linear-algebra

[broken hawk]

of high social status among people who identify with that word

[wintry steppe]

Failed to associate low temperature and linear algebra

[broken hawk]

that said, kay, not every matrix/transformation has a full set of eigenstuffs

especially if you’re not over an algebraically closed field

(ie ℂ in practical applications)

[half ice]

My life is a lie

[wintry steppe]

Not closed field?

[broken hawk]

algebraically closed = every polynomial has a root

over ℝ the characteristic poly may not have any roots ⇒ no eigenvalues

[winter reef]

OH YEAH

sry caps

[half ice]

We're likely working with subsets of ℂ here. Yes this gets weird for weird fields

[winter reef]

but I know what oyu mean

[broken hawk]

matrices over ℝ may not have any eigenvectors, and ones over ℂ may not have enough eigenvectors to diagonalize it

[wintry steppe]

Ah yup.

[broken hawk]

because just because the characteristic poly has a root of order, say, 3, does not mean there’s 3 independent eigenvectors associated to it

there’s at least 1, but that’s all you can guarantee

[winter reef]

ok so theres when the complex things come into lin algebra or wat?

[wintry steppe]

Duplicate roots are meh

[broken hawk]

pretty much; also in many applications using ℂ is more useful

e.g. physics is all about htose complex numbers

[wintry steppe]

Imo equality should be eliminated!

(Eliminates math)

[broken hawk]

you can probably do some interesting math by replacing equality with some weaker notion

just pick your favourite equivalence relation

and define that as your new “equals”

and see what happens

[wintry steppe]

Oh, weaker notion?

Like, can i have inequalities

<= in place of equality \o/

[broken hawk]

I mean you could see what happens when you work with the computer science equals: two numbers are considered equal if their difference is smaller than 2^-n for some given n

note: I doubt you’ll get a very interesting strucutre, and this weaker equals is not transitive

but at least it’s symmetric and reflexive, so that’s something

[wintry steppe]

Yeah

So, here is the thing:

Why am I even doing math

It seems like I'm delving into some nonsense no matter what field of math I learn

It is logical and sound, indeed, but what's the point

[broken hawk]

that’s a philosophical question

and there’s three answers to it, imo

[wintry steppe]

Oh, answers?

[broken hawk]

1. There may be practical applications you care about that you can appreciate or understand better with deeper knowledge
2. Just do it for the heck of it; math is fun and interesting, and often quite beautiful
3. Even if it may not have any immediate use, often mathematical research suddenly found unexpected applications; thus furthering human knowledge in mathematics is something worth doing for its own sake since you never know if it might not just solve world peace and cure cancer

I’m mostly in camp 2

[wintry steppe]

Ah. And..
4. It is actually nothing. No need to do that.

I'm leaning to 4

[broken hawk]

math is certainly not nonsense

it’s like music or art, but it also keeps satelites in the sky

[wintry steppe]

Uh

Do you mean calculation?

Or do you mean engineering

Also I'm not the only one on camp 4

Lots of people who dislike math are in there

[broken hawk]

I mean the fact that without math, we wouldn’t have general relativity, and without general relativity we couldn’t figure out why the heck our satelites are drifting off their paths, except we couldn’t even compute the (wrong) paths anyway so we couldn’t keep them flying autonomously anyway

[faint lintel]

Math is an art

[tidal nebula]

💯 .

[frosty vapor]

someone argued that math cant be art bc math is pragmatic

sad

[lean aspen]

@frosty vapor counterexample, math is useless.

[tidal nebula]

My savior 💙 .

[faint lintel]

[frosty vapor]

yea exactly katzen

[wintry steppe]

is there a practical version of gradient with quaternions?

[wintry steppe]

Sascha : General relativity uses math just as model. I think it could have used philosophical model instead

And calculate things based on that.

Like, 'There is no absolute time frame'

everything just uses math as a model, the universe could have something different in its source code

You can use math to model things, it's not the other way around.

In the same way, you can use logic and intuition to model things as well

Also lots of other variations we are yet to reach out

[lean aspen]

um

[sand hedge]

wtf is going on here

[wintry steppe]

the philosophy of linear algebra

[slow scroll]

I have a question: Is the rref of a matrix unique? It obviously is for invertible matrices as they all reduce to the identity, but idk how you would explain it for others.

[wintry steppe]

i still wanna know if i can easily sub quaternions for everything in vector calc

[lean aspen]

you don't want to

division rings are hell

do you know any algebra?

oh, vector calc

haven't seen that one in a while

thought u meant replace your field with quaternions

[wintry steppe]

that would be pretty crazy

[lean aspen]

(you can a lot of the time, it's not not so pretty)

[wintry steppe]

though i guess that would be something i would try if it seemed possible

[wintry steppe]

Quaternions for 3d vectors?

Hmmm

At least, it cannot represent all linear transformation of 3d vectors.

[trail knoll]

Have a super basic question about sets if someone can answer

B = {b1, b1, b2, b2, b3} C = {c1, C2, C1, C2, C3}

What is BC? Is it the same as union?

For example: {b1, b1, b2, b2, b3, c1, C2, C2, C3}

Trying to understand so I can figure this out: https://imgur.com/a/aEO7mdX

[dusky epoch]

how's BC defined in your text?

[pastel aspen]

let f:E-->F an injective linear application
then if G and H are two subvectorial spaces of E such that G + H is direct, then f(G) + f(H) is also direct

my question is, if f is not injective, on what condition on G and H can we find f(G) + f(H) to be direct

i can't find that condition

[brittle juniper]

Working in any dimension or finite dimension?

[pastel aspen]

any

[brittle juniper]

Also, do you want a necessary or sufficient or both condition?

[pastel aspen]

just sufficient

[brittle juniper]

mhmm if (G+H)∩Ker(f) = {0}, then the kernel of the restriction of f on G+H is {0}, so the restriction of f on G+H is injective

and of course you have $f(G)=f_{|G+H}(G)$

[stoic python]

Tuong:

[brittle juniper]

and the same, with H

because G is a subset of GUH, which is a subset of G+H

and so, I think you can use the result in which you have injectivity

[pastel aspen]

i was trying to find a condition to get the injectivity cuz thats the only way

i didnt think about this

nice ty

[brittle juniper]

You're welcome

[drowsy turret]

OH MY GOD THIS IS FUN

[lean aspen]

if T is diagonalisable/triangularisable/JNF-able/[insert decomp or form here] is f(T) also?
f is a poly over k, f(T) is the evaluation at T

@proper crescent

actually um

more actual question

[sour garden]

What is f

Yeah

Q=I

[lean aspen]

if T is diagonalisable/triangularisaible must there exist a automorphism Q such that QTQ⁻¹ is diagonal/triangle for some arbitrary fixed basis

[sour garden]

Lemme rephrase

[lean aspen]

pretty sure complex matrices are all triangularisable

but they're not all triangular

[sour garden]

Let Q be a diag/triangular matrix. Let B be a basis. Does there exist Q sucht that QTQ^-1 is diag/triangular in the basis B

[lean aspen]

something about change of basis matrices

are all invertible matrices triangularisable?

[sour garden]

No

Over R no

[lean aspen]

pretty sure no, consider the matrix in M2R corresponding to 90 degree rotation

[sour garden]

Take rotation matrix

Ye

[lean aspen]

are you saying we can always choose our automorphisms to be diag/triang-able?

I'm not sure you can do that

[steel cobalt]

ex: -5C^-1.. Do i multiply by 5 first then inverse? or inverse then multiply by 5? or does it not matter?

[slow scroll]

doesn't matter

[steel cobalt]

ty

[slow scroll]

np

[fervent totem]

Hi guys, is this statement true? A matrix is nilpotent iff it has 0 trace and determinant

[lean aspen]

what do u think

that doesn't seem like v convinving proof at least

[fervent totem]

if M is nilpotent then M^k=0 then use tr(AB)=tr(A)tr(B) and det(AB)=det(A)det(B). If matrix has 0 tr and det then its similar to an upper triangular matrix where the diagonal are all zeros(not sure about this direction)?

[jagged pendant]

@fervent totem trace is not multiplicative

nilpotent implies trace and determinant 0, that much I know is true

the converse I'll need to think about

are these matrices over C?

[fervent totem]

nilpotent matrices are similar to block diagonal matrices where each block is a jordan block of 0 on the diagonal

yeah

but im trying to think why thats true

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

[jagged pendant]

ah

because the eigenvalues exponent to 0

every complex matrix had a Jordan normal form

each block has its corresponding eigenvalue along the diagonal

and is upper triangular

when you multiply upper triangular matrices, you multiply entries on the diagonal

@fervent totem

[fervent totem]

ahh right I think get it

[jagged pendant]

but the converse doesn't make sense

just by using the same argument

trace 0 doesn't mean all the eigenvalues are 0

just that their sum is

[fervent totem]

if it also have 0 det doesnt it force all eigenvalue sto be 0?

[jagged pendant]

no

[fervent totem]

right I see lol

just need to have 1 zero

[jagged pendant]

yeah

[fervent totem]

Thanks!

[jagged pendant]

np

[livid crow]

If A is an mxn and B and C are two different vectors in R^m, with Ax= b and Ax= c both being consistent with solutions that are planes in R^n, why is it that these two planes might not intersect with specfic vectors b and c

Dont they live in the same space, so with certain B and C there is a combo where the planes can intersect?

[jagged pendant]

because if b is not equal to c

then for x in the intersection

you have Ax = b
and Ax = c
which implies b=c

@livid crow

[livid crow]

So they are essentially the same plane?

And therefore they cannot intersect each other?

@jagged pendant

[jagged pendant]

i assume b != c

and conclude b = c

which is a contradiction

[livid crow]

thus they are linear combinations

which is why they are different vectors but really the same vector

with one of them having a different weight

i still cant wrap my head around it i got this question and another one on my exam wrong and Im sitting here trying to figure out why, just not clicking

[jagged pendant]

matrices are functions of vectors

for a unique input, there is a unique output

it cannot equal two different things at the same time

therefore the solutions to Ax = b and Ax = c cannot overlap

as that would suggest Ax is equal to both b and c at the same time

alternatively, since they are linear, you can subtract them, and get

A(x-x) = A(0) = 0 = b-c

but b != c means b-c is nonzero

[slow scroll]

"for a unique input there is a unique output"
isn't that only true for an injective linear transformation?

If you have a mxn matrix where n>m, then I don't see why you couldn't have intersecting solutions?

[livid crow]

So with that explanation im getting the understanding that nothing can intersect ever?

[jagged pendant]

no you got it the other way around

for a unique output there is a unique input

it's like the vertical line test vs the horizontal line test

[livid crow]

Okay okay let me try again

Nevermind I cant get seem to understand why when b=c it means they wont intersect

[slow scroll]

you mean b!=c lol

[livid crow]

right right but looking through woogs work he said since ax=b and ax=c b=c

[jagged pendant]

like pick a specific solution

suppose A is a 3x3 matrix

[slow scroll]

okay using x as the input is confusing things i think
if Ax = b
and Ay = c
and b=c
then x=y

[jagged pendant]

and suppose that if v = (1,1,1), then Av = b, and Av = c

well, you can just multiply them together

and explicitly calculate what Av is

it cant be 2 different things at the same time

[livid crow]

so if it cant be two different things then they are one in the same no?

[slow scroll]

lets say i have x1v1 + x2v2 = b
and we know the values of x1 and x2 (scalars)
those same constants won't solve x1v1 + x2v2 = c
because x1v1 + x2v2 = b

thats all woog is tryna say

[livid crow]

Ohhhh so because they are different vectors that both are solutions to Ax there is no possible way for them to have the SAME solution so there is no intersection point

[slow scroll]

yea i think you got it

[livid crow]

sort of im bad with this conceptual stuff, gonna talk to proffessor about it to solidify it more

[slow scroll]

well kinda like woog was saying, we are dealing with functions that have a unique output for any input. It passes the "vertical line" test.

the only way for Ax = b and Ax = c to have the same solutions is if Ax somehow simultaneously equaled b and c where b!=c

[livid crow]

Which isnt possible cause they arent equal

thats where he said the contradiction comes in

ya ya its making more sense

If the columns of an nxn matrix are linearly independent they then automatically span Rn due to the fact that they can be written in the form c1v1 + c2v2....cnVn with all cs just being zero correct?

[slow scroll]

well, due to the fact that that you need at least n vectors to generate R^n.

err you got it mixed up i think

[livid crow]

on the exam it was specifically a 3x3 and R3

But im assuming it applies to all nxn

[slow scroll]

" c1v1 + c2v2....cnVn with all cs just being zero correct" means they are linearly independent. They are generating because you have at least n vectors and they are linearly independent. In other words, you are linearly independent because you have a pivot in every column of the matrix and generating because you have a pivot in every row of the matrix.

if you have a square matrix and every column has a pivot <=> every row has a pivot

[livid crow]

ohhh how did i forget about that

Gotcha

Thank you so much

[slow scroll]

np! c:

[glad abyss]

Can someone explain this to me?

Especially theFana

[proper crescent]

That's almost definitely a typo

This is Axler, right?

[glad abyss]

Not sure what Axler is

[slow scroll]

the guy who wrote "linear algebra done right"

[proper crescent]

Looked at a previous edition, he has nearly the same paragraph except he just writes "and replace the 2 or 3 with an arbitrary positive integer"

Basically you just take ordered n-tuples of elements of F

[glad abyss]

Yeah

Oh ok

Thanks

[proper crescent]

👍

[wintry steppe]

Oi I need some help yo so ima post something but before I do can i post a screenshot or nah?

SO Like in 35 I have this question and watch so I need to show
W = + or - || pq|| |project of F onto PQ|| ok?

wtf anyway

[hoary osprey]

this isnt linear algebra tho?

[wintry steppe]

it is.

its in my linear book

anyway

|| Project pq onto F|| = ||F|| ||cos(thetha)||

....

So I plug it back into the original equation and I get

• or - |PQ| | | | PQ || cos |

but since the cos is in the absolute value it will never be negative so how will I solve this 😐

explain where the signs ill be and all that

@me when anyone can help much appreciated and all that @wintry steppe miss you from your biggest fan

[mild narwhal]

@wintry steppe

you use -ve for the case on the left

+ve for the case on the right

[mystic goblet]

i got this problem for homework today

and i got 11a and 11b, but i couldnt do 11c

can someone help me through the steps?

[restive palm]

$\vb*{a}$ is parallel to $\vb*{b}$ \implies $\vb*{a} = k \vb*{b}$

[stoic python]

Cosmicrays:
Compile Error! Click the None reaction for details. (You may edit your message)

[restive palm]

For some k

[mystic goblet]

yes

[dusky epoch]

do you know how to find the angle between two vectors?

[bleak cairn]

do tri-diagonal matrices have to have the same value along a diagonal? for instance, 2 on the main diagonal?

[mystic goblet]

thats given in the question, its 60

[dusky epoch]

okay one of y'all will have to move.

[restive palm]

Oh wait I read the wrong question

[dusky epoch]

@mystic goblet that's not what i mean. what i mean is, if i gave you two vectors, would you be able to find the angle between them?

[mystic goblet]

yes

cos^-1 ((a.b)/|a||b|)

[dusky epoch]

well then.

what's stopping you from doing this in this problem?

you will express the angle in terms of m

and from that, find the value of m

[mystic goblet]

i tried that, but i dont get the same value

i tried to solve for m using a.b = |ab|cos∅

and then solving for m, but i dont get the right value

[dusky epoch]

then that means you fucked up in the algebra in one or both methods.

show your work

[mystic goblet]

yep

sorry for the bad quality, dont have a good camera

[dusky epoch]

ok so

fifth line

what did you do

[mystic goblet]

i squared both sides

[dusky epoch]

you didn't do that correctly on the right.

(a+b)^2 is not a^2 + b^2.

[mystic goblet]

oh right

[dusky epoch]

there goes your fuckup

[mystic goblet]

omg, thank you so much

im so dum

[frosty pewter]

is there a name for this D matrix? I don't really get what it is, can someone tell me what to look up or explain what all the O's and I's mean?

oh wait, I got it, nvm

[true zodiac]

Can you find a basis for a set in R3 that contains only the zero vector?

[slow scroll]

"a set in R3"
do you mean a subspace of R3 or all of R3?

[proper mirage]

Hey how is that possible

[true zodiac]

all of R3

[slow scroll]

no because the zero vector does not generate all of R3

[dusky epoch]

sure. an empty basis will do

the empty set is a basis for {0}

[proper mirage]

A basis has to have linearly independent vectors

How zero vector can be independent?

[dusky epoch]

i'm not saying {0} is a basis for {0}.

[slow scroll]

it is indepdent, but not spanning lol

[dusky epoch]

i'm saying {} is a basis for {0}.

if you think otherwise, give me a nontrivial linear combination of vectors from the empty set which sums to zero

well i mean ok

no

{} is a linearly independent set.

[proper mirage]

While any scalar multiple of zero vector is zero vector

Zero vector is dependent

Hence cannot be in a basis

[dusky epoch]

do you notice me putting it in my basis?

please read what i am saying

[proper mirage]

You imply that you can do so?

[dusky epoch]

i do not.

[proper mirage]

i'm saying {} is a basis for {0}.
This we know

[dusky epoch]

yes.

{} is a basis for {0}.

{0} is the subspace spanned by {}.

[proper mirage]

But this is the zero subspace

[dusky epoch]

it is

[proper mirage]

Well I cannot argue with you.

[hoary osprey]

[broken hawk]

the empty set is a basis for the zero subspace if you define the empty sum of vectors to be 0 (which is the only sensible definition, really)

[dusky epoch]

who doesn't define the empty sum to be 0?

[broken hawk]

idk but you do have to define it somehow

and I mean it’s ambiguous

if you sum no vectors, you get the 0 vector, but if you sum no numbers, you get 0 (the real number)

but 0 (number) ≠ 0 (vector)

[dusky epoch]

i mean

an empty sum still takes place in some known abelian group

and its value is then that abelian group's zero

[summer dagger]

Can a trivial solution to Ax = b also be considered a Unique solution

[slow scroll]

unique solution means there is only one input of x that outputs a b

[candid frost]

I've only heard of the trivial solution to Ax = 0, namely x is the zero vector

So it's a confusing question to me

[slow scroll]

yeah i don't really know what he was asking xd.

[candid frost]

Ah maybe there's a useful fact to know about a matrix A if the only solution to Ax = 0 is the trivial solution

[slow scroll]

yeah theres that

[candid frost]

What would that tell you about A?

[slow scroll]

are you asking me?

[candid frost]

Asking anyone

[slow scroll]

well it tells you that the solutions to Ax = b are unique for all b. i.e. the columns of A are linearly independent

[candid frost]

Ahh ok

[slow scroll]

but it doesn't tell you anything about the existence of solutions to Ax=b, so you could still have a value of b that doesn't have a solution because b is not necessarily in the span of the columns of A

[winter reef]

is matrix 2x2 sqrt(2) 0 in first row and in second 0 sqrt(3) diagonable in R and not diagonable in Q?

if diagonable is even a term since I couldnt find the word in english

[slow scroll]

diagonalizable is the correct word but idk anything about it

[broken hawk]

since sqrt(2) and sqrt(3) are not rational numbers, that is not a matrix over Q at all

but what I suspect you actually mean is whether a matrix which is similar to that one with rational numbers is diagonalizable

in which case, the answer is no

[winter reef]

oh true

[broken hawk]

because the diagonal entries are unique

there’s no way to get another diagonal matrix out of it (up to switching the order)

[winter reef]

but what would be the example of matrix 2x2 in Q that is diagonalizable in R but noit in Q?

ohh ok I know

I get it

[broken hawk]

uh, well, I mean the idea is good, you just have to find a matrix with rational coefficiants that is similar to that one

[winter reef]

when I get a characteristic polynomial and it has only compelx roots, right?

[broken hawk]

then it won’t be diagonalizable in R either

[winter reef]

okk

[broken hawk]

found one

the diagonal entries would just be the eigenvalues

which are not rational

done

(just guessed at random)

[winter reef]

wait

but to get characteristic polynomial I subtract lambda from diagonal

and then determinant

but the det of this will just be -1?

[broken hawk]

fine, if you want that

[winter reef]

nvm, got it thx

[broken hawk]

you have to do $\det \begin{pmatrix} 1 - \lambda & 1 \ 1 & -\lambda \end{pmatrix}$

[stoic python]

Sascha Baer:

[winter reef]

yeyeyeyeyeyeyey

I just cant subtract from zero

also I have a problem that Im not sure if its done or how to do: given matrix n x n if it has n different eigenvalues show that it is diagonalizable

so characteristic polynomial will be sth like (λ-a1)(λ -a2)...(λ -an)

each subspace V(a1) will just be lin of some alpha1

So I thought that basis of this matrix will be {alpha1,alpha2,...,alphan}

[broken hawk]

what conditions do you have that will let you conclude that the matrix is diagonalizable?

like what tools do you already have?

[winter reef]

similiar matrix

[broken hawk]

wdym?

[winter reef]

In linear algebra, two n-by-n matrices A and B are called similar if there exists an invertible n-by-n matrix P such that

${\displaystyle B=P^{-1}AP.} {\displaystyle B=P^{-1}AP.}$

[broken hawk]

like do you have any statements of the form “a matrix is diagonalizable, if…”

yea I know what similar matrices are

[stoic python]

dog:

[winter reef]

ohh

[broken hawk]

I’m asking you why you’re saying that word

[winter reef]

yeah so A is diagonalizable if and only if A is similiar to diagonal matrix D

[broken hawk]

that’s the definition, yea

[winter reef]

also something about basis of eigenvectors

[broken hawk]

eigenvectors

if anything

eigenvalues are just numbers

you can’t make a basis from them

[winter reef]

yes I know

[broken hawk]

okay, so, find that statement cause it’s important here

you’ll need it

[winter reef]

hmm so diagoalizable if matrix from standard basis of the endomorphism has basis made of eigenvectors

right?

[broken hawk]

which endomorphism

an endomorphism doesn’t have a basis

an endomorphism is a function

[winter reef]

I mean matrix of some transformation from standard basis to standard basis is diagonalizable if its basis is made of eigenvectors

[broken hawk]

“its basis” a matrix has no basis

no, I actually do not know what you mean

[winter reef]

dont tpype

[broken hawk]

I’m not being pedantic here

[winter reef]

sry sry

[broken hawk]

I’m trying to figure out what you’re saying

like, I know the theorem

but it’s not what you’re saying

[winter reef]

hmmm

[broken hawk]

you should look it up in your notes

but a transformation has no basis

that doesn’t make sense

[winter reef]

Ok, A is diagonalizable if and only if A is similiar to some diagonal D. Also, given endomorphism phi in K^n such that M(phi)st to st = A is diagonalizable, meaning K^n has basis made of eigenvectors

[broken hawk]

okay. so. can you find a basis of eigenvectors?

in your original problem

[winter reef]

I mean, yes, kinda, I typed it at the beginning but not sure iof its correct

[broken hawk]

(really what you should have is the theorem: if there exists a basis of Kⁿ of eigenvectors of A, then A is diagonalizable)

(that’s the one you need)

(you wrote the other direction)

then you can show that if the eigenvalues are distinct, then their eigenvectors are lin. indep. and therefore form a basis, qed

I’ll leave the details to you, I need to go to sleep

[winter reef]

ok gnight adn thx

[ruby topaz]

matrix multiplication
composition function🤔

[winter sage]

does anyone know how to do this problem

i was under the impression i had to get X and Y so the three matricies on the right multiplied out to be the matrix on the left

but apparently thats incorrect

[broken hawk]

uh yes, that’s what you have to do. and obviously I = identity and S = D - CA⁻¹B

[placid oracle]

how do u check these without examples?

[lean aspen]

elimination

[placid oracle]

is it C since BA is not accounted for

[lean aspen]

just go thru them one by one

[placid oracle]

yes but without plugging in a matrix how would i know

[slow scroll]

A^2 is invertible. Its inverse is just A^-1 applied twice

[placid oracle]

ik a and b are

but im confused as to how u would know it for AB and BA

[slow scroll]

umm is b?

d is invertible because A^2 + AB = A(A+B) and c is because A^2 + AB + BA + B^2 = (A+B)^2

[placid oracle]

oh

i see it

okay that makes a lot of sense

[slow scroll]

noice

[placid oracle]

this makes sense, right? since (AB)^T = B^TA^T and (B^TA^T)-1 = A^T^-1 B^T^-1 ?

[dusky epoch]

yeah

[placid oracle]

ok ty

[broken hawk]

hint: try to find a matrix whose square is -id

[summer dagger]

I'm lost as to how to set up a matrix to solve this

[restive palm]

Anyone got a clean proof showing upper traingular matrices are nilpotent. CBA to do induction on the ijth component of a matrix?

[restive palm]

Got it nvm

[dusky epoch]

upper triangular matrices with zero diagonals you mean?

[restive palm]

Yeah I should have specified strict upper triangular matrices

But I got the answer^^

[wintry steppe]

need help on a text reveiw

[wintry steppe]

what parts don't you understand?

[dull kettle]

Hello, is the answer to this question (III) only?

[spring wolf]

since I can think of a counterexample, nope

are we assuming u, v and w can't be 0? (if we aren't then ... this is a pretty boring question)

[half ice]

I think they're all true? Am I derping here? How did you rule out I and II?

Oh, I didn't think of the case where u, v, w may have been identical

[spring wolf]

Yep

[dull kettle]

if u, v and w equals 0, what does it imply? that it could be a linearly independent set

?

[half ice]

Or linearly dependent themselves

[spring wolf]

Also I) is a set of 4 vectors in R^3

[half ice]

So II may be wrong, but I is definitely right?

[jagged pendant]

yes

[spring wolf]

frick i misread

[jagged pendant]

I is correct, others arent

[half ice]

III is wrong?

[jagged pendant]

yes

[dull kettle]

why is III wrong?

[jagged pendant]

take v,w independent, u=0

wait

[spring wolf]

v and w dependent

[jagged pendant]

I guess that's still fine

[spring wolf]

and u independent of v and w

so take v=w

[jagged pendant]

ok take u,w independent, v=0

[spring wolf]

has solution

(0,1,-1)

[half ice]

u = (-b/a)v - (c/a)w

[spring wolf]

i.e (a,b,c) = (0,1,-1)

and then u is free to be anything you want

[jagged pendant]

like basically take u≠0, v=w=0

[half ice]

But then you don't have non trivial solutions

[jagged pendant]

yes you do

from u

[half ice]

OH you're right

[spring wolf]

explicitly, u= (0,0,1), v=w=(1,0,0)

has solution (0,1,-1)

but u is clearly independent

[half ice]

Right, I assumed you could divide by 0

[jagged pendant]

tsk tsk tsk

[spring wolf]

oof

[dull kettle]

erm, so is III the only answer?

[jagged pendant]

silly kaykay

[spring wolf]

nope III) is wrong we have counterexample

[half ice]

Only I

[spring wolf]

yep. do you see why, Contredict?

[dull kettle]

one sec

I don't understand 😭

[spring wolf]

Okay, so say you're checking one by one if the first three vectors are linearly independent

And say you're fine up to the third vector: i.e the first three vectors are linearly independent

what do you know happens if you have 3 linearly independent vectors in R^3

[dull kettle]

the entire set is linearly independent?

[spring wolf]

We aren't worried about the third vector just yet.

Hmm, have you come across the term basis before?

[dull kettle]

yes when it is linearly independent and it spans across a vector space

[spring wolf]

Yep. So we have three linearly independent vectors in R^3 - that means ... ?

well i mean I have a massive hint to the answer I'm looking for by asking about bases

[dull kettle]

the set is a basis for R^3? idk 😦

[spring wolf]

yep

[dull kettle]

but how do we know it spans R^3 which is the second condition

[spring wolf]

If n is the dimension of your space and you have n linearly independent vectors then they span

you can find proofs of that in a lot of places - it's a fairly standard result

[dull kettle]

okay but i'm trying to link back to what you said initially

[spring wolf]

if you're willing to accept it for the moment then we can move on

So we have three vectors which span R^3

if we add another, can it be linearly independent?

[dull kettle]

no

it becomes linearly dependent

[spring wolf]

Yep

so any set of 4 vectors is linearly dependent

(in R^3)

[dull kettle]

what I don't understand is the part on the trivial solution

if it has non-trivial solution, wouldn't it imply that it is a linearly dependent set and so at least one of the vector can be written as a linear combination

of the other

[spring wolf]

Yes, at least one
That one does not have to be u

[dull kettle]

oh right, but can you explain the workings you guys did above, something about assigning dependent and independent to u, v and w

[spring wolf]

So we want a non-trivial solution and for u to be independent from the other two

[dull kettle]

why do you want u to be independent from the other two?

[spring wolf]

For a counterexample

Since we're saying III) is not true

[dull kettle]

oh if it's dependent, would that imply it linear combination?

with v and w

[spring wolf]

Yes

As you said, at least one of the other vectors has to be a linear combination of the other. In this case, we're only left with v and w, so lets make our lives easy and pick v=w=(1,0,0). These are clearly linearly dependent and we can find a non-trivial solution regardless of what u is

[dull kettle]

am I right to say that v and w are linearly dependent because they are scalar multiple of each other?

[spring wolf]

yes

that's the only way for two vectors to be linear dependent

(to each other)

[dull kettle]

if u is a linear combination of v and w, would that mean there's a linear combination between v and w as well?

[spring wolf]

not necessarily

(1,1,0)=(1,0,0)+(0,1,0)

Also this is going a bit off topic: do you get how to find a counterexample to III)?

[dull kettle]

i get up to the point where v and w are linearly dependent

[spring wolf]

Great, so to find a counter example all you need to do is pick a u that's linearly independent of v and w

Because then a=0, b=1,c=-1 is a non trivial solution

(or you could just say one exists for the equation av + bw on its own since we know v and w are linearly dependent)

[dull kettle]

okay

[spring wolf]

did i make sense?

[dull kettle]

wait, how do I pick a u that's linearly independent of v and w

I can understand a = 0, b = 1, c= -1 is a non trivial solution

[spring wolf]

well we specifically chose v=w=(1,0,0)

you have a lot of choice for a linearly independent vector

[dull kettle]

you have a lot of choice because a = 0 right?

[spring wolf]

yes

literally any vector will do

[dull kettle]

okay, so how i'm understanding it is that, referring back to the original condition, we are trying to find a counterexample for u being dependent to v and w, and from the example where v = w = (1,0,0), we can establish the non-trivial solution as (0,1,-1) and this shows that u is independent to u and w

[spring wolf]

It doesn't show that u is

but it shows there's a non-trivial solution regardless of what u is

so we just pick a u independent of them

Because then we have a u that is not a linear combination of v and w, and we have a non-trivial solution which exactly what the statement says can't happen

Imma sleep soon but do you get how this works?

[dull kettle]

yes

thank you

[spring wolf]

Great

[dull kettle]

most patience teacher award goes to you

[spring wolf]

😊

i just answer questions here when I'm bored but too tired for other stuff

[slow scroll]

I'm not sure how to answer this using some argument involving dimension, but just to answer the hint, the dim(span(v1 v2 v3)) < 3. Thats all i know. Does that mean w1 w2 and w3 are linearly dependent as well, since this subspace cannot have 3 linearly independent vectors or something like that?

[spring wolf]

Hint before I sleep: consider where w_1, w_2 and w_3 are

[slow scroll]

they are linear combinations of v1, v2, v3, elements of a subspace with dimension < 3. so yeah i guess since the subspace has dimension less than 3, then there would be less than 3 vectors in any linearly independent set of vectors.

[misty quest]

UV=8 UW=-7 VW=6 -2U+6V=X what is WX can someone plz halp me I don't understand how to solve this

[placid oracle]

can someone explain how to do this problem?

[placid oracle]

<@&286206848099549185>

[wintry steppe]

@wintry steppe

[placid oracle]

@wintry steppe if this helps, these r the three conditions that make a subspace:

i think it should be b since a. c and d do not include the origin

which would account for the first property

[slow scroll]

lets take b. for example. Each vector of this subspace would look like
(x, -x/2) where x is in the field. Now to verify that this set is closed under addition.
let y be an element of the field
then (x, -x/2) + (y, -y/2) = (x+y, -(x+y)/2)
x+y is just another element of the field, and this vector still has the same form from before. so this is closed under addition. Check for multiplication now.

actually.... lol. You can still do this, its fine, but there is an even more basic thing to check: is the the zero vector an element of any of these subspaces?

[placid oracle]

@slow scroll thats what i was saying right above 😛

b is the only one with the zero vector in it

ur way works too

[slow scroll]

oops i didn't see that lol

[placid oracle]

haha nw i learnt another way to do it too

[slow scroll]

another way to think of it is that b is the only linear transformation there.

[placid oracle]

@slow scroll is this possible? because A in echelon form in the question, is there any way that a non-echelon version of that matrix would yield that as a basis for Col A?

i dont think it can

but im not sure

[slow scroll]

well the third column is a linear combination of the first two, but idk how you are supposed to know that the given set of vectors is a basis for A

[placid oracle]

well the pivot columns form it

so the first two

@slow scroll , but idk how to find out if a non-echelon version of that matrrix will yield those two matrices for the first two columns...

[slow scroll]

im not sure tbh

[sand hedge]

The basis of the column space will just be the pivot columns?

[dusky epoch]

the pivot columns will form a basis of Col(A)

[sand hedge]

[lean aspen]

@dusky epoch if you have a diagonalizable operator over a fdvs and its eigenvalues are nonzero, it'll be invertible and admit a basis of eigenvectors, right?

[dusky epoch]

that does not parse

please ask again

[lean aspen]

@dusky epoch

[dusky epoch]

yes.

[sand hedge]

Ann error detected

[wintry steppe]

lol

[hoary osprey]

lol

[kindred spruce]

determinant cant be 0

[tranquil schooner]

Hi I'm trying to remember a book I read

It was a linear algebra book

For undergraduates

And it was a yellow book

I think

And it had lots of images

[sage mauve]

how thick was it?

[rocky hill]

If there is a non-zero vector in the null space, how do I articulate that the columns of the matrix are linearly dependent? I mean in terms of language. Can I simply say "we have a non-zero vector in the null space, therefore the columns of A are linearly dependent" or is there some bridge I need to gap first?

[tranquil schooner]

@sage mauve it was a thin book

And extremely easy to understand

It was basically showing how eigenvalues had simple geometric interpretations

For example how translation and scaling work in linear algebra

So like something a college freshman would read

I can't find it but it was my favorite book because I liked looking at the images

[rigid cypress]

do you remember the edition of it?

[tranquil schooner]

I think it only had one edition

I read it 6 years ago

It's a small thin book

So not a regular sized textbook

Oof

[sage mauve]

the cover was something like this?

[rocky hill]

can somebody take a look at this?

https://imgur.com/a/i0nm2oq

[tranquil schooner]

@sage mauve no

But the form factor is similar

Just thinner

[wintry steppe]

axler?

but wait

[rocky hill]

Anybody?

[slow scroll]

@rocky hill What are you confused about? I'm confused about something too, but I'm curious whether its the same thing.

[sage mauve]

axler doesnt have lots of pictures doesnt it

is it linear algebra for dummies

i dont know if it has lots of pirctures but it sort of matches the form

oh but its pretty thick

[tranquil schooner]

It's not linear algebra for dummies

It's not linear algebra for dummies

[wintry steppe]

alright geez

[tranquil schooner]

It's probably linear algebra a geometric approach

[wintry steppe]

linear algebra: a linear algebraic approach

[frosty vapor]

math a math approach

[wintry steppe]

groundbreaking

[rigid cypress]

math: approaching

[frosty vapor]

limits

[flat pasture]

what is the difference between gaussian eliminatio nand gauss jordan

<@&286206848099549185>

thanks!!!

[half ice]

@rocky hill
This is late lol but yes that statement is strong

[rocky hill]

@half ice thanks. Turns out you can also actually use contradiction and then you don't even need to use matrix stuff.

[half ice]

Fair! Glad you got it

[rocky hill]

LA is the first math since algebra that I'm legitimately uncomfortable with lol

[half ice]

A lot of people are uncomfortable with it, it's a pretty different math for most people. If you understand your above statement, I would say you're doing pretty well!

[slow scroll]

@half ice I haven't actually learned rank theorem yet, but it seems like rank(A) + nullity(A) = n, since the vectors in the columns of A are vectors in R^n, not R^5?

[half ice]

That's rank nullity, yes

Why R⁵

[slow scroll]

https://i.imgur.com/WM0tHQjh.jpg

look at what blorgon posted. it says rank + nullity = 5, not n?

[half ice]

I was talking about this
If there is a non-zero vector in the null space, how do I articulate that the columns of the matrix are linearly dependent? I mean in terms of language. Can I simply say "we have a non-zero vector in the null space, therefore the columns of A are linearly dependent" or is there some bridge I need to gap first?
@rocky hill

I didn't see the other one lol

[slow scroll]

Oh. Is there a mistake in that picture though?

[half ice]

No need for "dim Null" as Nullity is already based on a dimension

[slow scroll]

well Nul is for null space. nullity := dim Null

[half ice]

Oh I see what you mean

Yeah sure

[slow scroll]

but it says rank + nullity = 5,
We are in R^n, not R^5

[half ice]

n = 5 here as there's 5 columns

[slow scroll]

rank + nullity = dimension of the vector space we're in i thought?

nvm i was mistaken

[half ice]

Rank + Nullity = size of the spanning set

I think lol

[slow scroll]

if A is an mxn matrix, then
rank A + Nulllity A = n

I thought it was = m, my bad

[rocky hill]

rank + nullity = columns

wait so dim Nul A is redundant?

[half ice]

Nullity = dim Null
Not redundant but irregular

[slow scroll]

makes sense i guess, sense rank = # of pivot columns and dim null = #free variables
pivot columns + #free variables = #of columns

I blame wikipedia for confusing me 😩

[flint sandal]

Oh I have a question. Since the set of all linear maps is a vector space. We can define the inner product of the space, we could then check how much two vectors of linear maps point in the direction of it. We could check something like if integration and differentiation are orthogonal to each other or if they point in the same direction. Would you us Hilbert space or something for that. I am just confused on the approach to it.

[maiden dagger]

"we could then check how much two vectors of linear maps point in the direction of it"

you're gonna have to make a little more sense in this part

[flint sandal]

oh define the inner product

and check for orthogonality i guess

like projection, but not

[lean aspen]

be more precise pls

[flint sandal]

uh that might be difficult

[maiden dagger]

so you just wanna put an inner product on some vector space on linear maps and then just evaluate it and see what happens?

[flint sandal]

i want to define the inner product on the vector space of all linear maps

[maiden dagger]

like ok, suppose you have an inner product on some function space, so what

[flint sandal]

and then see what happens

[maiden dagger]

and then what

[flint sandal]

so yeah

[maiden dagger]

what did you want to happen

[flint sandal]

i don't know

[maiden dagger]

like right now "suppose we have an inner product on L(V,W)"

okay, supposed

[flint sandal]

okay

i guess that is it

[wintry steppe]

Hey there, It's me, Bethesda's Todd Howard.

What's linear algebra like?

[maiden dagger]

linear

[slow scroll]

algebra

[wintry steppe]

Yes my child.

[maiden dagger]

it's pretty cut and dry

[wintry steppe]

I'm going to algebra 1 next year, and it has more than 200 endings now.

[lean aspen]

@flint sandal so might not be obvious but this doesn't always work

at least not in any interesting way

[flint sandal]

oh

i was just wondering how you would define the inner product on something like that

[lean aspen]

on L(V,W)?

[flint sandal]

yeah

[lean aspen]

so umm, try working with square matrices first

[flint sandal]

uhh

my teacher said use Hilbert spaces

[lean aspen]

🦋

[flint sandal]

but that is when i got confused

[lean aspen]

do know any analysis

[maiden dagger]

if V and W are finite dimensional and you pick a basis for each there are some naturalish inner products on the matrix representations

[lean aspen]

you can endow L² spaces

[maiden dagger]

i don't think in general there are cool inner products if V or W or both are themselves inner product spaces

[iron inlet]

I thought this channel was like highschool algebra 1 or 2. The name is deceiving.

[lean aspen]

it's literally linear algebra

[maiden dagger]

linear algebra is a standard first year college/university subject

[iron inlet]

ah ok

[maiden dagger]

if you want to talk hs algebra that's probably #prealg-and-algebra

[iron inlet]

Makes sense

[flint sandal]

so it would not be cool then

i was just thinking that checking if integration and differentiation were orthogonal was cool

[maiden dagger]

if you can put an inner product on a function space that has integration and differentiation type linear operators in it then you can try to ask but i can't really think of a natural way to do that

[flint sandal]

yeah that was the problem

and i wanted to see a generalized inner product on all linear maps

but that made no sense to me

[maiden dagger]

well remember you can't even just say "all linear maps", you need a domain and a codomain

[flint sandal]

oh

i see

[maiden dagger]

otherwise it's not even linear

what would adding a 5x3 matrix to a 6x24 matrix look like

it makes no sense

[flint sandal]

well wouldn't be infinite

or not

[maiden dagger]

like, L(V,W) (= the linear maps V -> W) has a natural vector space structure inherited from the vector space structure on W

f+g would be the linear map such that (f+g)(v) = f(v) + g(v), and kf would be the one where (kf)(v) = k*f(v)

[flint sandal]

oh i am starting to see

[maiden dagger]

no real way to transport an inner product off of W though

at least i can't think of one

[flint sandal]

thanks for the help

[maiden dagger]

thanks for a question that isn't a homework question 👍

[flint sandal]

oh i have anouther one

it is a less of a question though

and isn't fully thought out yet

i was wondering about the parallels between convex hulls and orbits. the center point and the centralizer are similar but tell different things and are for different data

[maiden dagger]

those have no relation

[flint sandal]

oh. so there is never any relation, except they kinda look similar

[maiden dagger]

"center" sounds like "centralizer"

as english words

[flint sandal]

i guess

well they do massively different things

and are for different types of data

i just thought they kinda looked similiar when they are finding the center and wondered if there was anything more behind it

but nope

thanks

sorry for the stupid questions i wished they taught more of it in school

[balmy bough]

Does anyone know what standard basis M31 is?

[tranquil schooner]

@balmy bough The standard basis in M31 is the standard basis for the 3x1 matrix

[dusky epoch]

is it just me or is this a stupid question bc the answer will be just X itself

[slow scroll]

maybe X isn't given relative to the standard basis in M31

[wintry steppe]

,w 2x +4y=15

[stoic python]

Display results online and refine query

Requested by Zenophius The Sin(pi)#0712

[winged pebble]

Hey does anyone have a good video that explains the process of finding the determinate of a 4x4 matrix

And one that explains how to find one of a 5x5

[broken hawk]

you really don’t wanna be doing that by hand, the caculations get really long really fast

unless there’s a lot of 0s

[winged pebble]

Gotta learn it for my test

[broken hawk]

but the procedure is pretty easily explained, one sec

[winged pebble]

Kk

[stoic python]

Sascha Baer:

[broken hawk]

this just for later

so, let’s say you have your 4x4 matrix (this’ll work exactly the same way for 5x5)

pick a row or column which has a lot of 0s

the more the better

and then you have to go along that row or column and calculate particular 3x3 determinants

let’s say you pick the first column

you would start at the top left

in your mind, remove first row and first column from the matrix

(the row and column containing the top left cell)

you’re left with a 3x3 matrix, right?

[winged pebble]

So you do this each time

[broken hawk]

you now have to compute the determinant of that, and multiply it by the number in the top left

then, since there’s a + in the top left, you add that

now you go to the next number

remove that row&column from the 4x4 matrix, calculate the determinant, multiply it with that number, and now subtract

[winged pebble]

So it’s kinda like a bunch of tiny 3x3 matrixes

[broken hawk]

yea, four of them

you can go along any row or column, but you have to pay attention to the signs

[winged pebble]

What are the tricks for 5 x 5

I know if A row repeats the determinate is 0?

Or something like that?

[broken hawk]

well, the brute force way would be to do it again like 4x4, but at this point you probably ahve to be clever

do you know gaussian elimination?

[winged pebble]

I do

[broken hawk]

good. gaussian elimination changes the determinant in predictable ways

specifically

any time you swap two rows, it gets multiplied by (-1), every time you multiply a row by some number λ, so does the determinant. if you add a mutliple of a row to another row, that doesn’t change the determinant at all

I’ll make an example afterwards of what I mean

now, the trick is this:

if you get your matrix to upper triangular form

then the determinant of that matrix is just the product of the numbers on the diagonal

and then, if you kept track of how you manipulated it, you can “undo” those changes to the determinant

for 5x5, this will definitely be quicker, even for 4x4 it probably is

again, unless there’s a lot of 0s

so, example time

(give me a moment, I’m drawing sth up)

[winged pebble]

Alrightyyyy

[broken hawk]

this is a 3x3 example cause like, effort, but it should show what I mean

so first, I do elimination to get it to upper triangular

then I can read the determinant of that by just multiplying the diagonal entries

then I go back

where I addedisubtracted rows to each other, I don’t change anything

but where I multiplied by some number, I now have to divide by that

and if I swapped two rows I’d have to do *(-1)

this is probably the most efficient way to do it in any even remotely large example

(there’s of course a reason why this works)

(it’s basically to do with how the determinant of the product of two matrices is the product of the determinants, you can represent each elementary row operation by a matrix multiplication if you so please; the things you’re doing here are just dividing out by those determinants again)

and yes, there’s some more trick
if you can tell that one of the following cases are true, then the determinant will be 0:
-one row/column is all 0s
-one row/column is a multiple of another one (including two rows/columns being the same, which is a multiple of 1)
-one row/column is the sum of (multiples of) other ones

[winged pebble]

Alrighty thanks a lot!

Imma try this method out!

Seems like it’s wayyy more realistic to do this in a test situation with limited time

Than just brute force

[broken hawk]

yea

for sure

[lean aspen]

Let $V = M_n(F), A \in M_n(F), T \in \End V$ such that $T(B) = AB$. Prove
$q_A = q_T$. \
%%%%%%%%%%%%%%%
Let $L_A: M_n(F) \to \End M_n(F), B \mapsto AB$ be the action on $\End V$ by left multiplication and fix $A$. For
$q_A(t) = \sum \beta_k t^k$. $L_A$ is linear and $L_A^k(B) = A^k B$, thus for all $B \in V$ we have
[(eval_{L_A} q_A)(B) =
\Bigg(\sum \beta_k (L_A)^k \Bigg) B =
\Bigg(\sum \beta_k A^k \Bigg) B = 0(B) =0
]
Therefore $(q_{L_A}) \subset (q_A)$.

In the other direction, let $q_{L_A}(t) = \sum \alpha_k t^k$.
Then for all $C \in M_n(F), \Bigg(eval_{L_A} q_{L_A}\Bigg)(C) = 0(C) = 0$. Thus $(q_{L_A}) \supset (q_A)$. \

$k[t]$ is PID and minimal polynomials are monic, therefore
$$q_{L_A}=q_A$$

[stoic python]

flimflam:

[lean aspen]

@dusky epoch @barren plank

not sure about second part

@wintry steppe ;__;

[barren plank]

what's q_...?

[lean aspen]

minimal poly

[barren plank]

I have no idea what that is

[lean aspen]

annihilates T

it's the largest ideal which does so

that is q_T(T) = 0, and for g(T) = 0, g ∈ (q_T)

get it together mniip

[barren plank]

are you like

considering End(V) as a ring?

cause Ab-cat?

[lean aspen]

well

you need to consider it as a ring for left multiplication lol

[barren plank]

not...really

[lean aspen]

since vector spaces don't have multiplication

[barren plank]

but M_n does

[lean aspen]

Not as a vector space

[brittle juniper]

so q_A and q_T are polynomials, but what meaning do you give to

$(q_{L_A}) \subset (q_A) $?

[stoic python]

Tuong:

[lean aspen]

q_A,q_T=q_{L_A} are polynomials in k[t]

so those are the ideals generated by respective poly

[brittle juniper]

Ah, ideals

[lean aspen]

well

basically the polys should just be equal

feel as though I skipped/missed a step at umm

Then for all $C \in M_n(F), \Bigg(eval_{L_A} q_{L_A}\Bigg)(C) = 0(C) = 0$.

Thus $(q_{L_A}) \supset (q_A)$.

[stoic python]

flimflam:

[lean aspen]

like maybe I should say
$eval_{L_A(C)}(q_{L_A})= ...$?

[stoic python]

flimflam:

[brittle juniper]

Mhmm, are $\sum\beta_k(L_A) ^k$ and $\sum\beta_k A^k$ supposed to be equal?

[stoic python]

Tuong:

[lean aspen]

ya

[brittle juniper]

But the first one is an application of M_n to End M_n, and the other is an element of M_n

They don't have the same type

[lean aspen]

umm

well properly the evaluation at B is equal to multiplying on the left by stuff

there's a B on the left

right

right

[brittle juniper]

if I follow your notations correctly, \ \
$\left(\sum\beta_k(L_A)^k\right)B\in\End M_n(F)$\
$\left(\sum\beta_k A^k\right)B\in M_n(F)$

[stoic python]

Tuong:

[lean aspen]

top should be (B) that is you should get a matrix back

oops

my bad

fixed

this is sorta messy sorry

[brittle juniper]

but top would still be the image of B by the application sum of the β_k(L_A)^k

which is an element of End M_n

[lean aspen]

that's just an endomorphism of M_n(F) tho

so if you evaluate at B in M_n(F), you get an element in M_n(F)

[brittle juniper]

$\left(\sum\beta_k(L_A)^k\right)$ is a an application of $M_n(F)\to\End M_n(F)$

[lean aspen]

oh

A is fixed

[stoic python]

Tuong:

[brittle juniper]

so when you evaluate at B, you get an element of End M_n(F), right?

[lean aspen]

L_A is in End M_n(F)

[brittle juniper]

Ah, now that's a bit different

but wouldn't it just be T then?

[lean aspen]

😰

ya

in retrospect that might have been easier

L_A = T

[slow scroll]

Ran A = column space of A

can someone explain the step I circled, It looks like A^T just gets factored out of the column space or something lol?

[sand hedge]

EA

[placid oracle]

having trouble here, they all seem false to me?

for c, doesn't Nul A need tobe = 0 according to the invertible matrix theorem

[slow scroll]

what about b?

@dame how is b false?

[placid oracle]

it isnt

[slow scroll]

ok then lol

[placid oracle]

wait, it is

A is the right answer

rankA+dimNulA = n

[slow scroll]

nono remember rank + nullity = rows

[placid oracle]

No, for answer choice b: 2+0 does not equal 3

and for answer choice c , 4+1 does not equal 4

[slow scroll]

um rows x columns

[placid oracle]

yes

wait

one sec

@slow scroll its not = rows, its = columns

[slow scroll]

oh wait yea oops

i had it mixed up yesterday to. I incorrectly corrected my correction from yesterday 🤦

[placid oracle]

haha good 😄

@slow scroll is there a nxn matrix where ColA would equal Nul A?

no right?

[slow scroll]

an nxn zero matrix would

[placid oracle]

oh yeah.. makes sense!

[slow scroll]

im not sure if there is anything less trivial tho

[placid oracle]

i think that would be the only one

[slow scroll]

yea i think so too.

[placid oracle]

@slow scroll srry mate, one more question: Can u explain why or why not this is true...

[slow scroll]

im pretty sure its true. They are linearly independent (not linear combinations of each other) and they have to span H because H is a 2D space. i.e. if they didn't span H, then there exists at least one vector that can be added to the basis, making it a >2d space @placid oracle

there is probably some better way to explain it though

[placid oracle]

ok so because theyre linearly independent, thery have to form a basis of a 2d subspace they lie in

[slow scroll]

yea

[placid oracle]

got it! ty

[slow scroll]

np

[frank warren]

Hi everyone!!
good to be here!
anyone here who is learning PCA?
I'd love to team up and learn/practice together!!

[wintry steppe]

what book you using?@frank warren

for PCA

I used D.C Lay, i have all worked out exercises

if yo can read my scribbles))

[frank warren]

i haven't started following a particular book yet

@wintry steppe

[wintry steppe]

have you started with linear algebra? excuse the question if you have

https://www.amazon.com/Linear-Algebra-its-applications-David-ebook/dp/B079H1WVK7/ref=sr_1_2?qid=1552147014&refinements=p_27%3ADavid+C.+Lay&s=books&sr=1-2&text=David+C.+Lay

let me find my notes, i scanned them in

some time ago, better pdf files then paper

[restive palm]

My initial thought for b is just 12000 - 12000/1.5 but I'm not sure why they asked to find the eigenvector in that case

[restive palm]

<@&286206848099549185>

[stiff surge]

so

prove that
1/4 + 2/8 + 3/16 + 4/32 ... = 1

directly

help 😭

[wintry steppe]

.

[brittle juniper]

Why is this in linear algebra??

[wintry steppe]

Don't ping helpers before 15 mins dummy

[stiff surge]

ahh sorry 😭

I don't know, this was given to me in lin algebra class 🤔

so i assumed this was lin algebra

[wintry steppe]

Did you even type what you meant to

[stiff surge]

yes

or technically, it's the sum of (n-1)/2^n from 1 to infinity

[wintry steppe]

That sum is equivalent to 1/2+3/16+1/8...=1, which seems sketch

[brittle juniper]

If it's linear algebra then, I suppose you have to do it through orthonomal total families of vectors and parseval?

[wintry steppe]

Oh I see

[stiff surge]

this was the example given that has fibbonacci sequence at top instead

what do you mean by sketch?

if it's not linear algebra, I apologize, please redirect me 😭 🙏

[brittle juniper]

That "add, matching like denominators" step, you can't do that until you've proven summability

[stiff surge]

summability?

[wintry steppe]

convergence

[brittle juniper]

you want to sum by grouping terms together

[stiff surge]

how would I do that?

[brittle juniper]

that's what the "proof" is doing

anyway, #calculus or any question channel is a better place for this

[stiff surge]

I'll move to there then thank you

[blazing cedar]

can someone explain what it means for columns to span Rn?

[ruby topaz]

S{α1,.......,αn}

[slow scroll]

ignore what i said im dumb.

[ruby topaz]

c1•α1+c2•α2+.......+cn•αn=O(O,zero vector)
c1=c2=....=cn=0(all scalar is 0.)
(α1,....,αn) basis
F1×n or R1×n

V is spanned by S.