linear-algebra

[wintry steppe]

I think why I don't understand it is because we haven't fully gone over it yet.

send it in

[misty ridge]

Recommendations for a good, basic linear algebra book?

[lavish jewel]

start by noticing that the original expression is convex, and so you can find a global minimizer by taking the gradient and equating it to 0

this gives you the answer to parts a and b

for part c just equate the given quantities and use the definition of the inverse

and for part d, use the given info from c, substitute it into the expression given in part b, and think about the pseudo inverse

[wintry steppe]

Wow dude thank you lol

[lavish jewel]

you were probably just asking in the wrong places

i don't think places where you pay to get your hw done are good

[spare crystal]

Does anyone know how to prove that a quadratic form is always equivalent to a diagonal form (over C, in R this just follows by diagonalizing the symmetric matrix I think)?

[lavish jewel]

the problem is also not very difficult, i recommend you review your notes (answering ost)

which definition of quadratic form are you using

[spare crystal]

Homogenous degree 2 polynomial in multiple variables

I think equivalently x^TAx for a symmetric matrix A

[wintry steppe]

It's not homework. How it works is here in norway is different. They don't teach nor give assignments that relate to the final exam. So we are given prior year exams with no solutions.

I do weekly problems, but none are like the previous years exams.

Our exam is 100% of our grade.

I'm essentially teaching myself with my friends. Please don't judge me so much.

Also thank you for your help, we're going over optimization of quadratic functions right now 😋

[lavish jewel]

i see. it worked similarly for me in germany, too

i didn't say that to be demeaning though, just that the problem SEEMS difficult, but it is actually made up of many small parts

[wintry steppe]

I did my bachelor's in the USA, was way different

[lavish jewel]

so try to break it down into small parts to not get overwhelmed

that was the message i meant to pass on

[wintry steppe]

Yeah I get it, your right

I'm bad at that, hence why I wanna prep

Get familiar if that makes sense

[lavish jewel]

yep

[wintry steppe]

Anyways, tyty have a great day

[lavish jewel]

you too

[wintry steppe]

any help with this? I know that the cross product will be (-1,1,0). I know the reflection on y-axis will be (-x,y,-z) but im not sure about the linear transformation. I found online that would be 2P-I. However I am not sure even if that is correct because how is that associated with the eigenvectors/values then?

[dusky epoch]

@wintry steppe are you familiar w/ the concept of specifying a linear map by what it does to a linearly independent spanning set?

[wintry steppe]

I think so, like after doing a linear transformation u get the spanning set as the linear combinations of the vectors u had

if that what u mean

[dusky epoch]

no

you're overcomplicating it and/or throwing around words you don't have full cognizance of the meanings of

[wintry steppe]

yes maybe I got some idea of it but not fully understood as i cant express it in my own words

[dusky epoch]

what i was going to say is:

for compactness of notation, let $\bd{u} = \bmqty{1\1\1}, \bd{v} = \bmqty{1\1\0}$ and $\bd{w} = \bd{u} \times \bd{v}$. then your linear map is uniquely determined by the conditions $T\bd{u} = \bd{u}, T\bd{v} = \bd{v}$ and $T\bd{w} = -\bd{w}$.

[stoic python]

Ann

[wintry steppe]

thanks but i dont get how to solve the conditions to find T

[lavish jewel]

if you want to turn ann's insight into a recipe, note that matrix multiplication of the form AB can be interpreted as applying the same transformation A to each column of the matrix B separately

so this gives you enough information to write something of the form AB = C, where you know B and C

then use your favorite method to solve for A

this is if you wanted to write it as a matrix, though, which isn't necessary

[ornate epoch]

Hi, when diagonalising a matrix, I need to find PDP^-1. When finding P, is there any order in which i should arrange the eigenvectors?

[wintry steppe]

is there a way with not finding matrices? bcs it seems hard for me

[lavish jewel]

the order should match the order of the eigenvalues in D, but otherwise it doesn't matter

[ornate epoch]

bless you, and thank you

[narrow iris]

If I had 2 planes that intersect, would the line of intersection of the 2 planes be parallel to the cross product of their normal?

[molten pilot]

yes

[narrow iris]

can you show me geometrically?

[molten pilot]

imagine one point of intersection

let's call the planes (A) and (B) and their respective normals $\vec{n}_A$ and $\vec{n}_B$

[stoic python]

DarQ

[molten pilot]

now if you were to go along any vector that is orthogonal $\vec{n}_A$

[stoic python]

DarQ

[molten pilot]

you'd still be in (A)

same for (B)

so if you were to go along a vector that is orthogonal to both $\vec{n}_A$ and $\vec{n}_B$ (i.e. orthogonal to $\vec{n}_A\times\vec{n}_B$)

[stoic python]

DarQ

[narrow iris]

yeah

just had to quickly google what orthogonal meant

[molten pilot]

you could've just asked me lol

[narrow iris]

oh right I knew that lmao

thanks heaps :D

[molten pilot]

[narrow iris]

@molten pilot can you check out my drawing to make sure I got everything ?

[dusky epoch]

it is the set of all even functions

...deleted

[fringe fjord]

Huh, where did the question go?

[wintry steppe]

I think I don't get it correct

[fringe fjord]

Impossible to tell now.

[wintry steppe]

{f ∈ R[x] | f(−k) = f(k) for all k ∈ R} I need to show is a subspace of R[x]

but I see it as the even function

[fringe fjord]

It is true that this is the set of even polynomials.

[wintry steppe]

but bcs is a subset of R[x] they must be polynomials

[fringe fjord]

That doesn't mean it's not a subspace.

Are you confused about whether non-polynomial even functions (such as the cosine) are in the set?

[wintry steppe]

tbh I started proving it and said The function 0 is the zero function defined by 0(k) = 0. It’s even since 0(−k) also equals 0. Thus the subset is non-empty/has the 0 vector. And then I got confused about the even function but I guess I am still correct even tho I got confused

[fringe fjord]

That sounds good so far. Can you describe what it is that confuses you now?

[wintry steppe]

I think I got it now thanks @fringe fjord

no wait mb that wasn't the q

(A,•,°) a ring
can all elements of A be written as a°b such as (a,b)€A²?

[fringe fjord]

If I understand you correctly, how about R[X]?

You can't write X as a product of two squares.

[wintry steppe]

they are written in terms of just f, but R[x] is a polynomial ring, so I guess is talking about finite polynomials

[fringe fjord]

(My latest posts were in reply to Sami).

[dusky epoch]

is A an associative ring with unity?

and is ° its multiplication?

[molten pilot]

is the restriction that v and w be bigger than 0 necessary?

[dusky epoch]

what does A>0 mean?

entrywise positive?

[molten pilot]

yeah

[dusky epoch]

right

then idt it's critical that v >= 0

at least from my pov

[molten pilot]

weird

[dusky epoch]

wait no hold on

oh wait no

it's not asserted that Av is nonneg

[molten pilot]

yeah

[molten pilot]

what does the highlighted text mean?

(I can provide more context but it's a bit long)

[dusky epoch]

do you know what 'restriction' means for functions

@molten pilot

[molten pilot]

I don't think I do

[dusky epoch]

given a function $f: X \to Y$ and a subset $X' \subseteq X$ (usually proper) the restriction of $f$ to $X'$ is a function $f\big|_{X'} : X' \to Y$ which sends every $x \in X'$ to $f(x)$

[stoic python]

Ann

[quiet salmon]

guys please

what did he do here?

apparently you can swap columns and revert the signal, I didn't know that

[hard drum]

Yeah, it's a general property of determinants of matrices that swapping two rows or two columns flips the sign of the determinant

[quiet salmon]

cool thanks

[wintry steppe]

thank you!

[wintry steppe]

it's its 2nd internal binary option

[wintry steppe]

no it wasn't given

how would that help tho I'm curious

[dusky epoch]

well if ° were multiplication and if it were given that A has a unity, i.e. a neutral element wrt multiplication,

then any element a ∈ A could be written as a°1

[wintry steppe]

oooooh

thank you!

[dusky epoch]

but you said now explicitly

[fringe fjord]

I think I misunderstood the notation in the question. Ann probably has the better answer.

[dusky epoch]

that A is NOT known to be a ring with unity

it's not even known to be associative!

it's not even known whether or not "2nd binary operation" is supposed to refer to multiplication!

[wintry steppe]

hold on

by multiplication u mean the one were used to?

[dusky epoch]

no, i mean multiplication in your ring

a ring comes with two binary operations

and the distributive law that governs them

[wintry steppe]

u gavee a counter example which was enough, Ann gave me the case when it's possible, which is equally appreciated

[fringe fjord]

A ring has two binary operations, but no matter what they actually are, they will be called "addition" and "multiplication" when talking about the ring.

[dusky epoch]

no i didn't

[wintry steppe]

ooooh

[dusky epoch]

i gave you a proof under a very strong assumption

which you said couldn't be made

[fringe fjord]

My counterexample was based on a misunderstanding of what you asked.

[dusky epoch]

anyway

[wintry steppe]

shit I'm sorry

[dusky epoch]

there are two binary operations in a ring. one (called multiplication) distributes over the other (called addition)

[wintry steppe]

ooo

then yrs

[dusky epoch]

are you explicitly banned from referring to them as such/

[wintry steppe]

° is the multiplication

[dusky epoch]

you have to refer to them as "the 1st operation" and "the 2nd operation" or face penalty?

[wintry steppe]

ye

yes*

[dusky epoch]

wow what the fuck

okay

[fringe fjord]

Ann, cut them a bit of slack. They're obviously unfamiliar with the usual terminology, and might well have a definition where all rings are unital without knowing that word.

[wintry steppe]

it's to remove confusion my teacher said

[dusky epoch]

is existence of unity (or NEUTRAL ELEMENT WITH RESPECT TO 2ND OPERATION) a ring axiom?

for you, that is

[wintry steppe]

hm

I'm sorry if I said anything wrong

[dusky epoch]

you didn't say anything wrong

[wintry steppe]

I'll look more into it

[fringe fjord]

Wow.

[dusky epoch]

i'm just trying to get through the conventions of your class

i don't know if you consider 2Z with the standard addition and multiplication a ring or not

2Z is the set of all even integers

or rather

2Z := {n ∈ Z | (∃m ∈ Z)(n = 2m)}

[wintry steppe]

nope we only considered Z

we did as examples lol

them*

[dusky epoch]

i'm not asking you what examples you did.

i'm asking you whether this, in your class, is a ring at all.

[fringe fjord]

Even if you only considered Z as an example, it's still possible that 2Z satisfies your axioms.

[wintry steppe]

Z is

[dusky epoch]

sigh.

[wintry steppe]

2Z isn't even discussed

[dusky epoch]

please show me the definition of a ring, exactly as your teacher gave it to you.

no alteration.

even if it's in another language.

[wintry steppe]

okay

[fringe fjord]

What Ann is getting at here is that there are several subtly different concepts that are all called "ring" by different books, and we're trying to find out which of them it is you're being taught.

[wintry steppe]

soit A un ensemble non vide muni de deux lois de compositions internes : • and °
on dit que A est un anneau si et seulement si :
-(A,•) est un groupe commutatif
-° est distributive par rapport à •
-si de plus ° admet un élément neutre, on dit que (A,•,°) est un anneau unitaire

• si ° est commutatif , on dit que (A,•,°) est un anneau commutatif

[wintry steppe]

oooo I see why now

[native rampart]

tfw you realise language doesn't stop you from reading definitions

[wintry steppe]

man thank you both for your time

means a lot

[dusky epoch]

aha.

[fringe fjord]

° admet un élément neutre
So you ring is unital and Ann's original answer works.

[dusky epoch]

-si de plus ° admet un élément neutre, on dit que (A,•,°) est un anneau unitaire

no

it says IF ° admits a neutral element...

[hard drum]

It says if

[fringe fjord]

Oh.

[dusky epoch]

if, in addition, ° admits a neutral element, we say (A,•,°) is a unital ring.

[wintry steppe]

yes

[dusky epoch]

sami, did the exercise say "anneau unitaire" or just "anneau"?

[wintry steppe]

ring

by itself

[dusky epoch]

okay

then the answer is in fact no.

there exist rings where not every element can be written as a product of two other elements.

[wintry steppe]

like the one Troposphere gave?

[dusky epoch]

no

take 2Z, the set of all even integers with the usual addition and multiplication of integers as the binary operations. the number 2 cannot be written as the product of two even numbers (as such a product would necessarily be divisible by 4, and 2 isn't divisible by 4)

[wintry steppe]

ooo I gotchu

thanks a lot and sorry for the misunderstanding 😅

[lone flume]

Btw, side question: A ring in measure theory has nothing to do with the above ring, right?

[fringe fjord]

It actually does.

[lone flume]

It does?

[dusky epoch]

you mean like a ring of sets?

[lone flume]

W.r.t. to unions and intersections?

Yes, ring of sets

[fringe fjord]

An algebra of subsets becomes a ring by taking "addition" to be symmetric difference of sets, and "multiplication" to mean intersection.

[dusky epoch]

not quite

oh

symmetric difference yes

[lone flume]

Only for an algebra?

[dusky epoch]

but in measure theory you do not really care that much about symdiff

[fringe fjord]

Having the addition be symmetric difference instead of union ensures that additive inverses exist -- namely, every subset becomes its own additive inverse.

[lone flume]

Oh wait

[fringe fjord]

It's right this is not particularly relevant in measure theory, but it explains where the word comes from.

[lone flume]

Algebras are rings

Okay

Sigma rings too?

[fringe fjord]

Yes, that just has extra properties.

There are a variety of things that are called "algebras" without actually being rings (according to the most common definitions), such as Lie algebras.

Or, indeed, Boolean algebras if we consider unions and intersections as the base operations.

This has partially historical background -- many of these were named before the concept of "algebra over a ring" became prominent.

[lone flume]

Alright, thanks for the info

[limber sierra]

and a lot of these structures were historically more like "vibes" rather than something with a concrete definition

there was over a century between the invention of groups and someone actually writing down group axioms

they were formalized mostly as subgroups of permutation groups in the meantime

(which is also a valid definition, to be clear, but not really what we think of as an "axiomatic definition")

[wintry steppe]

I was around in internet seeing some interesting lin alg problems and saw R^R (I guess presented as a vector space) and I would like to ask if someone can explain what is it just out of curiosity bcs I never saw it in my studies

[lone flume]

The set of functions from R to R

[hard drum]

Well it's the set of functions from R to R. Given two such functions f and g and some real c, we can define f+g and cf in the obvious ways to give R^R the structure of a vector space

We can similar consider C(R), the continuous functions R->R, and so forth

[zinc timber]

any algorithm to find the signs of eigen values without calculating the eigen values itself

matrix is not symmetric

also 3x3 if that makes it easier?

[fringe fjord]

Ignoring complex eigenvalues?

For 3×3 it ought to be feasible to calculate the characteristic polynomial explicitly and sketch its stationary points (if any) and y-intercept in a coordinate system. That's enough precision to see how many zero crossings it must have in each half of the x-axis.

[zinc timber]

thing is my entries are parameters

I just need the sign, not the actual values

this is my matrix

this is the char poly

so yeah

*sign of the real part

the imaginary part can be ignored

probably should ask in dynamical system... idk

[native rampart]

Count the number of elements in basis

[zinc timber]

number of vectors in the besis

[dusky epoch]

"and what if i'm unable to count"?

[native rampart]

Are you suggesting axiom of choice is false

[bold ermine]

can someone help me with this

[stuck tendon]

For the dimension of $M_5(\mathbb{R})$, consider the set of matrices $E_{ij}$, with a 1 in the ijth position and 0's everywhere else, for i,j between 1 and 5 inclusive. This forms a basis for the space.
For the subspace, I recommend trying to find a basis for it.

[stoic python]

1345631

[maiden star]

can someone help me out with this question?

[dusky epoch]

@maiden star what is your definition of an affine subspace?

[maiden star]

V is a vector space, U is a subspace
An affine subspace is
{v + U = V | v in V}

Something like that

P much an affine subspace is a point in a vector space

[dusky epoch]

as written {v + U = V | v in V} is nonsense

and no, "an affine subspace" and "a point in a vector space" are not the same thing

but despite your notation, it appears that your defn of an affine subspace is "a linear subspace shifted by a fixed vector"

[maiden star]

oh what

v + U | v In V is what I learned is an affine subspace

Where U is a subspace of V

Oh you meant like definition

[dusky epoch]

.....

[maiden star]

Mb mb LOL

[iron harbor]

while doing the gram schmidt finger exercise for these vectors a1 = (1,0,1), a2 = (1,1,0), and a3 = (0,0,1), I've reached a point where I'm unsure how to interpret the results:
I've constructed the first and second normalized orthogonal vectors, and lo and behold, the projection of the third vector is apparently the same as the projection of the second vector, which I can't really fit into my mental model of how I've been picturing this and seems like it must be a mistake

Try and ignore how horrific my handwriting is

guess it doesn't really make any difference because you can finish the algorithm without any trouble. but how can I picture that situation?

[lavish jewel]

the projection onto the vector u1, you mean?

[iron harbor]

Like this I guess, where x is the projection of a and of b in the direction of x?

[lavish jewel]

your drawing already captures it

except in 3d, the vector could lie anywhere on a cone surrounding the vector x

and it would have the same projection

this means there are infinitely many vectors with the same projection on x

[iron harbor]

got it, that explains it perfectly :)

[lavish jewel]

alternatively, if you consider that in the algorithm you explicitly split vectors into a component parallel to x and another perpendicular to it

you can fix the parallel component

and the perpendicular one can be anything you want

[hollow finch]

what signs are you looking for? there are some methods that can tell you specific cases and some general principles you may be able to use. also every time i say positive or negative i mean positive or negative real parts.

the most important fact: the trace is the sum of the eigenvalues and the determinant is the product of the eigenvalues.
this is true for any square matrix.

for 2x2s is pretty easy based on the trace and determinant. if det(A)>0 then the real part of both eigenvalues have the same sign, and if det(A)<0 then the eigenvalues are real and have differing signs.
something that applies to any sized matrix: if the trace is positive, you have at least one positive eigenvalue. similarly, if the trace is negative then there is at least one negative eigenvalue.

if the det of a 3x3 is negative, then either exactly one eigenvalue is negative, or all three are negative.

you can also look into the Routh–Hurwitz stability criterion. that tells you if the eigenvalues are all negative based on the coefficients of the characteristic polynomial. and it works for any square matrix.
you can also look at descartes' rule of signs.

another thing is that the characteristic polynomial of a 3x3 is
t^3-tr(A)t^2+(C11+C22+C33)t-det(A) (picture attached)
where Cii is the cofactor of the ii entry (the picture says minor but cofactor of a diagonal entry is just the minor so w/e). then you dont need to do the whole determinant formula for that mess of a matrix. but it looks like you got it already so idk

you might be able to use these if you know the signs of your parameters. but idk im looking at that matrix/polynomial and i dont really want to try applying any of them.

[wintry glen]

Hey, I have a proof that isn't the same as my textbook, so I'd appreciate if someone can check it for me.

Let H and K be subspaces of a vector space V. Show that dim(H ∩ K) ≤ dim H.
My proof: Let B be a basis for H. Then, B spans H, hence H ∩ K. Therefore, any basis for H ∩ K cannot contain more than dim H vectors, for they would be linearly dependent. Thus, dim(H ∩ K) ≤ dim H.

The argument my textbook gave was probably better, but I wanted to know if this is correct. The textbook essentially started with a basis of H ∩ K and said that it could be expanded, if necessary, to give a basis for H.

[native rampart]

It's correct

I will reformulate your statement a bit:
"if there's a basis for H \inter K containing more than dim H elements,see the set as a subset of H. Now,This has to be a Linearly independent set,but basis is the maximal independent set of H. So this set has to be linearly dependent and hence not a basis of H inter K"

[wintry glen]

Ah, that's much nicer phrasing. Gotcha now. Thanks so much!

[zinc timber]

yeah these are the things I have tried already. with RH I get F = det(A)-tr(J)*(tr(J)^2-tr(J^2))/2 > 0.
This is actually an epidemiological model where we have to find R0. I already have an expression for R0 but it's really long and I am supposed to show that R0 < 1 implies all roots of the matrix (real parts) are <0.
RH says it's enough to show that F<0 so I have to find something like F = K(R0-1)

really long expressions so I even want to bruteforce

picking it was a mistake

[void path]

Im trying to solve this. Im given the reduced row-echelon and i need to figure out the original A matirx.

My current thinking is "they have the same determinate, so UA must be a scaled version of A. Can i transpose UA, then multiply it by a constant so that it's first column = a1, then id have my answer?"

im not positive i can just transpose stuff freely though

or, maybe its the same as A but with different basis vectors technically? (ones that are scaled) so I need to do a change of basis? I cant figure out how to go about solving this

[lavish jewel]

why don't you do the row operations needed to turn columns 1 and 2 into the ones they have given you

wait they gave you rows, why did you write them as columns

yes, what you mentioned is correct

you could transpose the given mat and solve a system of equations

but what is a4 if there's no 4th row?

are you sure you wrote everything correctly?

omg those are 0s and not 8s

at any rate... you can thing of row reduction as a full rank matrix R that multiplies from the left

after transposing, this matrix multiplies from the right

you have that [a1 a2 a3 a4] = U(A)^T * R^T

[void path]

well, this is the point i dont quite get about linear algebra. If I have a matrix

1 1 0
1 0 1

then, that means "this is where your basis vectors are (each row is a 3D basis vector).

now, i have a collection of vectors I want to transform by that, i would put my vectors in a collection like this ```
2 1
2 2
3 4

where each COLUMN is a vector. 

Heres what i dont understand: if i have 2 collections of vectors (the second thing i sent), is that somehow different than what a transformation is? (the first thing i sent). Or could I treat any collections of vectors as a transformation as long as i transpose them before multiplication?

[lavish jewel]

this depends on the context in general

if you are explicitly told that one thing is a transformation and the other is a vector, and the transformation is supposed to act on the vector, sure

but normally one is careful to make it clear which things are "columns" and which ones are "rows"

and which ones are vectors or matrices, and which of them represent transformations

this is not implicit

[void path]

if i had a ball located at ```
1
1
1

and i wanted to know where it would be sent if i kicked it and there was wind, and i said the amount they move something by is
```        1          0
kick =  1   wind = 4
        1          3

would i need to Transpose the kick vector, then multiply it to figure out where it goes after the kick, then transpose the wind vector and multiply to figure out how the wind affects it?

Is a transformation merely "where your current vectors would go after these vectors (which are sideways) were to be added to the tips"?

I may be doing a terrible job at putting my question into words lol

[lavish jewel]

i'm struggling to follow, but i also just woke up

i think what would help is to think of vectors as matrices too

just size N x 1

then you see that addition is not defined for matrices that are not of the same size

[void path]

im asking if i can just traspose any collection of vectors to "apply" them to another group of vectors, or is there something fundementally different about a transformation?

[lavish jewel]

so you cannot add an N x 1 vector to a 1 x N one

which is what i understood you wanted to do in this example just now

separately from this, if you have column vectors, you transform them by multiplying them with a matrix from the left

e.g. Mx

if you have row vectors, you transform them by multiplying a matrix from the right, as in yP

[zinc timber]

given a matrix with non-negative entries, can I say we can find an eigen vector with max eigen value (abs) in the first quadrant?

I feel like it's true but don't want to prove it

any reference to cite will be helpful

[lavish jewel]

and for clarity, it is usually said that row vectors are the transpose of column vectors, so it should have been y^T P

[dusky epoch]

YTP

[lavish jewel]

what is ytp

i think doubly stochastic matrices might be helpful to look at, ryu

that's mostly a guess though

i know there are some results for their eigenvalues and all of their entries are >= 0

[dusky epoch]

youtube poop

i am being silly

[zinc timber]

I'll see

[bold ermine]

i didn’t clearly understand can u explain pls

[void path]

I answered my own question when i tried to visually explain it 😂

[lavish jewel]

that's not uncommon 😛 trying to put things clearly in words for others makes you fill in the gaps

[void path]

So, since A is a matrix, that is different than it being a collection of vectors, right? It means its side-ways?

So, are a1 and a2 column vectors, or are the row vectors?

[lavish jewel]

the "sideways" part doesn't matter

[void path]

im confused on "Transformations have their vectors as rows. A random collection of vectors have their stuff as columns. When you say that A is a matrix, are you saying that it is a transformation that is defining some sort of basis vectors, or are you saying it is an actual thing or collection of vectors, so its vectors are columns

[lavish jewel]

this problem only makes sense with more context, presumably given in your class and/or notes

[void path]

gotcha

the thing is, I think i have been able to do a lot of this math without actually learning this math

[lavish jewel]

linalg can do that, yeah

the way matrix multiplication works, you could treat matrices by looking at their rows or their columns

both interpretations are valid, but what is going on in each case is different

looking at rows, you're interpreting the transformation as taking dot products

as columns, you interpret it as taking linear combinations

[void path]

i think that makes sense. kinda like how an ascii and a digit are the same, its only different by how you interpret those bits?

im getting off topic, one sec let me word my actual question in a cohesive way

is a1 a scaled version of the blue, or the red?

how does a1 relate to U?

i know they have the same determinate, and that a1 could be a basis vector, but "sliding" a1 up and down would be confusing. U is telling me the sum-total change created when i increase a1

bc a3 and a4 probably undo a1 or a2 in some way, they arent helpful in our span

or, they can be undone BY sliding a1/a2 in some way

i just dont know if A1 would be a basis vector, or if the basis vector would be "the first digit of a1, the first digit of a2, the first digit of a3, and a4"

[lavish jewel]

no way to tell without further context. i would normally say column

you'll have to check the notation used in your notes

[void path]

actually, it would need to be colum right? Because we know that -3 turns into 1. meaning the row must have been divided by -3.

nvm, that cant be correct bc a2's first item is 4

[lavish jewel]

you need to look at the definitions given in your class

there is no way to know just from looking at this image

[low igloo]

For the linear algebra course in my school, we start with field then define VS on field we also used Zorn's lemma proving each VS has basis

what's a good approach to linear algebra actually

What are some elegant books on LA you have ever read

/lecture notes

[teal grotto]

what do you mean by this? did you not like something with the use of zorns lemma?

[low igloo]

like some people would start with determinant using gaussian elimination as the first principle for a more computational approach LA class

I want to know if there's any elegant approach to LA. Want to know it for fun

[native rampart]

Start with systems of linear equations

[teal grotto]

this is a pretty elegant approach imo

[native rampart]

Matrices come into picture because a Linear transform can be modelled like a system of equations(I mean linear combinations)

And we already developed theory to deal with matrices in context of systems of linear equations

[teal grotto]

conversely, you develop the theory for matrices by starting from fields when you want to study linear transformations between vector spaces. they just pop up naturally

[low igloo]

the LA course I taken in high school it was followed by linear equations, matrices, singurlarity, subspaces, null spaces, eigenvectors

I feel that I would like to have a LA course that also includes some abstract algebra. I feel like I would like it if the role of algebraic structures were stressed more

[teal grotto]

try hoffman and kunze linear algebra

[hushed hedge]

Aren't these two matrices equivalent to each other?

[teal grotto]

equivalent in what sense?

[hushed hedge]

if i don't want to work with a rational matrix can

can't i just multiply the second one with the denominator to get rid of the denominator? if that makes sense

[teal grotto]

if you mean to ask if those matrices are equal, then no, they are not

[hushed hedge]

to get the rid of 1/2 can't you just multiply B with 2 so you have A*2B or something

ok weird, i thought you could do that <.<

[teal grotto]

no, you can do what you just described

i guess i was misunderstanding what you were trying to say earlier

[hushed hedge]

yeah what i said didnt make any sense at first but

ok i did A*2B but for some reason i get the wrong answer

so

When i solve this i get this: but the answer is actually
4 -11
-6 15

i don't understand why it doesn't work, i get:
8 -20
-12 30 after i divide the last matrix with 1/2

I'm suppose to solve this equation AXB = AC

oh i need to divide it twice

[slate hatch]

Can I calculate the area of a triangle using vector algebra?

[teal grotto]

yea

[slate hatch]

Thanks

[icy blade]

In linear transformations what does T:R^n - > R^m mean

I just can't seem to get it, what exactly are we doing with matrix T

[teal grotto]

here is a possibly oversimplified explanation.

T is just a function that eats vectors in R^n and spits out vectors in R^m. we are just multiplying the matrix m x n matrix T by some n by 1 vector x to get Tx an m x 1 vector.

T is linear, so it has the property that T(x+y) = Tx + Ty and T(ax) = a(Tx) where x, y are vectors in R^n and a is a real number

[wintry steppe]

hi

when determining the eigenvectors of a matrix 5x5, i got solutions b=0, d=0, e=0

can anyone explain why the eigenvectors are (1,0,0,0,0) and (0,0,1,0,0) ?

and why not just the vector (1,0,1,0,0) ? is that because it's a sum of two linear independent vectors?

[lavish jewel]

you mean you get an eigenvector of the form (a,0,c,0,0)?

[wintry steppe]

yep

[lavish jewel]

if you don't have a=c, then the vector (1,0,1,0,0) does not cover all the possible vectors of the form (a,0,c,0,0)

e.g. if you take a = 2 and c = 5 (we can do this because a and c are free variables, so they should be able to take any value)

[wintry steppe]

i see, thank you very much!

[tidal flint]

Hi everyone! How would i go about doing this? "Is it true that for any invertible matrix A, we have tr(A^(−1)) = (tr(A))^(−1)?"

I thought I could prove using examples but it seems to me like I have to prove this by definition which I'm quite unsure of how to proceed

[lavish jewel]

do you know any relationships between the eigenvalues of A and the trace of A?

and also between the eigenvalues of A and those of A^-1

[viral olive]

I need to prove that the group (R^2,x) -> (G,x) an isomorphism. I proofed that (G,x) is an commutative Group. My problem is to assing G a basis to proof the homomorphis is surjective.

[wintry steppe]

or you can just show from the definition that it's surjective

[molten pilot]

is there a name for like, a subspace which contains vectors that're orthogonal to another subspace?

[lavish jewel]

orthogonal subspaces?

[twilit minnow]

why is it that if T is injective (with the corresponding matrix A being mxn) then A has a left inverse

[molten pilot]

I think it's called orthogonal complement

[lavish jewel]

it's only orthogonal complement if it contains ALL vectors orthogonal to the other subspace

[molten pilot]

yeah, that's what I meant

[lavish jewel]

i.e. their direct sum is the whole vector space

but they didn't say that was the case

[molten pilot]

who?

[lavish jewel]

oh lmao it was you

[molten pilot]

[lavish jewel]

anyway, it's only orthogonal complement if the two subspaces together form the whole vector space

but you didn't say that in your original post

[molten pilot]

yeah, sorry I wasn't clear enough

[reef gull]

So glad I found this discord!

[native breach]

so I had to prove that this set

was a basis

and now that I've done that I need to create the change of basis matrix from it to the standard basis (1,t,t^2,t^3)

but isn't that matrix just the coordinate matrix that I generated so I could row reduce it to prove it was a basis?

[lavish jewel]

sounds about right

e.g. it maps (1,0,0,0) to (-1, 0, 2, 1)

[native breach]

so this is my change of basis matrix then?

it's throwing me off cause I feel like I need to do math at it or something

[lavish jewel]

that looks correct

the question will probably go on to ask you for the other one

change from the standard basis to this one

[native breach]

and then for T_ab it's jsut that matrix ^-1

yeah

wait I thought this WAS the matrix to go from the standard basis to this one

[reef gull]

I am new to linear algebra, and haven't done math in 16 years lol can someone help to explain if this is linear or not?

[lavish jewel]

oh oops i misread your statement

no, this matrix takes coordinate vectors in this weird basis, and returns a coordinate vector in the standard one

you can see this by noticing that if you multiply this matrix by the vector (1,0,0,0) it gives you the polynomial 1 + 2t - t^3

[distant schooner]

\cal = ker, this is my proof of rank-nullity... i was wondering if it is complete/correct

any other feedback would also be appreciated

[lavish jewel]

have you tried using the definition of linearity?

[reef gull]

I know it needs to be a straight line

[lavish jewel]

that is not the case

[native breach]

I was initially tempted to claim that was exponential, but if you work out all the Lns I'm pretty sure it's just pi(2z)-1/2 y -2z =3 -x

which is a plane I'm pretty sure

[lavish jewel]

can you look up the definition of linearity used in your course or book?

[reef gull]

Looking now quick

[lavish jewel]

did you get why what you have is the opposite?

[reef gull]

all variables occur only to the first power and not appear as arguments of trig, log or exponential functions

[native breach]

does this server have a tex bot

[lavish jewel]

yes

[lavish jewel]

all right, and do you have any trig, log, or exponentials here?

[native breach]

so my understanding is that I"m trying to solve for $B\mathcal{B}= T\mathcal{A}$ where $\mathcal{B}$ is our new fucked up basis and A is the standard one that can be represented as the identity matrix

[stoic python]

mr.mseeks

[native breach]

so then, $\begin{bmatrix}
1 & 3& 0 & 4 \
2 & 1 & 0 & 0 \
0& 4 & 6 & -5 \
-1 & 0 & 2& 0 \
\end{bmatrix} = \begin{bmatrix}
1 & 3& 0 & 4 \
2 & 1 & 0 & 0 \
0& 4 & 6 & -5 \
-1 & 0 & 2& 0 \
\end{bmatrix} * I $

[reef gull]

inverse log In(e power 3)

[stoic python]

mr.mseeks

[lavish jewel]

not quite

[reef gull]

so sin cos tan will be trig right?

[lavish jewel]

yes

[reef gull]

and log then? or do we not look at inverse?

[lavish jewel]

so you want the change of basis matrix from the standard basis to this one

[reef gull]

exponent I see still the 2z

[native breach]

I want both, so if our wonky matrix is $B$ and that's the first matrix I want, then the other one is gonna be $B^{-1} right$

[lavish jewel]

what you originally have is $[x]\mathcal{A} = T^{\mathcal{B}}{\mathcal{A}} [y]_{\mathcal{B}}$

[stoic python]

Edd

mr.mseeks

[lavish jewel]

this vector y has coordinates using the weird polynomials as its basis B

e.g. it maps the coordinate (1,0,0,0) in the weird poly basis to (1 + 2t - t^3) in the standard poly basis, which corresponds to (1,2,0,-1)

it took it a coordinate 1 and it spat out a weird polynomial

[native breach]

right so that's changing our basis FROM A TO B

[lavish jewel]

no

it's saying that the weird poly has a coordinate of (1,0,0,0)

let's do this a different way

look at your matrix and consider its columns

let's say the matrix is M and we have a coordinate vector v

M has 4 columns, let's call them m_i

the product Mv is equivalent to writing m_1 v_1 + m_2 v_2 + ...

where v_i are scalars (the entries of v) and m_i are vectors (the columns of M)

you can see that the result of this operation is a linear combination of the columns of M

and the columns of M are the weird polys

i.e. you are using the weird polys as a basis

in this basis, the polynomial (1 + 2t - t^3) has coordinate (1,0,0,0)

in the standard basis, this polynomial has coordinate (1, 2, 0, -1)

there exists some OTHER matrix that provides the inverse mapping

we give in the coordinate (1,2,0,-1) , which corresponds to the weird poly, and it gives us its coordinate in the new basis

which would just be (1,0,0,0)

this would be the matrix that changes coordinate vectors from the standard basis A to the new one B

the columns of the matrix are the basis here

saying u = Mv is the same as saying "the vector u can alternatively be constructed by taking linear combinations of the columns of M, and the coefficients are the entries of v. this means v is the coordinate vector of u in the basis of the columns of M"

[reef gull]

okay I just saw that in(e^2) will be 2 for example?

so in(e^z) will be z

[native breach]

I think I kind of understand? I gtg but appreciate your help

[molten pilot]

I think I got a sufficient argument for (a)

[stoic python]

DarQ

DarQ

DarQ

DarQ

[teal grotto]

small note: you can just write, v1, v2, v1 u v2. a basis is already a set of vectors

[molten pilot]

that's just how my book denotes a basis

[stoic python]

DarQ

DarQ

[molten pilot]

(I can prove that lol, I'm just lazy)

[stoic python]

DarQ

DarQ

DarQ

[molten pilot]

does this work?

[twilit minnow]

how can you show that a linear transformation T that is injective always has at least one left inverse for its corresponding matrix

[molten pilot]

since T is injective, the columns of $[T]$ are linearly independent

[stoic python]

DarQ

[molten pilot]

hmmm

I can explain this better, give me a sec

[lavish jewel]

if you're going this route, why not gram schmidt

[molten pilot]

that means the rows of $[T]^T$ are linearly independent

[stoic python]

DarQ

[molten pilot]

which mean $[T]^T$ is onto

[stoic python]

DarQ

[molten pilot]

which means the equation $[T]^T \vec{x} = \vec{e}_1$ has a solution

[stoic python]

DarQ

[molten pilot]

now $\vec{x}$ is just the first row of the left inverse of $[T]$

just do this all the vectors of the standard basis and you'll construct a left inverse of [T]
does that make sense? @twilit minnow

[molten pilot]

coz I don't know what that is

[stoic python]

DarQ

[molten pilot]

screw my solution

it's ugly, I hate it

[molten pilot]

is there a better way of doing (a)?

[teal grotto]

if ker T2 is not a subset of ker T1 then there is some vector v in ker T2 not in ker T1. T1 v != 0 but T2v = 0. any linear transformation S has to take 0 to 0….

[molten pilot]

yeah, that direction is trivial

I meant the converse

[zinc timber]

[teal grotto]

oh did i do the wrong direction

[zinc timber]

[zinc timber]

what's the question

[molten pilot]

(a)

[zinc timber]

so you were able to show T1= S*T2 then ker T2 ⊆ ker T1?

or the other way?

[molten pilot]

the other way

[zinc timber]

ok

v ∈ ker T2 then S*T2(v)=0 => T1(v)=0 => 0 ∈ ker T1 so ker T2 ⊆ ker T1

Is this what you wanted?

@molten pilot

for converse use first isomorphism theorem, if not already done

I mean that's literally the first isomorphism theorem

[molten pilot]

but how do you know S exists?

wait

yeah, it exists by hypothesis lmao

[molten pilot]

what's the first iso theorem?

[zinc timber]

if ker T2 subseteq ker T1 then there is an unique map from Rⁿ/kerT2 to imT1

[twilit minnow]

that doesnt seem like a pretty way to do it

im looking at this now

how can he be so sure that transpose(A)*A will have an inverse

[molten pilot]

that's cheating

[twilit minnow]

i get that its a square matrix, but how do we know that the result will have full rank and be invertible

im assuming that the rank(A) = n here

[zinc timber]

it doesn't

but they have assumed that the cols are independent

otherwise use pseudoinverse

[twilit minnow]

im confused

cols are of what are independent

A?

[zinc timber]

yes

that's same as saying A'A has rank n

[twilit minnow]

okay but if the cols of A are independent, how do we know that transpose(A)*A will have an inverse

[zinc timber]

rank A = rank A'A

let me think of an easy proof

let v ∈ N(A) then Av=0. now <Av, Av> = 0 => <v, A'Av> = 0 for all v in N(V)

ok reverse this

I hope forst inclusion is clear, N(A) ⊆ N(A'A)

let v ∈ N(A'A) then <v, A'Av> = 0 since A'Av=0 so <Av, Av>=0 or ||Av||=0 so Av=0 or v ∈ N(A)

so N(A)=N(A'A) so null space have same dim, so does their col space and hence have same rank

[twilit minnow]

no lie chief

you confused me even more

[zinc timber]

ok assume A'A has rank n

ignore all that

that's what they are assuming anyway

[twilit minnow]

why is this the case though

[zinc timber]

Av= 0 mtlb A'Av=A'(Av)=A'0=0

[dawn lark]

Hello guys

I have a problem with a exercice

So i have this expression: $(u - a.Id_E) \circ (u - b.Id_E) = 0_{L(E)}$

[stoic python]

Pako

[dawn lark]

u is a linear funtion

I don’t know if you use L(E) in english to say that

And my first question is to prove that a or b is a eigenvalues

Then i have to proove that if u isn’t a homotety, a and b are eigen values of u

In this subject, a /= b

[zinc timber]

I doubt that you haven't written your question properly

[dawn lark]

Yes i know, i don’t speak English fluently sry

I will try to translate it with google

[zinc timber]

(U-aI)(U-bI)= 0 doesn't mean a and b are eigen values

[zinc timber]

wasn't talking abt that

[dawn lark]

It means that a or b are eigen values

I can proove that :

[zinc timber]

for example take U = aI then b is not an eigen value

yes a or b is correct

oh u said a or b my bad

[dawn lark]

Oh sry, i though that i said a or b

Ok^^

[zinc timber]

lol

[twilit minnow]

this makes no sense to me

can u explain

[zinc timber]

say (U-aI)≠0 then you have a vector v s.t.

(U-aI)v≠0

now assume (U-aI)v=u and look at (U-bI)v

[zinc timber]

did you get why N(A) = N(A'A)?

[twilit minnow]

no

[twilit minnow]

A'Av=0

how can u assert that

[zinc timber]

taking v in null space of A'A

[dawn lark]

Why take (u - aI) /= 0 ?

[twilit minnow]

and then how can <Av,Av> = 0

[dawn lark]

What is s.t. (Pls write with full name, all is different)

[twilit minnow]

where v is an element of N(A'A)

[zinc timber]

if (U-aI) is zero then you already got your Eigen vector

[twilit minnow]

s.t. = such that

[zinc timber]

[zinc timber]

properties of adjoint

<Av, w>=<v, A'w>

[dawn lark]

Okk i understand this point

[zinc timber]

so done?

[twilit minnow]

how do u apply that to what u said earlier

[twilit minnow]

with the <Av, Av>=0

[zinc timber]

well <v, A'Av> = 0

sine A'Av =0

now pull the A' back to first term

you get <Av,Av>

[twilit minnow]

this is the first time im hearing that u could do that o_O

[zinc timber]

that's the definition lol

not a property

[twilit minnow]

where can i read about this

[zinc timber]

which grade you are in?

[twilit minnow]

2nd year uni

[zinc timber]

ok not familiar with the metric

LADR is a good start

or you can just search adjoint operators

[twilit minnow]

what year r u in

[zinc timber]

2nd year masters (india)

[dawn lark]

In fact, I don’t understand

Because if u is bot a homothetie, u /= lambda .x

Are you agree with me ?

[dawn lark]

My country too 🙂

[dawn lark]

I dont understand from where come

The v

[clever prairie]

Hey can anyone tell me wether this is true?
(A + B)C = CA + CB

My friend says that it has to be

C(A+B) = CA + CB to be true

[dusky epoch]

are A, B and C matrices?

[clever prairie]

Yes sorry

just any m x n matrix

[dusky epoch]

if so then (A+B)C = AC+BC but not necessarily CA+CB

also you will need to be more careful when specifying sizes

but that is beside the point

[clever prairie]

My bad

Thank you though 😄

[dawn lark]

I have another plan

I suppose that b have not a eigen vector

But, i don’t know how to process then

[limpid frigate]

What is the difference between W⊥ and (W⊥)⊥?

W is and euclides subspace

[gleaming knot]

$W^\perp$ is the orthogonal complement of $W$ and $(W^{\perp})^\perp$ is the orthogonal complement of the orthogonal complement of $W$

[stoic python]

Icy001

[gleaming knot]

Does that answer your question?

[limpid frigate]

Ok i will fraze it diferently

Becouse i clearly didnt get something

W⊥ - means that every vector in this subspace is perpendicular to every other

Right?

(every other in this subspace)

So what is the point of orthogonalization of something that is already orthogonal

[gleaming knot]

No!

That would make no sense because if W is any nonzero subspace, it'll contain two vectors parallel to each other

just take some nonzero v in W and also take 2v

$W^\perp$ is another vector space, consisting of all vectors perpendicular to all vectors in $W$

[stoic python]

Icy001

[limpid frigate]

Oh so i missunderstood

W⊥ is not about all beeing perpendicular to each other
It's about beeing perpendicular to base subspace

Right?

[gleaming knot]

Yes

[limpid frigate]

Thanks, this makes more sense now

And any pair works, right?

[gleaming knot]

What?

Any pair of what?

[limpid frigate]

any pair of vectors is perpendicular

By pair i mean that they one comes from W and other from W⊥

[gleaming knot]

What do you mean by "any pair works"

[limpid frigate]

any pair of vectors is perpendicular

[gleaming knot]

I think you're missing some quantifiers in order for me to answer your question

Start from the definition of the orthogonal complement and write it as one whole sentence

[limpid frigate]

ok , so by perpendicularity i mean that the Dot products of 2 vectors is equal to 0

Now let there be a subspace of vectors W

And W⊥ - which is a orthogonal space of W

[fringe fjord]

By definition, an element of W^⊥ is perpendicular to every element of W.

[limpid frigate]

I just want to make sure i correctly understood that

[fringe fjord]

That's the definition of W^⊥.

[limpid frigate]

Sorry, I had only one class of this subject and dont fell quite good at it

Thanks guys, you really helped me out ❤️

[upper osprey]

A helping hand pls 😭 🤲

[hard drum]

Well what do you know as the statement of Lagrange's remainder theorem? what equation do you end up getting when you apply it to exp ?

You should precisely get that e - (1 + ... + 1/n!) = something you can estimate beneath 3/(n+1)! from that, lmk if you get stuck

[fringe fjord]

(This is pretty far from linear algebra, though).

[hardy inlet]

now he said theres not gonna be corrections for this one; but one of the proofs on this test was if T is a normal operator on Complex V with T^8 = T^9, prove that T is hermatian and idempotent.

I basically just said that from the complex spectral theorem, we'd have a diagonal matrix, and the only way the diagonal matrix = itself applied one more (8 to 9) is if everything was 1 or 0 on the diagonal, so any diagonal with 1 or 0 on the diagonal is hermatian and clearly idempotent?

i'm wondering how correct that is so i can estimate my grade

[fringe fjord]

Do you know V is finite-dimensional?

[hardy inlet]

i cant remember

it might have been

Also, another thing was in the "computation" section and it was something about a projection operator P_U on a proper subspace U of V, what were are the eigenvalues.
I just said 1 was the only eigenvalue because the only time a vector retains its form is for P_Uu=u, and hence lambda=1, otherwise the vector loses its direction (and hence isn't an eigenvector). But idk if thats what they were looking for or if its even correct

i guess 0 might have also been an EV since P_U(u perp) = 0

[fringe fjord]

Indeed.

[icy blade]

yo in linear algebra

whats the symbol for like saying a certain value in a matrix

lets say i want to show the value at row 1 column 2 of matrix T is 0.5

whats gonna be the symbol

[tribal willow]

$T_{1,2}$

[stoic python]

anamono

[icy blade]

alright thx my g

[tribal willow]

ye

[icy blade]

i have a question about the state vector in a markov chain

the thing u multiply by the transitional matrix

does it have to be like [ 0.25 0.25 0.5] so lets say 25% of students are in A, 25% are in B..

and totalling 1

or can it be like [ 100 200 250], 100 students are in A, 200 are in B...

does it have to be a proportion or can it just be like raw data

[icy blade]

nvm it has to be between 0 and 1 i got it

[zinc timber]

I don't get you, you can speak hindi

[iron harbor]

Test today 😬

[hushed hedge]

Is this formula familiar to anyone? It is used for a linear mapping problem, im wondering if it suppose to be inverse of S instead of transpose

[lavish jewel]

that depends on what structure S has

[zinc timber]

if A is symmetric then S is orthonormal

or "orthogonal" as some people say

[hushed hedge]

let me see

[lavish jewel]

as ryu says, some matrices are diagonalizable in an orthonormal basis, so that S is an orthogonal matrix

then S^T = S^-1

[hushed hedge]

" The imaging matrix in the new base is A = ST AS where S is the matrix whose
columns are the base vectors. So is (if we break out 1 / √6 from S), The image is a projection on the line through the origin with the direction vector "

[lavish jewel]

what structure does the original "imaging matrix" A have?

[zinc timber]

[hushed hedge]

one sec

and u_3 = 1/sqrt6(-1,1,2)

and this is the A matrix

ig you answered it up there

[lavish jewel]

looks symmetric to me

[hushed hedge]

yes 🙂

didnt read that carefully, ill keep that on my mind

[untold stag]

Can anyone help me here, I am trying to model inaccuracy in a WW2 RTS game.

Basically I have this base function that I want to scale non linear.
However I need multiple 'base' variants of this function which have a linear scaler, and I don't want to multiply the linear scaler by the exponential.
For reference. The base function is y=0.0266 * n * m where n is the position on the x axis and m is the linear multiplier of the function
The non linear scaler function is (1.08+0.015x)^n where x is a series of number 1-8.
I was just multiplying the two together until I realized its scaling the whole thing non linear
To put the idea in words; I'm creating an accuracy system based on minute of angle (0.0266m) and human misjudgment of range on first shot based on skill (x).
I want the MoA part to remain constant with addition of the misjudgement on non linear

[zinc timber]

'non linear' word doesn't belong here

[hushed hedge]

When choosing a direction vector (maybe it's called unit vector?) does it matter whether you pick the vector like $V=\overrightarrow{P_1P_2}$ or $V=\overrightarrow{P_2P_1}$?

[stoic python]

pesthaio

[hushed hedge]

or is there a rule that you should follow when creating a unit vector, (P_1 and P_2 are two points in the plane btw)

[halcyon spindle]

No rule, just depends on what your doing. Also you can obtain from from the other. V = p_1p_2 = -p_2p_1.

[lavish jewel]

just keep in mind that if the direction is reversed, your coefficients will have the opposite sign, as plegasus says

[hushed hedge]

alrighty

thanks

[fallen salmon]

can the determinant just be the diagonal of the matrix?

[dusky epoch]

the determinant of a triangular matrix is the product of its diagonal

[fallen salmon]

oooohhh yeah you're right

soooooooo if i have a question like this

i can just try making a triangular matrix

and get n! or n-1! or whatever it becomes afterwards

[umbral yew]

i mean that's a very specific example

im not sure how often you'll encounter a matrix that can be transformed so easily

it's not even possible for all matrices

[lavish jewel]

you can certainly row reduce any matrix

[fallen salmon]

how does my instructor get this on -1^

[lavish jewel]

look a the operations that are being done as elementary operations

scaling a column by -n multiplies the determinant by -n

[fallen salmon]

i understand that its n times -1 to the power of but i dont know why it's that power specifically

it's row + collumn

but thats n + n or 2n

[lavish jewel]

it's not clear to me what operation was done on the second row just from what is written

what is done to get from 1 to 2?

[fallen salmon]

laplace on n row

the all 0 and last one is n row

[lavish jewel]

hmm idk what you mean by that

let's see if someone else shows up

ah you mean laplace expansion of the determinant, so cofactors

[fallen salmon]

yeah that shit xd

im not a native english speaker i apologize

[lavish jewel]

so the sign changes depending on which row and column you are on

[fallen salmon]

yeah ii get it

it's -1 to the power of row + column right?

why is it 2n / n+n

[lavish jewel]

i don't think it's a division

it looks like a bracket

that was also throwing me off

[fallen salmon]

its in brackets i think

[lavish jewel]

what i mean is that what is written is

[fallen salmon]

oh okay

its ( 2n / n+n )

[lavish jewel]

not a division

[fallen salmon]

really?

[lavish jewel]

that's what it looks like to me

[fallen salmon]

okay i get that

and below that?

n-2 / 2

and i also notice that the columns are reversed

i think that has something to do with it?

[lavish jewel]

yeah

i guess they permuted the columns

it says something like permutation there, right?

[fallen salmon]

nope nothing says permutation

[lavish jewel]

but i'm not sure if they flipped only columns or also rows

[fallen salmon]

it says

we are flipping

switching

it says also

2nd row far right

its of format (n-1) X (n-1)

[lavish jewel]

what does it say on top and below the equal sign

[fallen salmon]

third row?

[lavish jewel]

yep

[fallen salmon]

we are switching

we are flipping

[lavish jewel]

[fallen salmon]

i see that the entire matrix is

inverted

instead of -1 ...

it's ... -1

[lavish jewel]

yeah

[fallen salmon]

how do you do that?

[lavish jewel]

ah i get how they formulated it

you take the first and last column, and swap them

this multiplies your determinant by -1

then you swap the second column with the second to last column

you get another -1

[fallen salmon]

oh i get it

and its n-2 times

[lavish jewel]

then as a toy example, consider that a matrix with 2 columns requires 1 flip

and a column with 3 columns also requires only 1 flip, the middle one isn't touched

[fallen salmon]

thats 1 * -1

[lavish jewel]

they have generalized it as, you have n-1 columns, we wish to flip half of them

and to make the expression work regardless of having n-1 odd or even, they have written this as floor((n-2)/2)

btw on the second row i would argue that it should be (-1)^((n-1)^2), but you still get the same result since this always yields 1 here

[fallen salmon]

umm

i kinda dont get it

sooo we have n-1

we want half of them

that's n-1 / 2

correct?

[lavish jewel]

you have to think about it a bit more carefully and see the pattern

if we have a 1x1 matrix, we need 0 flips

[fallen salmon]

2x2 its 1 flip

[lavish jewel]

for a 2x2 matrix, you need 1 flip

3x3 is also 1 flip

[fallen salmon]

oh i get it

[lavish jewel]

4x4 is 2 flips

[fallen salmon]

the middle still stays the same

on odds

[lavish jewel]

let's say the matrix has M columns now

[fallen salmon]

okayy

[lavish jewel]

let's see...

oh i think it's ceil, not floor. we round up

(1-1)/2 = 0

(2-1)/2 = 0.5, we round up to 1

(3-1)/2 = 1

(4-1)/2 = 1.5, we round up to 2

so in general for W columns, we do ceil((W - 1)/2) where ceil means rounding up

then we recall that for your matrix, there are W = n-1 columns

so ceil((n-2)/2)

and some people denote this rounding operation with brackets, so on your notes it reads as [(n-2)/2]

is that clear?

[fallen salmon]

im sorry im reading now hmm

oh so

n-1-1 is n-2

and then / 2

is the number of flips

sooo -1 to the power of n-2 / 2

i dont get the round up and ceil things you are saying

[lavish jewel]

the thing is that, for example, if n = 3

you get 1/2

you certainly don't expect the determinant to be imaginary

[fallen salmon]

yup

[lavish jewel]

there is no such thing as flipping half a column either

so you round it up

[fallen salmon]

soo you just round it up

does ceil mean rounding up?

[lavish jewel]

yes

it's short for "ceiling"

,w ceiling function

[fallen salmon]

oh okay i get it

[stoic python]

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[fallen salmon]

i get it

thanks fam <3

i just didn't know how flipping worked then

[granite wind]

Can someone help me w a question on determinants?

I had it on my test today and i remembered it i think i’m pretty sure i got it right but i wanna make sure for peace of mind

[celest ore]

Is it plausible to prove nxn matrix A has k^n*det(A) = det(kA) by associating the idea that A can be row reduced into REF, a triangle matrix, which has the property that the product of the elements on main diagonal equals det(A).

hope that makes sense

[zinc timber]

what kind of proof is that

also after REF you'll get 1's or 0's in the diagonal

not necessarily the det

[cunning viper]

i will not enter this discussion

nope cya

[zinc timber]

no body asked anyway

[granite wind]

DetA= 3 DetB=5 and it gives a) det(CA^3)=7 and u gotta find detC and b) det(2AB^-2)=? also it’s a 4x4 matrix

[zinc timber]

det(AB) = det(A) det(B)

[celest ore]

can I use this to prove that using k*identity matrix? Thought it might be circular logic

[zinc timber]

you can

$kA$ = \m{\dmat{k}{\ddots}{k}}A$

hint ^

[stoic python]

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

[twilit minnow]

ryu

is there a way to find the left-inverse with row operations

or is this really the best way

[zinc timber]

I like the 'tall matrix' term

[twilit minnow]

well you cant find it when n>m

only m=n or m>n

[zinc timber]

"left inverse" is kinda least square solution matrix

it has nothing to do with the classical inverse

[zinc timber]

in that case you have right inverse but I'll not get into it

these are just names, don't take the 'inverse' term too seriously

[twilit minnow]

okay assume we have to find the left-inverse of a tall matrix (m > n)

what would be the best way to do it

[zinc timber]

I mean they are 'pseudo inverses' but idk

[zinc timber]

the normal equation you are given

do you know least square?

say you have $Ax = y$ with $m>n$ then you can't find an solution of this system

[zinc timber]

but what you can do is find a solution $\hat{x}$ that minimizes the error, i.e. $A\hat{x}-y$

[twilit minnow]

x~ s.t. there is a soln right

[zinc timber]

no your system may not have a solution

x^ is a least square solution

i.e. one solution that minimizes the error

[twilit minnow]

oh, so its to avoid those cases

okay

[zinc timber]

yeah so it's like you can have a vector in a plane and you have to find one of such vector x^ s.t. the distance of Ax^ and y is the least

I hope you can see why it has to be perpendicular to our plane

actually let me send a video for you to follow

[twilit minnow]

that would be helpful

[zinc timber]

https://youtu.be/osh80YCg_GM

watch this

[lavish jewel]

they're equivalent when the mat is invertible

[zinc timber]

yes

[subtle gust]

I NEED TO KNOW IF LINEAR ALGEBRA GETS ANY EASIER

Cuz for the last couple of weeks classes have been absolute hell

Everyday we cover a topic that is even harder than the one before 💀

[zinc timber]

DEPENDS ON WHAT YOU ARE DOING

[lavish jewel]

you need to review the stuff until you understand it well, because every topic will build on the previous ones

i.e. it only gets worse

[zinc timber]

TENSOR PRODUCT OR MULTILINEAR STUFFS AREN'T EASY

[subtle gust]

Exactly!!

WHY ARE COVERING SO MUCH TOPICS LIKE TF?

[twilit minnow]

im looking at this review problem and i kinda get how to prove it but im getting stuck

so i know we need an iff proof here

rank(A) = n iff left inverse of A

[zinc timber]

BECAUSE THERE'S SO MUCH TO COVER

[twilit minnow]

so lets assume rank(A) = n

[subtle gust]

OK WELL IT SHOULDN'T ALL BE PUT IN ONE COURSE THEN

They can have a lin alg II or smthng

Like calc

[twilit minnow]

we can say that a transformation T for the corresponding matrix A will need to be injective

[lavish jewel]

it isn't put in only one course, you probably are only seeing a small part anyway

[twilit minnow]

^

in a proof how do we go from rank(A) = n to it being injective

[zinc timber]

ig you can say that

[subtle gust]

Ig so ...

Still think it's a lot of content tho ....

[zinc timber]

nullity = 0 means it has a left inverse

[twilit minnow]

in the proof how do i even get from rank(A) = n to injective

is it enough to say if rank(A) = n iff N(A) = {0}

[lavish jewel]

in the notation in your image, this isn't true

you need full column rank

so rank(A) = m

[twilit minnow]

rank(A) = n will give u full col rank though

A is m x n

[zinc timber]

m>n required

=

[lavish jewel]

only if you have some relationship between m and n

[twilit minnow]

assuming m != n

and m !< n

[subtle gust]

Wait if a matrix isn't full rank it wouldn't be invertible right?

[lavish jewel]

well, you have to state that

[twilit minnow]

my bad

forg0t

[lavish jewel]

because otherwise it's just not true

[zinc timber]

both sided invertible no, but one sides could be

[subtle gust]

For dquare matrices ofc