linear-algebra

[sonic osprey]

"An intersection that's something that's not just 0"

[wintry steppe]

Actually useless

[prisma kestrel]

non-trivial is always mentioned when theres one which works "always"

[sonic osprey]

Right

[prisma kestrel]

reminds me of linear combinations

[sonic osprey]

0 is always in a subspace, so the intersection of two subspaces will always have 0

[paper egret]

ah

[sonic osprey]

So non-trivial just means, not just 0

[paper egret]

omg i don't even know wtf i'm proving

holy shit

this is most likely wrong as hell, but that's what i got so far

wait nvm i have an idea

[sonic osprey]

V and W are linear transformations?

[paper egret]

wait shit

yup thats a mistake

B be a basis of U and C be a basis of
W

[sonic osprey]

Yeah that's pretty wrong

[paper egret]

ok yea big typo

lemme redo it

shit

[sonic osprey]

It should help you to think of actual examples

Nah it's more strong

Think of the R^2 example

Figure out an example in R^3 and see how that works

[paper egret]

i got an idea, i just represent vectors u and w by some linear combination of vectors, and then when I add them together, show that some vector v is also a linear combination of the linear combination of v and w

thus showing that B U C is a basis for V

[sonic osprey]

Try it

[paper egret]

yup it worked

goteeme

i think i overthought it too much

[prisma kestrel]

in what topic does construction of natural numbers etc fit in the best?

got a question on that

[sonic osprey]

#proofs-and-logic

[paper egret]

ok so i have to prove this corollary now, but i have one question

[prisma kestrel]

ty

[paper egret]

If V is the direct sum of U and W, then dim(V) = dim(U) + dim(W)

what happens if there is some identical vector in B and C, where B is the basis of U, and C is the basis of W

oh wait

nvm

that would violate the direct sum

nvm

im retarded

`Let U and W be subspaces of V. Then U + W is the subspace of V defined by

U + W = {u + w | u is in U, and w is in W}`

this is basically direct sum

but a little different

i think

[sonic osprey]

How so?

[paper egret]

oh this is a different question

this says nothing about uniqueness

given that definition

Prove this
`Let U and W be subspaces of V. Then V is the DIRECT SUM U and W if and only if

(a) V = U + W
(b) U intersect W = {0}`

ok yea this is kinda obvious, but idk how to prove this without being vague af

[clever cedar]

rearrange the condition

where u isolate a

[north sierra]

why do i solve for a?

@clever cedar

[clever cedar]

to find basis

after u do that

u know how to setup basis

which shows the subspace for R^3

[north sierra]

i haven't learned basis yet

can you explain with a little less terminology

[clever cedar]

right

are u familiar with the concept of span

?

[north sierra]

yeah

[clever cedar]

so the idea of basis jsut says

we can create the vector space R^3 (or R^n) with only a minimal amount of vectors

for example the basis of R^3 is {[1,0,0], [0,1,0], [0,0,1]}

because with ONLY those three vectors

we can create any other vector in the vector space R^3

[north sierra]

oh

[clever cedar]

so basically

what the question is asking u

[clever cedar]

is to set up a basis for that given set

[north sierra]

so in my case, a would be the only vector needed to create the vector space R^3?

[clever cedar]

there can be infinite basis

so ur question says let W be the set [a,b,c] where a - 3b - c = 0

thats a little hard to comprehend

so lets redefine it

let W be the set of vectors [a,b,c] where a = 3b + c

now we know that we can create this subspace with this given condition

as long as a = 3b + c

[north sierra]

ohhh i see what i was confused on earlier now

but wait

im sstill lost on why solving for b and c dont matter

[clever cedar]

sec sorry

multitasking

[north sierra]

aight

[clever cedar]

so we've set

let W be the set of vectors [a,b,c] where a = 3b + c

so in other words

we have to define W to be a subspace of R^3

where a = 3b + c

we can redefine this as

W = {[3b + c, b, c]}

now remember how a basis for R^3 is [1,0,0], [0,1,0] and [0,0,1]

?

was that clear to u

[north sierra]

ok wait

the question didn't set a = 3b+c

they just = 0

[clever cedar]

i can give u my book's example

[north sierra]

sure

[north sierra]

ok i get it now

thanks mike

[clever cedar]

np

[north sierra]

why is it true that Ax = v + w will also be consistent?

oh its cause colA is a vector space

[north sierra]

when it says "has a solution for each b in R^m" does it mean every vector in R^m or every entry in the vector b for that specific vector?

[slow scroll]

@north sierra the former

[north sierra]

thanks

[clever cedar]

i have a question

if x = 0

why cant i divide both sides by x

0/x = 0

so 1 = 0

[slow scroll]

because x = 0

[feral mountain]

or d/dx both sides :)

[lone quail]

whats the fastest way to find inverses of matrixes 4x4

im currently using the cofactor method

but i have to find 16 3x3 determinants which really can take a while in an exam

[rancid sonnet]

does a vector space have to be closed under the two operations?

I don't see any closure condition in the axioms

[prisma kestrel]

i only read for addition: ass law, com law, neutral element 0 and inverse element and for scalar mult: left dist law, right dist law, ass law and neutral element 1

and yes, it was fun to me to write ass law

[rancid sonnet]

so do these imply closure, then?

[prisma kestrel]

as concerns what i have read so far no since only neutral and inverse have to be element of the vector space

[dusky epoch]

if you don't have closure, nothing else even makes any sense in the first place.

[rancid sonnet]

yes, I can feel that intuitively

how do you rigorize that

[dusky epoch]

how can (x+y)+z = x+(y+z) make any sense if you don't know that x+y or y+z are still in your space

+: V×V -> V

*: K×V -> V

[prisma kestrel]

but do they have to picture in V?

[dusky epoch]

picture

what

[prisma kestrel]

idk

MAP

[dusky epoch]

yes of course they do where else would they map

[prisma kestrel]

english words will bring me into nuthouse one day

but why are they limited to map into V? is it a restriction coming from fields?

[rancid sonnet]

Isn't that what closure means?

[prisma kestrel]

afaik closure means that A x A maps into A

[dusky epoch]

but why are they limited to map into V? is it a restriction coming from fields?

where else could they possibly map into

if you add two vectors and get something that ISN'T, then maybe you aren't dealing with anything resembling a vector space at all

[prisma kestrel]

but where is it restricted in the axioms and definitions? thats what i am asking for

[rancid sonnet]

how can (x+y)+z = x+(y+z) make any sense if you don't know that x+y or y+z are still in your space

[prisma kestrel]

and so how is the vector addition defined? i have no clue since i dont know much about LA

[rancid sonnet]

could be defined any way you like as long as it fits the axioms

[mint sentinel]

if you have the characteristic equation (a+1)(a-2)(a-1), do you let the p1 be the matrix for when a = -1, the matrix p2 for when a= 2 etc. in that order?

[dusky epoch]

what is p1

what is p2

[mint sentinel]

bases for eigenspaces

for the eigenspaces*

[dusky epoch]

,,,,,,

,,,,,,,,,, no not even close then

[mint sentinel]

then what

is it in a different order

[dusky epoch]

nnnnnno you're just way off-base

[mint sentinel]

how so

[lone quail]

@mint sentinel i call then E-1 and E2 if thats what you mean

to find E-1 i would do:
1)sub a=-1 into the original matrix
2)set the product of that matrix times the general vector equal to 0.
3)create equations
4)solve equations.
5)use those values to create the bases for E-1
6)Make E-1 by adding the bases as column vectors

[mint sentinel]

can you give me an example? all im wondering is that, when im supposed to find a matrix P that diagonalizes a matrix A and get bases p_1,p_2 and p_3 for example for the eigenspaces, what order do i put them in to get the matrix P? or rather, how do i chose which of the bases are either p_1, p_2 or p_3?

[lone quail]

so basically you have to find the bases of E1, E2 and E3

which are the eigenspaces

then you put those bases in one whole matrix

thats your matrix P

(or at least i think so, i was taught eigenspaces 2 days ago)

[mint sentinel]

yes, as a matrix of column vectors right?

[lone quail]

yes

[mint sentinel]

yes, but if i get a vector E1, why not call it E2 and make the original vector E2 as E1? how do you know the order?

[lone quail]

E1 is the eigenspace

the vector is the base of the eigenspace

order doesnt matter

i think at lwast

I just started on this topic so i may be wrong

[mint sentinel]

alright, thx tho

[vast torrent]

@mint sentinel there's no "right" order for eigenvectors

if the choices of the matrix of the basis vectors give determinant +1 or -1, we generally prefer an order that gives +1

if the eigenvalues are all real, we often prefer them to be in ascending or descending order

but there's no "correct" order

@lone quail

[mint sentinel]

okay, thanks

[coarse rose]

Do i need any pre knowledge besides high school math to learn linear algebra?

[clever cedar]

Nope

[coarse rose]

thats great

[clever cedar]

U have to be able to think abstractly tho

Things get challenging when ur introduced to vector spaces in third dimension

[prisma kestrel]

THATS GREAT

from now i on i love you mike

#nohomo ofc

[clever cedar]

Love u too

[prisma kestrel]

xd

i mean im really interested in linear algebra and im not a math student so its great for me if i dont need pre knowledge

but i've got my lemme call it "semi-A level" so abstract thinking should be possible 😄

[clever cedar]

If ur interested in it and that means you wont give up at the first challenge then ull do fine

Try ur hardest and ask questions

Ull succeed

[coarse rose]

i dont have abstract thinking

cuz i havnt done math in 10 years

and i jumped into a engineer class

[clever cedar]

Believe in urself

If u wanna succeed then you will

Given u put in the effort

[prisma kestrel]

well i already constructed the natural and integer numbers

and i think thats quite hard if you havent done such an axiomatic construction before

AND YES IM HAPPY THAT IM ABLE TO CALCULATE THREE PLUS FOUR

and yes thats not linear algebra i think but kinda set theory if im not wrong?

---

on a german video about vector spaces, first comment: "vector spaces should burn in hell" 😄

[north sierra]

hello

[clever cedar]

Sup

[prisma kestrel]

which topic does stuff like equivalence transformations belong to? algebra i guess? but linear algebra as well?

[wintry steppe]

hi i need to multiply two matrices, B and A to get BA

B = 
[    1    0    ]
[    0    1    ]
[    -1    -1    ]

A = 
[    -3    2    -1    ]
[    0    1    1    ]

I managed to do AB, but BA is confusing because idk what to do when I reach the 3rd element of the first row of matrix A
Can someone help with this?

[clever cedar]

Can u take a picture of ur work

So i can see where ur confusion lies

[stoic python]

Marino:

Marino:

[clever cedar]

For BA

U have to make three copies of B

That are multiplied by each column of A

Respectively

[wintry steppe]

what do you want to see?
ok so first:
(1x-3)+(1x0) .... (0x2)+(0x1) ... then what do i multiply the last column of A by?

[clever cedar]

Its hard for me to type on my phone but if u give me 10 mins

I can write it out for u

So for example the first column of BA is

The entirety of B multiplied by the the column of A

[wintry steppe]

why is that so?

[prisma kestrel]

you can use my tex translations then and maybe copy it down

[clever cedar]

Ok ill write it on paper

[clever cedar]

Give mr 2mins

Or this bitch steals my thunder

[wintry steppe]

dude that's the answer xd

[clever cedar]

Ill show u the steps

[prisma kestrel]

i think the solution wasnt asked but the steps to get there :^)

[wintry steppe]

wait now i'll just try to figure out the steps

[north sierra]

oh lol

thought you wanted the answe

ok someone else could give you the steps

[wintry steppe]

ty @prisma kestrel for the tex

[prisma kestrel]

if im not wrong, matrix multiplication, it means you multiply col from first by row from second matrix

[wintry steppe]

yeah

[prisma kestrel]

so in B*A, m-th col, n-th row it means m-th col from B * n-th row from A

[wintry steppe]

but col of first is 1 short of row of 2nd

[wintry steppe]

can you take a closer pic please?

[clever cedar]

Open original

[wintry steppe]

zooming is blurry

[clever cedar]

But yh sec

[north sierra]

its row from 1st row matrix by column of the corresponding second matrix

[prisma kestrel]

like this?

[clever cedar]

Yh

Then expand

Notice those commas

Delimit columns

[wintry steppe]

yeah

[north sierra]

its easier to just do in your head imo

instead of writing all that

[clever cedar]

I mean sure but showing ur work proves u understand it on exam

[prisma kestrel]

if you understood you can do it in your head but if you havent understood it yet the steps are important 😉

[wintry steppe]

i don't understand why you are multiplying the whole of matrix B by each column

[clever cedar]

A matrix is just a set of vectors

Thats howby multiply a vector by a matrix

[wintry steppe]

what's the difference between a vector and a matrix?

[north sierra]

true marino i dont really understand the steps lol

[clever cedar]

A vector is a list of numbers

[dusky epoch]

set != array

[clever cedar]

Yh sorry disregard what i previiusly said about matrix

[lone quail]

@@vast torrent ok thanks

I was watching threeblue1brown

https://youtu.be/P2LTAUO1TdA

What is it doing at minute11?

How does it work?

[north sierra]

did you get it x14dev

[wintry steppe]

i still don't understand tbh why we multiplied the whole of B each time

no

did you get to your solution in the same way? @north sierra

[north sierra]

nope

ok so basically

ill write the steps

[wintry steppe]

ok

[prisma kestrel]

@dusky epoch whats an array in math for you?

[north sierra]

ok not really the steps but do you see whats happening ? @wintry steppe

one sec

i made an error

[clever cedar]

Does ur textbook teach that?

[north sierra]

im not sure

ok that should be good

so i take the first row of the first matrix and multiply it by the first column of the second matrix

then first row of the first matrix by the second column of the second matrix

then first row of the first matrix by the third column of the third matrix

(first row,first column) corresponds to (1,1) so you put the sum in (1,1)
(first row, second column) corresponds to (1,2) so you put the sum in (1,2) etc..

[prisma kestrel]

oh then i confused it, i thought it was col * row but its row * col oO

[clever cedar]

Its the same thing just without showing the steps

[north sierra]

yeah i dont really get the steps which is bad

lol

[celest bridge]

can someone please help me understand what the L_A matrix is?

[wintry steppe]

aha it makes sense now, so the entire row by each column

[north sierra]

yeah

[clever cedar]

????????? Its literally the same thing

How does his make more sense

??????:monkaS-1:

[north sierra]

i dont know

[celest bridge]

I don't understand why this isn't getting through my head, it seems like it should be real simple

[north sierra]

this way is more intuitive for me

[wintry steppe]

i have no idea how you got those smaller 1 column matrices

in the last step

@clever cedar

sorry not you

[north sierra]

kk

[wintry steppe]

thanks for the explanation too @north sierra

[north sierra]

np

[wintry steppe]

next time i will check the textbook too (i completely forgot to check it) uh

[prisma kestrel]

what the hell is an empty operation???

this thing

[wintry steppe]

that sign stands for an empty set, which contains no elements

but besides that idk

[gray glen]

just a function from $\emp^2$ to $\emp$

[stoic python]

CaptainLightning:

[gray glen]

although $\emp^2=\emp$ so this is actually just the empty function

[stoic python]

CaptainLightning:

[gray glen]

on the empty set

which if you consider its relation to be the function itself

is also the empty set

[half ice]

Can you have null×null?

[gray glen]

yes

[half ice]

Is that just the empty set?

[gray glen]

yes

[prisma kestrel]

yeah but

a map from A -> B means i can input elements from A and i get elements from B

but the empty set has literally no single element

so what should i input

and in reverse what should be the output???

[sonic osprey]

Look at the set theoretic definition of a function

[prisma kestrel]

this?

[sonic osprey]

that's not a definition

[prisma kestrel]

and i dont know where to find a definition

so that was my trial to find something in wiki 😄

[half ice]

It's the empty function. There's no mappings

[prisma kestrel]

dafuq is this now

you mean a function whose existence is clearly clueless?

[sonic osprey]

https://en.wikipedia.org/wiki/Function_(mathematics)#Relational_approach

[half ice]

One could think of a function A → B as the set of arrows starting at an element in A and ending at an element in B.

"No arrows at all" also counts as a function by this def

[prisma kestrel]

so this might be the definition

or rather that both criteria need to be fulfilled for a relation to be a function

[sonic osprey]

Yes

And you can check that the empty relation satisfies this definition

[half ice]

That second picture is a relation between sets X and Y

[sonic osprey]

Functions are just a special type of relation

[prisma kestrel]

yeah sure

left total and right unique if these are the english words idk

already heard about it 30 min ago 😄

but okay

so i first thought the 2nd condition isnt fulfilled

because for any given element from X (lets imagine X -> Y) there is no element in Y to be part of the relation

but okay there are no elements in X so the condition is still satisfied

[half ice]

There exists the empty function on A → empty as well

[prisma kestrel]

and the top one is always fulfilled because the lhs of the implications always yields false (because there doesnt exist any tuple)

X -> Y is our empty -> empty

A -> empty?

wait

but the bottom one says

for every a there has to exist min 1 y

but there exists no y

because y is element of empty

[half ice]

Oh true, mb there is no empty function on that

[prisma kestrel]

THANKS GOD

[half ice]

There's an empty relation though

[prisma kestrel]

because it didnt make sense to me otherwise

empty relation wdym

[half ice]

A relation between A and B is a subset of A×B. The empty set always counts

It's a set of arrows between A to B with no rules on the arrows

[prisma kestrel]

you mean that if R = subset of A x B and R = empty?

[half ice]

Ya ya

[prisma kestrel]

i mean

a ° b = false

xd

in computer science language talked

[half ice]

There could just be no arrows between A to B, that's an empty relation

[prisma kestrel]

with arrows you probably mean that if (x; y) is element of R, then x points an arrow to y

[half ice]

Ya ya

Sorry I can tell I'm not being clear here oop

[prisma kestrel]

its all good

i already guessed you meant that

[half ice]

You know, I don't know why I'm explaining this, you likely know these definitions oop

[prisma kestrel]

oop?

[half ice]

That's not a math thing just an expression I have

[prisma kestrel]

thanks god

what was the half group again? only associative?

collecting the stuff i learned in onenote so i wanted to summarize it if i once forgot it 😄

interesting...

so we could build a group of integers but only a monoid of natural numbers

understood that right? 😄

[half ice]

There's a semigroup, which only needs associativity

[prisma kestrel]

im sry

[half ice]

Closure is always assumed in a binary operator

[prisma kestrel]

trying to translate the german words for it

[half ice]

Ah IC

[prisma kestrel]

in german its "half group"

[half ice]

I've never used one for anything lol

[prisma kestrel]

a semigroup you mean?

[half ice]

Ya ya

[prisma kestrel]

well but its interesting how it builds up

so first ass law, then neutral, then inverse

no wait

[half ice]

"ass law" lol

[prisma kestrel]

yes

i loved to write that

if you might wanna ask

[half ice]

You're right, you need an identity before an inverse

[prisma kestrel]

identity?

[vast torrent]

xoe = eox = x

for the operation o, for any x

then e is an/the identity

[prisma kestrel]

in german its called neutral element

MATH

DONT FUCK WITH ME

[half ice]

That's probably a better name

I have no idea why we call it an identity

[gray glen]

really

no idea at all?

[prisma kestrel]

well neutral element sounds better to me as well

[vast torrent]

well the identity map is $x \mapsto x$

[stoic python]

gfauxpas:

[vast torrent]

because they're equal, identity implies things are equal

so I guess it's based on that

[prisma kestrel]

yeah i guessed the same

so i got a question

if we got a ring (M, +, *)

as concerns what i learned so far a ring has ass law, identity and inverse element for "+" and ass law + dist law of "*" whereas dist law is related to "+"

so a commutative ring restricts * to be commutative

and a unitary ring demands an identity element for *

but what about if we combine commutative and unitary?

does this have a name as well?

[dusky epoch]

commutative ring with unity

[prisma kestrel]

so it has no name, just two special properties

are operations in at least semigroups closed?

according to wiki they are, if wrong, just hit me up

is a general ring (R, +, *) commutative with respect to +?

because in my last video i've seen he didnt mention it but wiki tells otherwise

ok all comments say "+" needs to be commutative... 😄

[wintry steppe]

can this (B^T+A)C, be simplified to something other than CB^T + AC ? where B^T is B transposed

[prisma kestrel]

matrix?

[wintry steppe]

yeah

like is there a shortcut to solving it

[prisma kestrel]

maybe you should tell what C is

i cannot help you but those who can could be interested in

[wintry steppe]

C is a symmetric square matrix

i'll just do it this way, CB^T + AC

[prisma kestrel]

depends on the laws defined on matrices

distributivity in this case

if there is dist law or not

[dreamy depot]

To show indeed show that $\mathcal{F_{a}}$ is a subspace of $\mathcal{F}$ is it enough say that, $$(f + g)(a) = f_(a) + g_(a) = 0 \in \mathcal{F_{a}}$$ $$a(f(a)) = 0 \in \mathcal{F_{a}} $$ ?

[vast torrent]

what's $f_0$?

[stoic python]

gfauxpas:

[dreamy depot]

oh my bad typo it should be fixed now

[wintry steppe]

Yo if my line goes through (-3,-3) and x=0 how do I find the slope since I want to get a parallel line

[gray glen]

you need to check that all conditions are maintained under addition and multiplication by a real

[wintry steppe]

Wat?

[stoic python]

Zophike1:

[dreamy depot]

Ahh ok @gray glen

[wintry steppe]

O srry I thought u were talking to me, my bad

[gray glen]

I could say "what about af(1)"

also uh

[wintry steppe]

Yo if my line goes through (-3,-3) and x=0 how do I find the slope since I want to get a parallel line

[gray glen]

what are you trying to say here

[dreamy depot]

@gray glen let me relatex what I wrote

[gray glen]

you also can't say either of these things in general

[vast torrent]

the question is, is this suficient to prove the space is a subspcae

if you know these properties hold for all functions in a subset

[gray glen]

you can't say f(a)+g(a)=0 for all a in [0,1]

or that a(f(a))=0 for all a in [0,1]

even for f and g in the subset

[vast torrent]

they're saying, IF it's true

[wintry steppe]

im struggling to simplify this equation to find matrix C

[gray glen]

well it isn't much help to show that it's a subspace if you can't say either of those things for arbitrary elements of it

[vast torrent]

lightning, for example

[dreamy depot]

@gray glen the first part of the exercise was to show that it's a vector space so that's why i'm trying to show that it's a subspace

[gray glen]

I know

[dreamy depot]

Or wait is showing that it's a vector space implies that it's a subspace

a better way of going about it

[gray glen]

also what was the original space

[vast torrent]

go through the 3 step subspace test

nonempty, closed under scalar, closed under addition

the zero mapping is there

so it's not empty

[dreamy depot]

For the Subspace Test wouldn't I have to perform on the elements of my space

[vast torrent]

lightning, the original space is [0..1] integrable functions

[dreamy depot]

Like perform it on $f, g$ respectively ?

[stoic python]

Zophike1:

[vast torrent]

oh wait, specifically, continuous on [0..1], f(0) = 0, f(1) = 1

[dreamy depot]

No the condition is f(0) = 0, and f(1) = 0 for the context of the question that's why I tried doing f(a) = 0 for all in a

[vast torrent]

you have to show:

the 0 function is in the space or that it's non empty

that if f, g are in the space then so is the pointwise sum f+g

[dreamy depot]

ahhh ok

[vast torrent]

and if f is in the space and c is a scalar, the pointwise scaled function cf is in the space

[gray glen]

which is why I said you couldn't assume that

[vast torrent]

now, what is a

[dreamy depot]

a is some arbutiary constant right ?

[vast torrent]

I don't know, is it?

a is any number from 0 to 1?

[dreamy depot]

So the wise thing to do would be to consider $(f+g)(a) = f(a) + g(a) = (g + f)(a) $

[stoic python]

Zophike1:

[dreamy depot]

@vast torrent yes

[vast torrent]

well you dont need to check f+g=g+f, we know that's how pointwise addition of function works

[dreamy depot]

Oh my bad we would need to check that f(a) + g(a) is closed

[wintry steppe]

Yo if my line goes through (-3,-3) and x=0 how do I find the slope since I want to get a parallel line

Sorry for texting this for the third time it's just no one responded

[gray glen]

uh

this isn't really the channel for that

[wintry steppe]

O srry which one?

[gray glen]

#prealg-and-algebra

[wintry steppe]

can someone help me simplify this equation in a form that's easy to find C?

[dreamy depot]

@gray glen how would you apply the Subspace Test i'm a bit stuck 😦

[gray glen]

well

[dreamy depot]

https://proofwiki.org/wiki/Subspace_of_Real_Continuous_Functions I'm looking at the general proof for some insight

[gray glen]

given two elements f,g of the subset

show that their sum f+g has all of the properties necessary to still be in the subset

go through each property in turn

then do similarly for cf for an arbitrary real c

[dreamy depot]

so should I use the sum for continous functions ?

[gray glen]

for showing their sum is continuous in particular, you'd want that the sum of two continuous functions is continuous

(this is true)

would you be expected to prove this as well or would you take it as obvious?

it's definitely a very fast proof even if you need to

[dreamy depot]

Can't we take it as obvious as this case or would I have to explain it ?

[gray glen]

if you feel like it's obvious then you can probably just say it is

but still point it out

[dreamy depot]

but know I have a clearer idea on proving the second part of the exercise I was a bit confused since a lot of proofs applying the subspace test usually are with involve employing matrices

[vast torrent]

C⁰([0..1]) is uncountably infinite dimensional

So you can't use matrices

[dreamy depot]

fair point I'll spend more time playing with functions that form a vector space

[vast torrent]

Im not even sure you can do a "matrix" with countably many rows and columns for countably infinite spaces, actually . Idk if it's possible

But certainly not uncountably infinite dimensions

[wintry steppe]

can someone help me simplify this in a form where I can solve for C?

[dreamy depot]

@vast torrent Since indeed $f,g$ are continuous functions it's easy to see that $\mathcal{F_{0}}$ is indeed a subspace of $\mathcal{F}$ since one has to establish that $f+g \in \mathcal{F_{0}}$, and $a(f) \in F_{0}$. To show this one can note that

$$\lim_{x \rightarrow c} (f + g) (x) = \lim_{x \rightarrow c} (f(x)) + \lim_{x \rightarrow c}(g(x)) = (f +g) \in \mathcal{F_{0}}$$

as well as that,

$$\lim_{x \rightarrow c}(c(f(x)) = c\lim_{x \rightarrow c}(f(x)) = a(f) \in \mathcal{F_{0}}.$$

[stoic python]

Zophike1:

[dreamy depot]

^ I think this should be a good proof It's worded a little awakwardly but it should do the job

[gray glen]

on the left you have real numbers and on the right you have functions

it also ends with what you're trying to show stated (unjustified)

it seems like you're avoiding words in the later part and that might be the problem

you want to go like

(proof of property 1 for f+g)

[tame root]

what do these brackets around the vector mean? I've been seeing this a lot recently. The B is a basis for vector space V and x is an element of V

[gray glen]

"so we have property 1 for f+g"

(proof of property 2 for f+g)

[dreamy depot]

Ahh ok @gray glen

[gray glen]

"so we have property 2 for f+g"

until you get all the properties shown

and finally conclude f+g is in the subset

[dreamy depot]

also what was unjustified in my proof

?

[gray glen]

well to start you haven't said anything about (f+g)(c)

[dreamy depot]

ahhh ok

[gray glen]

and that's just talking about them being continuous

you said nothing about (f+g)(0) or (f+g)(1)

[dreamy depot]

So I would have to perform the same steps for (f+g)(c) and (f+g)(0) ?

[gray glen]

this is definitely a no go

well the steps are different really

[dreamy depot]

oh fair point

dang i'm really stuck on this 😦

[gray glen]

you should say like "for c in [0,1]" blah blah "as f and g are both continuous" blah blah "so as c was an arbitrary element of [0,1] (f+g) is continuous"

stuff to that effect

[dreamy depot]

so was my idea in the corrected solution wrong then ?

[gray glen]

I think the general idea was there but the execution was quite off yeah

[dreamy depot]

oh I just didn't state it rigorously ?

[gray glen]

I mean the picture I posted is just wrong

[dreamy depot]

ahh all right 😦

[gray glen]

oh and also show it's nonempty but that's a small thing

[prisma kestrel]

why is the 0 excluded in field's definition for the multiplication group being abelian?

so why for a field (K, +, *) is only (K \ {0}, *) an abelian group

[dreamy depot]

@gray glen I'm clueness at this point on how to show it I get how it forms a vector space and how prove that it forms a vector space but I don't get how to use the subspace test to show it

[empty copper]

0 doesn’t have a multiplicative inverse

[prisma kestrel]

wouldnt a commutative monoid work either?

[dreamy depot]

I think I know how to show it now give me a minute

[empty copper]

But that’s a monoid

[prisma kestrel]

well on the one hand we got (K, *) = comm monoid or (K \ {0}, *) = abelian group

[empty copper]

Not a group

[prisma kestrel]

no wait

we want the inverse for every element except for 0

🤦‍♂️

but it bothers me that we totally exclude it

because then we exclude the case where 0 wouldnt fulfill the ass law for example

really theoretically here

[slender yarrow]

can you really have an inverse for 0 though

that's the thing

[prisma kestrel]

well actually

is the 0 excluded because identity element of addition?

or where does the "0" symbol come from

i mean we're talking about a general idea of a field

[empty copper]

I’m pretty sure one can show that within any field, any additive neutral element cannot have a multiplicative inverse

The 0 symbol is usually reserved to denote the additive neutral element

[prisma kestrel]

but still we talk about any additive neutral element

[empty copper]

Also pretty sure it’s unique :p

[prisma kestrel]

so not specifically OUR neutral element of addition which has no inverse

[empty copper]

By virtue of it being an additive group

[prisma kestrel]

what is an additive group? you mean with operation = addition? 😄

[empty copper]

Yes

[prisma kestrel]

aight

well

but still can we say something about general stuff? where we havent got numbers etc?

[empty copper]

A field, by definition is an additive group that’s also a multiplicative monoid

[prisma kestrel]

wasnt addition and multiplication abelian in fields?

[sonic osprey]

multiplicative group, once you remove the zero

[empty copper]

Addition is almost unanimously deemed to be an abelian operation

I meant commutative

But multiplication is not

[prisma kestrel]

well but still in our general definition we only have binary operations which yield no idea of our usual addition etc

[empty copper]

Yes you could call them differently

How about this: a field, by definition, is a commutative group that’s also a monoid under another supplied operation

[prisma kestrel]

how can a field be a group? a group only has 1 op

[empty copper]

Ok, although there’s an addendum: only the neutral element of the commutative operation doesn’t have an inverse under the second mentioned operation

A field, restricted to the relevant operation :p

[prisma kestrel]

well actually

[dreamy depot]

@gray glen I coudn't get it in the end I ended up looking up the solution : https://yutsumura.com/subspaces-of-the-vector-space-of-all-real-valued-function-on-the-interval/

[prisma kestrel]

if i think about it correctly

actually its all just definition

which can be thought of as good or bad

and i think its imprecise

but in the end only K\{0} needs to fulfill the abelian stuff

[gray glen]

uh those are slightly different sets

[dreamy depot]

@gray glen the set for part (a) would be right for my quesiton

[gray glen]

that set doesn't have the continuous requirement

or that f(0)=0

[dreamy depot]

oh fair point but what was done in the solution for part (a) could work for the question i'm dealing with right ?

[gray glen]

uh I mean it's a bit different

but you could structure that part similarly

[dreamy depot]

@gray glen but I think I see how to solve my question know 🙂 sorry for being stupid 😢

[prisma kestrel]

#mathlearninginanutshell

#being5yearsdumbtobeintelligentasfuck

[dreamy depot]

That is a fair point but I get angry at myself for not being able to come up with things on my own

[vast torrent]

That comes with time

[dreamy depot]

I know

[half ice]

Math is too big to reinvent

[dreamy depot]

@gray glen here's my final solution

$\text{Proof}$

To show that $\mathcal{F_{0}}$ is indeed a subspace of $\mathcal{F}$ we first must define that,

$$\mathcal{F_{0}} = \big{ f \in C[0,1] , \big| f(0)=0, f(1) = 0 \big}.$$

Then we consider the addition $(f+g)(x)$ it's indeed easy to see that,

$$(f + g)(0) = f(0) + g(0) = 0 \in \mathcal{F_{0}}$$
$$(f + g)(1) = f(1) + g(1) = 0 \in \mathcal{F_{0}}$$

Thus it follows that $(f+g)(x) \in \mathcal{F_{0}} $

Since $f \in \mathcal{F_{0}}$ we have that $f(0) = f(1)$ the scalar multiple $(cf)(x)$ satisfies

$$(cf)(0) = c \cdot f(0) = c \cdot f(1) = (cf)(0)$$

Thus $(cf)(x) \in \mathcal{F_{0}}$

[stoic python]

Zophike1:

[gray glen]

where did you consider continuity

also this isn't really the idea

[dreamy depot]

oh my bad

[gray glen]

0 real number vs 0 function

(f+g) isn't necessarily the zero function

what you've done is show that (f+g)(0)=(f+g)(1)=0

which is good because that's one of the things necessary for f+g to be in the subset

but it isn't everything necessary for f+g to be in the subset

you still need that f+g is continuous

[dreamy depot]

I thought that for f+g to be in the subset it had to satisfy all the things within the subspace test

ahh ok to fix the proof should I just state that f,g are continous respectively

[gray glen]

and that this makes f+g continuous

make sure you don't get confused between the function f+g and the image of a certain point a under f+g

which is (f+g)(a)

[dreamy depot]

But wait I apologize for the dumb question but didn't I already define that f is indeed continuous earlier in the proof, by setting $\mathcal{F_{0}} = \big{ f \in C[0,1] , \big| f(0)=0, f(1) = 0 \big}.$ ?

[stoic python]

Zophike1:

[gray glen]

f is continuous and g is continuous

but you have to say something about f+g being continuous

[dreamy depot]

ooooh ok now I understand the issue

[gray glen]

and why it is

but yeah this is a dangerous thing to do

[dreamy depot]

So I have to also employ the property that it is continuous as well

[gray glen]

you have to say that it's continuous and why it's continuous yeah

[dreamy depot]

To show that $\mathcal{F_{0}}$ is indeed a subspace of $\mathcal{F}$ we first must define that,

$$\mathcal{F_{0}} = \big{ f \in C[0,1] , \big| f(0)=0, f(1) = 0 \big}.$$

Then we consider the addition $(f+g)(x)$ which produces a continuous function since,

$$D_{x}(f + g)(0) = D_{x}(f(0)) + D_{x}(g(0)) = D_{x}((f+g)(0))$$
$$D_{x}((f + g)(1)) = D_{x}(f(1) + g(1)) = D_{x}((f+g)(1))$$

Thus it follows that $(f+g) \in C^{(1)}(0, 1)$ and $C^{(1)}(0, 1)$ is closed under addition.

Since $f \in \mathcal{F_{0}}$ we have that $f(0) = f(1)$ the scalar multiple $(cf)(x)$ satisfies

$$D_{x}(cf)(0) = c \cdot D_{x}f(0) = c \cdot D_{x}f(1) = D_{x}((cf)(0))$$

Thus, $(af) \in C^{1}(0,1)$ and so $C^{1}(0,1)$ is closed under scalar multiplication.

[stoic python]

Zophike1:

[vast torrent]

If you prove f+g or cf you get the 0 function

Either 0f or f+(-f)

[dreamy depot]

Yeah is that a misstep on my part ?

[vast torrent]

What does Dx mean

We're not given tbat these functioms are diffable

[dreamy depot]

$\frac{d}{dx}$

[stoic python]

Zophike1:

[vast torrent]

How are you diff'ing if we don't know tje functions are differentiable

[dreamy depot]

But aren't continuous functions differentiable

[vast torrent]

No, differentiable functions are continuous

[dreamy depot]

ahh okay my bad

[vast torrent]

Consider |sin x|

[dreamy depot]

fair point now i'm confused how do I show that $f+g$ is continous

[stoic python]

Zophike1:

[vast torrent]

Continuous but not diffavle where sin x changes sign

Well there's more than one definition of continuous, which do you know?

[dreamy depot]

oh you would just take the limit

nvm then

[vast torrent]

that would work

write f(x+h)+g(x+h) = (f+g)(x+h) for |h| small and non zero

for epsilon != 0

maybe h would be better than epsilon because usually we use epsilon>0

changing it

there

note, however, that your work would be useful when analyzing the vector space of differentiable functions

something is still bothering me about this though

you say f(a) + g(a) = 0 in your space?

for every a?

[dreamy depot]

$$\mathcal{F_{0}} = \big{ f \in C[0,1] , \big| f(0)=0, f(1) = 0 \big}.$$

[stoic python]

Zophike1:

[dreamy depot]

Sorry posted the wrong thing eariler

[vast torrent]

yes but does that mean f(1/2)+g(1/2)=0 for all f, g in C[0,1]?

[dreamy depot]

yes

[vast torrent]

well that's a bizarre set

[dreamy depot]

Unless I'm misunderstanding the set

[vast torrent]

post the origina,l problem again

[vast torrent]

that doesnt say that f(a) +g(a) = 0

[dreamy depot]

oh shoot your right I must have misread the problem then

[vast torrent]

yeah, the problem is just to show this is a vector space

${ f \in C^0([0..1]): f(0) = f(1) = 0}$

[stoic python]

gfauxpas:

[dreamy depot]

ahhh okay I see where I got confused now 😦

[vast torrent]

so you need to prove two things

that it's closed under vector operations for continuity

and that the f(0)=f(1)=0 thing stays true for the vector operations

[dreamy depot]

I get how to show how it's closed but I don't get how to show that f(0) = f(1) = 0

[vast torrent]

You know that for both f(0)=f(1)=0 and g(0)=g(1)=0

Now just check (f+g) and cf

[dreamy depot]

oh ok

[vast torrent]

or just check (cf+g) in one step

[dreamy depot]

So to check $(f+g)$ I would just take that $$\lim_{h \rightarrow 0} (f(x+h) + g(x+h)) = f(0) + g(0) = f(1) + g(1) = 0 $$

[stoic python]

Zophike1:

[dreamy depot]

to conclude that indeed $(f+g)(x) \in \mathcal{F_{0}}$

[stoic python]

Zophike1:

[vast torrent]

You need to check closure under continuity seperately from checking closure under the boundary conditions

They're two different creatures

[dreamy depot]

ahhh ok I thought you wanted me to do it all at once

[vast torrent]

No, you can't

Or shouldn't

but instead of checking f+g and cf seperately you can check cf+g

Crossed it out bc people usually don't use that shortcut, at least not when you're taking your first la class

[dreamy depot]

That's fair i'm just trying to piece together the proof at this point I get some of the idea's but I can't put togther the bigger picture

[vast torrent]

Do you know what it means for a set to be closed under some operation?

[dreamy depot]

yeah when you add members of the set it produces members of the same set

[vast torrent]

"add" or perform whatever operation being considered

In linear algebra, the two operations you need to check for closure

Are addition, and multiplication by a scalar

[dreamy depot]

So I should do $(f + g)(x) = f(0) + g(0) = f(1) + g(1) = 0 $

[stoic python]

Zophike1:

[vast torrent]

Your set has two defining properties

Continuity and the bdry conditions

You need to check that both those things are closed under addition and scalar multiplication

[dreamy depot]

ahhh ok by boundary conditions you mean f(0) = f(1) = 0 ?

[vast torrent]

Yeah

You can't write f(x)=f(0)=0

That's not right

[dreamy depot]

At this point can I just see the proof ? I've been stuck on this for hours now

[vast torrent]

You're close

Do 4 things

Prove f+g is continuous

Prove cf is continuous

Prove (f+g)(0)=(f+g)(1)=0

Prove (cf)(0)=(cf)(1)=0

To prove a function is continuous, use this definition

$\lim_{h\to0}f(x+h)=f(x)$

[stoic python]

gfauxpas:

[vast torrent]

Go forth young man

[dreamy depot]

To prove the first point I noted that $(f+g)(0) = f(0) + g(0) = f(1) + g(1) = (f+g)(1)$

[stoic python]

Zophike1:

[dreamy depot]

^ Is this fine ?

[vast torrent]

Add =0

[dreamy depot]

all right thanks

Woah I got it thanks man @vast torrent 🙂 I see the proof and I get how and why it works

Also what was the shortcut you mentioned I shouldn't use ?

[clever cedar]

does anyone know if the cauchy swartz inequality holds for the cross product

[north sierra]

for 9, why is is that i can just make the equations = 0 and then that makes it a vector space?

[clever cedar]

wdym

the basis?

[half ice]

Are you talking about testing if the zero vector is in the set?

Well, is it?

[north sierra]

in the solution the guy just does thiis

[clever cedar]

are u familiar with subspace test

[north sierra]

ya

[clever cedar]

pretty much he's summarizing that the subspace given satisfies the subspace test

since u can make those vectors into a zero vector with being a suspace of R^n

as well as its closed under addition and under scalar multiplication

[north sierra]

but he doesnt show the addition and scalar multi part

[clever cedar]

slader is also student provided solutions so my guess is he didnt go into depth

but i could be wrong

[north sierra]

brb ill be back

idk

@half ice ?

[half ice]

Oh I see. He noticed that it is a null space of some matrix, which is a vector space

[dreamy depot]

Dumb question does a Cone form a vector space ?

[half ice]

Not a bad idea

[north sierra]

can you explain kaynex?

[half ice]

The null space of any matrix is a vector space.

[north sierra]

how did he know it was Nul space

[half ice]

Rearranged it so it was

Can you see the similarity between the matrix and the equations in the solution?

[north sierra]

ya

they’re the coefficients of the equations

but how does he know without row reducing

that theres a set of solutions to Ax=0

[half ice]

Remember that the null space is the set of vectors that, when put through the matrix, gives 0

[north sierra]

ya

[half ice]

We also have that a - 2b - 4c = 0, 2a - c - 3d = 0

What happens if we put [a,b,c,d] through that matrix?

[north sierra]

it should equal 0?

[half ice]

Ya ya. You'll get the zero vector. So [a,b,c,d] is in the nullspace of that matrix

[dreamy depot]

I'm struggling with the last part of this question i'm having trouble seeing that a cone is not vector space. I'm I missing something here because if it's closed under addition and mutiplication shoudn't the cone be a subspace or R^{3}. I'm I missing something here !!?

[half ice]

Here's a cheat, it's not a vector space if there's a constant in the equations. 7 and 8 are not.

[vast torrent]

@dreamy depot instead of testing f+g and cf in two steps you can test cf+g

[dreamy depot]

@vast torrent for the problem i'm facing or the shortcut I was asking about for the last problem

[vast torrent]

For testing closure under vector operations

Like if i have a map T and i want to see if its linear

I check T[x+y]=Tx+Ty

And T[cx]=cTx

[dreamy depot]

ahhh ok I see

[vast torrent]

But the shortcut is to check T[cx+y]=cTx+Ty

[half ice]

"multiplication by positive numbers"

There's other scalars that break this

What happens if your scalar is -1?

[dreamy depot]

ahhh I see so I would just have to throw some negative numbers at it

@half ice when you do multiply -\alpha (a) you would get atributary scales outside our side

so therefore $C$ cannot indeed be a vector space since the axiom closed under scalar multiplication is not satisfied

[stoic python]

Zophike1:

[dreamy depot]

@half ice ^ is this a good rigorous answer

[half ice]

Yeah

Just one counterexample is enough

[dreamy depot]

Sweet thanks man 🙂

[carmine terrace]

can anyone help me understand how they calculated cos for this

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

Dot product u*v makes it = to -1

and they use this formula

https://i.imgur.com/u8vifsV.png

to get -sqrt3/2

u*v = -1

is the absolute value u*v = 1?

[gray dust]

the bottom is magnitude u * magnitude v

[carmine terrace]

ah that's why

thanks, imma look up on how to get the magnitude

[clever cedar]

its a dot product property

u * v = ||u||||v||cosx @carmine terrace

u * v = |u||v|cosx

the bars should be double but discord sees that as a spoiler tag

its length of the vector

and * is dot product in this case

$\cdot$

[stoic python]

mike0x81:

[carmine terrace]

Thank you Mike, that's a good explanation.

[clever cedar]

anytime

[paper egret]

Let V be a finite-dimensional vector space. Let U and W be subspaces of V with dim(U) + dim(W) = dim(V).

The following are equivalent
V is the direct sum of U and W
V = U + W
U intersect W is the 0 vector

any hints

just want some sort of direction

dunno how to start this off

[clever cedar]

what is the question

find V?

[paper egret]

oh

prove the following are equivalent

mb

[clever cedar]

np

consider what dimension means in terms of its representation as a subspace

for example

if i say the set V has a dimension of 4

what vector space is it in

[paper egret]

no i know that, i just need some hints on how i should start this proof

whether its easier to start proving statement 1, or statement 3

kind of thing

[vast torrent]

what definition do you have for direct sum

I thought 2 + 3 are the definition of 1

[carmine terrace]

Hey guys I'm having problems with this problem

https://i.imgur.com/b2lh24T.png

How do i find L2?

For L1 I got

x = -3+-3t
y = 3+-3t
z = -1 + -t

but I'm not sure how to proceed for L2

I think this is L2
x= -18+-6s
y= 3+4s
z= 0 + 2s

then I just make em = to each other since they instersect

[north sierra]

kk thanks kaynex i get it

[clever cedar]

@carmine terrace you're suppose to use substution

so like for instance

you take one of the X's

and the other one isolate for T

then plug in the second one into the first one

then you should get a value for X

where if plugged into both equations satisfies

same thing for Y and Z,

[cloud cairn]

if i have 2 points in a plane. i know their points. (green points)
i have a 3rd point (blue) ALSO in the plane, i am trying to solve for.
i know the distance between each of the points. (3 lengths of a triangle given)
i have the equation for the plane (in picture top left)

how to go about solving the 3rd point?

i understand it will probably have 2 solutions. (cuz we could flip the triangle across the line formed by 2 green pts)
the blue point we want must satisfy the equation for plane

[clever cedar]

is the thrid point

on the plane?

[cloud cairn]

yes

all points on the triangle are in plane

[clever cedar]

interesting problem

let me think

wait

u know the distance between green points and blue point right?

[cloud cairn]

yes

L1, L2, L3 are all solved

[clever cedar]

why cant u just

solve for the vector equation

where a green point would be position vector

and direction vector would be distance

[cloud cairn]

i guess i havent covered direction vectors. or to say idk how to find vector that connects the blue dot to green

[clever cedar]

i see

would u mind giving me full problem?

or is that all of it

with the givens

[cloud cairn]

actyually i think i know what ur onto

the full problem is actualy something that preceded this. and i was trying to solve it usng this method

full problem something else 1 sec

[clever cedar]

ok

[cloud cairn]

given equation of plane. and then a point outside of the plane.

[clever cedar]

this is a simpler problem

i wish u presented to me this first

i could explain u it

[cloud cairn]

find the point on the plane closest to the point given

[clever cedar]

i can explain the full solution to u

if ud like

[cloud cairn]

ok

[clever cedar]

first what we want to do is find an arbitrary point on the plane

in order to do that

[cloud cairn]

right i know this part

[clever cedar]

we find [x,y,z] that satisfies the equation of the plane

ok

next

we want to find the vector from that point

to the point outside the plane

[cloud cairn]

right, red line on my picture

[clever cedar]

now

[cloud cairn]

er not red line

[clever cedar]

we want to project that vector onto the normal vector

[cloud cairn]

i mean one of the green lines

the projection is the red line

[clever cedar]

ok

now we take the length of that vector

using pyth theorem

[cloud cairn]

right

[clever cedar]

with that, we now know the distance from the point outside the plane

to a point on the plane

(shortest distance)

[cloud cairn]

yes

[clever cedar]

lastly

take the point outside the plane

and subtract it by the projected vector

and we get the point on the plane

[cloud cairn]

lol...

[clever cedar]

that is closed

closest

[cloud cairn]

"take the point outside the plane
and subtract it by the projected vector
and we get the point on the plane" this is too easy lol
i wish this popped immediately into my head.
im still working my head around vector addition, subtraction, how it all works. and i havent thought deep about it

[clever cedar]

lol

[cloud cairn]

even tho i know its one of the easier subjects of linear algebra

[clever cedar]

its crazy how abstract ideas can be solved with simple subtraction

i understand

hope that helped

[cloud cairn]

it did

and im facepalmin at how i missed that

[clever cedar]

lolol dw

[cloud cairn]

btw i was uisng the pythagorean theorm, everything you dud there

[clever cedar]

haha thats good

so u were done 9/10ths of the problem

[cloud cairn]

but then i went off into some bizaare tangent

[clever cedar]

happens

[cloud cairn]

ok question

you said to solve the length of the projection vector

and then you say subtract the projection vector from the point

when do we use the length?

[wintry steppe]

can someone help me simplify this further, to be able to solve for C?

[uneven bloom]

((AB^T)^(-1))^T=((B^(-1))^TA^(-1))^T=(A^(-1))^T*B^(-1), so the answer is ((A^(-1))^T)^T=A^(-1).
I’m using the properties:

1. (AB)^(-1)=B^(-1)A^(-1). This works since inverses are well-defined and since ABB^(-1)A^(-1)=B^(-1)A^(-1)AB=I.
2. (AB)^T=B^TA^T
   Element ij in (AB)^T is element ji in AB, which is sum a_(jk)b_(ki). But then element ij in B^TA^T is sum b_(ki)a_(jk), so these are equal.

[wintry steppe]

@uneven bloom ((B^(-1))^TA^(-1)) is it ^TA^-1 or ((B^(-1))^T x A^-1?

[uneven bloom]

T is applied to B^(-1) and separate from A^(-1)

[wintry steppe]

so C = A^-1 ?

[uneven bloom]

The RHS simplifies to A^(-1)

Solving for C is easy now

[wintry steppe]

ok

dang it this is so complicated

applying all these property rules

gonna take a bit till i can do that myself

could you just help me with this last one too please? @uneven bloom

[dusky epoch]

$D = \begin{bmatrix} 3 & 1 \ 5 & 2 \end{bmatrix}^{-1}$

[stoic python]

Ann:

[wintry steppe]

thanks @dusky epoch but how do i arrive to a diagonal matrix from that?

[dusky epoch]

what do you mean "arrive at a diagonal matrix"

[wintry steppe]

doesn't D have to be a matrix with 0s on each side of the diagonal?

[dusky epoch]

,,, where did you get that from?

just because a matrix is called D doesn't mean it has to be diagonal lmfao