linear-algebra

[abstract vale]

But would finishing the diagonal matrices that have trace 0 be similar to what I was doing at the beginning?

The 0 everywhere else but diagonal (1,-1,0000...)

[haughty berry]

Ah yes, but those are tuples, not matrices. If you meant matrices, then yeah, you were right

[abstract vale]

Thanks

[brave lintel]

Hey yall, this is my attempt at two of the exercises in axler. F denotes either R or C and the task Is to determine whether the subset (given at the top of each photo) is a subspace of f^3. Could anyone check my answers please because this is my first look into proof based maths and exercises. Thanks in advance :)

[tame mural]

Tiny grammar question -- do you say "a set of vectors are independent" or is independent?

[subtle walrus]

a set, a collection, a sequence, ... of vectors is linearly indepdendent
the vectors v_1, v_2, ... are linearly independent

[haughty berry]

Your proofs look correct

[brave lintel]

The strategy here is to prove that each of the subsets contain 0 , are closed under addition and under scalar multiplication, then subspace will naturally follow

[brave lintel]

that's reassuring because this is literally the 5th time I've ever tried to prove anything

lol

[subtle walrus]

the proofs are correct but

your counterexample in 1 should be more concrete

as concrete as possible, give actual elements

[brave lintel]

that's a better strategy

mhm

[subtle walrus]

i dont get what is happening here

or why

and the additive closure i would mark as insufficient (although i think your thought process is correct)

in general i would suggest adding more words and explaining what you do

[brave lintel]

A vector scaled by S satisfies the condition that x_1 + 2x_2 + 3x_3 = 0

[subtle walrus]

(and maybe less variables)

[brave lintel]

lol

Okay

I will take that into consideration with the next excercises

[subtle walrus]

yes but why the dividing by S

[haughty berry]

Your proof is a little overcomplicated, but it looks right

Also, the S is a bit confusing tbh. Id use a different letter for scalars

[brave lintel]

it's trying to show that S(a + 2b +3c) = a + 2b + 3c = 0, as 0/s = 0

[subtle walrus]

well

[brave lintel]

and hence that any scaled vector in the subset satisfies the condition

[subtle walrus]

you have (a+2b+3c) = 0

[haughty berry]

Yeah you know you could just say that since a+2b+3c=0, S(a+2b+3c)=S*0=0

[subtle walrus]

and S*0 = 0 hopefully

[brave lintel]

that's a lot less overcomplicated

[subtle walrus]

dividing by S is unnecessary and wrong if S=0

[brave lintel]

did not think of that

[haughty berry]

lmfao its fine. This is your first time, youll get used to these sorts of proofs

[subtle walrus]

(also the first "vector space" does not contain a zero vector, which is an easier proof)

[brave lintel]

that's also true

Idk the first thing that popped into my head was scalar multi. for whatever reason

Well, thanks guys

I'll do the next few and come back here to get torn apart

(jk Im very grateful)

[haughty berry]

lmfao

Dont worry, it just takes time and practice

[brave lintel]

Of course

Practice is what I'm lacking

but also what I'm doing

soon

I wil be the linalg king

lol

[haughty berry]

yeah, and good on you for that!

Are you selfteaching?

[brave lintel]

Yes, I'm 16

[haughty berry]

mmhmm

[brave lintel]

The whiteboard is from one of my classrooms which I used during my lunchbreak

[haughty berry]

Mm yeah i love using other class' whiteboards

makes you feel smart when other people walk in

lol

[brave lintel]

lol

[haughty berry]

Anyways good luck!

[brave lintel]

Thanks!

[abstract vale]

Need help with 4 ii)

This is the working out I have so far

[lavish jay]

Here, if i proof that IF ad - bc = 0 i would have infinite solutions will count as a proof that IF ad - bc =/=0 has a unique solution?

[quasi vale]

If ad-bc = 0, you can also have no solution

so you'll have to show either infinite or no solutions

better to just show unique solution

[lavish jay]

could you give me a hint ?

[quasi vale]

@abstract vale which book is that?

[lavish jay]

i know that it's the determinatn, but i've seen that in High School

not in the book yet

[quasi vale]

Okay so you haven't studied determinants or matrices yet?

in this course?

[lavish jay]

so i'm thinking that this is suposed to proof with row operations

[quasi vale]

like showing that the matrix of coefficients is invertible, hence a unique solution

?

[lavish jay]

with this book, yeah, i'm at the beginning

[lavish jay]

yeah, not yet

[quasi vale]

I think ur gonna have to work with some cases here

[lavish jay]

in the answer the book give a hint that use cases

[quasi vale]

ok so they are telling you to use the rank theorem

[lavish jay]

in the case that a = 0 i proof that

i think

i'll take a photo

[quasi vale]

Listen.

Since you have ad - bc not equal to 0

Take first case where ad is not 0 but bc is zero(which implies either b is 0, c is 0, or both. take any case u want)

then take second case where bc is not 0 but ad is 0(which implies either a is 0, d is 0 or both. again doesn't matter, use any case)

third case, where ad and bc are both not 0

using this, you can show that every case gives you rank 2

and then u use the theorem to conclude the proof

[lavish jay]

ok

i think i got ir

I use the same method in all the cases, but if ad =/= 0 and bc = 0 , b = 0
$\linebreak \begin{bmatrix} a & 0 \ c & d\end{bmatrix} \to^{\frac{R1}{a}} \begin{bmatrix} 1 & 0 \ c & d\end{bmatrix} \to^{R2-cR1} \begin{bmatrix} 1 & 0 \ 0 & d\end{bmatrix} \to^{\frac{R2}{d}} \begin{bmatrix} 1 & 0 \ 0 & 1\end{bmatrix}$ And the rank of this matrix is 2, so free variables equals 0, since a unique solution Q.E.D

[stoic python]

ᓇᘏᗢ Guilhotina ᓇᘏᗢ

[lavish jay]

@quasi vale is that right ?

[quasi vale]

Yes

You should also mention that since ad is not equal to 0, both a and d are not equal to 0

Hence the divisions by a and d are justified

[prime snow]

Could someone please explain Riesz representation theorem in layman terms, i've read the definition and the proof and I can see what everything means, but i'm struggling to understand what it's really doing. any help would be much appreciated cheers!

[lavish jay]

At this exercise i need to find the equation of Q, ok , so far i have that s -2t = a-1, s -t = b -2 and -s = c -3. t is the parametric of the equation that describes Q

[whole zodiac]

Is Gilbert Strang's "Introduction to Linear Algebra" better than LADW? I have been doing LADW and just looked at Strang today, Strang seems to have the same algebraic content but has a lot more on the geometry part as LADW does not have much geometry stuff.

[stone edge]

can someone help Use synthetic substitution to find f(4) for f(x) = x4+2x3-3x2-5x+7

[halcyon spindle]

#prealg-and-algebra

[raven badger]

I just gave it a quick glance; But I think you should start with making some sort of parameterization ala:

V *t= X = P + U * s.

Where t and s are parameters.

Then you have a system of linear equations, which you solve for s and t.

Set s and t to these values (or one of them,) if I remember correctly it should give a single point - unless they're parallel.

[solar steppe]

When you have a group of cointegrated serries. If you use the Engle Granger test you use a dependent variable. All Im trying to do is find out which one is it.

[nocturne jewel]

#prealg-and-algebra

[dapper gorge]

Let A be a matrix. There exists an n for which A^n=I. How are this matrices/maps... called?

If they are of any interest

[quartz compass]

I guess I've never seen them called roots of unity in the matrix setting

finite order matrices maybe

[dapper gorge]

they are pretty fun

I guess that name is fine

lol

[tacit pelican]

do yall think it's possible for me to start learning linear alg without a good understanding of proofs and not much experience of actual rigorous mathematics? ive done about almost half of calc 1 and i just want to get started even if it's just like a preview of the subject

[quartz compass]

yeah you'll be fine I think

[worldly bear]

i tool linear algebra the same semester i had an introductory proof class, can’t say the proof class helped my understanding of linear alg at all though

[nocturne jewel]

Proving things is just more common in LinAl compared to Calc1

But anyone can learn basic proof writing in a LinAl course (basic ideas of deductive logic should already be present kinda deal)

[halcyon spindle]

You can look at Loch’s intro proof which is 30 pages. It pinned in #proofs-and-logic

[silver cradle]

would this have to do with some rank nullity theorem stuff? i dont really know how to use it in this case, I think its # of variables = rank(A) + nullity(A) but what is rank(A)?

[wintry steppe]

dimension is rank plus nullity

rank can be seen as number of variables

nullity can be seen as number of equations

[wintry steppe]

whatcha think

[silver cradle]

if rank is the number of variables, then the nullity is 0

[wintry steppe]

nah

nullity is dimension of null space

null space is given as in the photo

rank nullity says

dim = rank + nullity

this is in reference to a given matrix

[silver cradle]

right, and what we want is nullity = rank - dim

[wintry steppe]

ye

dim - rank = nullity

[silver cradle]

im a bit rusty on linear algebra, is how do we determine the dim(A)?

you said earlier that the rank is number of variables

[wintry steppe]

um

a matrix is a linear transformation

so a 27 x 9 matrix

is a linear function from R^9 -> R^27

[silver cradle]

okay

yeah

so the dimension is 27?

[wintry steppe]

uh

rank is dimension of the span of the column space of a matrix

[wintry steppe]

i think so

[silver cradle]

so for A, its 23 - 23 = 0 as the nullity?

following that, wouldnt all of them have a 0 nullity

[wintry steppe]

where are you getting 23-23

[silver cradle]

whoops i mixed it up

so its a 7x23 matrix from R^23 to R^7

so dim(A) = 7

but if dim - rank = nullity and rank can be seen as the number of variables, then that would give a negative nullity

so i did something wrong

[wintry steppe]

uh

dimension is 23

rank nullity is stated like this

let V,W be vector spaces
Let T:V->W be a linear transformation
rank(T)+nullity(T) = dim(V)= n

If you see a 7 x 23 matrix

then we know that dimension is 23 since this matrix is a function R^ 23 -> R^7

what is left is to find nullity and rank

the question in your problem says choosing at random

so its likely that the column space will be linearly independent

my bad lol

there will be intersection

they wont all be linearly independent

Think about this

if you have a 5 dimensional vector space

What is max number of linearly independent vector in a set

[silver cradle]

5

[wintry steppe]

good

So you know matrix notation is Rows by Columns

meaning there are 23 column vectors

[silver cradle]

yeah

[wintry steppe]

now say we transport all of those vectors to a 7 dimensional vector space by a function

So we now have 23 transformed vectors in R^7

we know that this set isnt linearly independent

but by how much is a nicer question to ask

[silver cradle]

ah, 23-7 = 16

[wintry steppe]

by asking how much we are asking the nullity

yeah

[silver cradle]

thats a really good way to think about it, i think that clears it up

thanks 🙂

[wintry steppe]

i tried

[silver cradle]

much appreciated, you did great 😄

[torpid harbor]

Suppose that (V₁, V2, V3) is a linearly independent subset of R". Show that the set (v₁, V₁ + V2, V₁ + V2 + V3) is also linearly independent.

Pls explain

[brittle gyro]

You can use straight-forward definition of linear independence here. Say $a_j$ for $j=1,2,3$ are real constants such that $a_1v_1+a_2(v_1+v_2)+a_3(v_1+v_2+v_3)=0$. By the distributive rule this is the same as saying that $(a_1+a_2+a_3)v_1+(a_2+a_3)v_2+a_3v_3=0$. The hypothesis of ${v_1, v_2, v_3}$ being linearly independent means the constants $(a_1+a_2+a_3)$ , $(a_2+a_3)$ and $a_3$ are all zero. That implies $a_j=0$ for $j=1,2,3$.

[stoic python]

Sydd

[torpid harbor]

Oh right

[supple garden]

I don't understand what c) is asking

This is their solution but I don't really get it..

[brittle gyro]

Basically in this Z/2Z field every even number is a representative of the 0 element, so v1 would be 0v1'-0v2'=0 and therefore could not me the member of any basis

[supple garden]

Oh ok

I was wondering how you'd define e.g. 6*v1' when 6 isn't in the field

Would you say this is a strange/badly worded question

Or in general is it always the case that for Z/nZ we just assume to like reduce modulo n any integer that appears so that it's defined?

[subtle walrus]

in general there is a unique ring homomorphism from Z into any ring

so you will just interpret 6 as the image of 6 under that homomorphism

this basically turns into reducing elements of Z/nZ modulo n

but note that technically 0 or 1 isnt an element of Z/nZ either

and also that Z/nZ is only a field for n prime (since you are doing linear algebra, which requires fields)

[torpid harbor]

If A is a singular matrix

Then is it true to say that transpose of A should also be singular?

[nocturne jewel]

Yes

since det(A)=det(A^T)

[torpid harbor]

Oh

Right

Am so dumb

I literally forgot about that

@nocturne jewel thank u

[hearty rapids]

someone please help me make sense out of this

if she is doing -2R2 + R2

isn't that -2 + 1?

[tranquil steeple]

-2R3+R2

[hearty rapids]

yes thank you sm

i was driving myself insane

you can't multiply a row and add it to itself

[rugged trellis]

how to do i solve this

y = 4x + 2
and
y = 2x - 2

its a system

and idk how to graph it

[nocturne jewel]

if you're graphing lines, better suited for #prealg-and-algebra

cause I doubt you've learned matrices.

[rugged trellis]

why do u doubt it

fuck you

[edgy pecan]

You wouldn't need matrices for such simple set of linear equations

[rugged trellis]

its not simple

[nocturne jewel]

You typically never graph lines in LinAl

But sure, write it as an augmented matrix then invert the matrix.

or apply EROs

[lavish jewel]

why are you being rude to people trying to help you?

[whole zodiac]

set the equations equal to each other

[nocturne jewel]

They're in #prealg-and-algebra

[whole zodiac]

ic

[nocturne jewel]

Cause question is more on the pre-alg side of what LinAl you do in HS

[rugged trellis]

bc they being rude first

[sick sandal]

its quite reasonable to assume you haven't learned matrices with that question
dont take it the wrong way

[thin wing]

He didn't say anything mean, he just said they haven't taught you matrices yet. It's like if I said I doubt you know Homological Topology, it's a fact not an insult. This seemed like a weird misunderstanding.

[nocturne jewel]

Given you're commenting on graphing it to solve it, that is a HS way of solving stuff, usually ~grade 10. Grade 10 isn't 1st year LinAl.

[lavish jewel]

pixels reply was definitely unwarranted and out of proportion, but do try to be kinder/less negative with your wording, since that also egged them on (regardless of whether you were correct or not with your observation).

[frosty vapor]

the reply was not to you edd seems to have clicked on you by accident

im dumb i cant read

ignore me

[hearty rapids]

did he just curse at you

[nocturne jewel]

Yeah I was about to say Metal

[hearty rapids]

set them equal to each other

[gray dust]

nothing happened

[gray dust]

its moved to prealg, pls keep it there

[hearty rapids]

oh ok

[lavish jewel]

it's ok, pixels got help on another channel, don't summon them back here

[nocturne jewel]

People tried to help in pre-alg, however they gave up

[rugged trellis]

i gave up wdym

[nocturne jewel]

#prealg-and-algebra message

[thin wing]

He thought you meant they gave up, he was clarifying.

[main walrus]

#help-4 message continuing here the topic

[dusky wadi]

isn't this undefined

[nocturne jewel]

Nope

oh wait

yes

4x3 and 4x1 dont work

[dusky wadi]

ok

sneaky treil

[dusky wadi]

i found a different method to do matrix multiplication

if you have let's say

$\begin{bmatrix}a&b&c\d&e&f\end{bmatrix} \begin{bmatrix}x&u\y&v\z&w\end{bmatrix}$

[stoic python]

Alison40

[dusky wadi]

split the second matrix into $\begin{bmatrix}x\y\z\end{bmatrix}$ and $\begin{bmatrix}u\v\w\end{bmatrix}$

[stoic python]

Alison40

[dusky wadi]

then calculate $\begin{bmatrix}a&b&c\d&e&f\end{bmatrix} \begin{bmatrix}x\y\z\end{bmatrix}$ and $\begin{bmatrix}a&b&c\d&e&f\end{bmatrix} \begin{bmatrix}u\v\w\end{bmatrix}$

[stoic python]

Alison40

[dusky wadi]

and finally compose the resulting vectors back into a matrix

ending up with $\begin{bmatrix}ax+by+cz&au+bv+cw\dx+ey+fz&du+ev+fw\end{bmatrix}$

[stoic python]

Alison40

[dusky wadi]

is that a good method

for doing it

[nocturne jewel]

that's not a different method

[dusky wadi]

oh

nvm then

sorry

[wintry steppe]

matrix multiplication is only defined when the amount of rows in matrix 1 = amount of columns in matrix 2 and vice versa

Wait the messages didnt load for me rip

[lime zinc]

If A and B are two positive definite matrices , then what can we say about their product AB and ABA?

[zinc timber]

AB may not be symmetric (AB)'=B'A' =BA ≠? AB

if AB is symmetric then AB=BA => they are simultaneously diagonalizable => eigen values of AB are product of evs of A and B (in some order) => ev of AB are >0 => positive definite

@lime zinc

[lime zinc]

but in answer this is not positive definite.

[zinc timber]

in general, no they aren't

[lime zinc]

but when we talk about definitiness of matrix we consider that matrix is symmetric?

[zinc timber]

if matrix isn't symmetric, we can decompose it into symm part and anti symm part. for antisymm matrices x'Ax = 0, so only the symm part matters. that's why we take A to be symmetric to begin with

[lime zinc]

one more part how ,AB=BA?

[zinc timber]

if AB is symmetric, use (AB)'=AB

[lime zinc]

okk (AB)' =B'A' = BA= AB

[zinc timber]

yes

[lime zinc]

this one

[zinc timber]

I would have showed you an example but I don't have any

also g2g

[lime zinc]

its ok, so the main thing is If given AB is symmetric then we can say AB is positive definite , other wise we can't say anything?

[zinc timber]

@lime zinc

(notice that AB is not symmetric)

[lime zinc]

Thankyou

[rotund verge]

hi

can someone help me in this

[rotund verge]

<@&286206848099549185>

[zinc timber]

#odes-and-pdes

[jolly shale]

what's the requirement for matrix multiplication? was it "row size of first matrix must equal to column size of second matrix" ?

[zinc timber]

$m\times \fbox{\red{n\quad n}} \times k$

[jolly shale]

huge thanks

and the bot is very cool btw

[hushed hedge]

Just wondering what is the usage for matlab? Are you supposed to use it while solving problems or?

[lavish jewel]

you use it to solve large problems

where large is anything that would take you more than 30s ~1min on paper

[hushed hedge]

Alrighty

Can you use it while reading the text book? Are there any benefits ify what i mean

Like sort of complementary to the text book

[lavish jewel]

you can but it won't teach you anything that's in the text book

think of it as a fancy calculator

[hushed hedge]

Fair fair

[lavish jewel]

it can help you see if you understand procedures, if you code the procedures yourself

but you can do that in any programming language you like

the whole point is that matlab already has those algorithms built in

[hushed hedge]

The book i have is not so good since it's really old, i just realized that i can use matlab for some of the stuff they go through and use it to confirm if i understand something

[lavish jewel]

it won't confirm if you understood stuff

it can confirm if you got the result right

[hushed hedge]

yeah ig

[fickle citrus]

MatLab is terrible paid software. You can go for any FOSS-alternative without issue

[lavish jewel]

i wouldn't say it's terrible, but it's definitely expensive

[hushed hedge]

It's free for us

[lavish jewel]

the uni probably provided you licenses

[hushed hedge]

But it's really painful i have to say

[hushed hedge]

yea

Copy + paste for assignments

[lavish jewel]

matlab and python are both easier than proper programming languages though

and it's in your best interest to learn how to code at some point

[hushed hedge]

no way

[fickle citrus]

Don't rely on academic editions unless you think you'll work in a company that will pay for it, and it's even worse if you do not know what features MatLab has that alternatives do not which let you say you 'require' MatLab specifically than any alternative

[hushed hedge]

i find matlab super hard

i found java easier to learn

And i have zero experience in coding

[lavish jewel]

[hushed hedge]

actually

[fickle citrus]

Java is terrible

Though that is no longer #linear-algebra

[lavish jewel]

i disagree with everything shattered is saying

[fickle citrus]

Anyway, numpy will be sufficient for 99.98% of all cases

[hushed hedge]

matlab is easier but i find java a bit better idk

[fickle citrus]

Kotlin is better. Not sure what LinAlg algos are in Kotlin/Java though

I know most standard things are in the scipy numpy etc package

[lavish jewel]

the comparison doesn't even make sense to begin with, java is a programming language

matlab is for maths scripting

[hushed hedge]

They teach java to math students in my uni

• matlab

[fickle citrus]

Well I had to convert MatLab script to C++

[lavish jewel]

sure, but for programming, not to do math

[fickle citrus]

And it wasn't too bad

[hushed hedge]

yeah

[fickle citrus]

Tell your uni to update to modern things

If your professors aren't old they won't be that tied to MatLab/Java

[hushed hedge]

yeah no they won't

[fickle citrus]

Then ignore them and learn Julia

[hushed hedge]

shit uni lmao

[fickle citrus]

It's more than enough for standard LinAlg/any standard math

[lavish jewel]

matlab is still modern, tbh, it's just niche because it's used only by companies that pay for the expensive license

[fickle citrus]

The only thing I don't know if Julia can do is proofing/lean/coq but that's a separate thing by itself also

[hushed hedge]

haskell

[lavish jewel]

it's also faster than numpy

[hushed hedge]

that's where the fun is at

[lavish jewel]

but python has a much larger user base by now

[fickle citrus]

Python is always slow but that's just the way things go

[lavish jewel]

in spite of all its shortcomings

[fickle citrus]

The primary reason is that it is, to be laconic, child friendly

[hushed hedge]

Learn java then jump onto python

💯

[fickle citrus]

Everything that isn't C++ is just bad

[lavish jewel]

also note that the python notation for maths stuff is almost identical to matlab

so good luck

[fickle citrus]

They probably got inspired due to its influence

I think even Julia would be influenced

I mean MatLab is just very dominating

[lavish jewel]

yep, just pointing it out because pesthaio is shitting on matlab

but they can't escape it

[fickle citrus]

And it's difficult to convince people to jump if it is too different

[lavish jewel]

julia and python look the same

[hushed hedge]

i'm not shitting on matlab

[fickle citrus]

Julia is definitely more math-friendly than Python

[hushed hedge]

i just say i personally don't like it, i found it difficult, or not difficult but rather slow and unorganized.

[lavish jewel]

nothing is going to be as fast as typed, compiled languages

except assembly

[hushed hedge]

But i can definitely see that it has some uses. I quite liked some of labs we did, they were quite cool actually

[lavish jewel]

the idea is that it's higher level, in exchange for speed

so the stuff is already coded in

rather than coding a whole eigenvalue decomp yourself, you just eig(stuff)

and luckily for you, the matlab documentation is good

[hushed hedge]

i think the quite opposite in terms of speed, but perhaps because i don't know matlab that well

But it's a good program to use if you have a aight text book

[lavish jewel]

textbook for what?

[zinc timber]

octave

[lavish jewel]

for the language?

[zinc timber]

yes

[hushed hedge]

My text book skips over some steps because they think everything is 100% crystal clear, i can now use matlab to check if i understand what they mea n

[hushed hedge]

linear algebra book

[lavish jewel]

oh, well

matlab has nothing to do with your textbook, so

[zinc timber]

like I use matlab/julia to verify matrix related facts, like multiplying 2x2 matrices

[lavish jewel]

reading a linalg book won't teach to you to code in any language

that's on you

[hushed hedge]

no but i'm saying i can use it to confirm solutions

[lavish jewel]

you could, but usually solutions are also on the back of the book

if your issue is the book, the solution isn't matlab, it's using a different book

[zinc timber]

[fickle citrus]

Not a good enough replacement I think, or at least I tried it once and tried to run on .M and it failed spectacularly

But anyway I agree with everything said about books: if a book is not helping, try an alternative. There are way too many LinAlg books out there and there is high probability at least one is a good fit

[winged prairie]

what about julia

nvm already mentioned

[abstract vale]

Question 1a)

[dusky epoch]

what's $\varnothing(u,v)$?

[abstract vale]

How can I show this is a unique representation

[stoic python]

Ann

[abstract vale]

I linear form

[dusky epoch]

why are you denoting it with the empty set symbol

[abstract vale]

Bilninear*

Idek it was from part of an answer I found on the net

But ignoring that

How do I show its unique

[dusky epoch]

well you can let $\phi = s + a$ where $s$ is symmetric and $a$ is asymmetric

[stoic python]

Ann

[dusky epoch]

and by considering $\phi(u,v)$ and $\phi(v,u)$ you can show that $s$ and $a$ must be what you say they are just by simple algebra

[stoic python]

Ann

[abstract vale]

Oh are you going by my notation in what you wrote?

[dusky epoch]

well i can only assume that the empty set symbols are actually a fucked up phi

[abstract vale]

Yeah 🤣

[abstract vale]

This 1b) have I answered it correctly?

[lavish jewel]

,rotate 180

[abstract vale]

How can I improve my answer if possible

Sorry for the image pic 🥲

[lavish jewel]

wow wtf

[abstract vale]

Yep got it 🤣

[lavish jewel]

in what universe is that 180°

[abstract vale]

Crazy

Is the answer alright?

[nocturne jewel]

"<v,v>=0 if <v,v>=-<v,v>" reads more like you assumed skew symmetry then showed alternating, which is the opposite of what you did in that part of the iff proof

at least to me

[abstract vale]

But with the above formula above that doesn’t it show skew symmetry?

Is there anyway I should rewrite it then for more clarity

[gray dust]

doesnt look like an argument of alternating implies skew symmetric

[nocturne jewel]

The statement above doesn't say much to show that <u,v>=-<v,u>

[gray dust]

recall skew symmetric means <x,y>=-<y,x> for all x,y

[dapper gorge]

The main motivation of the determinant is to tell if a map/matrix is singular or nonsingular. Is this correct?

[zinc timber]

<,> is an iner product?

[nocturne jewel]

yes

[zinc timber]

in the above context

[gray dust]

its not the only one. for example we also want the oriented volume of the parallelepiped created by a set of vectors

[abstract vale]

How do I incorporate that from what I’ve already written

[nocturne jewel]

You can't from what you have written, cause what you have written only includes the vector v

[abstract vale]

Oh then what should I add

[nocturne jewel]

another vector in a smart way

hint: it's bilinear

[lavish jay]

I'm using three significant digits but i'm getting the right result not the wrong one

if the result of one calculation is -18795 should i round it to -187 or -188 ?

still i'm getting y = 1 and not 1.01

[tacit lava]

The ratio of wins to losses for a player in a tournament is 16:5. There
were 147 total games played. How many games did the player win?

[nocturne jewel]

#prealg-and-algebra

[tacit lava]

ok

[abstract vale]

I’m guessing <v,v>+<u,v> ?

[nocturne jewel]

aka <u+v,v>

[abstract vale]

Yeah I did that and equate to <u,v>

Is contradiction the path to go by ?

[lavish jay]

ok, i should also round things in the mid calculations, ok still don't get the right answer, but this distortion is caused by the rounding up i have to make when the last digit is 5 and because I'm missing information?

[abstract vale]

Or is there something much simpler

[nocturne jewel]

consider <u+v,u+v> and apply bilinearity

[lavish jay]

How can i prove that isn't just one operation, here the book is counting as an operation just multiplication and division??

[abstract vale]

I ended up with this after that hint

.rotate

[nocturne jewel]

what would that 1 operation be?

,rotate

[lavish jay]

dividing the first row by a1 , so i ended up with a leading one

[nocturne jewel]

yeah I think they mean operations not row operations

[nocturne jewel]

<u+v,u+v>=<u,u+v>+<v,u+v>=<u,u>+<u,v>+<v,u>+<v,v>

idk where all those 2s came from

[abstract vale]

Oh I thought I had to add <u+v,u>+<u+v,v>

[nocturne jewel]

that's applying linearity of one of the entries

however you need to do it in both

[abstract vale]

Ah I fumbled

I applied linearity twice

[abstract vale]

How did you know to consider <u+v,u+v> btw

Like what came to mind

[nocturne jewel]

cause it's bilinear so ik I can expand that to get stuff, and it's assumed alternating so ik it equals 0

[abstract vale]

Oh fair

Thanks

But would my if skew symmetric argument be fine?

The one I did before

[lavish jay]

i still don't understand what should i do then, since row operations is the only way i can think to get the pivoting one

[atomic kernel]

Hello! anyone can shed some light on the power of a jordan block with 2 distinct eigenvalues along its diagonal?

[brittle gyro]

a jordan block has only one eigenvalue per definition

[abstract vale]

Need help with the end of 6c

To show direct sum

Is showing that every v in v can be uniquely written as the sum of these two what I should aim for?

Or should I be doing something else

[gray dust]

sure. u can also show uniqueness by showing U cap U^perp={0}

[lavish jay]

so a spaning set of R^n has infinitely of "v" ?

[dusky epoch]

no

[lavish jay]

oooo, i think i understand now, i read it wrong

if span(S) = R^2 that means that we have R^2 of linear combinations of v1,v2... vk ?

[nocturne jewel]

yes

if span(S)=R^2, then any vector in R^2 can be written as a linear combination of vectors in S

[lavish jay]

R^2 is the Cartesian plane , right ?

only 2 dimensions

[nocturne jewel]

yes.

R^2 is a 2-dimensional space

[lavish jay]

Can i say by this definition that all set in R^n are linear depend of the span(e1,e2...en) where e1 = [ 1, 0 , 0 ,... 0] e2= [ 0, 1 , 0 ,... 0] and so on?

[lavish jewel]

hmm your wording is a bit off, but you have the right idea

they way i would write it is as "any vector in R^n is in the span of {e_i}" or "... can be written as a linear combination of the vectors e_i" or "any set of vectors {v, e1, ..., en}, with v and e_i in R^n, is linearly dependent"

[lavish jay]

ok thanks 💙

[pine shadow]

no clue

[oblique prairie]

this isn’t linear algebra

go to #prealg-and-algebra

[lavish jay]

In this matrix $\left[\begin{array}{ccc|c}
1 & 2 & 3& 4 \
5 & 6 & 7&8 \
9&10&11&12
\end{array}\right]$ in each collum i've the following pattern a2 = a1 + 4

[stoic python]

ᓇᘏᗢ Guilhotina ᓇᘏᗢ

[lavish jay]

where a2 is the second term of the collm

and a1 the first one

is there any matrix property that can help to transform this in RREF ?

wait that means that at row 2 and 3 are equal

and one is 0 0 0

since we consider row 1 as a1 in the progression

row 2 is 2 a1

and row 3 2 a3

is that correct?

[wintry steppe]

does this hold for all square matrices "A" or just 2x2 ones?

[nocturne jewel]

any matrix is a solution to it's own characteristic equation

but that's the char poly for 2x2 matrices

[wintry steppe]

oh oh right,, thank you

:D

[viscid lagoon]

$\text { Let } m \in \mathbb{N} \setminus{0}, n=2 m \text { and } \ U:=\left{u=\left(u_{1}, \ldots u_{n}\right) \in \mathbb{R}^{n} \mid \sum_{k=1}^{m} u_{2 k}=\sum_{k=1}^{m} u_{2 k-1}\right} \text {. }$ \
How do I find a basis for $U$?

[stoic python]

lewis

[zinc timber]

no

$A^2 = \tr(A)A - \det(A)I$

[dusky epoch]

$\tr AA$ \catthink

[stoic python]

Ann

[zinc timber]

how did u do \catthink

[dusky epoch]

it's part of texit now i think

try \catthink and \thonk they should work for you too

[zinc timber]

$.$ \catthink

[zinc timber]

nice

[dusky epoch]

you can do it inside mathmode too

$\catthink \thonk$ \catthink \thonk

[stoic python]

Ann

[zinc timber]

[quartz compass]

let $\thonk: \catthink \to \catthink$ be a continuous linear endomorphism of $K[X]$

[stoic python]

Merosity

[wintry steppe]

Is there a way that we can at least get a bit of idea of what will we obtain if we graph an equation in 4 variables and as well as the soln of system of some 4 variable equations?? like we know that in 3 variables an eqn is a plane in R^3 and the soln, is either a point or a line for the system of equations. So, what can we say about 4 variables?

[lavish jewel]

it follows the same pattern in higher dims, but there is no easy way to "visualize" it

the equation will be a "hyperplane", and solutions to systems of equations will form some sort of geometric object at the intersection of the hyperplanes

or maybe "flats" is preferred over hyperplanes, since they don't necessarily contain the origin

the pattern for hyperplanes is, a point on a line for 1D, a line on a plane for 2D, a plane in a cube for 3D, a cube in a ???? in 4D, a ???? in a ?????

and so on

[wintry steppe]

Alright, thanks

I have another doubt as well, how can I prove that there are only three conditions that could possibly result in the soln of a system of linear equations, that is either the solns are infinite or unique , or there is no solution at all?

[lavish jewel]

the case where there is no solution is simple enough

for the other two, you would use linearity and the null space

the null space of a linear operator (in your case a matrix) is the set of vectors x such that Ax = 0

[wintry steppe]

but how can we say that nothing else can happen to the solutions mathematically, instead of just proving it by common sense

[lavish jewel]

let's say now that there is a vector y such that Ay = b

by linearity, A(x+y) = Ax + Ay

and this is equal to 0 + b = b

similarly, A(cx) for some scalar c is equal to c(Ax) = c0 = 0

if any nonzero vector x exists, then Ay = b has infinitely many solutions

since A(y + cx) = b for any value of c

[lavish jewel]

idk what you even mean by this

i was giving you mathematical arguments

[wintry steppe]

we say that only 3 conditions are possible(as you explained), I can't understand that how we reached that conclusion that only 3 conditions are possible, how can we state that there is nothing else that can be a case other than the these 3.

[lavish jewel]

i just showed you how (in the cases where the underlying field is R or C, and when we're dealing with finite dimensional vector spaces)

as soon as there is more than one solution, there are infinitely many by linearity and homogeneity of scalar mult

[wintry steppe]

ohkkk

[wintry steppe]

finite dimensional vector space means?

[lavish jewel]

not infinite?

like R^10, for example

10 dimensions

[wintry steppe]

is it defined some other way for the infinite dimensions??

[lavish jewel]

don't worry about that for now

[wintry steppe]

Alright, thanks

[wintry steppe]

how do we prove that the no of solns won't be a finite no >1 ? (By proving that there are infinite and unique solns only?? and thus thinking of them as the only conditions, or there is some independent reasoning for this?)

[zinc timber]

say you have two solution x1, x2 of Ax=b, take x=2x1 - x2

[lavish jewel]

because you can just take a scalar multiple of the difference of the two solutions, and that will be in the null space

as ryu says

if Ax1 = b = Ax2, then Ax1 - Ax2 = 0, and A(x1 - x2) = 0 by linearity

now we let w = x1 - x2

A(x1 + w) = Ax1 + Aw = Ax1 + 0 = b + 0 = 0, this gives us one more solution

but A(cw) for any real c is equal to cAw, and Aw = 0, so Acw = 0

this means any vector cw is also in the null space

and c is any real number

so there are infinitely many vectors of the form cw

like 1w, 2w, 1.1w, 1.1111111111.... w

[dapper gorge]

If we consider rotations by an angle 0<x<pi in R^2 there are no real eigenvalues. However, if we allow eigenvalues to be complex, since C is an algebraically closed field, then there must exist (complex) eigenvalues. Can you interpret this geometrically?

what is the geometric interpretation then?

[lavish jewel]

it's very difficult to interpret geometrically, since visualizing vectors in C^2 requires 4 axes already

just abstractly, there are vectors in C^2 that won't "change direction", but only be scaled by a complex scalar, i.e. the corresponding eigenvalue

but scaling by a complex number is already a weird thing

[dapper gorge]

yeh

btw

[zinc timber]

like there's one way to interpret it, rotation matrices are of the form $\m{a & b\-b & a}$, there is an isomorphism of these matrices with the complex number, sending it to $a+ib$. since multiplying by complex number gives you rotation and scaling in complex plane, similar effect happens in $\mbb{R}^2$ as well

[zinc timber]

if you consider R^2 as the complex plane

[lavish jewel]

that's limited to rotations on real vectors tho

but yeah, for the real case, you can visualize it as turning it into an operation equivalent to multiplying 2 complex numbers

[dapper gorge]

what's the precise definition of rotation of vectors in R^n ? I suppose it means that a transformation is a rotation if you can fix a plane (2-dimensional) such that the transformation acts on all vectors in that plane just as a rotation would act in R^2. So similarity gives a nice form for such transformations.

Is this correct?

[zinc timber]

yes,

isn't it what they were asking?

[lavish jewel]

idk, i interpreted their question as wanting an interpretation for rotations in C^2

since they wanted to deal with the complex eigenvalues

[zinc timber]

I thought the question was rotation in R^2, allowing the eigen values to be complex (i.e. use C to find the roots of char poly)

[lavish jewel]

sure, but R^2 over C isn't a vector space

idk

maybe i'm overthinking it

[zinc timber]

for C^2 there's really no easy way to interpret the rotations, but maybe you can look at the primary decomposition then use the same argument, idk

[lavish jewel]

as for your other question, tybu, i think the usual way is to compose rotations on a plane along an axis

so 2D rotations

[dapper gorge]

ok

ty

ok this might sound stupid

Are there generalizations of rotations? In the sense that these rotations are 2 dimensional

maybe I'm not understanding idk

[lavish jewel]

you can yield an arbitrary rotation in many dimensions by doing several 2D rotations

beyond that, you can look at orthogonal groups

[dapper gorge]

cool

But, I suppose composition of 2d rotations yield a 2d rotation along a single axis

[lavish jewel]

no

[dapper gorge]

mb

I'll look into it

[lavish jewel]

well

yeah, the net effect, yes

[dapper gorge]

yeah

[lavish jewel]

but rotation is by definition given along an axis

[dapper gorge]

yeah, which defines a transformation in a plane

so a generalization being defining some motion on an n-surface which cannot be reduced to a 2d motion

in R^m

for simplicity

[lavish jewel]

ah, the overall transformation does not boil down to a single 2D rotation along one axis, that was my bad

but it is composed of many such 2D rotations

[dapper gorge]

oh cool

thank you for correction

[hushed hedge]

If you have a equation like in the picture, does this mean that the equation has a trivial solution or not?

I.e if you let 0 = t then you can solve for something like this:
x = t
y = -2t
z = t
is this considered a trivial solution?

Actually nvm, i found the answer in my notes :p

[dapper gorge]

It's a parameterization of the solutions

Usually, we call a trivial solution to be (x,y,z)=(0,0,0)

[hushed hedge]

yeah, i saw it in my notes, non-trivial solution is if you have for example 0 = 9 in the last step

[lavish jewel]

careful, that means there is no solution

[hushed hedge]

noted

[delicate jetty]

When we say that the set of matrices is over a field, is that field the codomain that the domain of the ordered tuples m x n are sent to through the matrix function M?

So for example if we had the identity matrix

The function would be M: [n] x [n] -> {0, 1}

So the matrix would be annotated as $M_{n \times n}( {0, 1 } )$?

[stoic python]

Forsaken

[lavish jewel]

the set {0,1} with usual addition and mult is not a field

[delicate jetty]

It isn't?

I think it is

[lavish jewel]

1+1 is not in {0,1}

[delicate jetty]

Well

You can assume 1+1 = 0

Right?

[lavish jewel]

that's why i said usual addition and mult

if you mean your field is Z/2Z, sure

it's not enough to just write {0,1}, because a field is tied to its operations

[delicate jetty]

I understand

I just want it for notation, nothing more than that

In the context of linear algebra this is assumed to be the case

Meaning, the identity matrix has as its field {0, 1}

[lavish jewel]

i guess one can argue that the finite fields with 2 elements are isomorphic though, so maybe it's ok?

if you really want to work with Z/2Z, yes

if you want to use the same identity matrix for M_nxn(R), then the field is R

[delicate jetty]

You mean that all tuple finite fields are isomorphic to each other therefore we could say that they construct Z/2Z or have I missed the point completely?

[lavish jewel]

the ones with 2 elements, yeah

[delicate jetty]

I see

Thank you both

[lavish jewel]

at any rate, the field would also depend on the other matrices in the set you're considering, and it will impact the operations you can do with the matrix

[zinc timber]

'both'?

[lavish jewel]

they knew you'd show up

[zinc timber]

[delicate jetty]

There was another person writing, didn't realize they had deleted what they initially sent here

[molten pilot]

Is there an intuitive way of thinking about this or do I just have to verify it computationally?

[wintry steppe]

det is alternating

when thought of as a function of 3 (in this case) vectors

in each equality you're making two swaps, so the minus signs you pick up cancel out

[hushed hedge]

x(1,2,1)+y(2,3,3)+z(-1,-2,-1)=x(0,0,0) are these 3 vectors linearly dependent or indepnedent? Looking at the picture we see that our equationsystem has a trivial solution, therefore it's dependent, we can let 0 = t, then we have: x = -3t, y = t, z = t.

In other words if the equationsystem has a **trivial **solution then it's dependent if not then independent, right?

[lavish jewel]

having a trivial solution to a homogeneous system says nothing about whether it's dependent or independent

[hushed hedge]

tbh i don't think i understand what trivial actually means anymore

what i basically mean is that

in our last step where we have 0 = 0

[lavish jewel]

that is not the same as trivial

trivial means that there is a solution of the form x = y = z = 0

[hushed hedge]

we can let 0 = t as we said before, then this means out equation system is dependent

[lavish jewel]

you're mixing up 2 different things

[hushed hedge]

meaning that x, y, z are the same number?

[lavish jewel]

not only that, but that they are equal to 0

it's called trivial because this is a solution to any linear homogeneous system

[hushed hedge]

yea, that's what i'm doing. But i can't seem to understand it now, i've overcomplicated things for myself

[lavish jewel]

your 3x4 matrix represents a system of equations

or a matrix equation of the form Ax = b

[hushed hedge]

ok and this is not where we let 0 equal to t?

[lavish jewel]

if the last column of the 3x4 matrix is 0, which is equivalent to letting b = 0, so that Ax = 0, we call this a "homogeneous system"

all systems of this form have a trivial solution: set all the variables to 0

[hushed hedge]

aha okok

[lavish jewel]

completely unrelated to this, you found out that, through row reduction, one of the rows turns into 0 0 0 0

which is equivalent to having 0 = 0

this means you have a "free parameter"

this happens when the rows of the matrix are linearly dependent

[hushed hedge]

yeah that's what i mean when i say 0 = t, go on 🙂

yup

gotchu

i gotchu

[lavish jewel]

you can then call the free parameter "t" if you like, this is common in books

[halcyon spindle]

If the trivial solution was the only solution to a homogeneous system then it would be linearly independent?

[lavish jewel]

yes plega

[halcyon spindle]

Ok thanks.

[lavish jewel]

when all the rows/equations are lin indep and the system is homogeneous

the system only has the trivial solution

when the eqs are lin dep, then you have infinitely many solutions

given by the parameters you end up with

it could be 1 or more

as you noted in your example, pesthaio, parameterizing with this new variable t, and then setting t = 0, is equivalent to setting all the variables to 0

at least in this case

that is in general not true

[hushed hedge]

yeah

but if we let t be 2

like so now we figured out that's it's lin dep

sorry for my english i cant write anymore

[lavish jewel]

no problem, you've been very clear so far

[hushed hedge]

we can describe it liek this x=-3t, y=t,z=t

yeah ok i understand 19eddy4, i can't really write anymore. I have been reading about this for 4 hours and my brain is not working anymore

But i understand, tyvm. i'll write it down in my notes 🙏

[lavish jewel]

right

and then the idea is that t can take values other than 0

in fact, t can be any real number

[hushed hedge]

yes exctatly

[lavish jewel]

so this system has infinitely many solutions

[hushed hedge]

i can actually show you what i mean instead of writing

[lavish jewel]

feel free to post it here, but i have to go eat :x maybe someone else can help you out further

[hushed hedge]

This is basically what i mean, this describes our vectors, like how they are dependent on each other

[hushed hedge]

enjoy, thank you for the help!

[zinc timber]

they are dependent because u can write one in terms of the other two

[hushed hedge]

yes exactly

[brave lintel]

Hey y'all, im doing axlers la rn, and I've been told that the first 5 chapters are the contents of a course. I'm teaching it to myself, would it be better to get through the entire textbook or just move on to a real analysis or algebra book

[wintry steppe]

depends on if you want to or have to read the rest

[brave lintel]

I don't know how useful the latter part of the book will be for other bits of maths

[wintry steppe]

that's the inner product stuff, right?

that's incredibly important

[brave lintel]

ill find the contents page

[wintry steppe]

inner products and jordan form stuff. also where he finally introduces the determinant

[brave lintel]

idk what happened there

lol

[wintry steppe]

they are

[brave lintel]

mkay

gonna spend like 5 months on this textbook then lols

ty tterra

I have no idea what the paragraph which I've put in brackets is trying to say. Can someone give a concrete example of what axler is trying to say?

,rotate

[wintry steppe]

is F = R or C for axler?

[brave lintel]

yea

[wintry steppe]

because what he's saying is false for F = Z/2Z

ok

he's saying that if $$a_nx^n + \cdots + a_1x + a_0 = b_m x^m + \cdots + b_1x + b_0$$ as polynomial functions $F\to F$, then $n = m$ and $a_k = b_k$ for all $k$. his argument is that their difference is a polynomial function which sends everything to zero, so its coefficients are also all zero. but its coefficients are just the differences $a_k - b_k$

[stoic python]

TTerra

[brave lintel]

that seems very obvious now lol, thanks

🙂

[wintry steppe]

the crucial fact is that a polynomial function which is zero as a function F -> F is also zero as a polynomial

[brave lintel]

mhm

[wintry steppe]

you can see it's true when F = R or C in a few ways

one way is to use calculus, just differentiate the polynomial a bit and get formulas for the coefficients

not reliable in fields other than R or C though

[hard drum]

e.g. nth degree poly has at most n roots so the difference must be identically 0

[wintry steppe]

if F is the field Z/2Z though, think about p(x) = x^2 + x.

[brave lintel]

I don't know what Z/2Z is

[wintry steppe]

that's fine

you can come back to this example when you do

just know that the thing about polynomial function = 0 -> polynomial = 0 is false for some fields

[brave lintel]

Okay, gotcha

[wintry steppe]

axler does functional analysis so he lives in a bubble where the only fields are R or C

[brave lintel]

Axler says that he proves it in chapter 4, so I'll wait until then because this is very intuitive for F = R or C

lol

thanks tterra

[molten pilot]

And why is that?

[wintry steppe]

definition

sometimes as "the unique alternating multilinear..."

in which it's built right in

no matter your definition of the determinant it will be true

and it's either defined that way, or a short computation away

[molten pilot]

I haven't reached that definition of determinates yet

I guess I should just be content with checking it computationally for now

[wintry steppe]

you can convince yourself of it geometrically too

for v and w in R^2, det(v, w) is the signed area of the parallelepiped they span. computing det(w, v) instead is like flipping the orientation and taking the signed area, so you should pick up a minus

something along those lines

(you're working on R^3 but i picked R^2 so i can type less)

[molten pilot]

Understandable, lol

Thanks for the insight!

[pastel brook]

What is e^I, where I is the identity matrix? I tried proving it using Maclaurin Series Expansion and got e^I = eI, but I feel as though I may be doing something wrong. Is this correct?

e 0 0
0 e 0
0 0 e

Wolfram alpha gives eI with the zeros replaced by ones:

e 1 1
1 e 1
1 1 e

any tips?

[lavish jewel]

how exactly did you do the expansion

[pastel brook]

here's what I've written so far:

[lavish jewel]

that seems all correct

[pastel brook]

I'm not sure if it's wrong, because

[lavish jewel]

my guess is wolfram took elementwise exponentiation

[pastel brook]

dang, wolfram alpha is wrong

I feel betrayed

[lavish jewel]

like so

[pastel brook]

Thanks for the help!

got it, thanks a bunch!

[lavish jewel]

👌

[wintry steppe]

to get WA to do the matrix exponential you usually need to type out the whole thing

matrix exponential [your matrix]

[pastel brook]

icic

[pastel brook]

yep, that did the trick!

[dapper gorge]

If t and g are two linear transformations defined on equal dimensional vector spaces with finite dimension and their minimal polynomial are equal, are t and g similar?

In the sense that their matrix representations are similar

[wintry steppe]

no, it's not true

$$\begin{pmatrix} 1 & 1 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 1 \end{pmatrix} \text{ and } \begin{pmatrix} 1 & 1 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix}$$ each have minimal polynomial $t^2$, but they are not similar since the geometric multiplicities of $\lambda = 1$ differ

[stoic python]

TTerra

[wintry steppe]

@dapper gorge

[dapper gorge]

their minimal polynomial is not t^2 tho

[wintry steppe]

(t - 1)^2, sorry

i had a slightly different example on paper that had t^2 instead

but switched it to 1's in typing so the jordan blocks would be clearer

[dapper gorge]

thank you

ok I see

[dark brook]

Would the dimension of an matrix N(A) not just be the number of linear independent vectors ?

[wintry steppe]

it'd be the largest number of linearly independent vectors from N(A)

[dark brook]

So if I had 4 pivotelements (4 linearly independent vectors) the dimension dim N(A) would be 4?

[dapper gorge]

N(A) is the null space of A or just a matrix?

I don't think dimension of matrices is defined ? but idk. You are probably talking about a vector space ?

[dark brook]

Yea its the null space N(A), and I want to find the dimension of N(A)

[dapper gorge]

TTerra already answered you mate

Is this proof correct?

I'm somewhat tired, so sorry if the writing is a bit dense

idk if it's just bullshit

The condition of the proposition is very strong obsiously, but with this we can easily proof stuff about cannonical forms of nilpotent maps

ok

I meant the dimension

not the null spaces

now fixed

Is this correct?

[zinc timber]

then it's correct yes, as said in the proposition

one easy way to verify this is N² sends all the jordan 2 blocks to zero, Nⁿ sends all the n blocks to zero

[zinc timber]

[dusky epoch]

why are you using the shitty geq

[rapid prism]

lmao i just call that the Russian geq

[dusky epoch]

yes and it sucks

[zinc timber]

$\ge \geq$

[zinc timber]

ik it sucks

,tex \rotatebox{90}{{\sffamily IV}}

[oblique prairie]

improvise adapt overcome

[viscid lagoon]

For $v=\left(-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3}\right) \in \mathbb{R}^{3}$ and the dot product $\langle\cdot, \cdot\rangle: \mathbb{R}^{3} \times \mathbb{R}^{3} \rightarrow \mathbb{R}$ such that $\left\langle\left(v_{1}, v_{2}, v_{3}\right),\left(w_{1}, w_{2}, w_{3}\right)\right\rangle:=v_{1} w_{1}+v_{2} w_{2}+v_{3} w_{3}$.
Define the linear mapping $f: \mathbb{R}^{3} \rightarrow \operatorname{span}(v), ; f(x)=\langle x, v\rangle v$.
What would the kernel of $f$ look like? What I have is: $\left{(-(x_2+x_3), x_2, x_3) \mid x_2, x_3 \in \mathbb{R}\right}$

[stoic python]

lewis

[viscid lagoon]

im just not sure if that's the best form of providing the solution, correct me if my solution is incorrect.

[lavish jewel]

i would go with something that makes the dimension easier to see at a glance

but that does seem fine

[viscid lagoon]

i mean

shouldn't the dimension be obvious

what would you have chosen here?

[lavish jewel]

sure, that's why i said easi"er"

i would've gone with span{[0,1,-1],[1,0,-1]}, probably

[viscid lagoon]

oh like that

yeah, ofc

[lavish jewel]

or something orthonormal

but what you had was already correct, too

[viscid lagoon]

yeah, ofc

span does seem easier

you're right

[viscid lagoon]

doesnt it have to be

span{[1,-1,0],[1,0,-1]}

or am i tripping

[lavish jewel]

there's infinitely many bases for the subspace

[viscid lagoon]

yeah, true

[lavish jewel]

there should be a transformation from the one i made up to yours

yeah

[viscid lagoon]

well i guess [0,1,-1] would just be in the span

i didnt check it

[lavish jewel]

[1,0,-1] - [0,1,-1] = [1, -1, 0]

so, same thing

[viscid lagoon]

so, yeah

yours works, too

[lavish jewel]

at the end of the day you should just go with whatever is more commonly used in your class/book. it anyway seems you have a good grasp on the stuff, so take your pick 😛

[viscid lagoon]

yeah

and one more, one of the questions is to represent the kernel as a linear equation

im not quite sure on what they mean by that

tbh

[lavish jewel]

maybe something like {x | <x,v> = 0}

[viscid lagoon]

is that a linear equation duh?

that would just lead to lke

something like $x_1 + x_2 + x_3 = 0$

[stoic python]

lewis

[viscid lagoon]

i guess that's a linear equation, but

the question seems weird to me

[lavish jewel]

i would have to read the original thing

but <.,.> is a linear operation, and you could also just write out a linear combination that generates vectors in the null space and do something with that

[viscid lagoon]

.

[viscid lagoon]

hmm

[lavish jewel]

idk, what i wrote is pretty much the definition of the kernel

the set of all vectors that satisfy that homogeneous eq

maybe someone has a different idea

[viscid lagoon]

could you expand on this

[lavish jewel]

which part

i mean something like

<a x1 + b x2, v> = 0

[viscid lagoon]

writing out a linear combination that generates vectors in the null space

[lavish jewel]

which is a linear equation

[viscid lagoon]

oh

like that

[lavish jewel]

and you pick some x1,x2 that are a basis for the null space

[viscid lagoon]

but i guess this would already give us our "linear equation"

[viscid lagoon]

this

[viscid lagoon]

that would lead tothis

i think

[lavish jewel]

v1x1 + v2x2 + v3x3 = 0, yeah

[viscid lagoon]

yeah

leading to x1 + x2 + x3 = 0

okay, yeah, thank you

[lavish jewel]

i don't see how you got the last part

ah

nvm

[viscid lagoon]

yeah

[lavish jewel]

all the entries in v are the same

[viscid lagoon]

yup

[exotic horizon]

hi! i'm trying to find the matrix associated to this application. The correct answer is (0 2, 3 0) but I don't know why. shouldn't it be a zero matrix?

This is what i’ve done

[amber bay]

uhhh

howd u get T(b_1) = (0,0)

[exotic horizon]

I just assumed it's (0x + 2y, 3x + 0y)

and then replaced x with 1 and y with 0

[amber bay]

ye

try doing it again slowly

[exotic horizon]

i really don't see what's wrong I'm sorry T.T 0*1 + 2 * 0 gives (0,0) no ?

[amber bay]

ye u got the first 0 right

but try 3x+0y again

[zinc timber]

why have taken vectors in R³?

[exotic horizon]

ohh i got it, I thought that 0*1 + 2 * 0 gives (0,0) but it only gives one 0, and i'm supposed to calculate the other T.T thank you!

[dapper gorge]

To show that a square matrix and its transpose have the same characteristic polynomial does it suffice to say det(A-xI)=det((A-xI)^t)=det(A^t-xI) ?

[tiny galleon]

It looks like it works to me

[wintry steppe]

Find a basis for $V = [(a, b, \frac{a}{b^2}, -b)], a, b \in \bR, b \neq 0 $. It would be easy without $\frac{a}{b^2}$. Could someone help me?

[stoic python]

mate

[haughty berry]

Thats not a vector space

Since $b\neq0$, $0\notin V$

[stoic python]

Slurp

[wintry steppe]

Find a basis for $V = [(a, b, \frac{a}{b^2}, -b)] \cup (0, 0, 0, 0), a, b \in \bR, b \neq 0 $. It would be easy without $\frac{a}{b^2}$. Could someone help me?

[stoic python]

mate

[wintry steppe]

Thank you, Slurp. Is it a vector space now?

[haughty berry]

Pretty sure its still not a vector space

Like (1,1,1,-1) is in V
but (2,2,2,-2) isnt

You can usually tell when something isnt a vector space. Like your criteria isnt linear

Exponents arent linear and neither is division

[wintry steppe]

Here [ ] means a span of a set.

[haughty berry]

Yep

Wait

Oh

Lol

Then my bad

It was originally a vector space

Spans are always a vector space

I thought you meant ${(a,b,\frac{a}{b^2},-b)\mid a,b\in\bR\land b\neq0}$

[wintry steppe]

Sorry, I will use span( ) next time.

Do you know how to find a basis?

[stoic python]

Slurp

[haughty berry]

Its a span of a single vector

So that vector is the basis

[wintry steppe]

Because we cannot rewrite $(a, b, \frac{a}{b^2}, -b)$ as a linear combination of $n$ vectors, $n \ge 2$?

[stoic python]

mate

[nocturne jewel]

what...?

@wintry steppe what does that have to do with anything?

span(S) has a basis S iff S is linearly independent

and since (a,b,a/b^2,-b) is not the 0 vector, it's an independent set