quartz compass]

eh?

I'd assume T_a has the matrix represetation a^T

erm that's nearly confusing

$[T_a]=a^T$

[stoic python]

Merosity:

[quartz compass]

as in this is the matrix representation

[elfin ingot]

idk whats a^t

[quartz compass]

transpose

[elfin ingot]

can u give me the def of anull space?

[quartz compass]

a being a column vector

[elfin ingot]

in ur words

[quartz compass]

no

[elfin ingot]

👍

[quartz compass]

I'm just suggesting an alternative guess as to what your notation could plausibly mean

you need to go back into your book and find what it actually means

there's no point in trying to guess notation and then reason off of that

[elfin ingot]

okay

1 more question

ppl say that the rank nullity theorem is analogous to first isomoprhism theorem

how is this true?

its just been given to me and i dont see it xd

[limber sierra]

its a corollary

[elfin ingot]

can u show me how?

[limber sierra]

it's immediate from the definition of quotient space that $\dim U/V = \dim U - \dim V$

[stoic python]

Namington:

[limber sierra]

first isomorphism theorem tells us that $V / \ker f \cong \text{image} f$

[stoic python]

Namington:

[elfin ingot]

okaaaaaay

[limber sierra]

consider dimensions

[elfin ingot]

yea

ker(f) is analogous to nullspace here

right?

[limber sierra]

$\dim V / \ker f = \dim V - \dim \ker f = \dim \text{image} f$

[stoic python]

Namington:

[elfin ingot]

yea

cool

i got it

tysjm

[limber sierra]

yes, the dimension of the kernel is the nullity.

[elfin ingot]

yea

okay the proof in the actual text is p trash ( for me )

can i use this as a legit proof?

proving the first iso theorem and then arriving at this just like u did>

[limber sierra]

sure, as long as you prove the $\dim U/V = \dim U - \dim V$ fact i mentioned

[stoic python]

Namington:

[limber sierra]

and as long as your proof of the first iso. theorem doesnt rely on rank-nullity but

im not aware of any proofs that do

[elfin ingot]

wait how does a proof of the first iso rely on rank nullity

[limber sierra]

i'd have no idea how

as i said

i'm not aware of any proofs

that would use it

[elfin ingot]

yea

i thought u meant ur not aware of an proofs that dont

cool

[brittle fog]

Is it possible to factor v = [w b] (w is mx1 and b is 1x1) out of (1 - y(w'x+b))?

[wintry steppe]

how do I take determinant of a 1 by 1 matrix?

By writing the number in the matrix

[dusky epoch]

garbage in garbage out

[brittle fog]

Thats not garabage thats just accurate

[cold topaz]

can u tell where is he getting the marked numbers from? Ive been able to get to the step right before it, but i dont know where he is getting thos marked numbers from...

[limber sierra]

are you given any other information

[cold topaz]

no

[limber sierra]

youre just asked to compute <2v - w, 3u + 2w> given

absolutely no information?

[cold topaz]

this is from my book

a)

[limber sierra]

huh

i... honestlly dont know what to say

are you using a notational convention where u, v, w denote the unit vectors or something? maybe?????

but then the numbers dont make sense

[cold topaz]

it might be from an older book

[dusky epoch]

are you SURE you aren't given what u, v and w are

[urban bough]

how do I solve an equation of the form u = e +Bu for some u

where e and B are fixed

[dusky epoch]

(I-B)u = e

[urban bough]

ah

and then u = (I-B) inverse e

right?

[dusky epoch]

yes, as long as 1 isn't an eigenvalue of B

[wintry steppe]

What did u mean by garbage in garbage out?

[dusky epoch]

what do you think i meant

[wintry steppe]

Do u think I'm garbage?

[dusky epoch]

no????

[wintry steppe]

so there's this theorem in linear algebra

that all of these must be true, or they are false

but i've found problems where matrices who do not fit the last condition are able to have a consistent Ax = b solution

[dusky epoch]

can you show an example of such a matrix

[wintry steppe]

for example this matrix

times x

equaling

the solution is

we dont have a pivot position in every row

yet Ax = b has a solution

[dusky epoch]

part (A) says Ax=b must have a solution for every b ∈ R^m (in your case, R^3)

but $A\bd{x} = \bd{b}$ fails to have a solution if $\bd{b} = \mat{8 \ 4 \ 2}$, for example.

[wintry steppe]

yeah it was able to have a solution for a b in R^m

R^3

[stoic python]

Ann:

[wintry steppe]

0 4 4

[dusky epoch]

so for the A that you showed, all the statements are false.

[wintry steppe]

How? set Ax = b with b being 0 4 4, and you get Ax = b being consistent

x_1 is 5/2 and x_2 is 3/2

A is true but d is false

because it doesnt have a pivot position in every row

b is in r^3

[dusky epoch]

bruh.

condition (a) says the system has to have a solution no matter what b is

[wintry steppe]

oh

for all b's?

[dusky epoch]

yes!!!

[wintry steppe]

btw what do they mean when they say the columns of A span R^m?

[dusky epoch]

they mean exactly what they say

when they say the cols of A span R^m they mean that the span of the cols of A is all of R^m

shocking right

[slow scroll]

wait ur doing RREF on the augmented matrix tho. Those conditions you posted a picture of are talking about the coefficient matrix.

[wintry steppe]

even if u do RREF on the coefficient it cannot have 3 pivot positions

[slow scroll]

true.

[dusky epoch]

a matrix with more rows than cols will never have a pivot in every row

no matter what

[wintry steppe]

i dont get what theyre saying when they say it spans R^m

thats very vague

what do they mean by that?

[gray dust]

did you learn the defn of span

[wintry steppe]

the set of all linear combinations?

[gray dust]

cool, so that's all there is

[wintry steppe]

R^n is the number of entries in?

i mean R^m

whats the m

how many entries in the vector?

[slow scroll]

the space of ordered n tuples, yea

[dusky epoch]

thats very vague
it is not

it is not vague at all

[wintry steppe]

can u show me a demonstration?

"the columns of A span R^m"

[gray dust]

A is an m by n matrix. m rows. R^m denotes the set of m-tuples with real entries

[wintry steppe]

i know, but what do they mean when they say that it spans R^3 for example

[dusky epoch]

let $A = \mat{1 & 5 & 11 \ 0 & 1 & -10 \ 0 & 0 & 7}$.

[stoic python]

Ann:

[dusky epoch]

the cols of $A$ are $\mat{1 \ 0 \ 0}, \mat{5 \ 1 \ 0}$ and $\mat{11 \ -10 \ 7}$.

[stoic python]

Ann:

[dusky epoch]

the span of $\curly{\mat{1 \ 0 \ 0}, \mat{5 \ 1 \ 0}, \mat{11 \ -10 \ 7}}$ is $\bR^3$ because every vector in $\bR^3$ can be written as a linear combination of $\mat{1 \ 0 \ 0}, \mat{5 \ 1 \ 0}$ and $\mat{11 \ -10 \ 7}$.

[stoic python]

Ann:

[wintry steppe]

every vector in R^3?

[dusky epoch]

yes, every vector in R^3.

[wintry steppe]

how is that even possible, lol

take vector 6 -9 20

how can u write that

[dusky epoch]

ok, one moment

[wintry steppe]

as a linear combination of those

oh i remember u do the RREF

and it should give u the values

[limber sierra]

solve $x + 5y + 11z = 6, y - 10z = -9, 7z = 20$

[stoic python]

Namington:

[gray dust]

,w rref [[1,5,11,6],[0,1,-10,-9],[0,0,7,-20]]

[wintry steppe]

yeah

[stoic python]

Display results online and refine query

Requested by RokettoJanpu#6002

[limber sierra]

this is fucking hellish

but it works

[gray dust]

there's your linear combo

[dusky epoch]

$\mat{6 \ -9 \ -20} = \frac{1577}{7} \mat{1 \ 0 \ 0} - \frac{263}{7} \mat{5 \ 1 \ 0} - \frac{20}{7} \mat{11 \ -10 \ 7}$

[stoic python]

Ann:

[wintry steppe]

wow

how do we know if a matrix is onto or one to one

do you mean a transformation?

[slow scroll]

onto iff the columns span everything, one to one if the columns are linearly independent

i mean, same thing

[wintry steppe]

yeah a transformation

nonzero determinant implies both onto and one to one

[slow scroll]

umm these aren't necessarily linear operators. They could be non-square matrices, so that doesn't work

[wintry steppe]

ok think I'm in the wrong realm

why is the inverse of a 2 by 2 matrix 1/det * switching the a and d and multiplying the c and b by -1?

is there some explanation?

[slow scroll]

its a special case of a complicated formula for the inverse of a matrix.

[gray dust]

that's just a shortcut they teach for inverting 2 by 2

[wintry steppe]

the formula is 1/det * adjA?

[slow scroll]

yea

[wintry steppe]

something like that

[gray dust]

1/det(A)*adj(A)

[wintry steppe]

but i tried doing that

on the 2 by 2

and it didnt work

[slow scroll]

adj(A)^T tho

[gray dust]

naw

[slow scroll]

or well, eh i think i forgot the def

[wintry steppe]

the adj thing doesnt work on the 2 by 2

[gray dust]

you're thinking cof(A)^T

[wintry steppe]

it explains why the b and c get multiplied by -1

but by that method, at the end, u transpose ur result

right?

so it doesnt equal the trick they teach us

[gray dust]

BEFORE transposing, that's the cofactor matrix

AFTER transposing, that's adjugate

[wintry steppe]

can u calculate the matrix of a 2 by 2 using the hard method?

[gray dust]

it's not hard. it's fast for 2 by 2 and tedious for all else

[wintry steppe]

if u transpose it doesnt quite get u the result u want

if its 2 by 2

[slow scroll]

ur probably just doing something wrong lol

[gray dust]

you're just getting smth wrong along the way

[wintry steppe]

a gets a positive, b gets a negative because (-1)^(2+1), c gets a negative because (-1)^(2+1), and d gets a positive

right?

ab are on the first row, cd on the second row

[gray dust]

ok walk through the steps slowly. form the matrix of minors first

[wintry steppe]

wdym by the matrix of minors in this case?

u cover up the first row and column, u get ONE number, how u get the determinant of that?

[gray dust]

det of 1x1 matrix is the sole entry

[wintry steppe]

oh i see

now i get it

[rugged wharf]

Hi, I have a question, in my exercise I have to find orthogonal projection of a vector on a subspace (vector u on subspace U). However, after trying to find a solution, I'm not able to find an orthonormalised basis.

The e2 length is not equal to 1, and I can't find the place where I made the mistake

lin is equal to span btw

[slow scroll]

there is no mistake. Just get an orthogonal basis and divide everything by their norms in the end. Does that make sense? @rugged wharf

so your orthogonal basis could be
(1,1,1,0), (-1,2,-1,6) (the second vector comes from the stuff you worked out already, i just multiplied by 3 to get integer coords), and then divide each of these by their norms to get the orthonormal basis

[rugged wharf]

hmm, so all I need is divide the second vector by it's norm?

because e1 is already of length 1

[slow scroll]

yeah. In other words, you can just get an orthonormal basis by taking some orthogonal basis and normalize all of the basis vectors. Everything is still orthogonal and linearly independent, so boom, orthonormal basis.

[rugged wharf]

Okey 😄

Thanks a lot

[slow scroll]

npnp

[hasty violet]

What is the difference between Orthogonal and Perpendicular?

[rugged wharf]

I think there is no difference

[hasty violet]

Why dont people just decide to use 1 of those words

[rugged wharf]

orthogonal is more universal?

[hasty violet]

oh

I think orthogonal also works in dimensions higher than 3

[subtle walrus]

yeah, orthogonal is a generalisation

[rugged wharf]

Yea

[hasty violet]

or something.

[rugged wharf]

exactly

[subtle walrus]

orthogonal depends on the specific dot product you use

and exists in every vector space with a dot product

perpendicular is a term from euclidean geometry

[hasty violet]

man learning math is always 👌

[latent marten]

If c is a value for which Bc is not invertible, can Bc be expressed as a product
of elementary matrices? Justify your answer.

i dont get this q

how do i prove it, ik that if Bc isnt invertible it cannot be represented as a product of elementary matricies

[quartz compass]

are elementary matrices invertible?

@latent marten

[latent marten]

yes

[quartz compass]

if you multiply two invertible matrices, is the result an invertible matrix?

[latent marten]

yes

[quartz compass]

alright good

[dreamy depot]

Can someone help me understand Example (4.4.4): https://math.libretexts.org/Bookshelves/Linear_Algebra/Book%3A_Linear_Algebra_(Schilling%2C_Nachtergaele_and_Lankham)/04%3A_Vector_spaces/4.04%3A_Sums_and_direct_sum

Does the two subspaces in Ex (4.4.4) form a direct sum form what i'm reading from there it dosen't but i'm not understanding how you can write (0,0,z) as (0,w,z)

[subtle walrus]

huh?

just set w=0

[dreamy depot]

ahhh ok that makes sense thanks

@subtle walrus could you do that also for Example (1.43)

[subtle walrus]

i have no idea how to navigate this site

[dreamy depot]

Oh I meant in the book LADR(Linear Algebra Done Right)

[subtle walrus]

what do you mean by "do that"?

[dreamy depot]

set one of the variables equal to 0

[subtle walrus]

sure

they variables are any field element

any field has a 0

[dreamy depot]

ahhh okay that makes sense so in general so if something is a direct sum in this situation you just plug in variables or numbers as test cases

[subtle walrus]

well, that will help you find out if something is not a direct sum

[dreamy depot]

ahh okay thx

[fossil quest]

So i got a question in linear algebra...
If x in Element of Z, how do i solve 7x + 3 = 4x + 1 (mod 10)

= is supposed to be the congruence symbol

[limber sierra]

rearrange as you would do a regular equation

(cancellation laws hold for modular congruence)

[quartz compass]

depends on what you mean by "cancellation"

division can break, so you gotta watch out for that

[limber sierra]

yeah true, let me clarify:

addition cancellation still holds

there's an analogue for multiplicative cancellation but it's "different"

since you're now working in Z/nZ

[fossil quest]

So Is it correct when I get 3x = -2 (mod 10)
And then x = -2/3 (mod 10)

Or can I add 20 on the right side to get it to 18 and then divide both sides by 3? Then I'd get x = 6 (mod 10)

[limber sierra]

well

-2/3 is not an integer

sooo probably not the best answer

but yes, you can verify that x = 6 works

7*6 + 3 = 42 + 3 = 45 = 5 (mod 10)

4*6 + 1 = 24 + 1 = 25 = 5 (mod 10)

the way i'd approach this is as you did, rearrange to 3x = -2 (mod 10)

and then 3x + 2 = 0 (mod 10)

and now just run through integer multiples

until the sum is a multiple of 10

ie:

3(1) + 2 = 5
3(2) + 2 = 8
3(3) + 2 = 11
3(4) + 2 = 14
3(5) + 2 = 17
3(6) + 2 = 20; bingo

this is easy to do in this case, since counting by 3s is "easy" and 10 is relatively "small"

[fossil quest]

Ok thx! But what is that you did to verify if 6 works?

Where does 7*6 + 3 come from?

[limber sierra]

we want to find an x such that

7x + 3 = 4x + 1 (mod 10)

so just check what each side is mod 10

when x = 6

to make sure 6 works

if they're the same (mod 10) then we're done

and indeed, they're both equal to 5 mod 10

[fossil quest]

Ohhh ok got it

But how do i even get a negative number like that -2 (mod10) ? That - in front of 2 kind of confuses me

How can i have a negative as a rest

[limber sierra]

-2 = 8 (mod 10)

think of it like this: each value that's equal to 8 (mod 10) is separated by a multiple of 10

that is

8, 18, 28, 38, 48, ... 108, 118, 128, etcetc

are all separated by 10

so if you go 10 backwards

18, 8, -2, -12, -22, -32, ...

all of these are also congruent to 8 (mod 10)

[fossil quest]

So 8 (mod 10) = -12, -2, 8, 18..... ?

[limber sierra]

yes, its congruent to all of those.

[fossil quest]

Ohh ok thanks now i understand

And how do i do these?

Whats F17 and [] around the number?

Kind of hard to solve questions without even having class bc of Corona :(

[shrewd mortar]

@fossil quest This would go in #elementary-number-theory

[sinful parcel]

How can I show that the transformation $T(ax^{3} + bx^{2} + cx + d) = cx + d$ is linear?

[stoic python]

Simula:

[limber sierra]

check the requirements

verifying linearity is very... formulaic

[sinful parcel]

i understand i need to show T(p+q) = T(p) + T(q) and that T(cx) = cT(x), just not sure how to do that with a polynomial

[half ice]

Let u and v be a vector in your space.

That is, they're each some cubic (or quadratic or linear or constant)

Can you compute T(u + v)?

[limber sierra]

i mean you can just literally compute T(p+q), T(p), T(q)

for arbitrary polynomials $p = ax^3 + bx^2 + cx + d, q = a'x^3 + b'x_2 + c'x + d'$

[stoic python]

Namington:

[limber sierra]

[or whatever notation you want to use to distinguish the coefficients of the polynomials]

[a_1 and a_2, or a and a_0, whatever]

whoops that should be x^2 not x_2

[sinful parcel]

im... so lost lol

[limber sierra]

let $p = ax^3 + bx^2 + cx + d, q = a'x^3 + b'x^2 + c'x + d'$

[stoic python]

Namington:

[limber sierra]

what is $T(p+q)$?

[stoic python]

Namington:

[sinful parcel]

cx+c'x+d+d' ?

[limber sierra]

sure

sooo what can you do with this

[half ice]

Nvm you nailed it haha

[limber sierra]

we want to show it's equal to $T(p) + T(q)$

[stoic python]

Namington:

[limber sierra]

so what is $T(p)$? what is $T(q)$?

[stoic python]

Namington:

[sinful parcel]

(cx+d)+(c'x+d')
T(p) = cx+d, T(q) = c'x+d'

[limber sierra]

right

[sinful parcel]

therefore T(p+q) = T(p) + T(q)

[limber sierra]

so T(p+q) = cx + c'x + d + d' = (cx + d) + (c'x + d') = T(p) + T(q)

that verifies additivity

now do the same thing but to show cT(p) = T(cp)

[sinful parcel]

alright. i can definitely do scalar mult on my own

thanks yall :)

[wintry steppe]

what do they mean in linear transformations when they say from R^3 to R^2

[sinful parcel]

(someone correct me if i'm wrong) it means that the transformation is a map that takes an input in the third dimension, and produces an output residing in the second dimension

[keen umbra]

yes but it should be more

[wintry steppe]

why is it R^n to R^m

what do they mean by that

[keen umbra]

those are just ur spaces youre working with

like R^3 and R^2

the reason why they write it arbitrary is for one to make it arbitrary and for two to make sure those indices matchup for whatever definition you use

so if I say a transformation from R^n to R^m, that would mean ur matrix representation for that transformation must be a m by n

Simula has the correct idea. using R^3 to R^2 as an example. That means if I add two vectors u and v in R^3 and i map the sum to R^2, that should be at the same location as if I have mapped u and v first then added them in R^2. Then same concept for scalar property

[half ice]

@wintry steppe
Rⁿ is the vector space of arrows in n-dimensional space. Or, lists of n numbers.

For example, (2,5,6,8) is a member of R⁴. Can also be thought of as an arrow in 4d space.

A function from R³ to R² takes every vector in R³, and returns a vector in R².

A function between vector spaces is linear if
f(u + v) = f(u) + f(v)
f(au) = af(u)

[cold topaz]

can you please explain how you would solve b)? I have the Slader open. But it is not helping at all.

[elfin ingot]

integrate from -1 to 1 x^2 dx ig

[cold topaz]

i did ||1-x||, => sqrt((1-x).(1-x)). then im stuck.

[gray dust]

@cold topaz why 1-x?

[cold topaz]

that's d(p,q)

[pallid rampart]

But q=x^2

[cold topaz]

ooooooooooooooooooooooo

[pallid rampart]

d(p,q)=||p-q||=||1-x^2||=sqrt(<1-x^2,1-x^2>)

[wintry steppe]

why is this not one to one???? The leading​ entries, denoted by box​, may have any nonzero value. The starred​ entries, denoted by star​, may have any value​ (including zero)

it has a pivot in each column

is it not linearly independent?

who says its not one to one

the problem

[eternal finch]

Do you know the relationship between one-to-oneness and the null space?

[wintry steppe]

no

maybe because there is one row that doesn't have a pivot? thats my guess

well i know that null space is

@wintry steppe thats not required for one to one

only for onto

and i know that for the equation Ax = 0

if its one to one, it has only the trivial sol

[eternal finch]

Good.

[wintry steppe]

thus being linearly independent

so does this have, here, only the trivial solution

make up numbers and see

[eternal finch]

Ok, yeah, that's pretty much what I was looking for. If you want to show one-to-oneness, you'd want to show that Ax = 0 only has the trivial solution.

[wintry steppe]

row 3 and row 4 is a bit confusing

how can x_3 have two solutions?

[eternal finch]

Well, the third row doesn't have to be nonzero, since it's a *.

[wintry steppe]

yeah exactly

which makes it confusing

so x_3 has two solutions???

how is that possible

[eternal finch]

If you get x_3 has two solutions, then there isn't a solution.

[wintry steppe]

then it's not one to one

[eternal finch]

However, that's not necessarily the case here.

[wintry steppe]

yeah it can be 0

or nonzero

if its nonzero, then x_3 has two solutions if the augmented vector is nonzero

[eternal finch]

Ok, let's just back up for a moment.

The goal is to prove the matrix is one-to-one.

[wintry steppe]

yh

[eternal finch]

The method is to prove that Ax = 0 only has a trivial solution.

[wintry steppe]

wait but i also read that

if its one to one

it has a pivot in every column

and that Ax = b, has either 1 or no solution

as well

the diagram has a pivot in every column so it should be one to one, but it's not, maybe that's a wrong assumption

[eternal finch]

if its one to one
it has a pivot in every column

You sure about this one?

it has a pivot in every column
and that Ax = b, has either 1 or no solution

This one is correct.

[wintry steppe]

thats what ive heard

[eternal finch]

I guess that would make sense.

A pivot in every column would mean full rank?

I don't remember much about pivots.

[slow scroll]

pivot in every row would correspond to being full rank. idk your matrix looks one-to-one to me

[eternal finch]

Ah, ok, every row. Gotcha.

[wintry steppe]

pivot in each row is for onto

[slow scroll]

onto means full rank

[wintry steppe]

@slow scroll exactly idk why they said its not one to one

Ax = 0, has only the trivial solution

right?

[slow scroll]

yeah. And you can tell just by looking that it does. The way they have the matrix drawn and pivots highlighted makes me think there is some kind of typo maybe? Since the third pivot belongs on the third row and the last row should be zero (if we're talking about row echelon form here). Either way, the REF of this matrix has a pivot in every column.

[wintry steppe]

what is an elementary matrix?

a matrix that differs from the identity matrix by a single row operation

[half ice]

The identity matrix with one "elementary transformation" done to it.

Elementary transformations are things you can do to systems of equations:

• Swap rows
• Multiply a row by a constant
• Add rows together

[wintry steppe]

wdtm by those?

wdym "wdym by those?"

what do they mean by those

statements

theorems

i understand the second one

but the first one

not really

so matrix E is obtained by preforming **the same **row operation on I, that was preformed on A to obtain EA

hmmm

i know how to get an elementary matrix, it was just explained to me

so basically u have an identity matrix, and u perform a row operation on it

correct

[clear vessel]

let T be the linear transformation from R^2 to itself so that:
if v is a vector on the line y = 2x, then T(v) = 2v
if v is a vector on the line y = -2x, then T(v) = 0
let A be the standard matrix for T, find the eigenvalues for A, and give a basis for each eigenspace
how do i approach dis

[hallow cliff]

a vector along y = 2x is an eigenvector so (1,2) is an eigenvector with eigenvalue 2

also a vector along y = -2x is in the nullspace as T(v) = 0, so the vector (1, -2) has an eigenvalue of 0

the basis for the eigenspace can just be the set of the eigenvectors

@clear vessel

[wintry steppe]

Hey guys I'm really stuck with this one could someone show me how to solve it?

[wintry steppe]

what do they mean when a basis spans H?

https://lmgtfy.com/?q=What+is+the+span+of+a+set+of+vectors%3F

lel

my question is not that easy

[eternal finch]

Usually, before you get to the concept of the basis, you learn about vector spaces, subspaces, span, and linear independence.

[sick dragon]

sounds like span but it's a linearly independent set?

i dont know basis yet either

[pallid rampart]

basis is not a span

It's a set of vectors that span H

and is also linearly independent

[wintry steppe]

@eternal finch i've learned about all of those

[eternal finch]

If you've learned those, then you should know what it means when a set spans a space.

[pallid rampart]

so a set of vectors span H if the span of the set of vectors is H

that's the definition

[wintry steppe]

how can we check if a set of vectors spans H?

[pallid rampart]

It heavily depends on the specific situation

[eternal finch]

In this case, they're talking about a basis of a subspace of R^n.

[pallid rampart]

Still

But

[torn hornet]

i mean it can still be p bad, if the cardinality of the spanning set isnt n for example

[pallid rampart]

Since you're introduced to basis, you'll probably learn some useful facts about them very soon

Like for example, a basis for R^n always have n elements

[eternal finch]

Right.

[pallid rampart]

So if a set has say, n-1, or n+1 number of elements, then it is definitely not a basis for R^n

[wintry steppe]

R^m or R^n

[pallid rampart]

A basis for R^n

[wintry steppe]

n is the number of columns?

[pallid rampart]

Um

[eternal finch]

R^m or R^n
When people say R^[some integer here], they mean all of the lists of length [that integer].

[pallid rampart]

Do you know what R^n means

[wintry steppe]

kind of, but sometimes i confuse it with the number of columns

[pallid rampart]

number of columns of what??

[wintry steppe]

of the matrix

[pallid rampart]

the matrix?

[wintry steppe]

n is the number of dimensions