linear-algebra

[wintry steppe]

I don't really know how to make it rigorous

[coral geyser]

determinants like this

what are the practical use for them

like finding coefficients in an arbitrary quadratic?

[nocturne jewel]

looks similar to the vandermonde

yes it's the 3x3 Vandermonde

https://en.wikipedia.org/wiki/Vandermonde_matrix

[coral geyser]

ooo

[wintry steppe]

Let S be a finite set of linearly independent vectors of F^infinity. For a vector v in F^infinity, call deg(v) to the highest term that is nonzero (can ignore the 0 vector). Then the set of those highest terms for elements of S is finite therefore bounded... so...

[faint dune]

Is it wrong to say the set of all functions form a vector space?

[quartz compass]

yeah I'd say so, you need to be more specific about what their domain is

for instance log(x) and sqrt(-x) are both functions but their sum I don't know if it's that useful to call it a function since their domains don't match up

[faint dune]

I had to proof that continuous functions over R form a vector space, so instead of proofing all axioms I just assumed the set of all functions Abb(R,R) is a vector space. So I only show subspace axioms

[quartz compass]

I wouldn't go about it that way, I think it's just simple enough to show directly it's a vector space

I'd start with f, g as continuous functions over R, then show that f+g is also a continuous function

[faint dune]

I wanted to keep the proof short by just showing 0 is part of it, f+g and a*f is part of it.

[coral geyser]

I have a question 🙂
So lets say X is a rectangular matrix
I have a left square inverse matrix Z
so ZX is gives identity
but XZ gives some random rectangular matrix

could XZ have another matrix that is its inverse

like left or right inverse

or nah

[quartz compass]

this doesn't make sense I'm afraid

maybe I've misread the problem actually

I assumed X=A

[coral geyser]

ok ok

sorry its pretty late here so i am just spewing things haha

if u have time could you explain if like

if u have a rectangular matrix right

it has one sided inverse

[quartz compass]

what is X

[coral geyser]

dw about that i mightve mistyped

😳

[quartz compass]

rectangular or square

[coral geyser]

oops

sorry for that

so sorry

[quartz compass]

lol

well if Z is square and X is rectangular then ZX and XZ can't both be defined unless X is actually square

[coral geyser]

yep

ok but if u have lets say ZX

right

[quartz compass]

don't delete past messages that just makes it more confusing

[coral geyser]

OH

sorry

i deleted cause i gave wrong info

[quartz compass]

so is X=A now or lol

[coral geyser]

yes 😦

[quartz compass]

I feel like we should start over

[coral geyser]

Yes please

ok so

we have one rectangular matrix

lets say it has a left hand inverse

[quartz compass]

wait gonna grab something to eat, you can take your time to formulate it again no rush

[coral geyser]

sure

and this left hand inverse is nxn and rectangular matrix is nxm

when they form a nxm matrix

(like the matrix is multiplied to the right, the square one to the rectangle one)

it forms that nxm matrix

will THAT matrix have any inverse

sorry if its really convoluted

[quartz compass]

hmm

I think how you were describing it before made more sense when you named the matrices

and showed the multiplication

A is nxn matrix and B is nxm matrix

are you saying A is the left inverse of B?

@coral geyser

[coral geyser]

yeah

[quartz compass]

that's not possible because AB=I means I is an nxm matrix, which means it's not the identity matrix

[coral geyser]

i cc

how would u do something like this btw

and like for these type of qtns

do u have any tips

[wintry steppe]

To check if a sum of two subspaces is direct, we check if we can write the zero vector in only one way. But why the zero vector? How do we know that, if we check that we can write the zero vector in only one way, then we can write all vectors in only one way?

What if it just happens so that we can write the zero vector in one way, but some other vector in more than one way?

[native rampart]

Let's say you have some arbitrary vector a which can be written in more than one way

Say a=u_1+v_1=u_2+v_2

Then (u_1-u_2)+(v_1-v_2)=0

Which means if there is a unique way of writing 0 as an element of U+V,u_1=u_2 and v_1=v_2

That is,there is exactly one way to write a as an element of U+V

[wintry steppe]

Thanks

[quartz compass]

try to see how they're similar

just start by multiplying by stuff like X or Y or X^-1 or Y^-1

the main thing is just try stuff and get started, then stuff has a way of popping out after a while

[coral geyser]

alright

btwwwww

u have a really nice pfp 😮

[quartz compass]

thanks lol

enjoy it while it lasts I think my nitro runs out like tomorrow

[coral geyser]

haha

[quartz compass]

here's another hint, multiply out MX and see if you can factor out N from that

[coral geyser]

wait btw

(A+B)^-1

i can distribute it right

or nah

[quartz compass]

check if can you do it for 1x1 matrices

in other words, just scalars

[coral geyser]

yeah

[quartz compass]

what's (1/2 + 1/2)^-1

[coral geyser]

gotcha

1

and not 4

yeahhh

[quartz compass]

👍

it's like trying to say (x+y)^2 = x^2 + y^2

[coral geyser]

yeahh

kinda stuckk ahaha

imma go sleep and try it when i wake up

[jagged granite]

Consider $n+1$ linear functionals $f_0,f_1,\cdots f_n$ on an arbitrary vector space V. I want to prove that if $\cap_{i=1}^{n} ker(f_i) \subset ker(f_0)$ , $f_0$ can be written as linear combination of rest of the linear functionals.

[stoic python]

bert

[rose cairn]

Is this equation linear or nonlinear? I think its nonlinear but how can i provide a rigorous and clear argument?

I want to mention linear combination somewhere in my proof

[limber sierra]

Linear in what sense? As in a linear ODE?

If so, why would linear combinations come up

[rose cairn]

Linear ODE

because we can apply linearity test to see if it is linear?

transform the function into a linear combination of output? and if it matches linear combination of input its linear?

I dont know...

@limber sierra

[wintry steppe]

https://en.wikipedia.org/wiki/Linear_differential_equation

[rose cairn]

uh ok..

[wintry steppe]

idk what you're about here tho, what function?

[limber sierra]

i feel like youre getting different meanings of "linear" confused

but i cant pinpoint exactly what the confusion is

linear in terms of ODEs has a different meaning than "linear" in a linear algebra sense

which has a different meaning than in calculus

which has a different meaning than in stats

its one of those annoyingly overloaded mathematical terms

[rose cairn]

ahhhh I see

this is for an ODE class

[steady fiber]

aren't the linear in ODEs and linear in lin alg kinda related

a linear ODE can be written as L y = b, where L is a linear operator

(on an infinite dimensional vector space)

[wintry steppe]

yea

[steady fiber]

although you really won't be using much finite dimensional vector space knowledge typically covered in a lin alg course in ODEs

[rose cairn]

I was only taught to determine linearity by using linear transformations

[limber sierra]

it is related but you have to look at the right vector space

and from the way theyve phrased their questions

[limber sierra]

i dont think thatll be feasible

[steady fiber]

I agree

[limber sierra]

ah i see what you mean

[steady fiber]

I was just saying that the "linear" terminology isn't entirely unrelated

thanks

[wintry steppe]

Why is the empty list linear independent?

I wouldn't say it's linearly dependent either before anyone says, I would just say it is undefined?

[faint lintel]

vacuous truth

A subset S {v_1, v_2, ...} of V is linearly dependent if there exists scalars a_1, a_2, etc , not all zero, where the sum of all a_i *v+i = 0

otherwise it is linearly independent

since S is empty

the statement is vacuously true @wintry steppe

[wintry steppe]

and it's useful for it to be linear independent. if you also consider the empty linear combination to be 0, then its a basis for the trivial space

[dreamy iron]

hi folks!

how do you guys define a finite dimensional vector space?

[wintry steppe]

It's a vector space that has a finite basis

[nocturne jewel]

basis has finite number of basis vectors

[dreamy iron]

yeah, i figured.

axler says its any vector space that is spanned by a finite list of vectors.

is he just being quirky?

[wintry steppe]

maybe would make more sense to call that finitely gerated but

[edgy kelp]

https://i.imgur.com/9iCKjQR.png

[wintry steppe]

can prove that if there's a finite spanning set then there's a finite basis

[nocturne jewel]

that's what we said

since a basis spans

[dreamy iron]

well....he seems to consider a spanning set to be more primitive than a linearly independent set, so sorta kinda maybe.

[nocturne jewel]

well yeah, spanning and independence arent the same thing

A basis of V is a set of vectors which are all independent and span V

[dreamy iron]

axler does give a two line proof of this, where the two lines refer to two other lemmas that do most of the heavy lifting.

[wintry steppe]

Equivalently it's a set of vectors which span V and that can write each v in V in unique way as linear combo of them

yea i guess it doesn't really matter

[dreamy iron]

okay, i wont worry about it too much then, thanks folks.

[remote gorge]

What is the number of independent complex entries of an antihermitian matrix?

[wintry steppe]

guys whats the pivot column here?

only the 1's right?

[nocturne jewel]

all 3 are pivots

[steady fiber]

non-zero leading 1 in a rref matrix <=> pivot element

[wintry steppe]

Is the 4th line also pivot

[hollow garnet]

what does pivot mean?

can you explain that to me first?

the answer is literally right above your message but it still looks like you don't know what it is.

[wintry steppe]

I know but I just wanted to make sure that I dont get it wrong

nevermind maybe the question wasnt necessary youre right

[wintry steppe]

For me its just something to figure out which vectors are linearly independent like you told me before on the defintion of the basis

[acoustic zodiac]

Does anyone have a link to read about notation used in linear algebra? I'm having a bit of a hard time adjusting to the notation because none of my profs have properly explained it. I'm talking about something like (if A is some linear map):

$$A(e_j)= a^i_j e_i$$

[stoic python]

rcatalang

[wintry steppe]

they're probably implicitly summing over i here

(einstein summation notation)

there's nothing more to it

if an index appears twice, once up and once down, it's being summed over

[acoustic zodiac]

And what about the j-index?

If we're summing over i, does the j stay static?

[remote gorge]

J is just a normal index (free)

[acoustic zodiac]

So it's $$a^1_j e^1 + a^2_j e^2 + ... + a^n_j e^n$$ for every column $$j$$?

[stoic python]

rcatalang

[remote gorge]

Yes

[jagged granite]

My qn got lost

[coral geyser]

can someone explain the first part

[coarse rain]

Suppose XYv = v, then Yv = X^-1v. Then, take w = X^-1v. Show that YXw = w.

[wintry steppe]

how would i do this

[shrewd shell]

Can annyone help me with linear functions

[dusky epoch]

you should post your question right away. don't ask to ask. just ask.

and also, check pins. your question might be more suited for #prealg-and-algebra.

[crude falcon]

Can someone name me two iterative methods for solving systems of lineal equations?

[lavish jewel]

gradient descent and jacobi method

that's implied

converges pretty quickly, too

[crude falcon]

Gradient descent is a first-order iterative optimization algorithm for finding a local minimum of a differentiable function

is this for systems of lineal equations too?

[lavish jewel]

sure

$\min_x \Vert Ax - b \Vert_2^2$ finds the x that brings Ax - b as close as possible to 0

[stoic python]

Edd

[lavish jewel]

i.e. Ax approx equal to b

if b is in the range of A, you can find one solution to Ax = b

you can do the minimization with gradient descent

[zealous junco]

the other jacobi method is cool

[lavish jewel]

which one?

[coral geyser]

If i take all the 1/3 out

i will have Q = 1/27 (whatever is inside)

My Transpose will it have 1/27 (whatever was inside transposed)

?

[lavish jewel]

where'd the 1/27 come from

[coral geyser]

uk inside u gave 1/3

i wanna take all of them out

cause uglyyy

[lavish jewel]

you're thinking determinants

[coral geyser]

1/3^3 (matrix)

i am dumb

ok

yeah

[lavish jewel]

all you have to take out is 1/3

[coral geyser]

thanks edd

Yes

LOL i am actually so dumb at times

thanks again so much

[wintry steppe]

$Ker T = {v \in Dom T : Tv = 0_{CoDom T} }$

[stoic python]

Carla_

[wind pasture]

how would i solve this?

[rare spade]

For $\lambda_2$ we have

$$
\left(\left[\begin{array}{lll}
2 & 0 & 0 \
0 & 1 & 1 \
0 & 1 & 1
\end{array}\right]-\left[\begin{array}{lll}
2 & 0 & 0 \
0 & 2 & 0 \
0 & 0 & 2
\end{array}\right]\right) \cdot\left[\begin{array}{l}
x \
y \
z
\end{array}\right]=\left[\begin{array}{c}
0 \
-y+z \
y-z
\end{array}\right]
$$

Since $x=0$ it implies that $x=a$ where $a \in \mathbb{R}$, we write the general form

\begin{align*}
x&=a \
y&=z \
z&=y
\end{align*}

$$\vec{v_2}=\begin{bmatrix}1 \ 0 \ 0\end{bmatrix}$$
$$\vec{v_3}=\begin{bmatrix}0 \ 1 \ 1\end{bmatrix}$$
$$\vec{v_4}=\begin{bmatrix}1 \ 1 \ 1\end{bmatrix}$$

[coral geyser]

Side question, for determinants can we change a row. Like JUST that row when doing row echelon operations

[stoic python]

Timur

[rare spade]

Is this all the eigenvectors which are linearly independent?

[visual hedge]

is the basis for the nullspace of A and determine the nullspace of A the same thing

[wind pasture]

@rare spade is that a solution to my question?

@rare spade im having a hard time understanding

[rare spade]

No it is not

[wind pasture]

oh

nvm

[visual hedge]

is the basis for the nullspace of A and determine the nullspace of A the same thing

[coral geyser]

can someone help me with this

this is what i got up to

[wintry steppe]

Is this a sufficient argument to show that (1,2) and (3,5) are basis vectors for F^2?

Let a(1,2) + b(3,5) = 0, which is true if a = b = 0, so they are linearly independent. We know that the length of a list of linearly independent vectors in a vector space ≤ length of a spanning vector list in that vector space. Thus, the spanning vector list is at least of length 2, and so we are done.

it needs to be true if and only if a=b=0 not just if. that doesnt show that any list of length 2 is a spanning list. you can use that any list of linearly independent can be extended to basis to prove that.

But why not? The spanning vector list has to be at least of length 2

We have a list of vectors of length two that are linearly independent

well you didnt say that any lin indep list of length 2 is a spanning list

What do you mean?

you say any spanning list is of lengh at least 2

but you didnt say that every linearly independent list of length at least 2 is a spanning list

Oh of course they have to be linearly independent sorry

a spanning list doesnt have to be lin indep

[visual hedge]

is the basis for the nullspace of A and determine the nullspace of A the same thing

[wintry steppe]

It doesn't, but we can use theorems to remove vectors that aren't linearly independent in that spanning list to get a spanning list that is no larger than a list of linearly independent vectors

With the same span

As adding or removing linearly dependent vectors won't change the span

yea but you need here to extend your list of linearly independent to a basis

it already is a basis but you have to show that

I just need to show that it spans F^2, which I think I've done? Because then I can say that any spanning list contains a basis

And a spanning list that is linearly independent is the basis, by definition

how do you know whether it is spanning tho?

Because the length of a list of linearly independent vectors in a vector space ≤ length of a spanning vector list in that vector space.

(1,0),(0,1) spans F^2

So that is the minimum length of a spanning list

So now that I have shown that they are linearly independent, I have two vectors that I can use instead of (1,0) and (0,1)

what you say doesnt immediatly imply that there arent lists of 2 linearly independent that dont span the space

you have to use that somehow

How?

you maybe know that every list of lin indep can be extended to a basis

Yes

and all basis of a space have the same number of elements

in this case seems you already know dimension of F² is 2

so the only way to extend your list to a basis is by doing nothing

Yeah

therefore it is already a basis

Yes, thanks

[visual hedge]

so like bois

whats the difference between determining the basis for a nullspace of a matrix and determining the nullspace of a matrix

[wintry steppe]

R² is a space, ((0,1), (0,1)) is a basis for that space

[visual hedge]

so the basis are just vectors

and nullspace is a linear combination of em?

[wintry steppe]

a space is generated by a basis of it

nullspace is not a linear combination. its the set of all linear combinations (of a basis of it)

[visual hedge]

so in your example above
the nullspace would be = k(0,1)

[wintry steppe]

to have a nullspace u need a matrix

[visual hedge]

ok

ty

[wind pasture]

can someone give me an example?

[zealous junco]

for diagonalizing using similar transforms like using givens rotation, I believe if u get a supercomputer u can rotate in parallel and it does 1 sweep of the entire matrix in O(n) time

[wind pasture]

is this true?

[native rampart]

Yes

[wind pasture]

how about this one?

[native rampart]

idk what a fundamental matrix is

[wind pasture]

is this one true?

[native rampart]

idts

If A was real,that would have been true

[dusky epoch]

take n=1

x' = ix

clearly has e^(it) as a solution

yet cos(t) and sin(t) are not solutions by themselves

[wind pasture]

how about this one @dusky epoch

[dusky epoch]

hold on

this is starting to sound more and more like a test

im not sure how productive itd be to just answer these t/f questions for you

[wind pasture]

its an online assignment, i understand if its true

but if its false can you give a counterexample

[wintry steppe]

if you re not sure there isnt a counterexample

then you dont understand if its true

can anyone help me with understanding hermitian adjoint

i don't even know how to think about it

perhaps some intuitive explanation or some examples of how they're useful would help me the most

i'm talking about this

[nocturne jewel]

Adjoint is useful cause they define isometries, which preserve norm

[wintry steppe]

how do you think about hermitian adjoints

[native rampart]

<Ah_1,h_2>=<h_1,Bh_2> then B is hermitian adjoint of A

for all h_1,h_2

In an orthonormal basis

[nocturne jewel]

$\langle T[v],v \rangle=\langle v,T^*[v] \rangle$

[stoic python]

moshill1

[spare crystal]

Is there a word to describe "nonlinear" combinations of vectors in an algebra (vector spaces with vector-vector multiplication, https://en.wikipedia.org/wiki/Algebra_over_a_field )? i.e., if a, b are elements of an algebra A, something like b^3 + ab^2 - 9ab would be a "nonlinear" combination of a and b. Basically a linear combination but allowing for multiplication

[stable kindle]

i don't know anything about anything but could it be this

https://en.wikipedia.org/wiki/Exterior_algebra

[native rampart]

It's an "A-linear combination"

[spare crystal]

when I google "A-linear combination" i get results for "a linear combination" :c

[steady fiber]

a tensor algebra maybe?

with tensor products being multiplication

it's the free algebra on a vector space, and "the largest" algebra over a vector space in some sense due to the universal property

and it basically acts as a noncommutative polynomial ring on basis vectors of a vector space

[wintry steppe]

a polynomial evaluation maybe?

[steady fiber]

idk what your goal exactly is so it's kinda hard to say things

[nocturne jewel]

how do you define a size of an algebra?

[steady fiber]

every vector space algebra is a quotient of the tensor algebra

due to the universal property

I believe

[spare crystal]

my goal is just to use this in my hw lol.. i might as well just define a term tbh and hope the grader doesn't cringe at whatever i choose

[steady fiber]

which is why I said "biggest"

gotta be more specific, you can define multiplication in any of a trillion ways

[spare crystal]

specifically im working with functions from [0, 1] -> R, multiplication is just pointwise multiplication

[gleaming storm]

How do you find volume of parallepiped?

in R3

[quartz compass]

take the determinant of a matrix with 3 vectors that define vertices adjacent to one at the origin

you can see in the case of a cube or rectangular prism it gives the right answer, and since the determinant is unchanged by shearing operations, it gets you the volume of arbitrary parallelepipeds as well

[barren blaze]

What does this tell me when looking for a basis for this vector space? I know how to do R^3 but R^1x3 I can't understand

[wintry steppe]

probably means 1 x 3 matrices with entries from F_2

[barren blaze]

I understand what I need in a basis of a R^n vector space. But when it comes to this I am struggling to understand what I need to change

Is the basis {(1),(1),(1)} ? Which makes no sense to me 😄

[quartz compass]

it might help if you give an example of an element of F_2^{1x3}, when you see what stuff looks like it might become clearer

[barren blaze]

For example (1 0 0) is an element of it right?

[quartz compass]

yeah good

so your question earlier you gave a set of stuff that looks like (1) not (1 0 0) so that can't be right

all your elements are of the form (x y z) so your basis must too

how comfortable do you feel with F_2 have you seen that before or is that totally new to you or does it make sense

[barren blaze]

It is kinda new to me tbh. The tutor mentioned it that it means this is a vector space from a "body" with two elements

But I am not sure what it means

I am familiar with the algebraic structure "body" (not sure what its called in english)

[quartz compass]

oh F_2 is a field

it's like R except instead of real numbers you just have 0 and 1

and so you can do addition and multiplication, it might help if you look up the field axioms

[barren blaze]

Yeah we call them Körper in German which directly translates to body. I guess it's called field in english

[quartz compass]

the rules are simple enough for F_2, you just have 0+0=0, 1+1=0 and 0+1=1 along with the regular multiplication you'd expect, 0*0=0, etc

so like here's an example [0 1 1] + [1 1 0] = [1 0 1]

just adding two vectors here, it works pretty much exactly the same

[barren blaze]

So does this mean I am used to working in F_R

And now we are restricting it to only two elements

[quartz compass]

sort of not quite but basically like that

R and F_2 are how you call them

a lot of the stuff you do in linear algebra doesn't actually depend on using R as your field though

like let's just pretend your problem was originally to find a basis for R^{1 x 3}

what would you write down

[barren blaze]

[1 0 0] [0 1 0] [ 0 0 1] ?

[quartz compass]

yeah

and that'll work as a basis for this too

there aren't that many possibilities for [x y z] because x,y,z can only be 0 or 1

[barren blaze]

What would have changed if this was in R_5 for example

[quartz compass]

F_5 would be how you write that

like F_5^{1x3} you'd write the same thing

[barren blaze]

Yeah sorry

[quartz compass]

although technically '1' and '0' are not the same things

F_2 is not in R and R is not in F_2

they're like entirely separate things

but we use the same symbols to represent them

[barren blaze]

What is F called?

[quartz compass]

these fields are often written slightly differently, you'll see F_5 written as Z/5Z or Z_5 sometimes

F_2 is the field with 2 elements

F_5 is the field with 5 elements

the F by itself doesn't mean anything

but often people might say F to represent a generic field, not referring to what it is like R or F_2 or anything in particular if the result doesn't depend on the field

[barren blaze]

Oh so the F stands for field. I remember the axioms etc. from Analysis but how it translates vectorspace is all new to me

[quartz compass]

vector space is basically taking a field but adding more structure on top of it

a field can be seen as a 1 dimensional vector space for instance

[barren blaze]

So am I correct when I say: when it comes to finding a basis the field's element count doesn't play much of a role but its dimensions do?

Dimensions may not be the right word for it

[quartz compass]

yeah that sounds good

the dimension is the number of basis elements

[barren blaze]

Yeah I know but in that case I meant R^{axb} and a and b play a role in the basis

That's why I wrote the dimension is the wrong word for it

[quartz compass]

yeah this has a*b basis elements for this vector space

you could write F^{axb} to mean for an arbitrary field

and it would hold

[barren blaze]

What is a and b called so that I can use it my future questions 😄

[quartz compass]

just the dimensions of your matrices

a is the number of rows, b the number of columns

so there are a*b entries in total

[barren blaze]

So it gets messy when in a sentence with a basis which also has a dimension, got it

[quartz compass]

you don't have to put your matrices in a rectangle box

you could put them all out in one single column or row of all a*b at once

to make it look more like a column vector

there's nothing particularly holy about how you write it down but it might make certain computations harder or weirder to carry out haha

if it makes you feel better you can relate the two by an invertible linear transformation

[barren blaze]

Yeah they almost go hand in hand so I am guessing I will learn a way to compute them together in near future

[quartz compass]

yeah probably

you'd really more encounter stuff like F_2 or F_5 in a number theory or abstract algebra course

I don't remember seeing them too much when I took linear algebra myself but I think they popped up a few times randomly and I didn't know what they were at the time either

if you're doing like physics/engineering/compsci you probably can get away with not knowing what they are too well tbh

[barren blaze]

One thing I still can't figure out how would I write this down as a set?

[barren blaze]

I am a comp sci student but many mentioned our program is very heavy on math and very abstract

I write {(1,0,0),(0,1,0),(0,0,1)} as a basis for F^{3x1}

[barren blaze]

How would I differentiate this one?

[quartz compass]

yeah same thing

I was writing [] the same as ()

it doesn't matter what kind of brackets you use to represent a vector or matrix

you can put commas or no commas too, whatever's clearer

[barren blaze]

I see, so is there no way to tell if a vector is in a F^3x1 space or F1x3 space or are they all essentially the same

[quartz compass]

well sort of

the difference is 3x1 means it'd be like $\begin{bmatrix} x \ y \ z\end{bmatrix}$ and in 1x3 it'd be like $\begin{bmatrix} x & y & z\end{bmatrix}$

[stoic python]

Merosity

[barren blaze]

Exactly, but I can't differentiate them when I am writing them in a set

[quartz compass]

but they're basically the same thing

well if you're writing them in a set you should write them that way too

being in a set doesn't change that

[stable kindle]

something something isomorphic

[quartz compass]

${ \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}, \begin{bmatrix}0 & 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1\end{bmatrix} }$ is not the same as ${ \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}, \begin{bmatrix}0 \ 1 \ 0 \end{bmatrix}, \begin{bmatrix} 0 \ 0 \ 1\end{bmatrix} }$

[stoic python]

Merosity

[quartz compass]

so if you wrote the second thing you'd be wrong, cause yours is the first thing

[barren blaze]

But on our scripts I have seen (1,0,0) used for a 3x1 vector

[quartz compass]

technically they should put (1, 0, 0)^T

[lavish jewel]

that's kinda cursed

why bother saying 3x1 when you'll write it wrong anyway

just say 3 then

[quartz compass]

but it's common not to write it if it's understood

[barren blaze]

Maybe they do and I didn't notice it before because this is my first time working with anything other than nx1 or nxn

[quartz compass]

because your problem is specifically writing F_2^{1x3} you should be careful about it

yeah, now that you're starting to get into linear algebra further, you have to start being a bit more careful about this sort of stuff

[barren blaze]

Thanks a lot for the help, I will now get back to battling the assignment. You really helped me open my eyes to how to think about fields

@quartz compass

[quartz compass]

yeah you're welcome 👍

[gleaming storm]

Oh ok thanks

[quartz compass]

you're welcome

[rocky hill]

Does an eigenvector have to stay "on its span" throughout the entire transformation (if you were to sort of "animate" the process of the linear transformation over time)?

I'm asking because to me, a rotation matrix acting on R2 of 180° or 360° would seem to mean that all vectors in the plane would be eigenvectors, but... that somehow doesn't seem right?

It's been a long time since I studied LA so I'm a little stuck in my head about it.

[wintry steppe]

most rotation matrices on R² have no eigenvectors

[quartz compass]

yeah but your rotation would be indistinguishable from a reflection and there's no unique way of interpolating from the identity transformation I to R

if you're trying to imagine like a 90 degree rotation as being partway through then you run into issues because it doesn't have real eigenvalues

[coral geyser]

is a nillpotent matrix commutative

[wintry steppe]

what is a commutative matrix?

[coral geyser]

like

AQC = ACQ

uk how matrix multiplication is not commutatitve

[wintry steppe]

ik it isnt

[coral geyser]

yeahhh

[wintry steppe]

still not clear what youre asking

[coral geyser]

You know how a nillpotent matrix is Q^2 = 0

if i have AQC which is basically 3 matrices multiplied

is that equivalent to me doing ACQ

A C and Q are matrices of same dimension

Q is nillpotent

[wintry steppe]

have you tried any simple examples by hand

[coral geyser]

its a qtn that just popped in my head

we havent really gone through them in class

[wintry steppe]

you dont need to go through something in class to investigate it

[coral geyser]

cause sometimes me duumbbb really dumb

and might miss out a case and all

yeah ❤️

ty for that

that is trye

true*

[wintry steppe]

if you try some simple examples youll find what to do

[rocky hill]

Right gotcha. So it doesn't matter "what happens in between" so to speak, since, well nothing does. I guess that's the problem with trying to visualize transformations...

[quartz compass]

it's not a problem since you can safely interpolate a rotation matrix, just write it in terms of an angle

$\begin{pmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{pmatrix}$

[stoic python]

Merosity

[quartz compass]

you can work this out and get the eigenvalues and eigenvectors, but they're complex numbers

this isn't shouldn't be surprising though, I wouldn't call this a problem, just freedom to take alternative pathways to get from one place to another is all it is

@rocky hill

[rocky hill]

Right, I guess I sort of meant that it doesn't really matter how you get there. The end result is -- like you said -- indistinguishable from a simple reflection (for some rotation matrices anyway).

And yeah, totally not surprising that the eigenvectors would be complex.

But I see what you're saying... I think.

by "interpolate a rotation matrix" I think you're talking about like, parameterizing theta as a function of time?

[quartz compass]

well the rotation matrix is parametrized by the angle theta

but you can then parametrize theta by time if you want too if that's how you want to think about it

although roots are not unique, you can fairly safely see that $R(\theta)^{1/n} = R(\theta/n)$ will get you the nth roots how you like

[stoic python]

Merosity

[quartz compass]

like if you want to see a snapshot in time of the matrix going through 3 frames starting from the identity I = R(0) and go to R(pi) you can look at n=2 and see R(0), R(pi/2), R(pi) and see these as being R(theta * k /2) for k=0, 1, 2

you can of course add in more I just am giving a simple example

[coral geyser]

does anyone have a good linear algebra book

this is for ug maths

first year ish

[faint lintel]

Dumb question

is the Adjoint of a unitary matrix the inverse of that matrix?

[native rampart]

UU*=I

For unitary

So yes

[faint lintel]

ok cool

I am stuck on this

So if it's upper triangular

say our matrix is A

A_ij = 0 then i > j

but then idk how to use the fact that it's unitary

[native rampart]

Inverse of a upper triangular matrix is upper triangular

[faint lintel]

OH

So

[native rampart]

You can kinda figure that out by writing the product out

[faint lintel]

A^-1 is upper triangular

but A^-1 = A*

since unitary

[native rampart]

Yes

[faint lintel]

so when i > j

(A*)_ij = 0

(A*)_ij = A_ji conjugate = conjugate of 0 = 0

or something like that the order is wrong

but you then when i > j and j > i, A_ij = 0

so diagonal

[native rampart]

Yes

[quartz compass]

I'd say if A is upper triangular then the adjoint is lower triangular, and since the inverse is upper triangular the only way you can be both upper and lower triangular is to be diagonal

[rocky hill]

linear algebra done right, sheldon axler

It starts with vector spaces, so it might help to have some basic abstract algebra under your belt first

for a more conventional approach... literally any university LA textbook is gonna be the same

[wintry steppe]

if you want a book that doesn't brush off determinants - which are incredibly important - try the book by the authors friedberg, insel, spence. same coverage, but doesn't ignore determinants

[coral geyser]

yeahh i need determinants too

[zinc lava]

wait what do you mean matrices?

oh i see. There is another nice way I think. Trying out Wolfram.

[zinc lava]

not possible i think (correcting myself)

5 5 6.5
7 5.5 4
4.5 6 6

hmm 🙂

[wintry steppe]

I want to show that if $v_1,v_2,v_3,v_4$ forms a basis of $V$ then so does $v_1+v_2,v_2+v_3,v_3+v_4,v_4$

[stoic python]

n/c

[wintry steppe]

First note that we can write any linear combination of that as $a_1v_1 + (a_1 + a_2)v_2 + (a_2+a_3)v_3 + (a_3+a_4)v_4$

[stoic python]

n/c

[wintry steppe]

We know that $c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4 = 0 \iff c_i = 0$

[stoic python]

n/c

[wintry steppe]

Since all of those coefficients are elements of the same field, it must be linearly independent

[nocturne jewel]

That's not the condition for independece

[wintry steppe]

It is, because v_1,v_2,v_3,v_4 form a basis

So it's in that form

Because v_1,v_2,v_3,v_4 are linearly independent

[nocturne jewel]

right, so c1 c2 c3 c4=0, but that means nothing about a1 a2 a3 and a4 as you've written

[wintry steppe]

How?

We have that $c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4 = 0$

[stoic python]

n/c

[nocturne jewel]

yes

[wintry steppe]

Just set $c_1 = a_1, c_2 = a_1 + a_2, \dots$

[stoic python]

n/c

[wintry steppe]

So it has to be linearly independent

[nocturne jewel]

yes, so say that explicitly since it's a proof

[wintry steppe]

Okay, well now by definition, $\mathrm{span}(v_1+v_2,v_2+v_3,v_3+v_4,v_4) = { k_1(v_1 + v_2) + k_2(v_2+v_3) + k_3(v_3 + v_4) + k_4v_4 : k_i \in \mathbb{F},v_i \in V }$

[stoic python]

n/c

[wintry steppe]

We can essentially do the same thing now, where we rearrange the coefficients to get them in that form that I wrote above

And so it must span V?

[nocturne jewel]

Is it always guaranteed I can pick k scalars because I can always pick c scalars to reach a vector in V?

[wintry steppe]

Well they're any number in F, so I don't see why we can't say that

[nocturne jewel]

$c_1=k_1\c_2=k_1+k_2\c_3=k_2+k_3\c_4=k_3+k_4$ Is there always a unique solution $(k_1,k_2,k_3,k_4)$?

[stoic python]

moshill1

[nocturne jewel]

answer yes, cause I can write the 4-tuple of k's in terms of c

which are already known scalars

[wintry steppe]

Ah okay, so we have $(k_1,k_2,k_3,k_4) = (c_1,c_2 - c_1, c_3 - c_2, c_4 - c_3)$

[stoic python]

n/c

[nocturne jewel]

yes, and that is the only solution

so the basis you're proving spans V

[wintry steppe]

It's proven now

Since they're linearly independent and spans V

[nocturne jewel]

yes

[wintry steppe]

So it is a basis

[zealous junco]

can some one explain how

R(P) orthogonal to N(P) and P^2 = P imply that P = P*

[lavish jewel]

what are R(P) and N(P)

[zealous junco]

range and nullspace

[lavish jewel]

oh

let's see

is an explanation using the svd ok?

[zealous junco]

ye

[native rampart]

Does the svd approach assume that the vector space is finite dimensional?

[lavish jewel]

i suppose it does

it assumes you can find a basis for col, row, null, and left null spaces, at least

[zealous junco]

yea i think its finite dim

[lavish jewel]

there's a missing step at the end

something something orthonormal columns

that should let T1* get rid of one of the sigma_1 matrices

i'm always prepared to be completely wrong tho, so do let me know if i'm peeing in the wrong bucket altogether

my argument there was orthogonal complements

oh, you were told that P^2 = P, so maybe let T1 = sigma1^-1

then you get that P = V1 T1 sigma1 V1* = V1 V1*, which has the hermitian symmetry you want ?

idk, something like that. i'm tired and i suck at math

it especially means that P was a projection matrix

is that what you were working with?

orthogonal projection matrix*

[desert wolf]

hi can I get some help with abstract algebra here

[lavish jewel]

there is an #groups-rings-fields channel

[desert wolf]

I can't type in that channel

[delicate jetty]

Go to the advanced access chat in the advanced mathematics section of the server and read the rules in order to gain access

#get-advanced-access here

[desert wolf]

thanks

[zealous junco]

yea sorry was in class

yep P was orthogonal proj and i was thinking to use things about inner product

I can show N(P) is a subset of N(P*)

[lavish jewel]

yeah then what i wrote above is what you wanted

[zealous junco]

thanks, i think i understand

[wintry steppe]

How do we check what we need to extend it by?

[dusky epoch]

?

what do you mean

[wintry steppe]

How do we know the list will be greater than length 4?

[dusky epoch]

if dim(U) were 4 then a basis of U would have to have 4 elements

but then appending anything else to it would make it into a list of more than 4 elements

which cannot be a basis of P_3(R)

[wintry steppe]

Okay, but why do we conclude that dim(U) is not 4?

[dusky epoch]

assuming dim(U) = 4 leads to a contradiction

[wintry steppe]

What impossibility?

[dusky epoch]

the impossibility of a basis of P_3(R) consisting of more than 4 elements.

[wintry steppe]

But there is no proof that dim U is not 4

[marble lance]

You could rephrase that part to say that if dim U= 4 then U =P_3(R) which we know isn't true. It's the same idea, but maybe you understand that better

[dusky epoch]

n/c

do you accept that proof by contradiction is logically valid

or are you going to pull the constructivist card on me

[wintry steppe]

@marble lance We could have a subspace with of a space with the basis of same length though?

[dusky epoch]

no, for finite-dimensional spaces all proper subspaces will have strictly lower dimension.

and P_3(R) is finite-dimensional, unless you reject that too.

[wintry steppe]

We've only proven that dim U ≤ dim V for U subspace V

[dusky epoch]

i find it a little odd how you just blurt out "there is no proof that dim U is not 4" immediately after i explain to you the proof that dim(U) is not 4.

[marble lance]

You have = when U = V. If U is a proper subset of V then you have <

[dusky epoch]

and you still have not answered my question of "do you accept proof by contradiction or are you a constructivist?"

[dusky epoch]

careful, this holds only when V is findim.

[wintry steppe]

Of course I accept proof by contradiction?

[dusky epoch]

so?

[wintry steppe]

I don't understand where we get the contradiction

We haven't proven the fact that you're using above

[dusky epoch]

assume dim(U) = 4.
then U has a basis of 4 elements.
extend this basis to a basis of P_3(R).
this basis of P_3(R) will be longer than 4 elements.
but dim(P_3(R)) = 4.

[wintry steppe]

Yes this line

this basis of P_3(R) will be longer than 4 elements.

How do you know that?

[marble lance]

Since U does not equal V, it can't just be the same basis. So you have to add something to it to get a basis for V.

[dusky epoch]

because the basis of U spans only U

and U is a proper subspace of V

[wintry steppe]

But again we've only proven that dim U ≤ dim V

Not dim U < dim V if U is a 'proper' subspace

[stable kindle]

that's a standard thing but ok

[wintry steppe]

So that kind of seems like a major part of a theorem to leave out

[stable kindle]

where's the proof...

[dusky epoch]

n/c, i am literally

proving

that dim(U) ≠ 4

for you

like right now

the basis of U spans only U

therefore extending it with something not from U will keep it linearly independent

[wintry steppe]

@stable kindle

[dusky epoch]

okay yknow what let's start over

suppose dim(U) = 4. then there exists a list v1, v2, v3, v4 which is linearly independent and whose span is equal to U.
agree or disagree?

[wintry steppe]

Agree

[dusky epoch]

U, as defined in your problem, is a proper subspace of V. in other words, there exist vectors in V which are not in U.

agree or disagree?

[wintry steppe]

Agree, like p(x) = 5x

[dusky epoch]

okay.

take one such vector, call it v5, and append it to our list. so we have a list v1, v2, v3, v4, v5 of five vectors in V.

(remember that V = P_3(R) here)

[wintry steppe]

No.

[dusky epoch]

what?

[wintry steppe]

We're trying to prove that dim(U) is not 4, not that is not dim(U) > 4

I understand that dim(U) is not 5 or 6 or 10

[dusky epoch]

you're not letting me finish.

[wintry steppe]

Okay sorry finish

[dusky epoch]

take one such vector, call it v5, and append it to our list. so we have a list v1, v2, v3, v4, v5 of five vectors in V.
(remember that V = P_3(R) here)

[wintry steppe]

Ah wait I see it now

Okay

So there exists vectors like p(x) = 5x

So that'd be the an example of a fifth vector that we can now reach after adding it on, but the basis of the P_3(R) is only 4

Got it

[dusky epoch]

since v5 ∉ span(v1, v2, v3, v4), and v1, v2, v3, v4 was linearly independent, the new list v1, v2, v3, v4, v5 is also linearly independent.
but the dimension of V is only 4, which forbids the existence of our five-element list.

hence the contradiction.

[wintry steppe]

And this new v_5 should allow us to span all of P_3(R), right?

[dusky epoch]

i said nothing of whether span(v1, v2, v3, v4, v5) = P_3(R)

because in fact i dont care

[wintry steppe]

Ah okay

Yeah you're right

Since the length of the list of linearly dependent vectors has to be no larger than the length of the basis

[dusky epoch]

yes.

that's the crux of my argument.

[wintry steppe]

Okay thanks got it now 😇

[wintry steppe]

How do I prove that the subspaces of R^2 are precisely {0}, R^2 and all lines in R^2 through the origin?

The word precisely is the tricky part here

[nocturne jewel]

Show that each is a subspace, so subspace / naive test 3 times

[wintry steppe]

That doesn't show there aren't more of them though

[marble lance]

Suppose that V is a proper subspace of R^2.

Then dim V = 0 or 1

If dim V= 0 then...

If dim V = 1 then...

[wintry steppe]

Okay I see thank you

[lapis osprey]

Can someone explain how to pivot a simplex tableau? It is confusing for me. How do I know when it is done?

[crude falcon]

if I have 2 linear maps, f and g, the matrix associated to its composition would be M(f o g) = M(f) * M(g)?

[dusky epoch]

yes

[crude falcon]

This is the matrix respect to a basis B associated to a linear map, f: R^3 -> R^3, if I have that B = {e1,e2,e3}, does this means that for example, e1 = (3,-2,4)?

[wintry steppe]

no

it means that Te1 = 3e1 -2e2 +4e3

[dusky epoch]

it means that Te1 = (3, -2, 4).

[gray dust]

the map is called f

[dreamy iron]

Hi Linear Algebra:

Can I get a scenario where I need to utilize the fact that **the set containing no vectors, i.e., the empty set of vectors, is linearly INDEPENDENT **?

[limber sierra]

you need that for the statement of the rank-nullity theorem to make sense

[dreamy iron]

hmmmm. What edge case are you hinting at?

[limber sierra]

since for a matrix with full rank, Rank(M) + Nullity(M) = dim(V) becomes dim(V) + Nullity(M) = dim(V)

and here the nullity of M is the zero space

(i.e. its basis is the empty set)

so for this statement to make sense here, we need dim({0}) = 0, i.e. we need the empty set of vectors to actually form a basis

if you said they were linearly dependent then {} would not be a basis for {0}

and so dimension of {0} would not be well-defined

[dreamy iron]

Thank you. That was less painful than I expected.

It actually makes perfect sense now.

[wintry steppe]

the empty set does form a basis cuz empty sum =0

[limber sierra]

well yeah it vacuously satisfies the definition of linear independence

i was just giving a "practical-ish" example of where that matters

rather than just being a "well, technically..."

[dreamy iron]

I really appreciate the concreteness of the example.

[wintry steppe]

another example is if you have a list of dependent vectors

theres an algorithm to reduce it to a independent one

[dreamy iron]

And I take nothing out??

If I take nothing out, it’s still dependent?

[wintry steppe]

well yea but

you want to get an independent one in the end

(e1, e2, e2) if you had this in R² u could remove e2 and get (e1,e1)

algorithm ends here

but if you had for example (0)

youd have to remove 0

if () wasnt linearly independent the algorithm wouldnt work for this case

[dreamy iron]

Ohhhhh

that’s almost magical!

[wintry steppe]

() could be seen as the basis case for the set of all linearly independent vectors ig

idk if thats ever useful tho

but if u wanted to prove something about all linearly independent vectors (of a space) ig you could prove for () and then prove for successors (and for limit ordinal length ones if infinite dimensional)

[dreamy iron]

Thank you! This was a most elucidating conversation. I’m so happy I asked this question.

[wintry steppe]

idk how to do it by trial and error but theres a way to represent complex numbers as 2x2 real entry matrices

[marble lance]

I did it by trial and error. I'll give you a hint by saying that you can do it using only 0 1 and -1 as entries. So just write out two matrices with no entries and think about what you have to multiply in order to get -I

So think like, I have to multiply this row with this column and get -1 so I have to use this here and here

[wintry steppe]

i like this way of thinking

[marble lance]

@wintry steppe

[ebon veldt]

If you want a less random approach than just guessing then look here

Matrix multiplication for 2x2 is the row of the first matrix are multiplied by the coloumns of the second one. So $$\begin{pmatrix} a & b \ c & d\end{pmatrix} \cdot \begin{pmatrix} a & b \ c & d\end{pmatrix}=\begin{pmatrix} a^2+b\cdot c & a\cdot b+b\cdot d \ c\cdot a+d\cdot b & c\cdot b+d^2\end{pmatrix}$$ So we want $$\begin{aligned}a^2+b\cdot c&=-1 \ a\cdot b+b\cdot d&=0
\ c\cdot a+d\cdot b&=0 \ c\cdot b+d^2&=-1\end{aligned}$$

Now either solve this system of linear equations or just guess from here (we quickly notice that $a^2,d^2\geq 0$), so setting those two equal to 0 is probably smart. Then we just have $b\cdot c=-1$ and $c\cdot b=-1$ left.

[stoic python]

ScapeProf

[heavy crown]

Hi. can someone explain to my why 6x is not a member of Z₇[X] ? isn't it supposed to be from 0 to 6?

that's what the teacher wrote but I don't understand why

[limber sierra]

im assuming they wanted you to rewrite that to use a coefficient between 0 and 6

-6 is not between 0 and 6

i.e. they probably wanted you to convert the -6x term to +x

[heavy crown]

oooh I think I understand you

so if we want -6 = (-1) * 6
while (-1) is actually 6 because 1 + 6 = 0

so (-6) would be 6 * 6 = 36, and in Z₇[X] would be +x

I guess

thank you 🙂

[native rampart]

I mean -(6) is clearly in Z_7[x]

[limber sierra]

yeah it seems like a silly nitpick if this wasnt clarified explicitly in the instructions

its a ring so every element has an additive inverse

including 6X

[slate siren]

how would I find all 2x2 matrices such that a^2=0?

ive started with a matrix with entries a,b,c,d and squared it

then equated that to the 0 2x2 matrix

but im not entirely sure how to solve the equations that are generated with that

[balmy phoenix]

In particular, you should have gotten a^2 = 0, b*c = c * b = 0, and d^2 = 0, if my mental math is right

ah, right, you have +s as well

a^2+cb=0, ba+db = 0, ca + dc = 0, cb+d^2 = 0, there we go

sorry, just woke up

[slate siren]

yeah

thats what i have

[balmy phoenix]

so yeah, you can conclude that a^2 = -cb, for example

that ba = -db

etc...

and that's basically solving the equations. Those are all of the matrices such that A^2 = 0 iirc

you can simplify this somewhat, for instance, a = -d

which happens twice

well, a = -d OR (b = 0 AND c = 0) I suppose

(but then a = d = 0 anyway, so...)

you see how the logic starts to pan out?

[slate siren]

yeah thats what I was sort of doing

I understand the case a = -d

but when a = d, why does that imply b=0, c=0

[balmy phoenix]

well, suppose a = d. Then ba + ba = 0, right? So then 2ba = 0, which implies either b = 0 OR a = 0

and a = d does not quite imply b = 0, c = 0, it implies one of them is 0, in particular

my logic was that either a = -d OR (b = 0 AND c = 0), since then ab + bd = 0 works out, and so does ac + cd = 0

if that makes sense

[slate siren]

But couldn't there be the case where d=0, which means a=0

then b and c could be anything

because 2ba=0 and 2dc=0

will work with a and d being 0

regardless of b and c

[balmy phoenix]

indeed

not quite though, you do require that b = -c iirc

[crude falcon]

the vectors that spans the image of a map are linearly independent right?

[slate siren]

oh, how do you come to that conclusion?

[balmy phoenix]

sorry, not b = -c, either b = 0 or c = 0

cause a^2 + bc = 0, so if a = 0, either b or c must also equal 0

[slate siren]

ah I see

thankyou for the help

[balmy phoenix]

not necessarily, though it must be the case that you have sufficient vectors to span the image. For example, [1, 0], [2, 0], and [0, 1] span the image of some R^2->R^2, but aren't linearly independent. Unless I'm mixing definitions, which is possible

Even this tutorial gives a set of vectors which are not linearly independent, if I'm reading it right https://www.projectrhea.org/rhea/index.php/Image_(linear_algebra)

[heavy crown]

is this how I find span?

because it seemed too simple idk

[faint lintel]

I would row reduce the matrix first @heavy crown

Because you're neglecting to check if those vectors are linearly independent

they very well may be

but you have to check

[heavy crown]

you mean I need to show this has singular solution to show its lineraly indepedant and then do what I did earlier?

but they just asked to find span why its necessary to say they're linearly independent?

[faint lintel]

Because if they aren't linearly independent

Then it could be the span of two vectors

Or 1

Your answer isn't wrong persay

But typically span is expressed in terms of linearly independent vectors

[heavy crown]

okay I understand thank you 🙂

[rare spade]

If a matrix A is invertible does it then make sense to raise it to powers like A^-0.1, A^-1291 etc?

[native rampart]

A^-1291 makes perfect sense

A^(-0.1) doesn't

[rare spade]

So only for i in Z, A^i makes sense

[native rampart]

Yes

[rare spade]

How do you write the A^-1291 for instance in product notation?

I have for $k \in N_0$ this $A^k=\prod_{i=0}^kA$

[stoic python]

Timur

[native rampart]

$A^{-k}=\prod_{i=1}^k{A^{-1}}$

[stoic python]

What's NDY?

[native rampart]

That product would be A^{k+1}

[rare spade]

What if I was given a negative k

This is the context btw

Task is to determine A^k for all k where A= V Lambda V^-1 but they do not specify what k is

🤔

[flint jackal]

@native rampart Actually A^(-0.1) makes sense. It's like A^(1/2) but it's much more difficult to calculate A^n power because it's not a diagonal matrix. If you do eigenvalue decomposition, then you can easily find A^n

[native rampart]

Well,There could be more than one square root of A

Like there are 2 matrices which satisify A^2+I=0(if A is 2x2)

[rare spade]

I'm very confused how to write it as a prod sum

[flint jackal]

That is true, but you just need to focus on the positive root. Because for instance, the decomposition for A = VΛ(V^-1), and if you want A^(1/2), we're equivalently saying B = V(Λ^(1/2))(V^-1) and B^2 = A, so you're going to be squaring the roots which makes it positive. @native rampart

[rare spade]

Idk if this makes any sense $$
A^k=
\left{\begin{array}{ll}
V\Lambda^kV^{-1} & k \geq 0 \
\prod_{i=1}^k{A^{-1}}=undefined & k<0
\end{array}\right.
$$

[stoic python]

Timur

[flint jackal]

I don't know what the product sum notation is, but if you do eigenvalue decomposition, where A^n = V(Λ^n)(V^-1), and use any n, you can find the power to any matrix

[rare spade]

Simply $\prod_{i=1}^5A^i=A\cdot A\cdot A\cdot A \cdot A=A^5$ for instance

[stoic python]

Timur

[rare spade]

but if the determinant of A is 0 then I can't find for n < 0 such that A^n?

[dusky epoch]

this is not true

$\prod_{i=1}^5 A^i = A^1 \cdot A^2 \cdot A^3 \cdot A^4 \cdot A^5 = A^{15}$

[stoic python]

Ann

[dusky epoch]

if you wanted the product of five copies of $A$ then that's $$\prod_{i=1}^5 A$$

[stoic python]

Ann

[rare spade]

Ah yes mb

[flint jackal]

A matrix should be invertible to be able to take an negative power

[rare spade]

Yes

I show that the matrix is non-invertible and therefore I do A^k=VLambda^kV^-1

but lets say it was invertible

how can I write it up lol

[flint jackal]

You have it backwards, an invertible matrix is represented by that equation

[rare spade]

Which equation are we talking about

[flint jackal]

A^k=VLambda^kV^-1

[rare spade]

Ahh

So I can write this $$
A^k=
\left{\begin{array}{ll}
V\Lambda^kV^{-1} & k \geq 0 \
V\Lambda^kV^{-1}=undefined & k<0
\end{array}\right.
$$

[stoic python]

Timur

[rare spade]

If I know A is non-invertible

[flint jackal]

Yeah, I believe it can be written like that

[rare spade]

I think it makes sense as well

Thanks @flint jackal

[sage ibex]

Magnitude as in? Are you putting a norm on the vector space?

[limber sierra]

just because they form a vector space doesnt mean they have a notion of "magnitude"

absolute value

(more generally the norm)

well your inner product will be the value of that integral, which will be a real number

so itll be the absolute value of that real number

[wintry steppe]

Hey guys this is what Ive to do: "Find a basis for the subspace of R[x] generated by x^2-1, x^2+x, 3x+1 and x^2-x+1"

and thats where I got but I dont know what to do next

I mean I know that the pivot columns are the 3 with the 1's so they are linear independent but what do I have to do next?

[hollow garnet]

What does basis mean?

and also ask yourself why you took rref of the matrix you created.

[wintry steppe]

a set of vectors in a vector space are called basis

[wintry steppe]

to look which vectors are linearly independent

[wintry steppe]

alright nevermind 😦

[hollow garnet]

Um, ok, no need to feel sad....

Anyway, let me give you a non-math example.

Let's say I wanted to paint something, and I wanted to purchase some colours, and I am tight on budget. What colours would I normally pick so I can make sure that I can create all possible colours, just from those colours that I buy.

Of course, white is a special case, but ignoring that.

[wintry steppe]

😄

Red green and blue. Some sources say its red yellow and blue and red

[hollow garnet]

yeah, so the point is that all colours can be created through (red, blue, yellow). Which means, we do not need to buy any extra colours. So if we purchased orange as well, this is unnecessary since orange = yellow + red. Similarly, if we only bought (red, blue), we would not be able to create all possible colours. Which means we require all three of those colours.

And if you notice, since red, blue and yellow are primary colours, we cannot create them through mixture of some colours like how we did for orange.

so we saw that (red, blue, yellow) are linearly independent and (red, blue, yellow) spans the colours space. Since (red, blue, yellow) satisfies these two condition, we can say that it is a basis.

So if we have some set $B = {\vb{v_1}, \vb{v_2}, \cdots, \vb{v_n}}$ which is a subset of some vector space $V$ of $\mathbb{F}^n$ and $B$ is linearly independent and spans $V$, then we say that $B$ is a basis.

[stoic python]

zslya

[hollow garnet]

which is why this definition is very wrong, hence the thonk face.

[wintry steppe]

OOOOh 😮

Thank you very much i appreciate the long explanation 🙂

I hope it did not take you long to write it though, for you are very busy for a couple of days ^^

[hollow garnet]

Ah that's because I am supposed to be updating my resume for possible internship.

Need to work on some projects to put in resume but, I am just lacking motivation....

[wintry steppe]

Yeah unfortunately motivation is not a good source to decide when to do something you dont want to do but have to do

[lilac stratus]

Huh quick question, I can't remember the proof I knew to show that the row rank and the column rank of a square matrix are equal

[warped cape]

dim(im(A)) = dim(ker(A^T)) = dim(im(A)^\perp) = n - dim(im(A)) = n - (n - dim(ker(A))) = dim(ker(A)) or sth

[limber sierra]

erm

cant you just follow the pivots

[lilac stratus]

oh huh yeah

thanks

[spiral star]

you gotta explain what you mean by vector multiplication

if you meant scalar multiplication then its easy to see from the inner product.

if f is in V and c is in R, then that integral for cf is given by <cf, cf> = c² <f, f> which is finite because of the inner product

:)

then you have your answer i guess

[hushed dock]

How do I determine the rank of a Jacobimatrix like this? p is a point in R^3

the p_1-1 i really tripping me up, long time since I had linear algebra

[dire thunder]

just do rref

it seems that rank is two

[tidal badger]

Could someone tell me, why the cross product works only in 3D and 7D? xD

[limber sierra]

see https://math.stackexchange.com/questions/706011/why-is-cross-product-only-defined-in-3-and-7-dimensions

tl;dr you get a structure compatible with this particular generalization of "Cross product" by taking a normed division algebra and restricting it to its imaginary dimensions

there are only 4 normed division algebras (standard fact of field theory https://en.wikipedia.org/wiki/Hurwitz's_theorem_(composition_algebras))

R, C, the quaternions, and the octonions

these have 0, 1, 3, 7 imaginary axes respectively

but cross products in the case of 0 and 1 are degenerate

(always 0)

so 3 and 7 are the only dimensions where its meaningful

[wintry steppe]

hey i have a linear algebra question in #help-3 thanks :3

A polynomial in one variable can be viewed as a list

But what about polynomials in multiple variables?

For example, (6,2,4) = 6 + 2x + 4x^2

But what about 6 + 2xy^2 + 4xy^4?

[native rampart]

You can still kinda see them as lists

There's a thing called monomial ordering

[dry tapir]

Can anyone help me with this

[wintry steppe]

ah yes

1.3.5

iterated decimals

[dry tapir]

Nope they show multiplication

[glossy sun]

1.3.5 I didn't get that, it's a representative of an operation?

[dry tapir]

In india it represents multiplication

[glossy sun]

I am Indian

But was not aware sorry

[dry tapir]

No issues

[glossy sun]

Oh thanks

[wintry steppe]

bad notation aside

this probably isn't a linear algebra problem; you may have better luck asking in e.g. #calculus

[dry tapir]

Ok thanks

[glossy sun]

Yeah I also advice that

[dry tapir]

Hmmm

[marble lance]

@wintry steppe you can also see a polynomial in x and y as a polynomial in y where the coefficient are polynomials in terms of x. Then you still have a list but the items in the list are themselves lists.

So in the example that you gave, the constant term has coefficient 6 which is (6) as a polynomial list, the y term has coefficient 0 which is (0), the y^2 term has coefficient 2x which is (0,2), the y^3 term has coefficient 0 which is (0), and the y^4 term has coefficient 4x which is (0,4). Then you can write it as

((6), (0), (0,2), (0), (0,4))

[wintry steppe]

Thanks Lunasong

That makes sense I suppose

[wintry steppe]

Anyone know how to prove this?

problem in inner product spaces

i think this is true more generally in normed spaces

you know how to prove it?

hint v = v-w+w

[lavish jewel]

i'm sure you can find several ways to prove it if you look it up by name, too

reverse triangle inequality

[wintry steppe]

so I used v = v-w+w

got what they wanted, but without the absolute

yea now try again but with

Could I just absolute both sides and remove the one on the right

w = w-v+v

Ok

Ok got the answer, thanks

had to combine both results in the end

[wintry steppe]

"Determine the row rank and column rank of the following matrix by choosing a maximum set of linearly independent rows and a maximum set of linearly independent columns.". I dont understand, I thought its only possible to determine one rank. So either the row or the column rank

presumably they want you to explicitly do computations to illustrate that row rank = column rank for this specific matrix

Do you mean like that, doing the row and column seperatly?

[flint jackal]

Honestly, I would have it in RREF first to determine the dimension of row space/column space

[wintry steppe]

Ok. But still how should I approach column and row rank at the same time?

Is it not the same?

[hollow garnet]

What is row rank?

And what is column rank?

[flint jackal]

@wintry steppe Determine the vectors that span row space and column space

[wintry steppe]

I understood it like that, the row rank is the rank from up to down and the column from left to right

[hollow garnet]

uhh, idk what that means but.

So explain what rank is then.

[north hedge]

how many linearly independent columns there are

so the dimension they span

[hollow garnet]

This question was intended for morethan2k

[north hedge]

oh sorry

lmao

[wintry steppe]

nevermind I was about to say the same 😉