linear-algebra

[sonic osprey]

To prove what?

To prove that this is the zero element, you want to show that p + 0 = p

Also, polynomials can only have finitely many coefficients, so it's kinda misleading to write p like that

[modern palm]

So then suppose P is the set of all polynomials, let $p \in P$ and set $p = a{0} + a{1}x + a_{2}x^{2} + ... + a_{n}x^{n}$, then use $p + 0 = p$ so $\bar{0} = p-p = 0$. Hence $\bar{0} = 0$.

[stoic python]

42_Toxic_Potatos

[sonic osprey]

I'm not really sure what you're trying to prove

[modern palm]

Im just trying to find the 0 vector of the set of all polynomials

so its not a proof i think

[sonic osprey]

Like I said, to prove that something is the 0 vector, you just need to show that p + 0 = p for all p

It seems that you're trying to prove that the 0 vector is unique for some reason

[modern palm]

yeah, ok so i dont even need to set p then bla bla. Just simply $ \bar{0} + p = p$ so $\bar{0} = 0$.

[stoic python]

42_Toxic_Potatos

[sonic osprey]

I'm still not really sure what you're doing

You can prove that 0 is the zero vector by showing that 0 + p = p for all p

Theres no need to deal with \bar{0} in any capacity

[modern palm]

Its a part of a bigger question where I have to prove U is a subspace of V. So i'm just trying to compute o vector of V, which happens to be the set of all polynomials

[sonic osprey]

This doesn't make any difference

[nocturne jewel]

0 vector of polynomials is p=0

[sonic osprey]

Wow thank you for repeating what I already said to them!!! Really helpful!!!

[nocturne jewel]

Ikr!!!

well you said f(x)=0, I said p=0 big difference \s

[modern palm]

ok, wydm

[sonic osprey]

The definition of the zero vector is a vector such that 0 + p = p for all p

So if you show that 0 satisfies this property, then you know that 0 is a zero vector

[modern palm]

i mean 0 is just a polynomial of $0 + 0x + 0x^2 + ...$

[stoic python]

42_Toxic_Potatos

[sonic osprey]

Yes

[modern palm]

ok, so i can group the coefficients with the corresponding x, then show that p is still p

[sonic osprey]

Yeah

[modern palm]

right. Was it wrong what I did before? As in $ \bar{0} + p = p$ so $\bar{0} + p + (-p) = p + (-p)$ so $\bar{0} = 0$?

[stoic python]

42_Toxic_Potatos

[lavish jewel]

what are you even trying to do there?

you proved -p is the additive inverse of p

[modern palm]

i see.

[noble swan]

This is a true or false question, but I want to understand the reason behind it. I don't really get it, so can someone help me understand, please?

[lavish jewel]

do you know how to get the equation of a plane?

[noble swan]

Errr, the r(t) = a(x - x_0) equation?

I know that's not the full equation, I just shortened it

[lavish jewel]

what does each thing mean there

[noble swan]

<a, b, c> is the normal vector

(x_0, y_0, z_0) is simply the intercepts of each axis

[lavish jewel]

the form that would help you here is n (x - x_0) = 0

[noble swan]

Okay, and what would I do with that?

[lavish jewel]

alternatively, just think of the dimensionality of the subspace spanned by u and v with the conditions they have given you

you could have some fun with cross products and build the plane eq explicitly

[noble swan]

Yeah, I thought of it as if x_2 & x_3 are free, you can call them 0, in which case you would get x = 0

So then it would go through the origin, making it true

But that's wrong, it's false

Soooo XD

[lavish jewel]

can you show the rest of the problem?

[noble swan]

Yeah, sure, the other questions don't help that much, though

Omg, my Snip & Sketch isn't working

[lavish jewel]

i'm kinda sleepy so i might be mistaken, but i guess they're asking for the solution space of the problem x = [u v] [x2 x3]^T

[lavish jewel]

and since u and v are lin indep, the matrix has full column rank

that'd mean the sol space is a single point instead of a plane

[noble swan]

I lost you at full column rank

[lavish jewel]

another way to put it is that it seems they're asking about x2 and x3, not about the vector x

[fast comet]

I'm working on a textbook question, but I'm not certain my answer falls within the spirit of the quesiton.

Q: There exists a matrix in reduced row echelon form that one column can be removed leaving a matrix that is also in reduced row echelon form. Find one or more examples of this phenomenon.
A: You can remove any column from any RREF matrix provided that column does not contain any pivots. Doing so will still result in a matrix in RREF.

[thorn yacht]

Could someone perhaps help in pointing out whether my solution is false or not:

Problem: "Suppose $v_1, \ldots, v_m$ is linearly dependent in $V$ and $w\in V$. Prove that $$\dim span (v_1 +w, \ldots, v_m +w) \geq m-1.$$"

[stoic python]

Kaishin

[thorn yacht]

I was thinking that if the list is linearly independent, then the dimension is m. If it is not, adding w to the list would create a span where the span of v1, ..., v_m is contained in and thus the dim of the new list must be greater than or equal to m-1, since span of v1, ..., v_m is a subspace of span of v1+w,...,v_m+w, and the dim of a subspace is less than or equal to the span of the space.

Does the argument hold?

[limber sierra]

you didnt find an example

[rough talon]

Does the cross product of 2 vectors typically only apply to R3? The only other example I can think of is if 2 vectors are antiparallel in R2 that could result in a vector perpendicular to both of them

[half ice]

Yes, cross product is strictly R3

[rough talon]

is there always some vector that is perpendicular to 2 vectors in R3?

[half ice]

It takes advantage of the idea that there's a unique* vector perpendicular to any two others

Where in R4 you get plane when you do the same thing

[limber sierra]

yes, but it wont necessarily be unique

if your two vectors are colinear it wont be

[rough talon]

in R3, does the direction of the vector (the result of the cross product) only matter as the dot product of the cross product to both (lets say vector a and b) have to be 0?

[half ice]

a×b returns some vector perpendicular to both a and b. Let's call that vector c.

a•c = 0, since they're perpendicular.
b•c = 0, since they're perpendicular.

[rough talon]

and if we took C and changed its direction (multiply by -1) that vector is still perpendicular to a and b right?

[half ice]

Ye

[rough talon]

a•c = a • -c?

[half ice]

a•(-c) = -(a•c)

Dot product associates like a regular multiplication, which is cool

[rough talon]

a•c = a • -(c) : but in the case of the cross product, this holds true

as they both result in 0

The direction of the cross product is simply defined?

From what I read, it seems the simplest answer is that subtraction is not commutative and when we switch the cross product from a X b to b X a that results in a sign flip

[hard drum]

Anyway, @ hismajesty... sure, so your question boils down to "how can we (efficiently) calculate A^n by diagonalising A"?

[cold ravine]

No I need to use diagonalization to prove that $F_{n} = \frac{\varphi^{n} - (1 - \varphi)^{n}}{\sqrt{5}}$

somehow

[stoic python]

HisMajestytheSquid

[hard drum]

Ye, which boils down to this

So we had an expression for F_n in terms of A^n

So if we can calculate the latter we have F_n

[cold ravine]

Oh okay

[hard drum]

Yee, do you know anything about calculating A^n given A is diagonalisable ( in general)?

[cold ravine]

Not really, it was sort of handwaved in class and then he moved on

Well..

[hard drum]

If so I could try to lead you there

*if not

Ah, sure

So the important thing is that if A is diagonasible, we may write it as A = P^-1DP for P an invertible matrix and D diagonal, agreed?

[cold ravine]

I guess I know how to diagonalize matrix. P = some matrix made up of eigenvectors D = some matrix with eigenvalues on the diagonal

[hard drum]

Sure

[hard drum]

Now, if you understand this, see what happens when you calculate A^n

like A^3 for example (in terms of P and D)

[cold ravine]

I'm reaching here because it's been awhile but I think A^3 = P^-1 D^3 P

[hard drum]

Correct

just because A^3 = P^-1 DPP^-1DPP^-1DP and we have some cancellations

[cold ravine]

Right, so in general, we have A^n = P^-1 D^n P

[hard drum]

Exactly. And D^n is very easy to compute as D is diagonal

does that give you enough stuff to work further on the problem?

[cold ravine]

Yeah, that helped a lot, thank you

[hard drum]

Np!

[wooden eagle]

hey guys so i'm doing this question and i've proved 7 axioms, and the last 1 left is to show that a multiplicative identity e exists in F.

but what if V and W have different multiplicative identities: as in if there is f such that f v_1 = v_1 and g w_1 = w_1, then how is it possible to have just 1 multiplicative identity e in F for the vector space Z

[sonic osprey]

How much do you know about fields?

It's true that theres only one multiplicative identity e in F

[limber sierra]

V and W having different multiplicative identities is irrelevant

i mean, you should account for that in your proof

but like

each element of Z consists of an element of V and an element of W

so if we want to find the identity of Z, what is your "first guess" as to what your element from V should be? your "first guess" as to what your element from W should be?

[wooden eagle]

well there's a corresponding identity for V and W, but we can't really combine those two identity scalars into one identity scalar for Z

[limber sierra]

oh fuck i brain farted

sorry i phrased that terribly

let me rephrase

the multiplicative identity of your scalars is necessarily the multiplicative identity of the underlying field.

V and W are vector spaces over the same field

[wooden eagle]

yeah but why's that

the requirement is only that there exists some c in F such that cv = v

and some c in F such that cw=w

[limber sierra]

compatibility

(ab)v = a(bv), right?

[wooden eagle]

yes

[limber sierra]

so suppose we have an identity, call it i

but its NOT the identity of the field

that means ib is NOT equal to b

since its not the identity of the field

i(bv) = bv since i is the identity of the vector space (the product bv is a vector)

[wooden eagle]

right i see

[limber sierra]

but compatibility says i(bv) = (ib)v

[wooden eagle]

thanks

makes sense now

[limber sierra]

this is... a problem

since that would imply ib = b

so yeah, necessarily the identity of vector-scalar multiplication is the identity of the underlying field

and that is unique

(do you know why?)

hence V and W have the same identity

[wooden eagle]

we can just use cancellation laws right

[cold moat]

3.The quotient of a number
and three is equal to twice
the difference of the square
root of the number and the
number squared.

PLWASE ANYONE

PLEASE

TRANSLATE INTO A ALGEBRAIC EQUATION

[limber sierra]

this isnt linear algebra

[cold moat]

WHAT IS IR

WHAT CHANNEL

[limber sierra]

please chill the caps

[cold moat]

Sorry

[limber sierra]

try #prealg-and-algebra or one of the 10 questions channels

[cold moat]

Ok

Thx

[limber sierra]

just dont interrupt anyone

[cold moat]

[reef sleet]

So he's just saying split up f as a sequence of functions in R^m?

[teal grotto]

not really a sequence, they are component functions. so for example, if $f$ represents left multiplication by the matrix
$$f=\begin{bmatrix}2 && 3\ 4 && 5\end{bmatrix}$$, then $$f_1=\begin{bmatrix}2&&3\end{bmatrix}$$ and $$f_2=\begin{bmatrix}4 && 5\end{bmatrix}$$ and you can check that
$f(x,y)=(f_1(x,y),f_2(x,y))$

@reef sleet

[stoic python]

c squared

[teal grotto]

more formally, for each $1\leq j\leq m$, $f_j$ in your text is defined as $f_j=\pi_j\circ f : \mathbb{R}^n\to\mathbb{R}$ where $\pi_j:\mathbb{R}^m\to\mathbb{R}$ is the $j$-th projection map defined as $\pi_j(y_1,\dots,y_j,\dots,y_m)=y_j$

[stoic python]

c squared

[reef sleet]

brain go owie man this is so much @.@

[teal grotto]

kinda symbol heavy, but i tried to give a simple example first lol

[reef sleet]

Yeah the example makes sense, don't worry. This is a multivariable calculus textbook I remember someone said yesterday that this stuff wouldn't be that useful for calculus or smth like that

[waxen jacinth]

How would I go about this?

Im stumped in how I should lay this down in writing

[lavish jewel]

just apply the definition of a vector space again

you will notice that a bunch of the properties are immediately satisfied (like the ones related to scalars)

then given that the sums and scalings of elements of W are in W, show that the other properties are also satisfied

[waxen jacinth]

Thanks, Ive completely forgotten linear lmao

[wintry steppe]

Can someone suggest an approach for #3?

[random axle]

What number can you multiply by itself to give i?

[lavish jewel]

putting it in polar coordinates might make it easier to see

[glad acorn]

Shouldn’t this be linearly dependent?

[dusky epoch]

why?

[digital gulch]

This is defination of linearly independent ,how can this be wrong

[reef sleet]

What am I doing wrong here?

https://cdn.discordapp.com/attachments/858954660513316875/887017157576261632/unknown.png

I already did the first part

But here's my work for the second problem

https://cdn.discordapp.com/attachments/858954660513316875/887017236919902238/unknown.png

The problem is x.u = -2, not 0

[reef sleet]

OMG

IM A FUCKING IDIOT

$\frac{2}{2} = 2$ btw guys

[stoic python]

feather

[reef sleet]

for those that didn't know : )

[wintry steppe]

🤨

[ionic laurel]

Does this set of vectors span r^3? Or is it a plane

When i check my work i see that there isn’t a pivot in each column

And b vector will not be consistent for all b

But i am getting a little bit confused

[deft apex]

what is the difference between a discrete basis and a continous basis

?

wtf is a continous basis

[dusky epoch]

@deft apex was there any context for that?

[deft apex]

no i am asking

i can't find the answer anywhere

[dusky epoch]

did you see the words "discrete basis" and "continuous basis" in a homework problem?

[deft apex]

in my qm text it says that we replace the sums by integrals if the basis is continous

so i am asking wth that even means

[dusky epoch]

can you show the text itself

i have a feeling it'll make things clearer if you show the exact words they use

[deft apex]

i think i misread something lmao

i'll come back to it if i don't understand it then

[ionic laurel]

@dusky epoch if you aren't busy could you help me out

[dusky epoch]

Does this set of two vectors span R^3?

no

[ionic laurel]

It spans a plane correct?

[dusky epoch]

no matter what those vectors are, there's not enough of them to span R^3.

yes, their span is a plane.

[ionic laurel]

Oh okay

Also one more question

[dusky epoch]

yes?

[dusky epoch]

,rccw

[ionic laurel]

This one would also be a plane?

[dusky epoch]

the span of the columns of $\bmqty{1 & 0 & 2 \ 0 & 1 & 3 \ 0 & 0 & 0}$?

[stoic python]

Ann

[dusky epoch]

yes

[ionic laurel]

ohhh okay

thank you very much Ann

[modern palm]

Is ${c_{0} + (5)^{1}c_{1} + (5)^{2}c_{2} + ... + (5)^{n}c_(n) = 0}$ the correct way to write the set of polynomials with $f(5) = 0$?

[stoic python]

42_Toxic_Potatos

[nocturne jewel]

Like... yes, but atypical imo

and it depends on what field the coefficients are from

[modern palm]

what is atypical

[nocturne jewel]

not typical

[modern palm]

ok

by coefficient you mean c_1, c_2, etc right? They are just real numbers

[nocturne jewel]

${p\in\mathbb{R}[t]_{\leq n}|p(5)=0}$

[stoic python]

Mosh

[zenith junco]

why

[modern palm]

okok, I kind of want a more clear visualization of the set since Im just beginning linear algebra. But that notation is better.

[nocturne jewel]

yeah, there's other notations for the polynomial space, I just use that one

Field[variable]_(=< degree) is set of polynomials in (variable) of degree at most (degree) with coefficients in (Field)

[modern palm]

also just as an extention, the polynomial f(x) = 1 is in the set right? By just setting for instance a_1 = 1 and the rest of the constants = 0

[nocturne jewel]

in this specific set or the general set?

[modern palm]

the specific set

[nocturne jewel]

no, cause 1 isnt 0

[modern palm]

ok so the polynomial f(x) = 1 would never be in the specific set since no constant in R can make that polynomial = 1

[nocturne jewel]

yes, it doesnt meet the requirement of having a root of x=5

it's in R[x]_(=<n) though

[zenith junco]

hi

how is it possible to have a basis

if a1 a2 a3 are linear combinations of each other

oh nvm they arent

oh wait

yeah they are because a1 = a3+a4

[manic blade]

note that a1, a2, a3, a4, a5 are not vectors they are elements of a vector in F^5

[lapis cosmos]

4 2

dim{x from V: Ax=0}= dimV-rank(A)

[zenith junco]

oh ye

thanks

[empty ibex]

if A is a 2 by 2 matrix with no entry being zero, and A**2= - A, How would I figure out A?

[teal grotto]

A is small enough to solve directly

just to make sure, it’s A^2 = -A, right? A**2 doesn’t mean something else?

[gleaming patio]

yea just take your matrix to be
a b
c d
and then solve the equations, you get, since the no of equations is lesser than the no of independent variables, multiple solutions exist, one of the trivial ones would be all of the entries being equal to -0.5

[empty ibex]

got it thanks!

[empty ibex]

nope ur correct

[empty ibex]

i cant seem to solve the equations, did u use elimination?

[gleaming patio]

Just to reconfirm, these are the equations you got right

1. a^2+bc=-a
2. a+d=-1
3. d^2+bc=-d
   So now since we have 3 equations, I just equated a=d and b=c, then everything became -0.5. You could assign a or any other variable a random value of your choice and still solve it, because there are infinite solutions possible

[teal grotto]

u need the set of solutions. i would substitute d = -(1 + a) and then subtract equation 2 from equation 1

should get a quadratic in a

[gleaming patio]

I just realised there are actually 2 free variables here, one would be either a or d and the other free variable would be either b or c . The reason b or c is also a free variable, because we are only concerned about the product bc, but there is no other constraint on either b or c, so I am not sure if we could get a set of solutions as such in terms of 2 free variables or 2 parameters

[teal grotto]

and i think ur missing one
ab + d^2 = -b

[gleaming patio]

how did you get that ? you will get -b only when you multiply the 1st row with 2nd column right, so that will be ab+bd=-b

[teal grotto]

? u need to solve for what values a,b,c,d can take on. not sure what ur talking about with free variables and stuff

[teal grotto]

multiply out A^2 and set it equal to -A

excuse me for being dumb
i mixed up my b and d. lol

[tranquil steeple]

[ a^2 + a + b*c, b + a*b + b*d]
[ c + a*c + c*d, d^2 + d + b*c]

solve for those four to be zero

[gleaming patio]

No problem, it happens 😆

[tranquil steeple]

first express a in terms of bc (top left), then d in terms of bc (bottom right)

[gleaming patio]

A free variable is one which can take any value and is not constrained by any equations, so for example if you have the equation ab=1, and you were to find out the set of solutions for this equation, then there is 1 free variable a which you can give any value you want, then based on that value of a you find out b.

[teal grotto]

yea bruh it’s late i need to go to bed and stop confusing ppl online

[empty ibex]

oh god I see what u mean....that was so dumb of me XD. thanks for all the help u guys!

[muted loom]

Is anyone able to explain how id go about moving this to echelon form and how that displays the constraints?

[teal grotto]

just row reduce the augmented matrix

a b | 1 0
c d | 0 1

until the left hand side is the identity matrix

[modern palm]

If I have $R^{3}$ with normal multiplication and new addition defined as $(a_{1}, a_{2}, a_{3}) + (b_{1}, b_{2}, b_{3}) = (a_{1} + b_{1} + a_{2}, a_{2} + b_{2}, 0)$. Does the 0 vector exist?

[stoic python]

42_Toxic_Potatos

[teal grotto]

u mean R^3?

[modern palm]

yes

the trouble Im running into the the last element.

one sec, let me process what i did

[lavish jewel]

only for elements with a3 =0?

this addition is a sort of projection onto the xy plane

[teal grotto]

if (a1, a2, a3) + (c1, c2, c3) = (a1 + c1 + a2, a2 + c2, 0) = (a1, a2, a3)
then a3 = 0, c2 = 0 and c1 = - a2. but thats not good, since c1 is fixed and a2 is variable

[modern palm]

I set $0 = (x, y, z)$. Then since $0+v = v = v + 0)$, I got two equations $(x,y,z) + (a_{1}, a_{2}, a_{3}) = (a_{1}, a_{2}, a_{3})$ and $(x,y,z) + (a_{1}, a_{2}, a_{3}) = (a_{1} + x + a_{2}, y + a_{2}, 0)$ Then I set $(a_{1}, a_{2}, a_{3}) = (a_{1} + x + a_{2}, y + a_{2}, 0)$.

[stoic python]

42_Toxic_Potatos

[lavish jewel]

oh i forgot to check commutativity lol but just looking at the 3rd elem alreafy shows that, at best, it wouldnt work on all of R3

[modern palm]

So $a_{1} = a_{1} + x + a_{2}, a_2 = y + a_{2}, a_3 = 0$

[stoic python]

42_Toxic_Potatos

[limber sierra]

i dont understand what youre trying to do

[teal grotto]

trying to show that the new addition on R^3 has no additive identity

[modern palm]

Trying to prove or disprove this is a vector space. Specifically, if a 0 vector exists

[limber sierra]

oh i see

thats what they meant by "zero vector"

was confused

[modern palm]

This is where Im having the trouble though. As there is nothing to solve for z.

[lavish jewel]

cuz its being projected on xy

[modern palm]

what does that mean

[lavish jewel]

that any nonzero z is changed to 0 by this addition

[modern palm]

So, essentially the third element of the 0 vector can be anything

[lavish jewel]

youd think that, but it actually means there is no 0 vector for vectors not on the xy plane

you cant add anything at all to a vectpr not on the xy plane, without making z be 0

[modern palm]

So that means that 0 vector doesnt exist unless its on xy plane

[lavish jewel]

you need a3 + b3 = a3 AND a3 + b3 = 0

thats what i said a while ago

[modern palm]

im just processing

oh ok

[modern palm]

that never happens

[lavish jewel]

it does if a3 = 0

amd b3 = 0

was the question to check if its a vevtpr space?

[modern palm]

yes

[lavish jewel]

i think there are also commutativity issues with this sum

[modern palm]

yeah, first thing I thought was the existence of 0 vector. So here we are lol

[north steeple]

Where can I ask about 2s complement representation of binary numbers?

[dusky epoch]

#discrete-math or any questions channel @north steeple

[thorn yacht]

Hey, I was looking at this problem, and was wondering whether my solution was valid, as it was quite different from the solution manual's, which seemed clean but I can't see how one would've come up with the idea.

My solution:

Nvm, just found a mistake

Actually, I'll just put it out there in case I'm wrong about being wrong...

It's clear that $null T \cap {au : a \in \textbf{F}} = {0}$. So we need to show that $V = null T + {au : a \in \textbf{F}}$. If $$T(u) \in {au : a \in \textbf{F}} \iff T(\lambda u) = \lambda T(u) \in {au : a \in \textbf{F}}$$ $$\iff (\lambda a)u \in {au : a \in \textbf{F}} \iff \lambda a \in \textbf{F} \iff \lambda \in \textbf{F}$$. So $$\forall v \in V: v = w + \lambda u, w\in null T$$ So $$V = null T \oplus {au : a \in \textbf{F}}$$

[stoic python]

Kaishin

[past violet]

I don't understand your second proof: you claim something about elements of a subspace of V, which turns out to be equivalent to a tautology? and that implies something on the whole V? 🤔 or else I didn't get who's lambda

I don't understand neither why the first proof is so wrong: the decomposition of v looks quite ok, as the denominator is not null, and it indeed seems to fit the needs @thorn yacht

[thorn yacht]

The first proof is right, it just isn't mine - felt it was a clever construction, that's all.
I was unsure whether the second proof holds (if lambda is a scalar on V).

[past violet]

the second proof looks reaaaally ambiguous. if I drop the details, the first long implication chain reads as: "T(u) is in a subset of V if, and only if, lambda (that I choose in F as I want) is in F"

(which is always true anyway, if I don't mistake)

so I don't see very well how you can infer something on any v in V, with this tautology

[thorn yacht]

I thought that if lambda is in F, then any element of V can be written as a w+lambda u, for w in nullT and lambda u for u not in nullT. If that is true and the intersection of nullT and the set {au : a in F} = 0, wouldn't that imply that V is the direct sum of nullT and {au : a in F}?

I'm probably wrong, I just wasn't sure why.

[past violet]

I think there is a small ambiguity in the quantifiers; you need to prove that for every v in V, you can find a lambda that matches the claim

here it seems you take it reverse: you start from any lambda and you want to say something

I think you're trying to postulate "there exists lambda such that for any v, [...]"

but that's false actually; the correct claim is: "for any v, there exists lambda such that [...]"

[thorn yacht]

I see. I'll go through it and try to rework it properly. Thanks for the clarification.

Oh, I think I meant for the long implication chain to show that the scalars on V are elements of F.

[lavish jewel]

just for a generic x

the given map is not pos def, right

[half ice]

@hexed plaza
You need to check non-negativity for all possible inputs.

However, if you break non-negativity by plugging in a specific input, then you've created a counter-example and you're done

Whoops I was not scrolled down all the way

Sorry about the ping

I see what you did above, and it looks good

[sonic osprey]

Maybe the Cayley Hamilton theorem is an example of the characteristic polynomial mattering outside of eigenthings

[wintry steppe]

not the values, but the coefficients of the characteristic polynomial are given by (up to sign) the traces of the exterior powers of your operator

maybe that'll help you find something

if $T\colon V \to V$ is a linear operator then for each $k$ it induces a linear operator $\wedge^kT\colon\wedge^kV\to\wedge^kV$ such that $$\wedge^kT(v_1\wedge\cdots\wedge v_k) = Tv_1\wedge\cdots\wedge Tv_k$$

[stoic python]

TTerra

[wintry steppe]

by the universal property of the exterior power or something

so then if your characteristic polynomial is $$p(t) = t^n + a_1t^{n-1} + \cdots + a_{n-1}t + a_n$$ you have $a_k = \pm\mathrm{trace} \wedge^kT$ (i do not remember the signs and it depends on your definition of $p$)

[stoic python]

TTerra

[wintry steppe]

sssssssomething like that

like you might know that for a 2x2 matrix, p(t) = t^2 - tr(T)t + det(T)

or something

this generalizes

now whether or not this gives you some cute interpretation of the values of the characteristic polynomial, i do not know

[wintry steppe]

no

[modern palm]

I dont understand how they showed span(u+2v, u-v) is a subset of span(u,v)

[lavish jewel]

show theorem 6.2.2.

what they're showing here is that you can write each set of vectors as linear combinations of the other vectors

[lavish jewel]

well

what they did is show that each set of vectors is in the span of the other, so you have 2 pairs of lin indep vectors in a 2D space

[modern palm]

i think it makes intuitive sense why span(u+2v, u-v) is a subset of span(u,v)

[lavish jewel]

the thing is that the span is the set of all linear combinations

if you can get the vectors w and z, you can also get scaled versions of them

and also sums of them. all due to linearity

[modern palm]

so is it correct to say span(A) will always be a subset of span(B), where B consists of all the vectors in A?

[lavish jewel]

yes, but idk if that is what you wanted to ask

[modern palm]

yeah, I understand why span(u+2v, u-v) is a subset of span(u,v).

Because like you said, span(u+2v, u-v) is just the set of all linear combinations of u+2v and u-v.

If we can form u+2v and u-v from span(B), that means we can take some linear combination of B to get to (u+2v, u-v)

But its a question asks to determine if span(1, x, x^2, 2x-2) = span(2, x - 1, x^2 + x, x^3), then the answer would obviously be no, right? Since span(1, x, x^2, 2x-2) doesnt contain x^3

[lavish jewel]

indeed

[modern palm]

Ok, but how would I *prove that? Since there are no counter examples

[lavish jewel]

wdym there are no counterexamples?

you just gave one

x^3 is in the span of one set and not in the other

[modern palm]

okok, thank you

[wintry steppe]

help

[modern palm]

how can we show this?

[teal grotto]

given a linear combination a + bx + cx^2, show that it can be written as d(1+x) + e(x + x^2) + f(1+x^2) for some some coefficients d, e, f

[lavish jewel]

you can also follow the procedure from the previous problem and show that you can express 1, x, and x^2 in terms of the other basis

[modern palm]

sth like this?

[teal grotto]

ye. u can now solve for d e and f in terms of a b and c.

not sure what previous procedure was, so sorry if this was off track

[lavish jewel]

nah it's the same thing

[modern palm]

okok,

[lavish jewel]

this is the same as expressing the canonical basis in terms of the new one

which is what was done in the previous task

[modern palm]

what if the questions is $span(1, x, x^{2}) \subset span(1+x, x+x^{2}, 1+ x^{2}, 2x-1}$?

[stoic python]

42_Toxic_Potatos
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[modern palm]

Using the same way, this would produce 3 equations with 4 variables

[lavish jewel]

why 4 vars

you're still looking for the coeffs of a + bx + cx^2

the equations for each are longer tho

[modern palm]

doesnt this produce 4 var? d,e,f,g

or do i need a break 😑

[lavish jewel]

the issue is that the sets have different numbers of elements, so your options will either be infinitely many sols or no sols

[empty ibex]

I was able to show it mathematically like this...but how do i prove it in words?

[past violet]

you need to explain; so for example you could say:

$f_\theta \circ f_\phi$ corresponds to first applying $f_\phi$ (rotate by $\phi$ radians) and then applying $f_\theta$ (rotate by $\theta$ radians). As a result, the composed transformation is a rotation by $\theta+\phi$ radians, so $f_\theta \circ f_\phi = f_{\theta+\phi}$. Since linear map composition translates as matrix composition, this yields the desired identity

[stoic python]

judekeyser

[empty ibex]

why are we rotating by phi radians first tho? is it because of the matrix composition?

[past violet]

yeah in your exercise it's phi first

I'd just explain with geometric content (so talking about rotations in the plane) that what we are proving, is basically: when I rotate first of phi, then of theta, it's the same as rotating only once, of theta+phi

[empty ibex]

wait so if I rotate by theta first and then phi, the answer would still be the same as rotating once by theta+phi?

[random axle]

I have solved part a) of this problem, by finding the determinant of the coefficients of the system of equation and equating that to be 0. Therfore, the system of equations only have a unique solution when t = 1 or t = -1/8. I am now stuck on how to go about part b) What relationship between must be true in order for the system of equations to be consistent? I.e. for the system of equations to have itleast 1 solution for t = 1.

[past violet]

indeed

in the plane, rotations are commutative: you can apply them in the ordering you want

[empty ibex]

I see.. this makes it so much clearer. Thanks a lot!

[rigid sorrel]

Let A,D, and P be nxn matrices satisfying AP=PD. Assuem that P is nonsingular and solve this equation for A.

This is what I did

we have AP=PD

we right multiply with P^-1

by Doing so we have APP^-1=PDP^-1

on the right side P*P^-1=1

Therefore, we have A=PDP^-1

However, apparently my answer is wrong, and I don't understand why.

Can anyone tell me where I messed up?

<@&286206848099549185> Can I get some help please

[hard drum]

I mean that is correct

Could you send the original problem?

[rigid sorrel]

Sure

[warm rock]

guys

[rigid sorrel]

Here it is

[hard drum]

Bruh

I agree with you lol, odd

[rigid sorrel]

i guess maybe it's just something wrong with the software then?

[fiery silo]

Oh

Maybe you actually have to put "A ="?

"A = PDP^-1"

Not sure how many attempts you have, but that might work

[rigid sorrel]

let me try that

[rigid sorrel]

Thank you so much! That worked!

[fiery silo]

Np, glad that worked

[naive crest]

hey

can anyone

help

w

some

angebra

if anyones

free

[empty ibex]

I did it this way but im unsure if this is the answer for 9

[zinc copper]

Asking again since there was no one available yesterday but would someone who can read French quickly look over my first semester Lin alg lecture notes and tell me if they seem standard?

[opal mortar]

hey guys,, can the result of the determinant be a decimal or a fraction?

[limber sierra]

certainly.

not if the matrix itself consists only of integers, though.

but consider, say, an identity matrix except you replace one of the 1s with a pi

the determinant of that matrix would be pi.

(since the determinant of triangular matrices is the product of their diagonal entries)

use 3/4 instead, and its determinant is 3/4

etc

[plucky crypt]

Hello, given a normed vector space over $\mathbb{C}$ and two linear independent vectors $v,w$ how can I show (if true) that $\inf{|v+zw|: z\in \mathbb{C}}>0$?

[stoic python]

Maldor

[opal mortar]

would you check it out? where am I wrong in the calculation?

[dusky epoch]

what is this symbol

it looks like some kind of weird backwards &

[zinc copper]

Prolly an 8

[zinc copper]

While you’re here Ann do you think you’d have the time to give this a quick look?

[dusky epoch]

uh

i can certainly try

may i have your notes then

[zinc copper]

Yeh ima fetch the@

Them*

[opal mortar]

28/5

[dusky epoch]

so that's an eight...

in any case idk maybe you screwed up some copying or some arithmetic somewhere

i would suggest avoiding fractions to whatever extent possible

[opal mortar]

I don't know where I went wrong, after looking so far. the determinant is 173.6 ... so is it possible if the determinant is decimal ?

[dusky epoch]

you started with a matrix made entirely of integers, so no, in your case that couldn't have happened.

[limber sierra]

your headache comes from trying to put this in rref

rather than just ref

all you need is a triangular matrix to compute the determinant quickly

[zinc copper]

There are definitely other errors but the very first manipulation misses a minus sign

[dusky epoch]

are we explicitly required to do this with elementary row ops

[opal mortar]

the question command is asked to search with ERO

I looked it up in the matrix calculator on computer and it turned out to be 958

[plucky crypt]

<@&286206848099549185>

[opal mortar]

Well, I think I made a mistake in calculating it, but I don't know where the error lies

[topaz moon]

hi everyone, I was wondering if I could receive some help on this question

I believe A,D are correct. I think b is wrong and I'm not really sure about c

[lavish jewel]

a and d seems right. you can check C by building A^T A

this would be V S U^T U S V^T. U^T U is an identity mat, since U is orthonormal

also S^T = S, since S is diagonal

so you get V S^T S V^T = V S^2 V^T

that is of the form Q D Q^-1, an eigendecomp with eigenvalues D

you can see what's what from there

[topaz moon]

Oh I see

Thanks, I'll work that out and see what I get

[lavish jewel]

you can do that same procedure for all of A,B, and C, btw

D is just a definition that you can look up in your notes or on wikipedia 😛

[topaz moon]

yeah that's where i found it!

[wintry steppe]

is there a well defined algo to check if N matrices commute?

[lavish jewel]

a quick check that is necessary but not sufficient is to see if you can diagonalize all the matrices simultaneously (or make them all triangular)

at least over C

[zinc timber]

I have a question from Lp space

Given $| f | = |g | = 1$ and $f_t = tf+(1-t)g$ show that $|f_t| < 1 $ unless $f=g$ a.e.

[stoic python]

Ryuzaki

[zinc timber]

how do I show that if $f\neq g$ then equality cannot hold

[stoic python]

Ryuzaki

[dusky epoch]

you're working in $L^p$ with $1 < p < \infty$, yes?

[stoic python]

Ann

[dusky epoch]

@zinc timber

[zinc timber]

yes

[dusky epoch]

ok nevermind i lost my idea

[zinc timber]

[lavish jewel]

couldn't you just use triangle ineq? i'm hypoglycemic atm so i might be way off

(i'm seriouly hypoglycemic)

[dusky epoch]

me too

[zinc timber]

How triangle ineq's gonna help?

[lavish jewel]

yeah idk lol, it ends up with the same problem of having to show that you strictly have < instead of <=

[zinc timber]

yeah

for R2, it's kind of easy

geometrically, but for Lp.....

[lavish jewel]

how about saying g = f + q

and see what happens to q

[zinc timber]

There's also another problem I'm stuck with, given $l:X\to \mathbb{R}$ be a linear functional, show that $l$ is continuous iff $ker(l)$ is closed in $X$

[lavish jewel]

i'm literally making shit up, it'll take my blood glucose like 30 mins to normalize, so feel free to ignore me

[stoic python]

Ryuzaki

[zinc timber]

bruh no one

[zealous junco]

hm are you stuck on showing continuous if kernel is closed

[zinc timber]

yeah, that one is easy

if stuck with 'if kernel is closed then l is continuous'

[zealous junco]

yea ic, so

[zinc timber]

lol I misread yr statement

[zealous junco]

i think you should work with sequential def, pick some x_n -> x and l(x) = l(x_n) + l(v_n) where v_n -> 0

in the norm space

then you can write v = norm(u)*u where u is a unit vector, and the goal is now to show the image of unit ball under l is bounded if kernel of l is closed

once that is done, the rest is just by noting l(x) - l(x_n) = l(v_n), but now you know l(v_n) -> 0

[zinc timber]

let me see

[mystic dome]

Hi. I have some issues understanding this question. Please do not reply with the answer 🙂 but just with clarification/suggestions . To me, that would just be H @ H. But it is not the answer accepted

[zinc timber]

those parts are already clear to me, but how do I show that unit ball is bounded

[zinc timber]

just expand H² out, u don't have to multiply,

these matrix has a special name 'Housholder Reflection matrix'

[mystic dome]

Great, thanks a lot. Is there any reason why one should be interested in H^2? or it is just an exercise? Edit: uh, now i see 🙂

[zealous junco]

ok if it were not, then you would find a sequence {u_n} unit vectors s.t. l(u_n) > n right? then choose an x s.t. l(x) = 1, and the sequence x_n = x - u_n/l(u_n) from the kernel space converges to x, meaning l(x) = 0

because kernel is closed

[zinc timber]

oh yeah nice tx

[zinc timber]

Not really H² but H matrix is often used to find the eigen values/vector of a symmetric matrix. there might be other applications in physics but idk

[mystic dome]

Well. At the end H^2 = H. So it was just an exercise. Thanks for the explaination!

[zinc timber]

oh wait my bad it's not Housholder,, there'll be a 2 in front of 1/n. it's simply a projection matrix. Sorry about that

[empty ibex]

can someone review this and see if my answer makes sense?

[chilly jackal]

I'm lost

somebody get me a map

[mystic dome]

f: X -> Y
it is a simple map, but that should do the trick

[chilly jackal]

Isn't that a function?

Oh and thanks

I found the buried treasure

[zealous torrent]

Say I have a dataset of a couple thousand nxn matrix pairs, (X and X'), and I want to solve the following "A X_i ~= X'_i" with a best fit approach, how would I go about that?

I guess I am trying to find the A that minimises this function:

[stoic python]

V4

[lavish jewel]

so minimizing the frobenius norm

let's see

a single A, right?

[zealous torrent]

Yup, that's right

[lavish jewel]

lemme see. you can either directly use matrix calculus to compute gradients, or vectorize the matrices X and X' and find an equivalent map B that can be reshaped into the A you want

i think there's a straightforward way to compute the frobenius norm using traces, so that'd be my suggestion

yes, sqrt (trace(M^T M)) if your matrices are real. hermitian transpose if they're complex

[zealous torrent]

I'm really rusty on my linalg, say I use that way to calculate the frobenius norm, that simplifies the summation. I'd still have to calculate the gradient of the function, right?

[lavish jewel]

mhm

[zealous torrent]

Alright, I'll report back in a bit. I don't have pen and paper atm

[lavish jewel]

does the cost function have to be that one?

[zealous torrent]

Nope, what did you have in mind?

[lavish jewel]

square the frobenius norms to get rid of the roots and make the gradient nicer

[zealous torrent]

Oh right, and it should still give the same results if my understanding is correct

[lavish jewel]

i'm too sleepy to even think about why, but i think that squaring each term should preserve the argmin since the numbers are positive and x^2 is monotonically increasing

[zealous torrent]

So here's what I got

$\sum_{i}Tr\left(\left(AX_{i}-X'{i}\right)\left(AX{i}-X'_{i}\right)^{T}\right)$

[stoic python]

V4

[zealous torrent]

$\sum_{i}Tr\left(\left(AX_{i}-X'{i}\right)\left(X{i}^{T}A^{T}-X'_{i}^{T}\right)^{T}\right)$

$\sum_{i}Tr\left(AX_{i}X_{i}^{T}A^{T}\right)+Tr\left(AX_{i}X'{i}^{T}\right)+Tr\left(X'{i}X_{i}^{T}A^{T}\right)+Tr\left(X'{i}X'{i}^{T}\right)$

[stoic python]

V4
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[zealous torrent]

Gradient;

$\sum_{i}2AX_{i}X_{i}^{T}-2X'{i}X{i}^{T}=0$

[stoic python]

V4

[zealous torrent]

Which gives

$A\sum_{i}X_{i}=\sum_{i}X_{i}'$

[stoic python]

V4

[zealous torrent]

At this point I'm not sure anymore, that can't be the answer right?

[still lodge]

can someone help me define transition, translation and transformation matrices and their differences pls

i think the last two are interchangeable but i know there's some subtlety with transition

[short elm]

What else would be in the range?

[nocturne jewel]

do you mean the 0 mapping?

if so then the image is just 0

[still lodge]

"this is the gun and these are my bullets" - my prof explaining linear transformations

[empty ibex]

Anybody?

[nocturne jewel]

yeah so A is the 0 transformation

[empty ibex]

Anyone can help figure out 3? I can't understand the question

[zenith junco]

could someone please explain this to me

[nocturne jewel]

so trivially they are independent sets, so you need to check if they span V

[zenith junco]

i dont understand how it proves this

[nocturne jewel]

you have an independent spanning set

that's the definition of a basis

[zenith junco]

and then

but then how is (U+v, au) also a basis for v

[nocturne jewel]

cause it spans V

[zenith junco]

how do we know its lin indep

how do we know adding vectors from a basis still makes it span and is lin indep

also i dont understand the soln at all for showing it

😔

[nocturne jewel]

You consider the span of the set

$x(u+v)+y(au) \ =(x+ay)u+xv$

[stoic python]

Mosh

[nocturne jewel]

setting this equal to a generic vector in V: $$(x+ay)u+xv=cu+dv$$

[stoic python]

Mosh

[zenith junco]

what is x and y ? scalars?

so x+ay are scalars?

[nocturne jewel]

so you need to solve the system $x+ay=c$ and $x=d$ which in matrix form is $\begin{bmatrix}1&a\1&0\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}c\d\end{bmatrix}$

[stoic python]

Mosh

[nocturne jewel]

and want this system to have a unique solution for all c,d in K, whatever the scalar field of V is

[zenith junco]

why where is this comming from...

where is x, y, c, d from

im sorry lol

[nocturne jewel]

need scalars.

[zenith junco]

where did you get that matrix from

[nocturne jewel]

read

I compared the scalar on u and v

and got a system for x and y

[zenith junco]

why did you multiply them by x and y

[nocturne jewel]

...

[zenith junco]

ok look

i know im dumb ok

[nocturne jewel]

cause I considered the span of u+v and au

[zenith junco]

so x and y are like c1, c2?

do you have a yt vid i could watch to understand this fully

[nocturne jewel]

organic chem probably has stuff on bases

or khan

[zenith junco]

ok thanks

[jolly tapir]

idk know which one of these is wrong

can someone help

[zinc timber]

@jolly tapir 4 T?

[jolly tapir]

oh yes

[still lodge]

will ask again but what's the difference between a transition matrix and just thinking about a matrix as a transformation

[nocturne jewel]

Transition matrix I inherently a concept of Markov chains, matrix representation of a linear transformation is what it says on the tin

[sonic osprey]

Do you have some condition on f? This isn't true unless you assume continuity or something

[dire thunder]

it is enough to have set of nonzero points to be of measure 0

[sonic osprey]

lmao

[gray dust]

last q i saw em ask explicitly mentioned C[0,1]. i think they're still working in this space but failed to mention it

[sonic osprey]

yeah, so one direction is clear, if f is the zero function, then the inner product will be zero

Then, its easiest to try to show that if f isnt the zero function, then the inner product isnt zero

Think about a picture, how would you show that the integral of a continuous non-zero and non-negative function is not zero?

Hm, I'm not really sure how that would help

I'm not really sure what you're trying to show

[empty ibex]

cant figure this out

need help

[lavish jewel]

what kind of condition? off the top of my head, you could require sin(theta) > 0 and -sin(theta) < 0

[empty ibex]

not sure.. thats where m struggling

[lavish jewel]

this is simple and it works lol

[flat narwhal]

https://youtube.com/playlist?list=PLbMVogVj5nJQ2vsW_hmyvVfO4GYWaaPp7

Anyone please tell me prerequisites for learning this Playlist?

Calculus enough?

[zinc timber]

actually u don't even need calculus for this

[flat narwhal]

What is needed? I just don't want to start and stop in the middle

[zinc timber]

metrics and determinants will be enough I think

what grade u r in?

[flat narwhal]

12th

Metrics?

Or matrices?

[zinc timber]

probably u shouldn't watch it then

it's for UG/PG studs

u don't need to go that advanced

[flat narwhal]

😭😭

[zinc timber]

I'm familiar with yr syllabus (indian) lol

[flat narwhal]

😭😭

[still lodge]

go watch the 3blue1brown series

at least the first 8 videos are fairly intelligble, if maybe a bit dense in terms of content presented in a short period of time

and it'll give you a really good foundation

[still lodge]

https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=1

[modern palm]

if i try doing it this way, is there a way to calculate what the coefficients are? Or is it a guess and check thing

[modern palm]

How do I know whether this has infinitely many solutions or no solutions

[tranquil steeple]

it is quite a big difference between the two, just start counting 🙂

[reef sleet]

If a system has an infinite number of solutions and the problem says to "solve" it, will just picking one solution work? Or should I just leave it in parametric form?

[winter harbor]

When someone asks you to solve a system of equations Ax = b for it usually means to find all possible x such that Ax = b.

So if the system doesn't have a unique solution

You will still have to specify all of them

In the case of linear system of equations

Via a parametric equation

[gray dust]

it's like only giving x=2 when told to solve x^2=4

[reef sleet]

Yeah it sounded wrong to me too but I wasn't sure. I left it parameterized

[gray dust]

when put this way it sounds quite silly to only give one of possibly many sols

[winter harbor]

If you have a function f : X -> Y, solving the equation f(x) = y_0 is pretty completely specifying the set f^{-1}(y_0). Which is called the fiber of y_0.

If this function is not injective

There's no reason f^{-1}(y_0) consists of a single element

[reef sleet]

This is cool cause I actually remember this stuff 😁

[versed yew]

Hey guys can someone teach me about REF's and RREF's i am having a hard time to adapt in a class rb

[flat narwhal]

Thanks @still lodge 🌹

[winter harbor]

There's some content on reduced row echelon form and systems of linear equations on Khan Academy.

I mean, this stuff is pretty much computational.

You will learn it by doing tons of examples.

Teaching something like this over discord is quite impossible because of its algorithmic flavor.

But if you have an specific question, then sure just let us know.

[zenith junco]

omg

plz

can someone expalin to me whats happening

[limber sierra]

the idea is, we know matrices correspond to linear transformations

but how do you find this correspondence?

i.e. if i give you a matrix, how do you determine what linear transformation it corresponds to?

it turns out that it suffices to look at what the matrix does to basis vectors

so the example looks at what M does to (1 0 0), (0 1 0), and (0 0 1) [which each correspond to a column of M]

and uses this to reason what the corresponding transformation T must be.

[zenith junco]

ok ok so

i see that it takes the (1 00) vectors right

but i dont understand how we end up with 1e1 + 4e2

what is e1 and e2

[limber sierra]

e_1 and e_2 are the standard basis vectors of ℝ²

i.e. (1 0) and (0 1)

[zenith junco]

okok

and then once we have those equations

wait what how do we use

[limber sierra]

you take the sum, as in the next line

[limber sierra]

given any of the three basis vectors in ℝ³, we can express its image under T as the given sums of basis vectors from ℝ²

so given any vector from ℝ³, we can express its image using this

by writing the vector as a sum of ℝ³ basis vectors and translating those into their images under T

you might notice a "pattern" in your result

[zenith junco]

OFMOGOMGOG

I SEE IT

the e1 and e2 corresponds

ok so

which way

do i use

to solve this type of question i nthe future

[limber sierra]

you mimic this process:

• Determine the images of the basis of V under M.
• Write the above images in terms of the basis of W.
• For any arbitrary vector in V, write it as a sum of basis vectors (we know this is possible because that's what makes a basis a basis), and compute its image by looking at the images of each basis vector (exploit linearity).

[zenith junco]

im so sorry but i lowkey dont understand what you wrote

also what is this saying ;-;

do you have maybe a yt video i could watch on this topic

idek what this topic is

[nocturne jewel]

it says that every transformation from V to W has a unique matrix representation T[v]=Mv, then vice-versa, every matrix M represents a unique transformation from V to W

[torn stag]

@zenith junco You have bases $\beta = {v_1, \dots, v_n}$ of $V$ and $\gamma = {w_1, \dots, w_m}$ of $W$ and a linear map $T \colon V \to W$. You encode the map as a matrix $[T]\beta^\gamma$ by setting the $j$th column of $[T]\beta^\gamma$ to be the coordinates of $Tv_j$ in the basis $\gamma$.

[stoic python]

IlIIllIIIlllIIIIllll

[zenith junco]

ok what does "setting the kth column of matrix M to be the coordinates Tv in basis r " mean

[torn stag]

$v_j$ is in $V$

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

$Tv_j$ is in $W$

[stoic python]

IlIIllIIIlllIIIIllll

[zenith junco]

oooo

so map the entire column vj by doing t(vj) = W

what is this process called

[torn stag]

since $\gamma$ is a basis of $W$, there are unique $c_1, \dots, c_m \in \mathbb{C}$ such that $Tv_j = c_1 w_1 + \dots + c_m w_m$.

[stoic python]

IlIIllIIIlllIIIIllll

[zenith junco]

WAIT SO

[torn stag]

The coordinates of $Tv_j$ are $(c_1, \dots, c_m) \in \mathbb{C}^m$.

[stoic python]

IlIIllIIIlllIIIIllll

[zenith junco]

I USE MATRIX M to transform any vertex in V into a vertex in W?

im so confused

[torn stag]

So $(c_1, \dots, c_m)$ is the $j$th column of $[T]_\beta^\gamma$

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

What is a vertex? You are simply encoding a linear map $T$ into a matrix

[stoic python]

IlIIllIIIlllIIIIllll

[zenith junco]

like a vector *

mb

VECTOR****

[torn stag]

The matrix transforms $\beta$ coordinates of a vector $v \in V$ into $\gamma$ coordinates of $Tv \in W$.

[stoic python]

IlIIllIIIlllIIIIllll

[zenith junco]

oo

[torn stag]

Meaning if $x \in \mathbb{C}^n$ are the coordinates of $v \in V$ in basis $\beta$, then $[T]_\beta^\gamma x$ are the coordinates of $Tv$ in the basis $\gamma$.

[stoic python]

IlIIllIIIlllIIIIllll

[zenith junco]

thank u

[torn stag]

Yeah I guess this is the ultimate motivation behind this way of defining $[T]_\beta^\gamma$

[stoic python]

IlIIllIIIlllIIIIllll

[limber sierra]

this is kind of the ultimate motivation for bases in general, honestly

a basis is the first example an algebra student sees of a "generating set"

which are valuable not just because they let us think about the structure (in this case the vector space) in terms of the basis (a "coordinate system"), but also because they make it really easy to tell what homomorphisms ("linear maps") will do to that structure

[zenith junco]

thank you

[zenith junco]

OMG IHAVE A QUESTION

beta and gama are just pases like

bases*

[1, x, x^2, x^3, ...]

[versed yew]

is this linear?

[nocturne jewel]

yes, they're bases

[torn stag]

@zenith junco $[1, x, x^2, \dots]$ is a basis of $\mathbb{C}[x]$, the space of formal polynomials with coefficients in $\mathbb{C}$. But in the case of $\beta$ and $\gamma$, it is necessary that $V$ and $W$ are finite dimensional to get a matrix representation of $T \colon V \to W$.

[stoic python]

IlIIllIIIlllIIIIllll

[lavish jewel]

what is this M(f) supposed to be

try showing the full problem. i speak spanish and know a little italian

i'm not sure what they're doing, since that can't even be multiplied by the vectors

hmm wait

this is very weird lmao

i'm used to placing transformations on the left

like take vectors V and transform them as MV, with some matrix M

what this person did

they put the vectors in a matrix V that is 4 x 3

and defined the transformation from the right

so you'd take VM = V' to get the output vectors

idk if this is the way you're doing it in class

this is the same as taking a transformation matrix M' that is 4x4, and first converting the vectors with some sort of projection with 4x3 matrices

if you put your 4x1 vectors as cols of a matrix V that is 4x3, you can express the transformation as a 3x3 matrix

i honestly would not have done it this way, but ok

the other would be to make a matrix M' that is 4x4

and make the transformation as M'V

i don't think that's necessary

i guess finding the 4th vector does make it easier (or maybe it was necessary in the end lol)

a vector (-1,1,-1,0) is orthogonal to the other 4. you can use that info

what one could do is make a matrix W with the 3 vectors from V, and a 4th column containing the vector that is orthogonal to all of those 3 (the one i gave above). then we take the inverse of this matrix

in the middle goes a 4x4 matrix that does the transformations we want. call this one T

the important part is to take f(v_i) and express them in the basis of the columns of W

and then the full matrix is W T W^-1

since it's an endomorphism V->V, i'd expect the matrix T to be block diagonal

all right, all good then

what i said works in general for this type of problem tho

just for future reference 😛

aight cool

[flat narwhal]

Anyone please reccomend me some Playlist covering these?

Advance thanks

*youtube

Some of the topics are missing in khan academy

Or khan academy Playlist enough for this?

On linear algebra

[limber sierra]

i have yet to find a youtube series that covers rigorous linear algebra to my satisfaction

mind, i havent looked very hard

but i think textbooks will invariably be a more time-efficient way to learn

[flat narwhal]

Maybe I will follow khan academy

Only thing missing is Cayley Hamilton theorem

I guess

[limber sierra]

it might talk about it at some point, and just not have a section called it

if you dont need the proof, the result itself is pretty simple

every square matrix satisfies its own characteristic equation

[flat narwhal]

Got it, thanks!

[wanton spoke]

There are probably some recorded uni courses on linalg on youtube

Just gotta look 👀

Mit puts a lot of videos so can be a good idea to check out if they have some

[vocal prairie]

Axler has a playlist for his book, which is basically him reading out all the important theorems/results sectionwise. He walks through the more involved proofs, and refers to his book for the rest.

[flat narwhal]

@chrome basin thanks

[flat narwhal]

Sure thanks❤️

[wintry steppe]

Does anybody know if calculus is a prerequisite for linear algebra?

[thorn yacht]

It's not, and I know people who strongly believe linear algebra should be taught before multivariable calculus.

[wintry steppe]

so i dont even need single variable calculus or elementary calculus?

[hard drum]

No you don't - it's just that most people learn calculus beforehand anyway

[wintry steppe]

great

[wintry steppe]

ill start linear algebra now

[primal schooner]

hi can someone help me?

[wintry steppe]

if you post a question, then maybe someone might

[west saddle]

What does the subspace of Psub3(x) mean?

does that specifically imply each row or column is to an additional power? ie [x,x^2,x^3]

and the matrix is a 3x3

[wintry steppe]

what's "Psub3(x)"?

[west saddle]

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

[wintry steppe]

so P_3(R) you mean

it tells you right there what the subspace you care about is

[west saddle]

and its a 3x3

[wintry steppe]

i don't know what that means

what's 3x3?

[west saddle]

a matrix?

[wintry steppe]

no

this consists of polynomials

not matrices

[west saddle]

ok, that makes sense then

[wintry steppe]

they are degree 3 polynomials though (the elements of P_3(R))

[west saddle]

but if it was M_3

thats a 3x3 matrix

[wintry steppe]

sure

[west saddle]

ok got it

[wintry steppe]

that would be the case

[west saddle]

teacher didn't really go over P_3 in class

he implied that there was a problem in the homework but that was all

that helps my understanding a bunch

[wintry steppe]

$$P_3(\bR) = {ax^3 + bx^2 + cx + d : a,b,c,d\in\bR}$$

[stoic python]

TTerra

[west saddle]

Yes and that explains chegg a lot better

[wintry steppe]

chegg

[west saddle]

ya ya i know 😦

this is the only non matrix problem

lol

anyways thanks for the help

[slim knot]

Consider a collection $X$ of pairwise orthogonal unit vectors in $\mathbb{R}^n$. Then for each $i \in [n]$, $\sum_{x \in X} x_i^2 \leq 1$. There is a hint which says this follows from Cauchy-Schwarz but I don't see it yet.

[stoic python]

geekotechy

[zinc timber]

@slim knot isn't is supposed to be equality?

or this one is a practice problem

[slim knot]

Equality may not hold if there are fewer than n vectors

[zinc timber]

hmm didn't consider that,

[zinc timber]

@slim knot one approach I came up with which doesn't use CSw ineq is

extend X to orthonormal basis

the extra terms will act like slack variables and you can remove the

[still lodge]

notational question but what is M_2(R) supposed to be

[zinc timber]

given that $\sum x_i^2 = 1$

[stoic python]

Ryuzaki

[zinc timber]

2x2 matrices on real numbers

[still lodge]

so a vector in M_2(R) is a 2x2 matrix

[zinc timber]

I really don't see the cauchy swarts approach

[zinc timber]

ya

[pine lion]

Hello! I'm trying to show that V, the set of all real numbers x (such that x is greater than or equal to 0) is a vector space, where addition is defined as
x + y = xy+1
Now I'm trying to show that this is associative. So far I've let u=x, v=y and w=z and I'm trying to write out
u + (v + w) = u + (yz+1) but im not sure how to proceed next
I'm not sure if it'd be xyz + 2 or xyz + x + 1

[still lodge]

so if my prof is asking about idempotent matrices in M_2(R), is he asking about vectors in M_2(R) or matrices of matrices?

[wintry steppe]

just elements

vectors in M_2(R)

[wintry steppe]

x(yz + 1) + 1 = xyz + x + 1

[zinc timber]

@still lodge all matrices

[pine lion]

Thank you :) that does make more sense

[wintry steppe]

you might wanna use a different symbol for the addition in your space

[still lodge]

wha-

[pine lion]

Yeah on paper I do but idk how to get a circleplus here

[wintry steppe]

you can use the latex bot

$\oplus$

[stoic python]

TTerra

[wintry steppe]

but that has a learning curve

[pine lion]

but thanks 👍

I know a tiny bit of latex so I'll try next time I have a question

probably will butcher it at first though lol

[wintry steppe]

call up the bot by putting single dollar signs around your equation/symbol/etc., double dollar signs if you want it displayed

you can test in #latex-testing, ask for advice in #latex-help

[pine lion]

Oh okay cool

[wintry steppe]

here's an inline equation: $x^2 + y^2 = z^2$. here's a displayed equation: $$\frac{d}{dx} e^x = e^x$$

for example

[stoic python]

TTerra

[wintry steppe]

once you know that you can pretty much figure it out from there

[pine lion]

Okay displayed being centered and away from all other words?

[wintry steppe]

right

nice and big in the center

you might want to use it if you have something with \frac, since inline fractions can be squished and look bad

[pine lion]

good to know, thanks

[stoic python]

ADevPlayer

[charred dome]

can u help me?

;-;

[slim knot]

Nice

[zinc timber]

didn't see the cauchy inequality approach tho

@charred dome isn't Ann itself is a component, not a tensor

[charred dome]

Yes, I have already solved it.
Anyway, thanks

[zenith junco]

hi

can i please have a hint to approaching this problme

i have the solutions but i didnt look at it yet

like i dont even know what this question is even saying

[sonic osprey]

@zenith junco what do you not understand?

[zenith junco]

the first sentence

like i read the question and idk what to do

plz some guidance

imassuming im doing the subspace thing where i check for

1. 0 vector
2. closed under scalar mult n addition

[sonic osprey]

What are you not understanding about the first sentence?

[zenith junco]

i dont understand what M is

theres some M number??? where the absolute value of the function f(t) is always less than it

[sonic osprey]

I mean, yeah, there exists an M

[zenith junco]

idk how to do this basically

idk how to start

[sonic osprey]

They're telling you what it means for a function to be bounded

[zenith junco]

oh

lol

so

how do i prove this B(R) is bounded and since its bounded its a subspace of f(r,r)?

btw F represents the field of functions or?? and is B(__) also a function?

[sonic osprey]

It doesn't make any sense to ask that B(R) is bounded?

[zenith junco]

um

[sonic osprey]

Read the first sentence again. It tells you

[zenith junco]

oh its the set of bounded functions

ok ok ok

soo

i kind of get now what its saying

prove that bounded functions are a subspace of all functions

[sonic osprey]

Right

[zenith junco]

ok

imma attempt n show u my work in like 1min

thanks

oh no

theres a problme

f(ct) < M

but

cf(t) is that < M ???

cf(t) < cM ??

omg

[nocturne jewel]

you're proving that 1) 0 is bounded (trivial), 2) if f and g are both bounded, f+g is bounded and 3) if f is bounded and c is a real number, then cf is bounded

[zenith junco]

nvm iget it now

thaknkns

yo im so confused

what is this

[winter harbor]

What exactly do you not understand?