linear-algebra

[nocturne jewel]

so clearly you require $2=q_1-q_2 \ 2=-2q_1$

[stoic python]

Mosh

[summer sinew]

interesting!!!!

wow, wait so

[nocturne jewel]

which is like I said... you just solve the system

[summer sinew]

ok

This is really interesting like the theory behind it

[nocturne jewel]

...

[summer sinew]

im so confused rn

[nocturne jewel]

I had said the theory behind it already, it's the sets ability to equal 0 via a linear combination

[summer sinew]

im fuckin LOST rn bro

your doing your part

its just all new for me

nvm i think i get it

is it 2 = q1b - q2c

[nocturne jewel]

no

2 is a scalar and b and c are vectors

so that is nonsensical

[summer sinew]

damn man

i wish i knew it how you knew it

Not clicking with me

I get that q1 and q2 are units

[nocturne jewel]

scalars

not units

[summer sinew]

scalars are essentially units of measure right?

[nocturne jewel]

scalars are objects which, for all intensive purposes, arent vectors

they just scale vectors

if I work over a R space, then the scalars are real numbers, work of a C space, scalars are complex numbers

[summer sinew]

Ok

Thank you for your time

[crude sedge]

i solved this pretty quickly with simple logic and geometry, but i can't figure out how to do it using dot products. anyone have some insight?

pretty sure the angle is 120 degrees, btw

[teal grotto]

@crude sedge
v • u = cos(t)

1 = |u + v|^2 = (u + v) • (u + v) = ||u • u + 2 u • v + v • v = 1 + 2 u•v + 1||

and i’m getting 120 degrees as well

[hard drum]

Equilateral triangle basically yes

[crude sedge]

ohhh ok thank you!

[wintry steppe]

How many pivot columns must a 5 X 7 matrix have if its
columns span R5? Why?

is it 2?

<@&286206848099549185>

anyoone there

[lavish jewel]

5

look up "rank"

[wintry steppe]

so what does pivot columns have to do with rank

they're equal?

[abstract prawn]

ye

Trying to do a least squares solution for this with the normal equation, I'm confused on what to do once I do the row reduction.
I dont really know what x should look like because theres a free variable and I dont really know how to present an answer because its in 3 dimensions.
Can anyone help?

[nocturne jewel]

well if they're saying A*A isn't invertible, you're gonna have a subspace of solutions, so either a plane or line in R^3

so it should be a span of either 2 or 1 vector(s)

[abstract prawn]

kk ill have another crack at it with that in mind

tyvm!

[frail ember]

hey positive semi-definite matrices have determinants >= 0 right? why exactly?

wait nvm i get it

one way you can say this is that positive semi-definite matrices have non-neg eigenvalues, and the determinant is a product of those so

[empty grotto]

anyone able to explain what to do here to me

im sorta lost

[kind cobalt]

you have Ax = b, which is equivalent to LUx = b with L being the left matrix part of A and U the right one, and try to set Ux = y, so the equation becomes Ly = b, find y then solve Ux = y

[empty grotto]

didnt I do that?

[kind cobalt]

Dont know i was looking at your notes upside down so i didnt pay attention i thought you were asking if we could explain the method
But if you solved Ly = b and Ux = y then yes this is the right way to do it

[empty grotto]

i should flip the image just realised lol

[half ice]

,rccw

[half ice]

Yeah it should work

[empty grotto]

but i dud that i used the upper triangular matrix and the vector and set that equal to y

[empty grotto]

idk how to continue from here tho

[half ice]

It looks like you actually set that equal to b

Either way, the bottom row reveals no solution

[empty grotto]

yeah

[empty grotto]

wdym by no solution

[half ice]

Bottom row implies 0 = 963

[empty grotto]

yeah

i see that

but it wants a vector with 3 components

do i just solve for it ignoring the bottom row?

[half ice]

Well, it can't have one

[empty grotto]

pain

ik that but is there any way i can get like a partial answer

bc the math program my teacher uses to grade hw doesnt take dne as an answer

[half ice]

I'm checking your answer. y2 is off

[empty grotto]

am i solving for it wrong then?

[half ice]

I don't know what that is

I'm getting:
y1 = -153
y2 = -2
y3 = 12
y4 = 0

[empty grotto]

how?

[half ice]

How did you do it?

[empty grotto]

for the first ros isnt it 1y1 = -153

and for the second row

3y1 + y2 = 457

etc

[half ice]

-3y1 + y2 = 457

[empty grotto]

ic

i may be unable to write down negative signs

thanks a ton for taking a look kaynex

┬─┬ ノ( ゜-゜ノ)

[rugged reef]

Does there always exist a 3x3 invertible matrix A such that $BA^T=A^{-1}C$ where B and C are any symmetric 3x3 matrices?

[stoic python]

ALBERTO BALSAM

[rugged reef]

I posted the above in the questions thread but it seemed no one could answer it there...

[lavish jewel]

off the top of my head, this is a similarity transformation with an orthogonal matrix A

i wouldn't think symmetric matrices are always orthogonally similar

[half ice]

Consider taking the determinant of both sides:
ba = c/a
a² = c/b
Which gives a necessary condition on A. Noticibly, A doesn't exist for some zero values of b or c

@rugged reef

[rugged reef]

If B and C are non-zero will there always exist a A which satisfies the equation though?

[half ice]

I'm using b and c as the determinant of B and C

If C is not invertible, and B is invertible, A doesn't exist

Like Dawdle of Doubt is saying, this is a type of similarity, so B and C will have to have some properties in common

[rugged reef]

Ok thanks.

[summer sinew]

Good morning,

how can I find the values of q1 and q2 that allow us to write a = q1b+q2c

[lavish jewel]

notice that what you have written is a linear combination of b and c

this is also equivalent to putting b and c as columns of a matrix M = [b c]

so that if we make a vector x = [q1; q2], we now have that a = Mx

and you wanna solve for x

you can do this with your favorite method

gaussian elimination, gauss jordan, inverting M, etc

[summer sinew]

Thanks, so what is q1 and q2 mean?

and what is the equation mean

[lavish jewel]

q1 and q2 are the coordinates of the vector a in the basis {b, c}

that's also what the equation means

[summer sinew]

Not sure I understand let me just re read and process this

so I thnk you explained the theory

but I dont understand what the equation means

and how to get values q1 or q2

even if I were to get the values I dont understand like what the problem is saying/meaning

should I focus on the meaning first or the how to

[lavish jewel]

on the meaning, i would say

you can read up on what a basis means and what a change of basis means

[summer sinew]

Like I still can't figure out how to get q1 and q2 w the explanation

[lavish jewel]

you invert M

[summer sinew]

I know a basis is a set of n vectors that is not linear combination

[lavish jewel]

or use gaussian elimination

[summer sinew]

What is M

or gaussian elimination

[lavish jewel]

oof

[summer sinew]

I just want q1 and q2

[lavish jewel]

M is a matrix whose columns are the vectors b and c

you should already know how to solve linear systems in some way by now

you should at least know elimination and substitution

[summer sinew]

I know how to solve systems...

[lavish jewel]

mhm

so solve the system

[summer sinew]

I just dont know how to compile systems from the question

to solve it

[lavish jewel]

a = Mx is the system of equations

in matrix-vector form

do you know how to solve a system in matrix-vector form?

[summer sinew]

just the algebraic form

[lavish jewel]

ok

[summer sinew]

so firstly I get the vector values of each line right?

[lavish jewel]

let $\boldsymbol{a} = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$

[stoic python]

Ephemeral Dawdle of Doubt

[summer sinew]

so a = [2,2] b = [1,-2] c = [-1,0]

[lavish jewel]

actually, let me take a step back

yeah

so far so good

[summer sinew]

great

[lavish jewel]

and you want a = q1 b + q2 c

[summer sinew]

I want q1 and q2

[lavish jewel]

do you remember how scalar multiplication works?

if you multiply a constant by a vector

[summer sinew]

Is that when you multiple each iteration by the other vector iteration and add each one

so like a1 * b1 + a2 * b2

[lavish jewel]

no

q1 is a scalar

and b is a vector [1, -2]

so q1 b = [q1, -2q1]

[summer sinew]

Hmmm

i have no clue what you just did there

So how do I get q1?

[lavish jewel]

sorry, i can't help you

[summer sinew]

Damn, its ok

[lavish jewel]

you'll have to go review basic operations with vectors

[summer sinew]

I know its a hard problem

most ppl havent been able to solve

[lavish jewel]

lol it isn't

[summer sinew]

I dont blame u bro

[lavish jewel]

you just don't know stuff, either because you haven't learned it or you aren't keeping up with the content

so you'll have to go review by yourself

good luck

[summer sinew]

Thanks

I appreciate you trying!

hopefully someone else can get it cuz this one is interestingly weird

[lavish jewel]

i'll just mute you for trolling now

[sonic beacon]

does it make sense to ask whether addition and scalar multiplication on the set hom(V,W) (where V,W are vector spaces) is a linear map itself?

[limber sierra]

they arent functions on hom(V,W), theyre functions on hom(V,W)²

[sonic beacon]

what is hom(V,W)²

[limber sierra]

like, hom(V,W) × hom(V,W)

because addition takes two inputs

[sonic beacon]

oh i see.

[limber sierra]

correction: scalar multiplication is a function on F × hom(V,W) where F is your underlying field

[sonic beacon]

so i guess does it make sense to ask if + and * hom(V,W)² is a linear map

[limber sierra]

but yes, it does make sense to ask that

and you can check the properties by hand, but it is indeed linear

[meager steppe]

hi i'm not really sure what to do after I get the identity matrix after row reduction

[winter harbor]

You supposedly finished the row reduction.

This implies that the matrix has full rank.

And so the kernel of such a matrix consists solely of the 0 vector.

And a basis (indeed the unique basis) of the vector space {0} is the empty set ∅

So the basis for this solution space would be ∅

[lavish jewel]

(with dim 0)

[meager steppe]

i dont understand what you mean by this

[winter harbor]

Rank is defined as the dimension of the collumn space/number of linearly independent columns.

When a matrix has full rank

This means that that the only solutions to Ax = 0 are x=0

[meager steppe]

ok that makes sense now, thank you

[meager steppe]

and what would you refer to this as

[winter harbor]

What do you mean?

[meager steppe]

like its literally just nothihng?

like an empty set

does it have a name

[winter harbor]

Yeah, it's called the empty set.

And it is a basis for the vector space {0}

The solution set of that linear system of equations

[meager steppe]

am i understanding correctly that the span of a vector space is all the linear combinations of the vectors in the vector space? And the difference between the span and the basis is that the basis of a vector space must not only span the vector space but also be linearly independent (so 0 would not ever be a basis because it is always linearly dependent, but it could span a vector space)

[winter harbor]

That's perfect

The reason why ∅ is a basis for {0} is because ∅ is vacuously linearly independent

And also span(∅) = {0}

The span of a subset S in a vector space V may also be characterized as the smallest subspace of V that contains S.

With this characterization

It is easy to see why span(∅) = {0} on a vector space V

{0} is the smallest subspace of V that contains ∅

[meager steppe]

technically every every single subspace includes the empty set right

but 0 is the smallest?

[winter harbor]

Yeah

Smallest in the following sense

If you take any other subspace that also contains ∅

Then this subspace is going to contain {0} as well

[meager steppe]

alright that makes sense, thank you for the help

[winter harbor]

Let $V$ be a vector space, and $S \subset V$ a subset of $V$. Let us denote by:
$$
\mathcal{S} = { W \subset V , \vert , S \subset W , \text{and W is a subspace of V} , }
$$
Then, we have the following characterizations of $\text{span}(S)$
\
\
$\text{span}(S) = \bigcap\limits_{W \in \mathcal{S}} W$
\
\
$\text{span}(S) = W$, where $W \subset V$ is a subspace of $V$ for which $S \subset W$ and such that for any $W' \subset V$ subspace of $V$ with $S \subset W'$ we have $W \subset W'$.

[stoic python]

MisterSystem

[winter harbor]

Proving that all these are equivalent characterizations of the span is a neat exercise.

[meager steppe]

for this question I think my answers wrong because it is in R^4 (?) but I'm not sure what I did wrong

[winter harbor]

No, that is correct

Ah

It was prolly a typo btw

I think whoever wrote down this exercise sheet meant R^4.

[meager steppe]

ok thx

[acoustic crescent]

For solving a upper triangular matrix you can use gaussian elimination, but what about lower triangular matrix? <@&286206848099549185>

[cosmic jacinth]

what would be a great book to supplement my schools textbook? axler's?

[nocturne jewel]

You... still do gauss jordan

Gauss Jordan is just a general process/algorithm for solving any linear system

[acoustic crescent]

im not sure i understand the difference between LU decomposition and Gauss jordan, dont we get the same result?

[nocturne jewel]

one's a factorization... one's solving the system

this is like asking the difference between factor x^2+5x+6 and evaluate x^2+5x+6 at x=2

[acoustic crescent]

okay, so to solve a upper triangular matrix we can use gauss elimination then back subsitution right

wait sorry upper triangular is already in correct form so no need to gauss

[nocturne jewel]

if you have $[A|b]$ then you solve it with gauss jordan no matter what A is

[stoic python]

Mosh

[acoustic crescent]

if we want to solve the lower triangular we do forward subsitution?

[nocturne jewel]

you just perform gauss jordan

and row reduce

[acoustic crescent]

its because my book want me to do them separately

"Solve upper triangular system using forward substitution" and likewise "Solve lower triangular system using forward substitution."

and i was wondering why i couldnt just do gauss jordan cause that solves the whole system

[acoustic crescent]

so i guess im asking is why you would ever solve for L and U seperately during LU decomposition instead of just doing gauss jordan, but i guess LU is more efficient in certain circumstances?

[nocturne jewel]

#prealg-and-algebra #precalculus this isnt linear algebra

[cosmic jacinth]

in back substitution, is it always -2 you multiple the first equation to when adding to second?

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

[meager steppe]

is R^n the literal definition of a vector space

[teal grotto]

no

[meager steppe]

except n defines the number of n tuples inside each vector?

i phrased that poorly, i should say is R^n a vector space where n is the number of n numbers/values in each vector

[teal grotto]

each vector is an n-tuple in R^n, since R^n is R x R x ... x R (n times)

[meager steppe]

i understand that

[meager steppe]

I thought n denoted how many values there are in each n tuple

[teal grotto]

you changed your wording when you edited your last comment, it seemed like you were suggesting something else

[meager steppe]

oh yeah i wrote it wrong, sorry. I originally wrote n is the number of n tuples

but is R^n a vector space with n number of values for each vector? or is this statement incorrect in any way

[teal grotto]

yes that is correct

[meager steppe]

thx

[nocturne jewel]

n number of values from R

[wintry steppe]

myy brain is melting

like they want them to be in terms of i, j, k, does that mean taking s*cos(alpha) and phi cos(theta)?

so they're all parallel to the x, y, z axis and can satisfy those equations shown below

but then we're only taking their x-components

[wintry steppe]

update if anyone will chime in later, i got the coordinates of phi and s, and basically dot producted both of them and said the result = 0, to find alpha, and ended up with tan(alpha)*tan(alpha+90) = -1, so i messed something up

[lavish jewel]

this is one of the most misleading depictions of cylindrical coords i've ever seen

[wintry steppe]

shots fired

[lavish jewel]

to make it easier, i'd take k = z, then note s points to the projection of P onto the xy plane

and finally get phi with a x product

[wintry steppe]

oh interesting

though i thought you can't say k = z, since k = 1 and z is infinite pretty much?

[lavish jewel]

what

[wintry steppe]

isn't z an axis?

that goes from 0 to infinity?

[lavish jewel]

sigh

[wintry steppe]

i just got digitally sighed

[lavish jewel]

ok, note the 2 coord systems use the same k vector

[wintry steppe]

you mean cartesian and this chunky boy?

[lavish jewel]

mhm

[wintry steppe]

yeah

[lavish jewel]

then s = i + j

you can get phi from those

either with a cross product or by rotating s 90 def with your favorute method

[wintry steppe]

S is the projection of P, so Pcos(whatever), and P is a given vector that's kind of a function of S and Phi?

those are the trash connections my mind is trying to conjure up the solution with

kill me

[lavish jewel]

P is a function of k, s, and phi

[wintry steppe]

oh but k is constant?

ANYWAY

this is diverging from the question

the real question is why does k being in two coordinate systems relate to s = i + j?

[wintry steppe]

i have no idea how you came to the conclusion that s = i + j, from this

[lavish jewel]

hmm yeah now that i look at it lol

[wintry steppe]

i was thinking about it and maybe you meant I should be imagining i and j superimposed onto x and y?

then S would constitute their addition?

[lavish jewel]

kinda, yeah

[wintry steppe]

i shouldnt have used superimposed, it probably means something else in maths

[lavish jewel]

call x y and z the scalars for i j k

[wintry steppe]

i get it

i dont get it as in i understand the world, but i can swallow that for thiis question

[lavish jewel]

let's try this a different way

[wintry steppe]

i mean im thinking of this as x, y, and z are variables that can be whatever scalars depending on a function

i mean S iis moving according to the question

[lavish jewel]

so you know you can represent a point in 3D space in terms of i, j, k, yeah? the vectors [1,0,0], etc

[wintry steppe]

yuep

[lavish jewel]

but you have to scale them depending on the point

ok

[wintry steppe]

yuoop

[wintry steppe]

also what do you mean by "def"

[lavish jewel]

oops that was a typo, meant "deg" or degrees

[wintry steppe]

o

[odd prism]

anyone wanna do my ixl

i would rather kill myself than do it

[wintry steppe]

ima try the cross product since i have a framework for what that means, since i know k is probably the answer since its orthogonal

[lavish jewel]

let's try it this way

more geometrically

you can project the lengths of i and j onto s, since you know the angle

[wintry steppe]

do yoyu mean i can project x and y onto s?

i just projected i onto S and did it to find S using the dot product equation, then learned S = 1 and that i was a clown

[lavish jewel]

lol

the thing is x and y are scalars

[wintry steppe]

oh right

[lavish jewel]

the vectors are x i and y j

[wintry steppe]

ii understan that

[wintry steppe]

i tried doing a cross multiplication, with the result being k = 2xy, orthogonal to them in the +z direction

cross multiplication of s against phi

[lavish jewel]

you're using the scalar definitions

those only give length

use the vector ones

[wintry steppe]

oh right

[lavish jewel]

the x prod needs a matrix determinant, and the vector projection is a dot product multiplied by a unit direction vector

[wintry steppe]

a dot product of what

[lavish jewel]

e.g. x i dot s, which is arguably not that useful

if you wanna do it with the scalars to save yourself some heartache

note that you have a right triangle with x, y, and the length of s. let's call that r

[wintry steppe]

the hypotenuse is r?

[lavish jewel]

mhm

[wintry steppe]

ii mean i now have magnitude of S = sqrt(x^2+y^2), that could potentially hypothetically probably maybe be used against the k = 2xy

[lavish jewel]

that 2 there makes it seem like you assumed an angle somewhere though

and k really shouldn't depend on x and y at all, btw

[wintry steppe]

i have no idea what that means, I got that 2 from doing that determinant cross product method, with S = xi + yj and Phi = -xi+yj

[lavish jewel]

it's orthogonal to them

[wintry steppe]

LMAO

it has 0 of it

[lavish jewel]

wait ugh i keep going back and forth between the scalars and the vectors, wth

[wintry steppe]

kill me im losing hours of sleep

[lavish jewel]

let's do it this way, maybe this is more intuitive

we already started with something like S = xi + yj

[wintry steppe]

mhm

[lavish jewel]

but we want unit vectors, ideally

and as you pointed out, the length of this vector is sqrt(x^2 + y^2)

so let's call the unit vector s (lower case now), and it's s = 1/sqrt(x^2 + y^2) (x i + y j)

this is already pretty good, cuz if you go back to the drawing i made

this has a geometric meaning

if we distribute the sqrt

we get that s = x / sqrt(x^2 + y^2) i + y / sqrt(x^2 + y^2) j

[wintry steppe]

please wait

[wintry steppe]

why does s = 1/S

[lavish jewel]

no, s and S are vectors

rather, |s| = 1

and s = S / |S|

[wintry steppe]

is that 1/S? 1/magnitude * vector expression

[lavish jewel]

yes

that's why i explicitly wrote i and j there, to make it clear it's a vector

more clearly yet,

[wintry steppe]

AHHHHH

[lavish jewel]

$s = \begin{bmatrix} \frac{x}{\sqrt{x^2 + y2}} \\ \frac{y}{\sqrt{x^2 + y2}} \\ 0 \end{bmatrix}$

[stoic python]

Ephemeral Dawdle of Doubt

[wintry steppe]

though WAIT why is there a xi+yj in |S|?

[lavish jewel]

S is a vector

|S| = sqrt(x^2 + y^2)

[wintry steppe]

yep

why multiplyy by xi+yj

[lavish jewel]

cuz we want a vector s

not it's magnitude

we already know it's magnitude

[wintry steppe]

wait

you're multiiplying the entire expression or just the bottom part?

[lavish jewel]

what

[wintry steppe]

S/|S| = 1/sqrt(x^2+y^2)

the sqrt part is the bottom part of the divsion

it has a name

[lavish jewel]

no what

[wintry steppe]

denominator

[lavish jewel]

no

S/|S| = 1/sqrt(x^2+y^2) * (xi + yj)

[wintry steppe]

are you multiplying xi+yj by the nominator or denominator

[lavish jewel]

use PEMDAS

the numerator is (xi + yj)

[wintry steppe]

its hard to see in computer notation

[lavish jewel]

that's why i wrote this

it's literally the same thing

[wintry steppe]

oh so you're multiplying by the nominator

NUMERATOR

[lavish jewel]

as pemdas implies

[wintry steppe]

well i couldnt tell

i thought perhaps you put the space there for no reason

usually iin written stuff that xi+yj would be in the center and big

ANYAWY

i'm startiing to rrealize something

if S is a unit vector, how are we getting another unit vector (s) out of it?

[lavish jewel]

S isn't a unit vector

S/|S| is a unit vector

[wintry steppe]

i'm so happy you replied

[lavish jewel]

as soon as sqrt(x^2 + y^2) is different from 1, S is not a unit vector

ez example

set x = 2, y = 0

S is not a unit vector

s = S/|S| is a unit vector

i do have to go sleep tho

,ti

[stoic python]

The current time for Edd is 02:36 AM (CEST) on Fri, 01/10/2021.

[wintry steppe]

WAIT

please just show me how to solve this unsolvable problem

ii need to sleep too

[lavish jewel]

look at this drawing and realize that the vector s = S/|S| is written in terms of sines and cosines of alpha mutliplied by x and y

and then use the property you were given that k x s = phi

or rotate s on the xy plane by 90 deg

[wintry steppe]

that's it?

[lavish jewel]

that's it, as i said from the beginning :p

[wintry steppe]

well then, i'll head to bed now. and hope it all makes sense when i wake up

[wintry steppe]

also i asked if S was a unit vector because it has a hat iin the original picture

[lavish jewel]

i tried to distinguish that by using upper and lowercase

[lavish jewel]

since it's a pain in the ass to keep doing $\hat{s}$

[stoic python]

Ephemeral Dawdle of Doubt

[wintry steppe]

wait what capital S were we talking about then?

[desert palm]

Does it have a minimum x value ? I think yes

[lavish jewel]

is this an exam?

[desert palm]

No

[lavish jewel]

anyway, wrong channel

try #calculus

[desert palm]

Ok

[lavish jewel]

i made the substitution $s \to S$ and $\hat{s} \to s$

[stoic python]

Ephemeral Dawdle of Doubt

[lavish jewel]

upper case for non unit norm, lower for unit norm

anyway, nite

[noble swan]

Awww, dang, are you going to sleep, too?

[wintry steppe]

fuck my life, thanks so much @lavish jewel , and i hope when i read and try out the solution it works

[noble swan]

RIP, I was waiting for help, too

[desert palm]

Ephemeral you know that thing I sent

[wintry steppe]

it worked, i got the solution, beautiful @lavish jewel : )

s = cos(alpha)i + sin(alpha)j, phi = sin(alpha)i - cos(alpha)j

goodnight

[cosmic jacinth]

https://i.imgur.com/x6X8IN2.png how in the fuck

when I do r2-2*r1, and r3-[3(r1)]

I cannot get to their answer

[undone axle]

How do you get the inverse of f(x) = x/(x-2)

[nocturne jewel]

not linear algebra

[undone axle]

I also graph it though?

Which sub channel is best then

[nocturne jewel]

#precalculus

[undone axle]

Thanks!

[glad acorn]

would we use the vector notation on the zero vector

[zinc timber]

ya

you can just use $0$ but I've seen ppl use $\vec{0}$ and even $\text{O}$

[stoic python]

Ryuzaki

[glad acorn]

i think O should for matrix

instead of column vector

[dusky epoch]

does it really matter if everyone knows what's meant

[past flax]

Let $V$ be an n-dimensional vector space, $(v_i)$ be a list of n vectors, and $A=[A^i_j]$ be an invertible matrix. How do I conclude from $$\sum_i A^i_j v_i = 0$$ for all $1 \leq j \leq n$ that $v_i=0?$

[stoic python]

Brian485

[past flax]

Is "multiplying both sides by A^-1" allowed?

[lavish jewel]

from A being invertible

sure, you can do that

you can also just say A is full rank

[past flax]

is it necessary to transform v_i into a column vector first by some choice of basis before multiplying by A^-1?

[lavish jewel]

in what you wrote, v = [v_i] is already a column vector

presumably in the basis A is acting on

[past flax]

oh no i mean v_i is a vector itself for each i

[lavish jewel]

your summation notation is a bit confusing then

it doesnt convey that

[past flax]

right sorry

Ah so im asking whether the matrix A thought of as a function A: V^n -> V^n by "matrix multiplication" of n vectors is injective

wait it has inverse A^-1

ok haha thanks

[empty ibex]

if a 2 by 2 matrix has one unkown value k in the a11 position, how do we know for what value of k is it diagonalizable?

[dusky epoch]

@empty ibex do you still need help w/ this?

[empty ibex]

Yeah

[dusky epoch]

okay so like, you know how to find if a matrix is diagonalizable, right

like if all of its entries are known

do you know what to do

[empty ibex]

Yeah AM=MD

Find eigen vectors and values

[dusky epoch]

yes the eigenvalues are key

if the eigenvalues are different, the matrix is diagonalizable always

if there are repeated eigenvalues there might be issues - however for 2 by 2 matrices it is much more clear cut: a matrix with two repeated eigenvalues is not diagonalizable unless it is already diagonal (i.e. [λ 0; 0 λ])

so you'll want the discriminant of the charpoly of your matrix

see when it's positive

and when it's zero

can i see your matrix btw?

[empty ibex]

Yeah sure

[dusky epoch]

splendid

let's calculate its charpoly

$(\lambda - k)(\lambda - 3) - 1 \cdot (-1) = \lambda^2 - (3+k)\lambda + (3k+1)$

[stoic python]

Ann

[empty ibex]

Yeah I did this and got a root of lambada in terms of k

But im stuck from there on

[dusky epoch]

no need

just find the discriminant

D = (3+k)^2 - 4(3k+1) = k^2 + 6k + 9 - 12k - 4 = k^2 - 6k + 5 = (k-1)(k-5)

[empty ibex]

Ohhhhh

[dusky epoch]

so for k not in [1,5] we have two distinct real eigenvalues and the matrix is diagonalizable

[empty ibex]

Damn it was so straightforward

Ty so much!

[dusky epoch]

yes that it was

obviously it would be just the same if there were multiple k's in there

[empty ibex]

I see

[ocean sequoia]

Can you invert an infinite dimensional matrix

[winter harbor]

It depends

For instance, matrix multiplication is usually defined for row/column finite matrices only.

And the reason is as follows

Suppose that we have a linear map $T : V \rightarrow W$ between two (infinite dimensional) vector spaces over a field $\mathbb{K}$, and we have $\mathcal{B}$ a basis for $V$ with $|\mathcal{B}| = J$ and $\mathcal{B}'$ a basis for $W$ with $|\mathcal{B}'| = I$.
\
\
Then, $\forall b_{j} \in \mathcal{B}$ we have that $\exists n \in \mathbb{N}$ and $b'{k{1}}, \cdots, b'{k{n}} \in \mathcal{B}'$ and $\alpha_{k_{1},j}, \cdots, \alpha{k_{n},j} \in \mathbb{K}$ such that:
$$
T(b_{j}) = \sum\limits_{i=1}^{n} \alpha_{k_{i},j} b'{k{i}}
$$
Then, we have that for each linear map
$$
T : V \rightarrow W
$$
Between two (infinite dimensional) vector spaces and ordered basis $\mathcal{B}, \mathcal{B}'$ we can associate a column finite matrix
$$
[T]^{\mathcal{B}}{\mathcal{B}'} = (\alpha{i,j})_{i \in I, j \in J}
$$

[stoic python]

MisterSystem

[winter harbor]

Analogously, for each finite column matrix we can associate to it a linear map

And so multiplication of finite column matrices is well defined

And we define it the same way as we do for matrices usually

Multiplication of matrices representing composition of linear maps

If we are dealing with matrices that do not have finite column or finite rows.

This concept is usually ill defined, since in general there's no way we can associate to such a matrix a linear map and define multiplication the usual way.

Another way to see this is that we would have to be "summing over" infinite non zero terms, which is generally an ill defined thing to do.

So we can talk about matrix multiplication as long as they have finite columns/finite rows

And the way to define invertibility is analogous

A matrix with finite columns is invertible iff the associated linear map is an invertible linear map.

If the field we are working over is a topological field, I think we can probably ask weaker conditions than column finiteness under when we can perform matrix multiplication.

But I have never seen that come up.

[ocean sequoia]

huh ok let me think about this for a second

thank you for the indepth answer

never done anything in infinite dimensions and so I wasnt sure about how think about it

seriously this is a great answer

[scenic lichen]

For 3x3 matrix A, If Ax=b has no solution is it possible for Ax=y to have a unique solution?

[lavish jewel]

no. Ax=b having no sol means A is rank-deficient

[scenic lichen]

Ahh ty. I’m reading about ranks online but have yet to get there in class

[empty grotto]

Does anyone know what the additive inverse here implies

[cosmic jacinth]

Hey

in system A, if the associated homogenous system is B & it has infinite solutions, then A will have infinite solutions too right?

[winter harbor]

There are two cases

Either A has not a solution

Or if A has a solution

Then it is indeed not unique

and there are infinitely many

The reason is as follows.

Suppose $T : V \rightarrow W$ is a linear map between finite dimensional vector spaces over the reals and we have that $\ker(T) \neq {0}$. Suppose now that for $b \in W$ we have that the system $T(x) = b$ has a solution for some $x \in V$. Then, $\forall y \in \text{ker}(T)$ we have that:
$$
T(x+y) = T(x)+T(y) = T(x) = b
$$

[stoic python]

MisterSystem

[winter harbor]

So this means that if I sum any element y of the kernel to a solution x

we have that x+y is still a solution of the system

and since ker(T) is infinite in this case

we have that there are infinitely many y such that T(x+y) = b

and the system has infinitely many solutions

Just remember that we have to make sure that at least one solution exists for the argument to apply

there can be cases where no solutions exist at all

but if at least one exist, then there are infinitely many

[cosmic jacinth]

ah!

It was actually false

thank you

after doing r2-3r1 my system became this:

1 -1 -2 2 | 0
0 5 -7 -9 | 0

do I divide everything by 9?

or 5

wait, 5 so that it becomes a 1, right

[winter harbor]

5r_1 + r_2 -> r_2 should do the work

[cosmic jacinth]

sorry, can you explain that in different words

I'm not sure what you mean by 5r_1 or -> r_2

[winter harbor]

multiply the first row by 5

then sum it to the second row

and then you substitute that into your second row

[winter harbor]

Do you still need some help with it?

[empty grotto]

Yeah it's due next week and I am doing it during class at my HS so haven't made any more progress

[cosmic jacinth]

I have no sweet fucking clue what's going on

[winter harbor]

Yeah, if we have a vector space $V$ with an operator of sum $+$, the additive inverse of a vector $x$ is the unique vector $y$ for which $x+y=0$

[stoic python]

MisterSystem

[winter harbor]

In this case, we have that $(-3,2)$ is the additive inverse of the operation $\boxplus$ and for a pair $(x,y) \in \mathbb{R}^{2}$ we want to find an element $(x',y') \in \mathbb{R}^{2}$ for which
$$
(x,y)\boxplus(x',y') = (x+x'+3,y+y'-2) = (-3,2)
$$

[stoic python]

MisterSystem

[winter harbor]

So we want that
$$
x+x'+3 = -3
$$
and
$$
y+y'-2 = 2
$$
We then have that $x' = -x-6$ and $y'=-y+4$ so the additive inverse of $(x,y)$ under $\boxplus$ is $(-x-6,-y+4)$

[stoic python]

MisterSystem

[winter harbor]

I encourage you to convince yourseld that (-3,2) is the neutral element of $\boxplus$ and convince yourself of this argument.

[stoic python]

MisterSystem

[median ocean]

can anyone help me with 1?

[winter harbor]

what have you tried so far?

[median ocean]

i have no idea how to start it

so is v1[s, s, 4s] and v2[t, -t, -3t]

[winter harbor]

no

(s+t,s-t,4s-3t) = (s,s,4s) + (t,-t,-3t) = (1,1,4)s + (1,-1,-3)t

so we may take v_1 = (1,1,4) and v_2 = (1,-1,-3)

and it is easy to see they indeed span W

[median ocean]

so part a is v_1 = (1,1,4) and v_2 = (1,-1,-3)?

[winter harbor]

yes, and verify they indeed span W...

[median ocean]

how do i verify it

the work?

[winter harbor]

you just have to verify that any vector in W can be written as a linear combination of v_1 and v_2

[dire wharf]

For T:Rm to Rn linear transformation, can I tell whether T is one to one, onto or both/neither depending on if m or n is greater?

[median ocean]

@winter harbor why is W a subspace of R^3

[winter harbor]

I mean, it is the span of two vectors, so it has to be a subspace of R^3

I recommend you to review this stuff

[median ocean]

yes ik

[winter harbor]

You can get some information on that.

More specifically

we can know wheter or not a map can't be surjective/can't be injective/can't be bijective

With the following lemma

Which can be proved via rank nullity

Let $V,W$ be finite dimensional vector spaces such that $\text{dim}(V) < \text{dim}(W)$, then no map $T : V \rightarrow W$ is surjective.
\
\
\textbf{Proof}
\
\
Indeed, suppose $T$ is surjective, then we would have that $\text{Im}(T) \cong W$ the rank-nullity theorem we would have $\text{dim}(V) = \text{dim}(\text{Im}(T)) + \text{dim}(\text{ker}(T))$ this would imply then that $\text{dim}(V) = \text{dim}(W) + \text{dim}(\text{ker}(T))$. Which is a contradiction, since then we would have $\text{dim}(V) - \text{dim}(W) = \text{dim}(\kernel{T}) \geq 0$ which is not the case since $\text{dim}(V) - \text{dim}(W) > 0$ ; $\square$

[stoic python]

MisterSystem
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[winter harbor]

Analogously we also have that for $V,W$ finite dimensional vector spaces with dim(V) $>$ dim(W), no linear map $T : V \rightarrow W$ is injective

[median ocean]

is there other ways to explain now W is a subspace of R^3

how*

[stoic python]

MisterSystem

[winter harbor]

So the thing is that we can use dimension to infer when a map is not injective/surjective/bijective

but usually not the prove that it is indeed injective, surjective/bijective, etc...

[median ocean]

3 coloms

columns

[winter harbor]

It is useful for the negation, let's put it in this way

[winter harbor]

Using the definition.

take two vectors in W

prove that W is closed under linear combinations

and that 0 is in W

[median ocean]

so even if you sum it up or multiplay the vectors it will still be in W

so then any vectors of W will be in the subspace of R^3

[winter harbor]

That's a weird way to phrase, but yeah, all you have to show is that 0 is in W (equivalently W is non empty) and that any linear combinations of vectors in W is still in W.

[median ocean]

how do i show that 0 is in W

so like s+s+4s=0 and t-t-3t=0 and 0=0

[teal grotto]

this should read dim V = dim W + dim ker(T)

[median ocean]

how do i know if [v1, v2] is a basis of W

[winter harbor]

Let $V,W$ be finite dimensional vector spaces such that $\text{dim}(V) < \text{dim}(W)$, then no map $T : V \rightarrow W$ is surjective.
\
\
\textbf{Proof}
\
\
Indeed, suppose $T$ is surjective, then we would have that
$$
\text{Im}(T) \cong W
$$
and by the rank-nullity theorem we would have
$$
\text{dim}(V) = \text{dim}(\text{Im}(T)) + \text{dim}(\text{ker}(T))
$$
this would imply then that
$$
\text{dim}(V) = \text{dim}(W) + \text{dim}(\text{ker}(T))
$$
Which is a contradiction, since then we would have
$$
\text{dim}(V) - \text{dim}(W) = \text{dim}(\kernel{T}) \geq 0
$$
Which is not the case since $\text{dim}(V) < \text{dim}(W)$ ; $\square$

[winter harbor]

Yeah, corrected it tho.

[stoic python]

MisterSystem
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[median ocean]

@winter harbor how do i know if [v1, v2] is a basis of W

[winter harbor]

You have to show they are lineraly independent

you have already shown they span W on exercise (a)

now show they are linearly independent

[median ocean]

if vecotr is linearly inde to W right

[cosmic jacinth]

Yo

[cosmic jacinth]

oh lagrange

[cosmic jacinth]

oh god

[astral abyss]

yo lets say i have matrix A and C, and we want a matrix B such that AB=C. are there many Bs?

[slow scroll]

Not always, but there could be, yes

[gritty furnace]

can u help me

[slow scroll]

wrong channel. Your question belongs in #precalculus #prealg-and-algebra or possibly #competition-math

[gritty furnace]

oh ok sorry i'm newbie

[upbeat elbow]

"prealg

probably compmath

[astral abyss]

thanks

[slow scroll]

np

[hot swallow]

quick question, if there are more columns than rows then is it linearly dependent?

[slow scroll]

if a matrix has more columns than rows then the columns would have to be linearly dependent

the rows could be linearly independent though

@hot swallow

[hot swallow]

ahh i get it, ty

[slow scroll]

np

[jolly tapir]

does anybody understand why this is wrong?

[astral abyss]

use parenthesis for the point instead of <>

[jolly tapir]

oh i got it 😂

thanks

[median ocean]

how do you do part c

[winter harbor]

You have to solve the system of equations Ax = 0

[silver heath]

Could someone explain to me why g(0) must be zero?

The explanaition of the answer is confusing for me

nvm im a moron

[wintry steppe]

even though it has a proof

i can't understand the proof because i am dumb

please, can someone explain it

no wait

first tell me why this is even useful

or what it's trying to even do

[scenic spoke]

u need help?

[wintry steppe]

yes please

[scenic spoke]

u have talked to the wrong person

[wintry steppe]

[scenic spoke]

idk why im in here

[silver heath]

why would you say that if u cant help

[wintry steppe]

it's okay

[scenic spoke]

cuz im very smart guy

[winter harbor]

Oh ok

You are probably familiar with the fact that linear maps between finite dimensional vector spaces are associated to matrices, given a choice of ordered basis for the domain and codomain.

And analogously, for each matrix we can associate a linear map.

[wintry steppe]

yes

[winter harbor]

Now

You are probably also familiar with the transpose of a matrix.

What this theorem tells us

[wintry steppe]

yes i just want to understand the proof

[winter harbor]

Is that if we have a linear map T : V -> W and ordered basis B for V and B' for W, then in fact the induced map T* : W* -> V* corresponds to the transpose of T when we view T* as the matrix associated to it on the dual basis B* and B'*

And at least I find that quite nice, because we have a basis free way to define the transpose of a linear operator.

[wintry steppe]

yeah

[winter harbor]

Now for the proof

It is pretty straightfoward, it is the standard kinds of proofs we do in linear algebra. Maybe the notation is a bit messy.

[wintry steppe]

ok

[winter harbor]

But ok, suppose that we have finite dimensional vector spaces $V,W$ and a linear map $T : V \rightarrow W$. Suppose that $V$ has ordered basis $\mathcal{B} = {e_{1}, \cdots, e_{n}}$ and $W$ has ordered basis $\mathcal{B}' = {e'{1}, \cdots, e'{m}}$
\
\
We then have the dual basis $\mathcal{B}^{\ast} = {e^{\ast}{1}, \cdots, e^{\ast}{n}}$ and $\mathcal{B'}^{\ast} = {(e')^{\ast}{1}, \cdots, (e')^{\ast}{m} }$
\
\
Notice the following, $\forall j \in {1, \cdots, n}$, we have that
$$
T(e_{j}) = \sum\limits_{i=1}^{m} \alpha_{i,j} e'{i}
$$
Where $\alpha{1,j}, \cdots, \alpha_{m,j} \in \mathbb{R}$ are the coefficients of $T(e_{j})$ in the ordered basis $\mathcal{B}'$.
\
\
Notice precisely that $\alpha_{i,j} = (e')^{\ast}{i}(T(e{j}))$.

[stoic python]

MisterSystem

[winter harbor]

Can you understand the proof up until now?

I will continue if you don't have any questions

[wintry steppe]

let me see

by the way really appreciate it

not many people are willing to spend time to help so thanks alot

[winter harbor]

Np

[wintry steppe]

ok i understand

[winter harbor]

nice

This then means that the matrix
$$
[T]^{\mathcal{B}}{\mathcal{B}'}
$$
has entries
$$
(\alpha{i,j}){i \leq m, j \leq n} = ( , (e')^{\ast}{i}(T(e{j})) ,){i \leq m, j \leq n}
$$
So its transpose matrix $([T]^{\mathcal{B}}{\mathcal{B}'})^{t}$ has to have, by definition, its (i,j)-th entries given by
$$
(\alpha{j,i}){j \leq n, i \leq m} = ( , (e')^{\ast}{j}(T(e{i})) , )
$$

[stoic python]

MisterSystem

[wintry steppe]

let me think

[winter harbor]

Yeah, the idea is that we will show T* : W* -> V* has these entries when we consider the dual basis

and like

I think that the main idea we use here

is that the dual basis of a given basis gives the coordinates of a vector

[wintry steppe]

i see

wow

[winter harbor]

For instance, if we have a vector space $V$ and a basis $\mathcal{B} = {e_{1}, \cdots, e_{n}}$
We can write any vector $v \in V$ as
$$
v = \sum\limits_{i=1}^{n} \alpha_{i} e_{i}
$$
For some coeffiecients, what we have is that $e^{\ast}{i}(v) = \alpha{i}$ gives exactly the i-th coefficient and we can write
$$
v = \sum\limits_{i=1}^{n} e^{\ast}{i}(v) e{i}
$$
Instead

[stoic python]

MisterSystem

[winter harbor]

We will use this idea again

Ok, so we have the induced map
$$
T^{\ast} : W^{\ast} \rightarrow V^{\ast}
$$
Such that $\forall \varphi \in W^{\ast}$, $T^{\ast}(\varphi) = \varphi \circ T$.
\
\
Let us find its matrix in the ordered basis $\mathcal{B}^{\ast}$ and $\mathcal{B'}^{\ast}$ given by the dual basis.
\
\
We have that $\forall j \in {1, \cdots, m}$,
$$
T^{\ast}((e')^{\ast}{j}) = \sum\limits{k=1}^{n} \beta_{k,j} e^{\ast}{i}
$$
Let us find these coefficients.
\
\
We notice that $T^{\ast}((e')^{\ast}{j}) = (e')^{\ast}{j} \circ T$
\
\
This implies that
$$
(e')^{\ast}{j} \circ T = \sum\limits_{k=1}^{n} \beta_{k,j} e^{\ast}{i}
$$
Notice that if we apply $e{i}$ on both sides, we have
$$
(e')^{\ast}{j}(T(e{i})) = \beta_{i,j}
$$
But notice that
$$
(e')^{\ast}{j}(T(e{i})) = \alpha_{j,i}
$$
And so the coeffiecients of the linear operator $T^{\ast}$ with respect to the ordered dual basis are indeed given by $(\alpha_{j,i}){j \leq n, i \leq m}$ and so
$$
([T]^{\mathcal{B}}{\mathcal{B'}})^{t} = [T^{\ast}]^{\mathcal{B'}^{\ast}}_{\mathcal{B}^{\ast}}
$$
As we wanted to prove.

[stoic python]

MisterSystem

[winter harbor]

And that's the proof

I could have used some better notation instead of prime for the other basis tbh lmao

So yeah, the dual map T* corresponds (in the dual basis) to the transpose

this map is also called pullback too

And we haven't used much, just the fact that the dual basis are precisely what some people call the coordinate functions.

and the the definition of the pullback/dual map ofc

[wintry steppe]

wow

ok thanks alot

this was much clearer

i'll have to look at it a few more times to get it but i have the big picture now

thanks alot really really really appreciate your help

[winter harbor]

Np

[gusty elbow]

Let V be the space of all functions from R into R which are continuous.And we define a LT as,
$$(Tf)(x)=\int^x_{0} f(t)dt$$ . What would be the Range space ? It won't be V right because no function will integrate to a constant.

[stoic python]

negative_epsilon

[lavish jewel]

0 integrates to a constant, and both are continuous

but where did you get the "integrate to a constant" bit anyway

[gusty elbow]

Hehe, I thought that but what I meant was nonzero constant. And I didn't get it from anywhere.

Wait is this a hint?

[dusky epoch]

g ∈ ran(T) iff g(0) = 0 and g is C^1, i think.

[lavish jewel]

ann's answer is precisely what i would've thought, using the fundamental thm of calc

you get a once continuously diff func

and you need it to follow the def of linearly, which imposes conditions on F(0)

(with F(x) the antiderivative of f(x))

[gusty elbow]

Wait

Why should F(0) be 0?

[lavish jewel]

so that a zero element gets mapped to another zero element

i think this is required for this transformation to satisfy "homogeneity"

or maybe it's enough to set the constant of integration to 0

[gusty elbow]

Constant of integration should not necessarily be zero I think.
Like if
f(x)= sin(x) then
F(x)=-cos(x) and
g(x) = (Tf)(x)= -cos(x)+1
so it belongs to Range of T but here F(0) isn't 0 but g(0) is.

[hard drum]

Wht "constant of integration"? If you simply plug in 0 into the equation we find (Tf) (0) = 0

[lavish jewel]

that's what i mean

[gusty elbow]

Yes that I can understand

So That's the only condition

[solid smelt]

Looks like some quotient space of C^1

C^1/R

Or something like that

[winged prairie]

yo is kerT (null space) many-to-one or one-to-many

[hard drum]

it is not a function, so neither

unless i'm missing what you mean

[nocturne jewel]

Nope, you're right based on what they said

[hard drum]

(also maybe this isn't that strict but to me, and apparently according to some sources i've found online too, null space usually refers to matrices whilst kernel is more for linear maps)

[shell kindle]

Chegg is wrong with this one, right?

[nocturne jewel]

yeah, typically

but all LT's can be written as a matrix wrt some basis/bases

so

[shell kindle]

This is the problem, btw

[hard drum]

eh there's still a distinction as a linear transformation can be written as a map v -> Av for a matrix, A isn't the map in itself ig

c is a spanning set im pretty sure (1 = x+2 - (x+1), x = 2(x+1) - (x+2), x^2 = (x^2 - 1) + (x+2) - (x+1)

[shell kindle]

Yeah you right, just noticed what they did was solve for a1, a2 and a3. Thanks!

[median ocean]

@winter harbor so would the 2 procedures be RREF and then solve the system of equations Ax = 0?

[hard drum]

Row operations don't (necessarily) preserve column space though ig

like (11 / 11) has rref (1 1 / 0 0) which has a different col space

[winter harbor]

Oh yeah, true.

Oh

Ok makes sense

I didn't actually read question (c)

It is asking to find the null space of A

Yeah, so the procedure would be to solve Ax = 0 after row reducing A.

Corrected it

[median ocean]

ok so i did it correcrt then?

[winter harbor]

Yup

[median ocean]

ok thanks

[midnight trout]

To find right-hand values for b that make the system solvable, I choose any value for u, v, and w and then let that determine b.

But what about for the second part of the question? I originally assumed that, because solvable b's must lie on a plane, that you could just shift a solvable b along any single axis to get a value for b that isn't solvable. But, after thinking about this, this doesn't work in cases where the solvable-b-plane is orthogonal to any axis.

So, how do I approach this then?

[zinc timber]

notice that if u write yr equation in matrix form Ax=b, u see that A. rank deficient

so for this equation to have solution b must lie in the range space of A

can u find the range space? just pick 2 vector from it

[midnight trout]

the range space would be column 1 and 2

or a combination of any of the three because the vectors are on a plane

[zinc timber]

since one of the basis is degenerate, you can pick any 2 or even 3

won't matter

look for trivial cases, like picking u,v,w yourself

[versed yew]

can someone help me with this question

[quartz compass]

what have you tried?

[versed yew]

wdym

[turbid aspen]

As in what have you tried while attempting the question, so Merosity can know where to step in

[versed yew]

not yet

[turbid aspen]

Now just so you can be able to firmly grasp the solution to a question, it is highly advisable that you have attempted it, you may then post it if u are stuck somewhere or something like that.

I have provided a link below, it will help you attempt the question.

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-equations/x2f8bb11595b61c86:number-of-solutions-to-systems-of-equations/a/number-of-solutions-to-system-of-equations-review

[versed yew]

thanks!!

[winged prairie]

can anybody explain this corollary

[turbid aspen]

Okay but I am not good with latex, so I will be describing some symbols here rather than displaying them

[turbid aspen]

Now u see that curvy minus sign, it is usually a notation for something called an 'equivalence relation'. Can I go on?

[half ice]

What's it a corollary of?

"corollary" means "easily deduced from previous theorem"

[turbid aspen]

An equivalence relation on a set by the way is any relation that is reflexive, symmetric and transitive. That is it satisfies the rules stated below:

Reflexive: an element x of the set can be related to itself

Symmetric: if an element x is related to an element y then y is related to x.

Transitive: if an element x is related to y and y is related to z, then x is related to z

With this you can see that the = sign you are familiar with is an equivalence relation

x = x (Reflexive: any number x is always equal to or related to itself)

Symmetric: if we say that x = y, then we can be sure that y = x. I think this is part of the definition of the = relation.

Transitive: ___________. Can you fill the gap?

[wintry steppe]

please help

[wispy pewter]

is there a word for two if then statments joined by an and?

For example, I know an if then statmente is a conditional statement

but if you link two conditional statements with an and does that have a name

[winged prairie]

yes

[winged prairie]

if x = y and y = z then x = z

[half ice]

~ is "like an equal sign" except where "equal-like" is given by a specific rule.

[x] is a set. We call it an "equivalence class". It means "all elements equivalent to x". By transitivity, all of these elements are also equivalent to eachother.

The previous theorem is a big deal, that the equivalence relation partitions the set. That is every element goes into some "equivalence class", and no element goes into multiple "equivalence classes".

.
If we write that idea using our new notation, we get that:
x ~ y ⇔ [x] = [y]
(If two elements are equivalent, then they belong to the same equivalence class)

Either [x] = [y] or [x] ∩ [y] = empty
(Any two equivalence classes are the same, or they have nothing in common. This is what it means to partition)

@winged prairie

[winged prairie]

Thanks for your answer

imma need a few mins to process this

[half ice]

That's fair, this idea is huge and abstract. Very useful to get it down though, it's used everywhere

[turbid aspen]

Two statements with an 'and' is called a conjunction, with an 'or' is called a disjunction

[wispy pewter]

I want to say that if X =6 then X + 3 = 9 and if X = Y then Y + 3 = 9

is that a conjunction

@turbid aspen

[turbid aspen]

Yes it is a conjunction

You know a conditional statement is still a statement and there are two of that combined with an 'and'

SOMETHING TOTALLY UNRELATED
Hi everyone,
How do I draw a locus satisfying certain conditions when there can be an infinite number of points, satisfying any condition

[midnight trout]

21: do you think they mean to ask "when equation 1 and 2 are replaced by the sum of equations 1 and 2" ?

[turbid aspen]

Don't understand

[midnight trout]

I don't understand whether they want me to examine a system with three equations or if they want me to replace the second equation with the sum of the first two equations.

[turbid aspen]

Did you upload a picture because all I see here is blank, it might be a bug, give me a sec

[midnight trout]

ok

[turbid aspen]

I see it now

[midnight trout]

I answered both interpretations. Found that the combination version gives a 3x4 matrix that looks like
1 2 2 2 6
0 0 1 1 3
0 0 0 1 2

reduced it to
1 2 0 0 0
0 0 1 0 1
0 0 0 1 2

and said everything changes: the row image, the column image, obviously the coefficient matrix, and there are now infinitely many solutions to the system

[turbid aspen]

I have very little idea about this

[wintry steppe]

hey can anyone help me
thank youuuu

[winter harbor]

I think this is best suited to for #prealg-and-algebra

[ionic belfry]

Hey could anybody help me with linear algebra

[winter harbor]

You just ask

[ionic belfry]

Can I ask later because I am at work

[limber sierra]

No one will stop you

[torpid wedge]

if you use gaussian elimination to check if a matrix is dependant or not

you cant 'take out' a row 2x right?

like you cant use the same row twice for the transformation

[surreal wadi]

can someone help me with these?

[cerulean garden]

These transformations are such that any combination of these are another transformation

[surreal wadi]

and this one

im stuck on them and dont really understnad how to prove them

[nocturne jewel]

Looks like you're missing a lot of information

namely are you given A?

[cerulean garden]

for example, substracting 2 times the row is the same as substracting this row twice

So you can use the same row any amount of times as long as you are doing it right

[torpid wedge]

I see, thank you! @cerulean garden

[midnight trout]

does eq3=eq1 or something?

[vivid bane]

How do I prove this with matrices?

[winter harbor]

Do you still need help with that?

[vivid bane]

yes please

[vivid field]

How would I set this question up to begin to solve it ?

[winter harbor]

Ok so, notice that via a completing the square kind of argument, we can reduce the equation of a circle to:
$$
(x-x_{0})^{2}+(y-y_{0})^{2} = r^{2}
$$

[stoic python]

MisterSystem

[winter harbor]

For some x_0, y_0 and r reals

I think you should be able to do this

This makes things simpler

Because the problem now boils down to

Given points $(a_{i},b_{i}) \in \mathbb{R}^{2}$ for $i \in {1, 2,3}$ not colinear, we have to prove that the system of equations:
$$
(a_{i}-x_{0})^{2}+(b_{i}-y_{0})^{2} = c, i \in {1,2, 3 }
$$
Which is a system 3 unknowns, namely $x_{0}, y_{0}, c \in \mathbb{R}$, and 3 equations has a solution.

And how do we prove this?

Well

[vivid bane]

we can now make a matrix?

if it equals to a homogeneous matrix, then there has to be anontrival solution?

[winter harbor]

Not really, because this system of equations is not linear, yet.

But we can use like

One of equations

To reduce the other two to a system of linear equations.

In the following way

[stoic python]

MisterSystem

[winter harbor]

For the sake of simplicity I changed r^2 to c

But anyways

Notice that given
$$
(a_{1}-x_{0})^{2} + (b_{1}-y_{0})^{2} = c
$$
We can set
$$
(a_{1}-x_{0})^{2} + (b_{1}-y_{0})^{2} = (a_{2} - x_{0})^{2} + (b_{2}-y_{0})^{2}
$$
Simplify this down and you will get a linear equations on $x_{0}$ and $y_{0}$
\
\
We can also set
$$
(a_{1}-x_{0})^{2} + (b_{1}-y_{0})^{2} =
(a_{3}-x_{0})^{2} + (b_{3}-y_{0})^{2}
$$
And this simplifies down to a linear equation.
With these simplifications, we get a 2×2 system of linear equations on the variables $x_{0}$ and $y_{0}$.
\
\
By using the hypothesis, notice that we can then conclude this system of linear equations has a (unique) solution.

[stoic python]

MisterSystem

[winter harbor]

So after you are done solving for x_0 and y_0

You can plug these values back in

And solve for c

[vivid bane]

when i simplify, i was suppose to expand then collect like terms to one side?

[winter harbor]

Yup

Notice that x_0^2 and y_0^2 will cancel out

And you will a linear equation on x_0 and y_0

And you can apply linear algebra to study the solutions

Notice that you are given a 2×2 system of equations

So you can now like

Prove that a unique solution exists via soma determinant argument

Or some other method to prove that the system is non singular.

[vivid bane]

ok thank you very much

[winter harbor]

Yeah, tbh you don't even need to complete the square.

But it prolly makes things easier

Because you reduce the number of variables

So I find it worth

[vivid field]

If B is in the Col(A)....

Can B only be unique ?

Can someone explain this ?

[limber sierra]

b is "prechosen" by the equation Ax = b

it doesnt make sense to ask whether b's unique

since it's defined as part of the problem

its like if i have the equation 3x = 6 and ask "is 6 unique?"

[vivid field]

Thanks.

Let V = {(x, y) : −3 ≤ x ≤ 3, −2 ≤ y ≤ 2}.

How would I give a geometric description of this subset of R^2

And how would I determine and prove that V is or is not a subspace of R^2 ?

I am a little confused

[limber sierra]

can you draw it?

[vivid field]

Draw it ?

How would I draw it ?

?

[limber sierra]

yeah, thats what i meant

its a rectangle with corners (3, 2), (3, -2), (-3, -2), (-3, 2)

[meager steppe]

Can somebody help me figure out 4b? I’m not sure if it’s correct

[limber sierra]

now, do you think its a subspace or not?

what do we need to do to check whether somethings a subspace?

[vivid field]

How would I go about determining that ?

[vivid field]

I am not sure

[limber sierra]

a subset of a vector space is a subspace iff:

• it is nonempty, and
• it is closed under addition (so adding two vectors in the subset always gives another element in the subset), and
• it is closed under scalar multiplication (so multiplying any vector in the subset by any scalar always gives another element in the subset)

[vivid field]

I am aware of the properties. I just dont know how I would apply it

[limber sierra]

try adding some vectors in V together then

is the result always in V?

[vivid field]

would the vectors in V be any numbers in between −3 ≤ x ≤ 3, −2 ≤ y ≤ 2 ?

[limber sierra]

theyd be of the form (x, y) for −3 ≤ x ≤ 3, −2 ≤ y ≤ 2

so (2.5, -1.5) is in V

but (2.5, -2.5) is not

since y = -2.5 does not satisfy −2 ≤ y ≤ 2

[vivid field]

No the result is not always in V

[limber sierra]

right

(2, 2) + (2, 0) is not in V for example

since the sum is (4, 2)

but 4 is greater than 3

[vivid field]

Yeah!

[limber sierra]

so V is not a subspace

in general, subspaces of ℝ^n have to "extend infinitely"

they can be a line (1 dimensional), a plane (2 dimensional), or a hyperplane (more than 2 dimensions)

but they cant have any boundaries

(well, theres also the 0 dimensional subspace which is just the zero vector)

(but thats just a single technicality)

[vivid field]

Got it! Thanks!

I have another quick question which I may be just over-tthinking. I found the spanning set of nul(C)

How would I Provide 3 non-trivial, explicit examples of vectors in Null(C) ?

Maybe I am just confused on the wording but I am not sure

Here is what I have

How would I Provide 3 non-trivial, explicit examples of vectors in Null(C)

Any ideas ?

[limber sierra]

well you already have 2 examples

the vectors in your basis

take any linear combination to get your 3rd

(as long as it isnt trivial, i.e. isnt 0)

[vivid field]

Can You explain further. Sorry I must be forgetting stuff or something.

Are they not supposed to equal to the o vector

Do you mean any linear combination of (3,0,1,0) (-1,-2,0, 1) ??

[wild fulcrum]

ye

[runic oar]

any recommendations for the best yt channels for vids for a linear algebra+diff equations course? Taking it async rn with a textbook that isn't working great for me so another resource would be helpful. Have used Khan for some which has been helpful but there seem to be a lot of gaps in what they have stuff for as compared to what I need for the course

[dusky epoch]

@calm matrix do you still need help with this?

[stoic python]

EmilD

EmilD

EmilD

[wispy pewter]

is there a trick for this

[calm matrix]

I think if u can help me that's much better haha

[wispy pewter]

I mean I guess I found them by guessing

is that all you can do?

[calm matrix]

okie. I have another ques..

[half ice]

Which one?

[torn leaf]

Hi can anyone please help

<@&286206848099549185>

[half ice]

See the rules on helper pings

[torn leaf]

Oh mb

[half ice]

,w row reduce {{1,-2,2,3,-1},{-3,6,-1,1,-7},{2,-4,5,8,-4}}

[stoic python]

Results provided by WolframAlpha

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Requested by Kaynex#7520

[lavish jewel]

is this an exam/quiz?

[torn leaf]

No

Hw

[half ice]

With that row reduction, you can read most things off the matrix

[torn leaf]

Hm not getting it

[wild fulcrum]

did you learn row reduction in class

[torn leaf]

Yes