linear-algebra

[plush canyon]

is it possible to do this in latex

to make it more readable?

[lavish jewel]

it'll look the same

lets use c instead of lambda

[plush canyon]

Ok

1/c

[lavish jewel]

1/c_i

i'll do one eigenvalue at a time and draw the conclusion they're all the same

[plush canyon]

ok starting here 1/ci

I'm usnure of this

[lavish jewel]

A^-1 has an eigenvalue 1/c_i, since A has an eigenvalue c_i

[plush canyon]

is that from our matrix defintion?

[lavish jewel]

oh, you'll have to look up the proof for that, then

none of the observations i'm using are trivial

[plush canyon]

Oh ok

well I'll learn ur method

[lavish jewel]

the first thing would actually be to show that -A(A+2I) = I means A is invertible

[plush canyon]

ticked

got that

[lavish jewel]

so you need a proof that for square matrices, the left and right inverse are the same

(here it's easier because the matrix product commutes)

then you need a proof that the eigenvalues of A^-1 are 1/ the eigenvalues of A

and you also need a proof that the eigenvalues of A + cI are the eigenvalues of A + c

[plush canyon]

ok

the last 2 characteristics

[lavish jewel]

now, since when we showed A was invertible, we immediately got that the inverse is -(A + 2I)

[plush canyon]

yes

[lavish jewel]

-(A + 2I) = A^-1

[plush canyon]

yes

[lavish jewel]

and we can equate their eigenvalues

[vernal yew]

final answers

[plush canyon]

Yes

-1

[lavish jewel]

what

[plush canyon]

by eqauting

[lavish jewel]

wait

-(c_i +2) = 1/ci

you need to find the roots of that polynomial

there are 2

[plush canyon]

OHHHH

[lavish jewel]

then you need to show which one works

so you see the problem is not trivial

[plush canyon]

I can find the roots

[lavish jewel]

you need proofs for all of these things along the way

the roots are +1 and -1

[plush canyon]

how would Ik which one works

[lavish jewel]

you substitute them back in and test

[plush canyon]

by subbing back in??

and it works because it's = to the inverse?

[lavish jewel]

no, you substitute them back into the original poly equation

[plush canyon]

YES

LOL

THANKS

OMG

I'm stupid

[lavish jewel]

then you note that A^2 has eigenvalues c_i^2

which is another thing that needs proof

[plush canyon]

wait

Why do I have to prove that last bit

If Ik what the eignvalue is

[lavish jewel]

because the polynomial has A^2 in it

we found candidate eigenvalues for A

not A^2

the poly is A^2 + 2A + I

[plush canyon]

oh

wait

we need the eigen value for A

that's what the questions asking

[lavish jewel]

yes

[plush canyon]

so we stop here

we don't need to prove the last bit

[lavish jewel]

you haven't shown what the eigenvalue is yet

you found two candidates

only 1 works

you have to find which

we're not done yet

[viral olive]

wait

[plush canyon]

Ok

[vernal yew]

The orthonormal basis for N(A) = {[-1/sqrt3 , 1/sqrt3 , 1/sqrt 3 ]} with dimension 1 and the orthonormal basis for R(A) = {[1/sqrt2 . 1/sqrt2 , 0] , [sqrt2 / 2 , - sqrt2 / 2 , 0 ]} with dimension 2

[viral olive]

Edd did you we use the characteristical Polynomial to find out the possible candidates?

[plush canyon]

yes we did

[viral olive]

so you guys have two candidates right?

[lavish jewel]

char poly of what

i didn't

[viral olive]

Of A? Or how did you guys found possible eigenvalues?

[lavish jewel]

you can scroll up and check

[plush canyon]

we solved the poly

[lavish jewel]

we don't know what size A is to begin with

[plush canyon]

giving 1 and -1

[lavish jewel]

we used properties of eigenvalues to construct a poly whose roots are possible eigenvalues, for each eigenvalue

[plush canyon]

and - 1 works because we sub it back in get the original poly

[lavish jewel]

that's wrong

we still haven't shown -1 works

[plush canyon]

ok now we need to do that

what do we do here

[lavish jewel]

we go back to the original poly and rearrange it as when we proved A was invertible

[viral olive]

Now makes sense

[lavish jewel]

so we have -A(A + 2I) = I

we observe that A and A+2I have the same eigenvectors

(technically we need to use jordan cannonical forms in general here, which i won't do for simplicity. instead i assume the matrix is not defective, which would need to be proven)

so that multiplying both sides of the eq by v_i, the eigenvector of c_i, we get that

-c_i(c_i + 2)v = v -> -c_i(c_i + 2) = 1

now we test the possibilites

we set c_i = 1 and find that -(3) = 1 does not work

however c_i = -1 yields -(-1)(-1+2) = 1

[plush canyon]

kk

[lavish jewel]

so the eigvals are all -1

[plush canyon]

AHHH nice

[lavish jewel]

you need about 5 proofs along the way. it depends on your teacher whether they accept these statements without proof

[plush canyon]

now I personally find latex more helpful. any chance u are willing it to write it in latex

[lavish jewel]

no

[plush canyon]

probably will accept it wothout proof

But I will do proof for my understanding

Thanks edd. 10/10

[vernal yew]

so um, is mine correct? @lavish jewel

[lavish jewel]

yes, those work

[vernal yew]

really?

[lavish jewel]

yes

[vernal yew]

first try?

wowzers

so i might pass today after all?

[lavish jewel]

i would've personally used [1,0,0] and [0,1,0] as a basis for the col space

it's already orthonormal and saves you a gram schmidt

[vernal yew]

how did u get that basis? i just used the columns with leading ones (the only way i know how) . did you do it the transpose way?

[lavish jewel]

i just looked at it and went "oh shit"

but i guess yeah, you can do it with the vectors as rows instead

as long as you don't get mixed up

[vernal yew]

how do you just look at something and know the basis lol. a skill i will never have

[lavish jewel]

i've only been giving you examples one can do in their head without much effort

cuz i didn't wanna use paper

[vernal yew]

maybe you can

i cannot

it took me like twenty minutes to solve that

[bleak folio]

Let $\Pi_S$ be a projection onto the subspace $S$. Define ${s_1, ..., s_d}$ as an orthonormal basis for $S$. Can I then write $\Pi_S = \sum_{1}^{d} s_i s_i^{T}$?

[lavish jewel]

[stoic python]

Kas.st

[lavish jewel]

yes

[bleak folio]

thanks 👍

[vernal yew]

but everything i did is right? im having trouble believing that

[lavish jewel]

looks ok to me

stuff is orthogonal and unit norm

dims are ok

i'd just ask you to write the general form of a vector in the col space, and the general form of a vector in the null space

to drive the idea home

[vernal yew]

the what now

[lavish jewel]

the same thing i've been asking you all along whenever i say "write a generic linear combination"

which you have not yet understood 😛

[vernal yew]

no idea

[lavish jewel]

scroll back through our convo

[vernal yew]

not a braincell up there

u want me to write the orthogonal bases ive found as linear combinations?

[lavish jewel]

i want you to use the definition of basis, span, and linear combination to show how any vector in the col space would look like

[vernal yew]

a[1/sqrt2 . 1/sqrt2 , 0] + b[sqrt2 / 2 , - sqrt2 / 2 , 0 ]

where a and b are any scalar

[lavish jewel]

a and b, but yes

there we go

how about in the null space

[vernal yew]

its just a single guy

a[-1/sqrt3 , 1/sqrt3 , 1/sqrt 3 ]

where a is any scalar

[lavish jewel]

wait

now that i think about it

your null space basis is wrong

hmm wait

nvm

i always mix that part up

[vernal yew]

Orthogonal basis for N(A) Part 1

Orthogonal basis for N(A) Part 2

[lavish jewel]

what were the original vectors?

[1,0,0] [0,1,0] and [1,1,0]?

[vernal yew]

given {[1,1,0], [1,0,0], [0,1,0]}, place these vectors as columns of a matrix M. find the dimension and an orthonormal basis for the column and null space of this matrix.

my top left is rref of given

[lavish jewel]

ok all good then

[vernal yew]

scared me for a sec

[lavish jewel]

now for the last part

geometric intruition

[vernal yew]

i think we covered everything on the test but linear transformations

what

[lavish jewel]

describe the the column space and the null space geometrically

[vernal yew]

ahaha

one sec

[lavish jewel]

which geometric figures do they form

1~2 words suffice

[vernal yew]

the null space is a line in r3 and the range is a plane in r3

[lavish jewel]

byotiful

[vernal yew]

!

i did it?

letsss goooooo

[lavish jewel]

now

[vernal yew]

so all x that make Ax = 0 nontrivial form a line?

[lavish jewel]

yes

[vernal yew]

woot woot

[lavish jewel]

now, since you said linear transformations

[vernal yew]

and all x that satisfy Ax make a plane

or all solutions to Ax make a plane

[lavish jewel]

show that f(v) = av, where a is a scalar, is a linear transformation

[vernal yew]

okay

[lavish jewel]

similarly, show that g(v) = v + [1,1,1] is not a linear transformation

(v is in R^3)

[vernal yew]

one at a time man

for f(v) = av we must check for closed addition (i) and closed scalar multiplication (ii)

[lavish jewel]

no

[vernal yew]

no?

[lavish jewel]

linear transformations are ones that satisfy

[vernal yew]

yeah, u said show that it is a linear transformation..aka does it meet i and ii

im not there yet

[lavish jewel]

ah then nvm

[vernal yew]

that was my next thing

i aint even started yet

for (i) f(v+u) = f(v) + f(u)

[plush canyon]

Edd quick

[vernal yew]

so a(v+u) = av + au and it does, so check on i

[plush canyon]

Why did u get 1 and -1

I got -1 twice

[vernal yew]

for ii f(cv) must equal cf(v) so acv = cav and it does, so check

[lavish jewel]

oh maybe you're right, i might have factored the poly wrong

i wasn't using paper

[vernal yew]

it is a linear transformation

[lavish jewel]

you might be correct, and then the problem is easier

[plush canyon]

Yeah

[lavish jewel]

(no need to test the +1 and -1, so it ends at that step)

[plush canyon]

Thanks

[lavish jewel]

if you're sure you did it right, then that's all 😛 i might've messed up

[plush canyon]

So if I get - 1 twice

I don't need to prove it?

[lavish jewel]

yeah cuz there is only one option

[plush canyon]

Kk ty

[lavish jewel]

right. that was just a quick, simple check to warm up

[vernal yew]

ye

and u stopped me short haha

i aint just see stuff like you. your a genius with this

thanks again for helping me

[lavish jewel]

now try the other one

[vernal yew]

okay, one sec

[vernal yew]

for i g(v+u) = v+u+[1,1,1] must equal g(v) + g(u) = v + [1,1,1] + u + [1,1,1] X not a linear transformation cuz its not closed under addition. do not need to check for closed under scalar cuz i already proved its not a linear transformation

[lavish jewel]

good

don't forget the "linear", it IS a transformation of some sort

[vernal yew]

i guess i got this part down pat

[lavish jewel]

just not linear

[vernal yew]

good spot

whats a linear transformation anyway

[lavish jewel]

let's get a bit creative, because why not

hmm?

[vernal yew]

like intuitively?

what is it. i know how to check for it but idk what it is

[lavish jewel]

in this setting, a transformation that preserves straight lines and also the origin

[vernal yew]

so the usual definition of linear

[lavish jewel]

it takes in straight lines and spits out new straight lines after scaling and rotating

[vernal yew]

right right

okay

[lavish jewel]

the nice thing though, is that it generalizes

which is why we use axioms and abstract properties

have you taken differential calculus?

[vernal yew]

in it now

[lavish jewel]

then

using the properties you know of derivatives

show that differentiation w.r.t. x of univariate functions is a linear transformation

[vernal yew]

if u take the derivative of a linear function then u still get a linear function.

its only nonlinear if it starts out that way

[lavish jewel]

you're looking at it the wrong way

the definition is the same as before

[vernal yew]

and a linear function will eventually differentiate down to a constant

[lavish jewel]

f(av) = af(v)

[vernal yew]

um

[lavish jewel]

f(u + v) = f(u) + f(v)

[vernal yew]

thats cool

that is also cool

you can differentiate terms individually and you can pull out constants

check check

[lavish jewel]

yes

[vernal yew]

ye, my linear algebra prof mentioned this

[lavish jewel]

i.e. differentiation of functions is linear

[vernal yew]

ye

[lavish jewel]

integration is linear, too

[vernal yew]

ye

[lavish jewel]

and here the "vectors" are any function of x that has a derivative

[vernal yew]

what

[lavish jewel]

doesn't matter if these functions are nonlinear or linear

[vernal yew]

oh

weird

[lavish jewel]

look again at the definition of linear transformation

a transformation acts on a vector

so here, differentiation is a (linear) transformation

[vernal yew]

so derivatives is a linear transformation regardless of input

[lavish jewel]

differentiable functions are vectors

mhm

[vernal yew]

even with nonlinear inputs it doesnt change what derivatives and integrals are

[lavish jewel]

this is the reason why the definitions given in linalg are so abstract

[vernal yew]

ye

[lavish jewel]

sure, you mainly work with matrices and vectors in R^n in your class

but it applies to many other things

[vernal yew]

so i keep hearing haha

so i should be able to handle any problem within these topics now?

[lavish jewel]

hopefully lol

[vernal yew]

hahah

more than before

[lavish jewel]

at any rate, i must now disappear

good luck in your exam

[vernal yew]

GG bro

ur a saint

want me to let you know how i do?

[lavish jewel]

sure, why not

[vernal yew]

sweeet, have a great day!

[lavish jewel]

you too

[burnt timber]

Hey could anyone explain to me what linear spanning and a list of vectors being spanning means? Ive got the definitions but I still dont really understand it based off that, is there any easy way of understanding it

[fringe fjord]

Can you try to state the best understanding you currently have of the definition? Then it will be easier to figure out what you're missing.

[burnt timber]

Im going off these definitions, my understanding is that a spanning list basically means that any vector in the vector space can be made by a linear combination of the vectors in the spanning list

If the list spans the vector space

So I think i understand what it means for a list to be spanning, but I dont really understand Definition 1.3.1

Is it that the span of a list S is all the vectors it can create using a linear combination of the elements in it?

[spare widget]

yes

but S is not generally a list

It is a subset

The list of finitely many vectors there is a special case

That's why you have the other definition there

[burnt timber]

Ah okay so for the subset its just the vectors that can be made by a linear combo of vectors in that subset?

[spare widget]

If you are given S = {v1, ..., vn} then span(S) = {\sum_i \lambda_i v_i}

[burnt timber]

Okay and where do the lambda i's come from?

[spare widget]

Any real numbers

e.g. span(v) = {lambda v }

[burnt timber]

So the span of v is a set of v's with different coefficients?

[spare widget]

if I pick lambda = 0 -> 0, if I pick lambda = 1 -> v if I pick lambda = 1/2 -> 1/2 v etc

You get a line

For two vectors you get a 2d plane

[burnt timber]

Okay I think that makes sense

[spare widget]

There are some technicalities if S has infinitely many elements

But the first definition (with the intersection) will work in that case

[burnt timber]

I think ive probably learnt about it, weve done it a while ago ive just forgotten a lot of it

[oblique prairie]

does someone mind explaining this proof to me? my book seems to be trash at presenting proofs of diagonalization theorems

specifically the last sentence in the first image

how they turn the first sum into the second sum

because the way they said it, i don’t get it

[dusky epoch]

$(A - \lambda_r I)\sum_{k=1}^r c_k \bd{v}k = \sum{k=1}^r c_k (A - \lambda_rI)\bd{v}k \ = \sum{k=1}^r c_k (A\bd{v}_k - \lambda_r\bd{v}k) \ = \sum{k=1}^r c_k (\lambda_k\bd{v}_k - \lambda_r \bd{v}k) \ = \sum{k=1}^r c_k (\lambda_k - \lambda_r)\bd{v}_k$

[stoic python]

Ann

[dusky epoch]

@oblique prairie it's just linearity and defn of eigenvectors nothing more

[oblique prairie]

ok thanks

just wasn’t obvious to me

wait a second

the second summation goes to r-1

not r

oh god the last term is 0

i see it lol

thanks

[tribal willow]

is the minimal polynomial a canonical form?

[viral olive]

Are you asking if the minimal polynomial is its easiest form to show or if there is another form to make it more simple?

[halcyon spindle]

My book asked if readers is interested to showevery subfield of the field of complex number(denoted by C) must contain every rational number.
I am having trouble showing it. I feel like going the indirect way is not the right way,
That is Suppose S is a subfield of C such that there exist a rational p not in S.
I can't seem to get a contradiction with this.

I guess my problem is, I don't have much information on how S is formed.

Except it being a subfield of C.

Any hints?

[teal grotto]

S has 1 and zero so S has a copy of the naturals in it, and consequently a copy of the integers.

try now to build a copy of the rationals inside of S

[halcyon spindle]

Wow, thank you. I try that.

[tribal willow]

former

[pulsar wedge]

Hey guys

Can I ask a question

Can a matrix in RREF with the same column values have a different value in the original matrix

[dusky epoch]

this question appears not to make any sense

[pulsar wedge]

Okay

Let’s say

U have a row reduced echelon matrix

Okay

And two of the columns in that matrix are identical

[dusky epoch]

okay, sure

are you asking if identical columns in the RREF necessarily imply the same columns should have been identical in the original matrix?

[pulsar wedge]

Yes

[dusky epoch]

the answer is in fact yes

[pulsar wedge]

Oh really

Damn okay

There’s no counter example

[dusky epoch]

it's not very hard to prove

a matrix with two identical columns has a vector that looks like [0, ..., 0, 1, -1, 0, ..., 0] in its nullspace

where the 1 and -1 are at the same positions as your duplicate cols

[pulsar wedge]

Oh dang

So there isn’t a counter

[dusky epoch]

"counter"?

[pulsar wedge]

Like

Is there any case

Where that’s false

[dusky epoch]

where what is false?

[pulsar wedge]

are you asking if identical columns in the RREF necessarily imply the same columns should have been identical in the original matrix?

[dusky epoch]

would have appreciated if you didn't copy my messages word for word so blithely

but yes, if you want me to repeat myself a third time, yes, identical columns in original <=> identical columns in rref.

[pulsar wedge]

Oh I see

[dusky epoch]

is there any reason why you want that not to be true so badly?

[pulsar wedge]

I was just thinking about it

That’s all if there was any case where that’s not true

[vernal yew]

idk how to thank you @lavish jewel , i crushed the test today (high 90's) i was able to do everything and verify my answers. there was only one minor thing i couldnt recall (the cosine relationship formula for dot product) but i only lost like 2-5 points there on a 60 point exam. I didnt think to review it but i looked it up right after and its ((length a )(length b )cos ( theta ) )= dot product of a and b

[lavish jewel]

nice

congrats

[vernal yew]

all thanks to you buddy. your a saint!

[zinc timber]

[viral olive]

So a question i have to get the basis of a kernel from the field C^3. So i have by gauß elimination s element of C. (ixs ; s; -ixs - sx3) = (x;y;z). While there is only one variable i do know in C the basis vectors arent just simply real numbers but also complex numbers. My guess here is that the basis would only be one vector sx(i;1;i-3). Is that a correct thought process or am i forgetting something?

[golden reef]

if i have 2 matrices A and B. Can i rewrite this (A+B)^-1 as (B^-1 + A^-1)

or would that just be (A^-1 + B^-1)

[spare widget]

no

https://en.m.wikipedia.org/wiki/Woodbury_matrix_identity

If you scroll down to the special cases they have an expression for (A+B)^{-1}

But it's not as simple as what you wrote

[spark hornet]

Any idea how I can go about proving this?

<@&286206848099549185>

[wanton night]

Look at the definition of the L2 norm: It’s the max of the eigen values of A.

[spark hornet]

Hmm, yes I managed to now get the right part of the inequality 🙂

[devout raft]

Hey fellas
I'm currently taking linear algebra in uni. I've been trying to develop a strong intuition towards subjects and I've managed to do that to a pretty good degree until my last couple of lessons.
I was wondering if anybody's got good resources that would help a student develop the time of imagination needed to "see" equations in their head in a clearer manner, and to make more sense of statements instead of just memorizing them.
I watched 3blue1brown's first 3 videos of "Essence of Linear algebra" and those were god-sent at the time, but the material I'm currently studying isn't entirely relevant to the rest of the videos(Fields, Vectors, Matrices, and Determinants)
Pardon any faulty-translations(And this big wall of text) English isn't my native language 🙂

*tl;dr: Need some resources to comprehend the material, would love some recommendations

[subtle gust]

Hey y'all

I have a question that's kinda dumb 💀

Or it's not idk

Why is it that when we check if a set of vectors spans R^n

We consider the case where the resulting system of equations has infinitely many solutions

Wouldn't this mean that the vectors have infinitely many coordinates

When they should be unique

Am i overthinking this or is there smthng i'm not understanding 💀

Ay @devout raft you're answering my q or are u typing another q? 💀

[devout raft]

Answering you, it relates to my own question haha

[subtle gust]

Oh

I'll wait then lol

[devout raft]

A set of vectors that spans R^n means that we can 'reach' any vector within the field(by adjusting the vectors scalers, that is).
The field R, as you probably know, is an infinite field, and has an infinite amount of vectors accordingly.
So we have a system that can 'reach' any vector within infinite field.

To sum it up:
The span of R^n can reach any vector within R^n(by definition)
The vectors do indeed have infinitely many 'coordinates', but we can reach them all when we adjust every vectors scaler
Their representation is not necessarily unique, they're only unique if you have an n amount of vectors

Hopefully that's clear enough ^^

[subtle gust]

The "infinitely many coordinates" i mean

[devout raft]

I mentioned 'The Essence of Linear Algebra' in my question, I suggest you google it. You'll probably understand what vectors are all about with the first two videos

[subtle gust]

Is that you can get the same vector by an infinite number of linear combination

Which shouldn't be possible i think

[subtle gust]

I watched it

Like a trillion times since tjis sem started lol

If we have a set {V1,V2,V3}

And we have a vector (x,y,z) in R^3

And the linear system
(X,y,z)=aV1+bV2+cV3

Has infinitely many solutions

This means that we can get the same (x,y,z) using different a,b and c right?

[devout raft]

You joined a vector with a vector? I'm not entirely sure what you did there.
(x,y,z) would be the solution to the equation. It could be a column vector, and in that sense you're multiplying {V1,V2,V3} with b=(x,y,z)

[subtle gust]

So (a,b,c) the coordinates are not unique

[devout raft]

Hm, let me see if I can find a real example

[subtle gust]

Exactly

And we need this system to be consistent

To prove that any vector in R^3 (which we represented by x,y,z) in within our grasp

[devout raft]

Okay so if we were to draw out the following:
1 0 0
0 1 0
0 0 1
How exactly would you go about representing the vector (3,2,1)?

V1 is the first column, V2 is the second..

[subtle gust]

The solution would be something like:

a=some expression in x y and z
b= same
c= same

[devout raft]

x=3, y=2, z=1 In this case. Those are the scalers in our equation

[subtle gust]

They're not? 💀

Aren't those

[devout raft]

You said it yourself, aV1 + bV2 + cV3

so 3V1 + 2V2 + 1V3

[subtle gust]

Yeah but a b and c wouldn't be equal to x y and z

[devout raft]

Huh?

[subtle gust]

That would be true if V1 V2 and V3 are the standard basis vectors

[devout raft]

Oh, you represented... (x y z) as their sum?

I'm not entirely sure what you mean

There's the 'system', there's the 'solution to the system' and there's the vector that you get when combining the two

[subtle gust]

Ok so we have a set of vectors {V1,V2,V3}

And we want to check if, using those three vectors, we can represent any vector in R^3

So we take a generic vector in R^3 (x,y,z)

[devout raft]

Simplify it, don't take arbitrary vectors. Take the simple ones first and then branch out your knowledge
Stick to the following system in the mean time, where each column represents a vector accordingly:
1 0 0
0 1 0
0 0 1

[devout raft]

Mhm mhm

[subtle gust]

And we set it to be equal to a linear combination of V1,V2,V3

So

[devout raft]

Right, by definition

[subtle gust]

(X,Y,Z)=aV1+bV2+cV3

If we found that this system has infinitely many solutions

[devout raft]

Oh okay, I see what you're saying
(a, b, c) as scalers
V1 V2 V3 as our vectors
X Y Z as the result of the two

[subtle gust]

Yup

X Y Z are the resulting vectors

A b c are the coefficients

[devout raft]

That's impossible, it can't both scale R^3 and have an infinite amount of solutions

Not in this case, as you only have 3 vectors

Over R^3 that is

It can, at best - span R^3 with a single solution to each vector (X Y Z) as you describe it

[subtle gust]

Why is it defined to span R^3 if the system is consistent then

[devout raft]

Huh, it has to be consistent in order to span R^3

As each vector has a single representation in this case

[subtle gust]

Consistent includes infinitely many solutions

[devout raft]

That's not true, consistent just means the system works with a given solution(Not even all solutions, just a solution)

[subtle gust]

I was told a system is consistent if it has a unique sol or infinitely many solutions 💀

[devout raft]

So you don't get something like the following:
a b c | 1
s t v | 2
0 0 0 | 1
Obviously, 0 times 0 times 0 can never be 1. That's inconsistency

[subtle gust]

Yeah

But if we get
0 0 0 | 0 we would have infinitely many solutions

What if that were the case here

[devout raft]

No, consistency just means the system has a solution at a given state

[devout raft]

Alright, with 3 vectors you mean?

Let's see with the last vector, for example

[subtle gust]

Wdym a system having infinitely many solutions exactly follows the def u just typed

[devout raft]

1 0 0
0 1 0
0 0 0

If we would want the vector (X, Y, Z) = (2, 2, 3), how would we achieve that?

[devout raft]

Having an infinite amount of solutions doesn't mean an infinite amount of solutions to every single vector

[subtle gust]

X= some expression in x,y and a free variable z
Y= same
Z= the free variable

I wanna cry

[devout raft]

Well yes, Z is indeed a free variable. But you cannot obtain the vector I just described, as you would simply be multiplying by 0 every time

[subtle gust]

Wdym multiplying it by 0

Multiplying what by zero 💀

[devout raft]

Let's check for the solution (a,b,c) = (2,2,100)
so you get
2 0 0
0 2 0
0 0 0
^ ^ ^
2 2 100
(Because 100 * 0, is still 0!)

[devout raft]

The vector silly

[subtle gust]

If we pick z=3 we can generate the vector u described

[devout raft]

Each column I just described here is a vector

Nu uh, column
Look closely, if we were to look at the original matrix again that* I described:
{(1,0,0), (0,1,0), (0,0,1)}
V1 V2 V3

I just put them next to one another, in a column

[subtle gust]

I feel so lost rn

[devout raft]

Have you learned about matrices? Maybe I'm speaking gibberish to you

[subtle gust]

I have yeah

[devout raft]

Alright, have you ever represented vectors as a matrix

[subtle gust]

Yeah each col is a vector

[devout raft]

Right

[subtle gust]

So what was your point cuz i didn't get it

[devout raft]

And each column, has a scaler next to it

[subtle gust]

Oh yeah yeah

So a linear combination

With each col being a vector in our set

[devout raft]

Good! so lets look at the 0 vector again
1 0 0
0 1 0
0 0 0
You called them (a,b,c) in this example

each row is x, y, z

[subtle gust]

True

[devout raft]

so you have:

x  1  0  0
y  0  1  0 
z  0  0  0
   a  b  c

Something like that

Is this clearer?

[subtle gust]

Yeah makes sense

[devout raft]

Alright

[subtle gust]

Which brings us back to the fact that z is a free var

[devout raft]

That's not wrong, but you need to acknowledge something

No matter which 'c' scaler you choose, you will never, ever be able to change 'z' to be anything besides 0

Is this understandable?

[subtle gust]

No tbh

Cuz

Wouldn't x and y be dependent on z

So if we choose any number for z

It would generate values for x and y

[devout raft]

I'm not sure what dependent would mean, like a representation?
I think you need to understand each one is a different axis entirely. There is not combination of x dimension and y dimension that would add value to the z dimension

[subtle gust]

I need to realte this to the linear combination somehow but i just don't understand...

[devout raft]

I think I might be able to understand the confusion

[subtle gust]

Like

[devout raft]

Wait, let us narrow this down to 2 dimensions alright?

[subtle gust]

The values of x and y would be expression of z

[devout raft]

Wait wait

[subtle gust]

Alr

[devout raft]

I have a good analogy

Okay so, X and Y, right?

[subtle gust]

Yeah

[devout raft]

So let us say, that in order to reach the point of (1,1), I need to give X a value of 1, and Y a value of 1

[subtle gust]

Makes sense

[devout raft]

For (3,2)
X a value of 3
And Y a value of 2

So far so good?

[subtle gust]

Yeah

[devout raft]

Awesome

[subtle gust]

That's assuming we choose the standard basis vectors ofc ...

[devout raft]

Let us put this into an arbitrary matrix

x  1 0
y  0 1
   a b

the first column is responsible for the value of X
second column is responsible for the value of Y

a and b are the multipliers

So we have two vectors, (1,0) and (0,1) alright?

[subtle gust]

Yeah

Oh

[devout raft]

Okay, so if we would want (3,2)
We would need to go:
a * (1,0) + b * (0,1) = (3,2)
What is a and b in this case?

[subtle gust]

A=3 and b=2

[devout raft]

Very good!

Okay, now let me do what we did to z in the last matrix

x  1 0
y  0 0
   a b

a * (1,0) + b * (0,0) = (3,2)
a = ?, b = ?

[subtle gust]

You can't

The system is inconsistent

[devout raft]

It is not inconsistent.
Well, it is in this case, you're right

But it is not inconsistent for values such as:

x  1 0
y  0 0
   a b

a * (1,0) + b * (0,0) = (3,0)
a = ?, b = ?

a = 3, b = 0. The system is consistent again

You see?

[subtle gust]

Yeah but that would mean it doesn't span R^2

[devout raft]

Correct!

[subtle gust]

Yeah

[devout raft]

Right, so let's try going back to our original example

[subtle gust]

Your point is

[devout raft]

x  1  0  0
y  0  1  0 
z  0  0  0
   a  b  c

[subtle gust]

No matter what value of c we pick

We will end up with a z coordinate of zero

[devout raft]

Yep!

[subtle gust]

And therefore we wouldn't span the whole R^3

[devout raft]

Absolutely, as you can agree that we cannot reach any vector we want

[subtle gust]

Yeah

But like i don't get how we reached this result considering that we started from 3 generic vectors

[devout raft]

Another interesting example if you want:

x  1  2  3
y  2  4  6
z  0  0  1

This system doesn't span R^3 either

[subtle gust]

I mean it's obvious

Since col 2 is 2 col 1

[devout raft]

Yeah

[subtle gust]

The set is linearly depen

[devout raft]

Correct

So what's the confusion again?

[subtle gust]

The confusion is abt one point

When we have a set of vectors and we want to determine if it spans R^n

[devout raft]

Mhm

[subtle gust]

What do we do exactly? (Just to reach to my point)

[devout raft]

Oh, that's simple

I think.

[subtle gust]

Let's work in R^2

[devout raft]

I'm at my first month of Uni so the following statement might be a little shaky

[spare widget]

How many vectors?

You need n<= vectors to span R^n

[subtle gust]

Let's say we have a set of 2 vectors

[devout raft]

"If we can bring down the matrix into a canonic form, and every single variable has a representation, it spans R^3"

[subtle gust]

And we want to determine if it spans R^2

[subtle gust]

R^2 u mean?

[spare widget]

A set of 2 vectors spans R^2 if they arelinearly independent

[devout raft]

Oh I think criver might have more knowledge than me

[devout raft]

Any R, quite frankly

[spare widget]

Similarly n lin indep vectors span R^n

[subtle gust]

Wait this differs if we work in R^3 right

[spare widget]

inwhat sense?

[subtle gust]

They have to be linearly independent and span R^3

Both conditions have to be satisfied

[devout raft]

So

1  0  0  3  3489  -12348
0  1  0  31  423  234
0  0  1  13  312  42308

Should still span R^3? I think

[spare widget]

If you're in R^3you need 3 lin indep

[subtle gust]

Well that wasn't my point anyways...

My original question was the following

[spare widget]

For a n-dim space you need n linearly independent vectors to span it

[subtle gust]

Every vector in R^3 has a unique coordinate right?

[spare widget]

If you pick a basis then yes

[subtle gust]

Ok so when we have a set of 3 vectors and we want to determine if it spans R^3

We consider the case where the resulting system of equations

Has infinitely many solutions

[spare widget]

then just check that they are lin indep

[devout raft]

A span doesn't have to be linearly independent though

[subtle gust]

Even though that would mean the vectors have infinitely many coordinates

[devout raft]

Oh wait, sorry

[spare widget]

If the system has infinitely many solutions then they are lindep

[devout raft]

Yeah in the case of 3 vectors, they would have to be linearly independent. I suggest you look at other examples too though

[subtle gust]

Huh

[devout raft]

That's just a "law" for spans

[subtle gust]

If it has inf many solutions the system is lin dep ?

[spare widget]

Kernel is trivial for full set of lin indep vectors

That is for a nxn matrix A

Ax = 0, det(A)!=0 <-> x = 0

[subtle gust]

I thought the system is consistent => the set spans R^3 hmm

[spare widget]

no

You can have singular consistent systems

[subtle gust]

It's not AX=0 tho

It's AX=B

[spare widget]

what is B?

[subtle gust]

Cuz we are trying to determine if we can represent and vector (x,y,z) as a lin comb of the vectors in our set

So the procedure we follow is this

For a set {V1,V2,V3}

If we want to determine if this set spans R^3

[spare widget]

ok I got what your b is

[subtle gust]

We do this
(X,y,z)= aV1+bV2+cV3

[spare widget]

Then sure Ax = b must be consistent for all b

[subtle gust]

And if this system is consistent the set spans R^3

now if this system has inf many solutions

[spare widget]

it must be consistent for all b

[subtle gust]

Wouldn't this mean that the vectors have infinitely many coordinates

As in

There are multiple coordinates that produce the same (x,y,z

[spare widget]

Why would it have infinitely many solutions?

[subtle gust]

Cuz it might?

Consistent includes both a unique sol and inf many sol

[spare widget]

A system has infinitely many solutions when it is not full rank

But then the kernel is non-trivial

And thus it is inconsistent for some b

-> contradiction with the assumption

Assume A in R^{n x n}, then if A is full rank -> Ax = b is consistent for any b

More importantly A^{-1} exists, so x = A^{-1}b

Now assume A is not full rank

Then its kernel is non-trivial

that is, there are multiple x for which Ax = 0

[devout raft]

Sorry to budge in I just have to go for the day - I'd still appreciate any input if someone would care to provide one 🙏

[spare widget]

Try using several books simultaneously, e.g. axler + friedberg + hoffman + halmos +shiffrin

If you don't get a concept explained in one of those then look for it in another

[subtle gust]

Test

Dafaq

Are u getting my msgs?

Discord crashed

[spare widget]

But your question is too general to be able to get a meaningful answer

[subtle gust]

And i couldn't get it to work

[spare widget]

Yeah discord did some update of things

[subtle gust]

Anyways, we still haven't reached rank, nullity and kernel

Maybe i'll understand stuff better when we reach it

[spare widget]

anyways A not full rank -> range != R^n -> there are b for which the system is inconsistent

[subtle gust]

Idk what rank is sooo ....

[spare widget]

The number of linearly independent columns or rows in the matrix

[subtle gust]

Oh

If V1,V2 and V3 are linearly independent

[spare widget]

if A is full rank then it is invertible

[subtle gust]

Wouldn't that mean they're full rank?

[spare widget]

If it is invertible then x = A^{-1}b

Not only do you get a unique solution but the above tells you how to compute it

[subtle gust]

Yeah

So here A is full rank

And we have a unique solution

A being [V1,V2,V3]

[spare widget]

If it is not full rank then you can see how you can get multiple solutions

e.g. say y is from the kernel

Then Ay = 0

Let x be a solution for Ax = b

But then also (x+y) satisfies A(x+y) = b

[devout raft]

I havent emphasized this: English isnt my native language so books are a little hard to understand. Visual aids really help me understand terminologies and such. Would any of these books stand out in particular in terms of drawings or something?

[subtle gust]

The fact that the existence of multiple solutions doesn't violate the rule that every vector MUST have a unique coordinate is what's confusing me

[spare widget]

Then you're probably better off finding multiple books in your mother tongue

[spare widget]

having multiple solutions meand that the rank is not full

So it violates it

I just showed you an example

[subtle gust]

Yeah

[spare widget]

Ax=b and A(x+y) = b

[devout raft]

Oh. Are you from the US or some place else? If you dont mind me asking

[spare widget]

The solution is not unique

[spare widget]

I am Bulgarian, though idk how that's relevant

[subtle gust]

Yeah i think i got it

I have one last question tho

To prove that a set of vectors span R^n

[spare widget]

Not only that but there exists a b' such that Ax = b' is inconsistent

[devout raft]

Oh, the way you phrased yourself made me think we might be from the same country. Apologies
Anyway I really do have to kick off this time. Thanks criver

[subtle gust]

Is it enough to show that the det(A)=/0

[devout raft]

And good luck @subtle gust hope I managed to help 🙂

[subtle gust]

You def did!

Tysmmmm

Truly appreciate your help

[devout raft]

👍

[subtle gust]

@spare widget too did a great job!!!

[spare widget]

Yes

[subtle gust]

I appreciate y'all effort

[subtle gust]

Good then

Tysmmm @spare widget

[spare widget]

You can just do ref though

If you get no zero rows

Then it's lin indep

[subtle gust]

Wouldn't have understood it without u guys @weak ivy @devout raft

Yeah computing the det is much faster tho

[spare widget]

computing det for large matrices is a pain

I'd suggest ref

[subtle gust]

Row reduction goes purrrr

I mean the max we get on tests is 4×4

[spare widget]

Try row reduction for a 5x5 matrix, it's not fun

[subtle gust]

So i should be good

[spare widget]

A 5x5 ref isususlly simpler

[subtle gust]

Yeah prob

[spare widget]

Then it's fine

[subtle gust]

We don't get such huge matrices on tests tho

[spare widget]

But yeah, I recommend ref

You have to do it anyways for a lot of stuff

[subtle gust]

I hate G.J to my core

[subtle gust]

True

[spare widget]

idk what gj is

[subtle gust]

Gauss jordan

It's the process where u reduce a matrix to RREF

Row reduction basically lol

[spare widget]

oh ok, they didn't used to add the jordan part

we justcalled it gaussian elimination

eitherway, you'll use it a lot

Also for matrix inversion, even though you could compute it with dets too, but it's more effort

[hushed hedge]

Not sure how i can get this latex to work but i hope it's readable. The question is decide the rank for the matrix and find the basis in the null space (i believe it's called). I don't understand the last step, where do they get that x_1 = -3t, x:_2=5t, x_3=7t, any ideas?

The last step makes no sense, where do they get the values for t?

[vague crane]

it's from the matrix right before it

look at the numbers and then the coefficients of the equation

and also just substitution

[hushed hedge]

like -7x_2 = -5x_3

?

[vague crane]

yeah

[hushed hedge]

oh let me try that

[vague crane]

so x_3 = 7/5 * x_2

well i'm 90% sure

idak linalg lmfao

[spare widget]

x2 = 5/7 x3, since ghey don't want to work with fractions they substitute x3=7t

Then x2 = 5/7 * 7t = 5t

[hushed hedge]

aha

wow

[spare widget]

Finally x1 = x3-2x2 = -3t

[hushed hedge]

i'm guessing that a fraction would be equally correct?

[spare widget]

yes

This has infinitely many solutions

[hushed hedge]

perfect, thank you both!

[spare widget]

Since 2 eq 3 unknowns

Also the two matrices are not equal

That's abuse of notation

[hushed hedge]

yeah, i understood that. But had no idea how they got the values for t, i tried sub before but since i got a fraction i assumed it was wrong. Figured perhaps you need to transpose the matrix or something

[hushed hedge]

straight up copied from my book

<.<

[spare widget]

yes, your book abuses notation

[hushed hedge]

yikes

[spare widget]

Try picking some other books to go along with it

[hushed hedge]

It's a exercise book, even though they make mistakes like that it is easy to understand

[wanton night]

So if you can prove 1/max(λ) <= |λ| then you have it

[wanton night]

Which would be proving: max(λ) >= 1

[robust flicker]

can someone help me with getting the kernel of problem c??

im not sure if im doing the free variable correctly, since there's a row of all zeroes

[wintry steppe]

assuming you row reduced correctly

write out the solution more neatly

like $\begin{bmatrix}x\y\z\end{bmatrix}=\ldots$

[stoic python]

jswatj

[wintry steppe]

you're free variable is correct

(again, assuming you row-reduced correctly)

[robust flicker]

with what u told me this is what i got

[wintry steppe]

yeah so the kernel would be the spanning set containing [-1, 0, 1]

[robust flicker]

i think i row reduced correctly hehe

alrighttt

the free variable topic was confusing to me ty for clarifying :))

[wintry steppe]

Let K be an algebraically closed field, then there exists infinitely many A ∈ M2×2(K) such that malg(A, λ_A) = 2 and mgeom(A, λ_A) = 1 for some λ_A ∈ K (where the element λ_A may vary depending on A). Can someone help me to identify if that is true or false?

[dusky epoch]

is λA meant to be λ_A, ie an eigenvalue of A?

[wintry steppe]

yes sorry I should put a subscript on A

[dusky epoch]

i have a feeling your statement is a little ambiguous as worded

is λ fixed?

or do you just want to know whether there's infinitely many matrices with an eigenvalue whose geometric mult is strictly smaller than the algebraic mult

bc if so then i think yes?

just take any conjugate of [λ, 1; 0, λ]

theres gonna be infinitely many of those since an alg closed field is guaranteed infinite i think

[wintry steppe]

I think λ is not fixed it depends on A

[dusky epoch]

do you have the original problem exactly as stated

i want to ensure nothing is getting lost in translation or anything like that

[wintry steppe]

We need to prove if its wrong or true

[dusky epoch]

hrm

well ok like

whether there's infinitely many matrices with an eigenvalue whose geometric mult is strictly smaller than the algebraic mult

again im pretty sure the answer is yes but its even simpler now

just take 2x2 jordan blocks

[wintry steppe]

thanks but what are jordan blocks sorry never seen them before

[dusky epoch]

[λ, 1; 0, λ]

a jordan block is like a scalar matrix except just above the diagonal there are 1s

[wintry steppe]

oh oke I just looked it online now yes, but how do I prove there are infinetely matrices that the statement is True?

[dusky epoch]

i just gave you a family of such matrices where there's one matrix for every scalar in your field

surely you can take for granted that an algebraically closed field is necessarily infinite

[wintry steppe]

oke thank @dusky epoch

[round granite]

A nice and quick way proof of this is (just in case @wintry steppe hasn't seen it)

Suppose a_1, a_2, ... a_n were all the elements of a finite field F that was algebraically closed
The polynomial p(x) = 1 + prod (x-a_i) is a polynomial in F that can't have a root in F

[wintry steppe]

Yes I get that why alg closed field is infinite. I just don't get how to show there are infinite many matrices on M 2x2 (K)

[dusky epoch]

.

you were literally given a family that's the same size as your field???

yet you still doubt the existence of infinitely many such matrices?

[round granite]

it's okay, so the thing to notice is that the type of matrix you're looking for is

[dusky epoch]

literally what i said earlier

[stoic python]

CreativeMath

[round granite]

and you can verify that the eigenvector for this must be in the span of [1,0]

this works for any lambda in K

so each lambda gives u such a matrix

your field K is infinite, thus you can get infinitely many such matrices

(by just letting lambda be any of the infinite values in your field K), do you see why there's infinitely many matrices like that in M_2x2(K) now?

[wintry steppe]

yes yes I get it thanks a lot @round granite

[fiery rune]

I somehow accurately calculated the hypotenuse's length using gradient and one side length.
Why do I never hear about this?

[round granite]

Great, now try to generalize this for n×n

(and use higher dimensional analogues of the matrix above, so same eigenvalue repeated a few times on the diagonal but with a 1 above it)

You'll encounter a matrix that has dim(V) eigenvalues, but not an "enough" dimension of eigenvectors

Infact, this will precisely be the reason why some matrices aren't diagonalizable. You've just stumbled on Jordan blocks, and there's a theorem that says every matrix can be "diagonalized" with the diagonals being Jordan blocks

So that's some good motivation to study Jordan normal form

[misty pine]

Hello everyone. I'm from Georgia living in Poland. Want to start coding, AI and machine learning is what I want to enter. But need help with math, like guiding. Can't understand simple staff to build up on that. If anyone willing to help me with basic stuff and guide me through the book "Mathematics for Machine Learning" by Mark Peter Deisenroth, I'll be very grateful.
Or if I can't find such people, hope I can post here screenshots with questions.

[iron harbor]

Could you be more specific about which topics you need help with?

that is, which basic topics do you struggle with?

I'm no math expert, but I can give you some advice that took me a long time to learn by myself:

1. it's ok not to understand something when you've just started
2. you have to work really hard to understand stuff sometimes
3. it's absolutely worth it

[misty pine]

So, for example here:
Where does elements c(ij) came from, if later stated that i=1,...m and j=1,...k.

[lavish jewel]

this is common notation where the first index is a row in a matrix, and the second is a column

[misty pine]

Or in sum notation, why under sum is written l=1

[lavish jewel]

it's telling you to evaluate l at every integer value starting at 1 and ending at n

[misty pine]

Yes, but why to change those variables. It's really confusing. Randomly giving new letters

[lavish jewel]

$\sum_{l = 1}^n a_{il}b_{lj} = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots$

[stoic python]

Edd

[lavish jewel]

why not do it?

the letters themselves mean nothing

could've used x and y if they wanted

[misty pine]

So I can swap any letter for convenience? And get same result?

[lavish jewel]

that's the idea

otherwise there wouldn't be enough letters and symbols to represent everything

[iron harbor]

it's also not random; what this is saying is that you add each number from each matrix (where the numbers are at the same position in both matrices) together:

| 0, 1 | + | 3, 4 |
| 2, 1 | + | 5, 6 |
=> 
| 0 + 3, 1 + 4 |
| 2 + 5, 1 + 6 |

those letters are saying something like: for each number of the first column in the first matrix, add the number at the same index from the second matrix. this results in a new matrix

[misty pine]

So, here x1, x2, x3. Is it same variable? How can it be same if for example x(1) = 4, that means x(2) equals 4 too?

[lavish jewel]

nope, each one is a different variable

could've used x,y,z instead of x_1, x_2, x_3

[misty pine]

Aha

[lavish jewel]

subindex notation is powerful because it lets you refer to different things by juxtaposing a letter and a number

otherwise a 5x5 matrix would need 25 letters to represent

as opposed to a single letter and 2 subindices

e.g. x_1,1, x_1,2, etc

[misty pine]

Got it, thanks. So I can rely on you guys? Later I'll start from the beginning and will write anything down asking by the way questions like this

[iron harbor]

it won't always be the same people in the channel who can answer you; I can't answer many questions because I'm learning this topic myself too

but there is almost always someone here who will help

[lavish jewel]

for most of these questions, google will be your best friend

or simply reviewing the book's notation, which should be in some glossary either per chapter or at the start or end of the book

[iron harbor]

not always though: when I first started I didn't know what to ask in order to find out what a sigma notation means because I didn't know what sigma notation is; being able to ask by posting a screenshot of the textbook and saying "what is this weird letter with all these other letters?" is absolutely helpful

[misty pine]

Yeah, I have same problem, don't know what to ask. Or how to ask

[lavish jewel]

fair enough

[misty pine]

I understand concepts but fail in that simple things

[iron harbor]

@misty pine this page is long but sometimes you can figure out what something means by scrolling through here until you find it. https://en.wikipedia.org/wiki/List_of_mathematical_symbols_by_subject

for instance

[misty pine]

Okay, thanks. I'll do my best. Thank you guys.

[iron harbor]

👍 you can do this :)

[plush canyon]

@lavish jewel r u happy now.

[lavish jewel]

yes, this is a good alternative approach

[plush canyon]

is it all correct?

[lavish jewel]

would be good to note something like A^2 + 2A + I = (A + I)^2 and doing a substitution like W = A + I so that you can apply the proof you did above more properly

this is the same as saying A + I is nilpotent

[plush canyon]

does it make sense

what I said tho

[lavish jewel]

it's missing steps

[plush canyon]

figured.

what's the missing step?

I didn't feel like there was

[lavish jewel]

you made the assumption that A is diagonalizable

but as part of your proof you included nilpotent matrices, which are not diagonalizable (other than the 0 matrix)

so you need the extra stuff i mentioned above

[plush canyon]

do I NEED to show a substitution? or is that just good maths

[lavish jewel]

a lot of your steps are not well justified, it looks like you just wrote random stuff

[plush canyon]

ok

[lavish jewel]

that's how i would put it

[plush canyon]

Thanks for your advice, I'll take it into consideration.

[lavish jewel]

you can do something a little more careful by first considering something like matrices W^n = 0 for finite n. Then W^n x = 0 for any x, and in particular, this includes all eigenvectors of x. regardless of algebraic multiplicity of the eigenvalues, you get that W^n x = lambda ^n x = 0, so lambda = 0 for all of the eigenvalues. this means nilpotent matrices have eigenvalues all 0.

now consider the given matrix equation A^2 + 2A + I = 0 and factor it as (A + I)^2 = 0

we can let W = A + I and n = 2, and so W has all eigenvalues = 0

then note that the eigenvalues of A + cI are the eigenvalues of A + c. here, c = 1, so lambda + 1 = 0

so lambda = -1

[plush canyon]

thanks Edd while you're here, do u mind jumping to ODE's to help me with one final thing. I really appreciated your help here and It was useful

[lavish jewel]

i'm rusty on ODE, better wait for someone better at it

[plush canyon]

kk

thanks

how are u with Maclaurin series

I just need to verify one small thing

👍 TY

[zinc timber]

#calculus

[plush canyon]

Does it not have it's own seperate topic

sequnces/series?

kk

[zinc timber]

a separate channel for maclurin series?

[plush canyon]

nvm

[zinc timber]

what was your question?

[plush canyon]

ok

so what do I do here

if u can help me on matlab great]

if u cannot use matlab could u pls explain

[spare widget]

What do you not understand

They explain everything as far as I can tell

[plush canyon]

What the task wants me to do

Find the e value/vector of matrix a

And then apply power method to A?

It's difficult cause the teacher never showed us what the power method was for this he said. Do ur on research

☠️

[spare widget]

They have given you the algorithm in the exercise

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

[plush canyon]

criver are u goot at matlab?

[spare widget]

I will not write the code for you

look up

https://de.mathworks.com/help/matlab/ref/mtimes.html

https://de.mathworks.com/help/matlab/ref/transpose.html

https://de.mathworks.com/help/matlab/ref/norm.html

this ought to be enough to implement the algorithm

you'll need to make a for loop too obviously

[plush canyon]

wait

Is this the correct?

I applied the power method to a basic matrix

@spare widget sorry

[spare widget]

idk, I can't really tell from that, and I am in a meeting at the moment

[plush canyon]

My bad

[crisp minnow]

Is there a quick "trick" for getting the eigenvalues for a 3x3 matrix like there is for 2x2 matrices or do I just have to do the long expanded triangle formula for determinant and then find roots?

[plush canyon]

Lol

I'd like to know too

[spare widget]

what is the trick for 2x2?

[lavish jewel]

you can use sarrus for the det of a 3x3

[crisp minnow]

$\lambda_{1} \lambda_{2} = m \pm \sqrt{m^2 - p}$ where $m = \frac{a+d}{2}$ and $p = det(A) $

[stoic python]

Joachim

[crisp minnow]

https://www.youtube.com/watch?v=e50Bj7jn9IQ

[crisp minnow]

ouu that looks at least somewhat faster, thanks ❤️

[zinc timber]

[spare widget]

so a reformulation of finding roots of a quadratic polynomial

[crisp minnow]

Yes

[lavish jewel]

still, finding the roots of a cubic is not always easy

so in general you'd fall back on numerics

[crisp minnow]

yeah spent most of a page on one now

[spare widget]

cubic and quartic are solvable, but you really don't want to see the explicit solution for quartic, so yeah numerics at some point are just nicer

[zinc timber]

lol I've been stuck with this for 3 days now

welp

[spare widget]

physics?

[urban magnet]

Quick question. If I have a question asking if a matrix is diagonalizable over R or over C how would I do that exactly? I am studying minimal polynomial so I assumed we use the fact that it is diagonalizable if we minimal polynomial is a product of distinct monic linear factors but how does that differ over R and C?

[iron harbor]

When checking for linear independence, if you can work one of these out to a form like k1 = -k2, does that immediately imply that the vectors are not linearly independent?

[spare widget]

I don't think so

k1 + k2 = 0
k2 = c

(1 1) and (0 1) are linearly independent

you can either compute the det and show that it is 0

or do ref

and show that you get a 0 row

[iron harbor]

the way they showed it in the reference solution is more difficult for me to follow than I would like; they do it with operations like subtracting a multiple of one of the entire equations from one of the others

[spare widget]

yes you would have to learn gaussian elimination

I suggest learning it with matrices though

see the first chapters of hoffman and kunze

[iron harbor]

nice, thanks for the reference. I've been looking for a good book on these topics written in english

[spare widget]

hoffman and kunze, friedberg et al, strang, axler, halmos, shifrin, etc

pick your poison

[iron harbor]

this certainly does feel like poison some days :P

[lavish jewel]

that's disgusting

[spare widget]

consume it gradually and you become resistant

[zinc timber]

no maths, epidemic modelling

[zinc timber]

u still need help?

[vague crane]

isn't hoffman and kunze not recommended for beginners?

[spare widget]

maple can't group things and solve the cubic?

[zinc timber]

wait till I show u 6x6 one

[spare widget]

seemed pretty beginner-friendly tbh compared to halmos

Roman is not beginners friendly

[vague crane]

huh

[zinc timber]

it's already solved, it's just ugly to look at

[vague crane]

maybe i'm just dumber than i thought

[zinc timber]

no absolutely not

[iron harbor]

ah, right. now I see the connection between how doing it the matrix way and this system of equation way works. I'll try to figure out the matrix way after doing it this way a few more times

we had a lecture on how to solve these with matrices on monday but I need to work them out myself to feel like I really understand what's happening