linear-algebra

[spiral star]

since (v1, v2) are independent, the map defined by [v1, v2] is injective. hence, there is a unique inverse from its image to its domain

you can always invert injective functions if you restrict the codomain to the image

but inverting the matrix is the same as doing elimination

so you dont gain anything from it when you only need to do this for a single w

[hollow finch]

aha that makes perfect sense

tyvm

[spiral star]

@hollow finch oh i just realized that you were only talking about spaces spanned by 2 vectors right?

then you dont need elimination for the inverse

if $A = \mqty[v_1 & v_2]$ then $(A^\top A) \in \bR^{2 \times 2}$

[stoic python]

Flow:

[spiral star]

so you can invert that square matrix without elimination

and then your inverse becomes $U = (A^\top A)^{-1} A^\top$

[stoic python]

Flow:

[spiral star]

$(A^\top A)^{-1}$ is 2x2 so you can just invert it immediately by a formula

[stoic python]

Flow:

[spiral star]

this might be quite fast to compute

ah crap that was the wrong inverse... nvm then

lol

then it means you would have to do elimination, and i have to learn the difference between left and right

[hollow finch]

worked out for my particular example

[spiral star]

lol

[hollow finch]

i think that U works

[spiral star]

i thought i did it wrong

[hollow finch]

i mean it algebraically checks out

[spiral star]

oh

yea... right

[hollow finch]

((A^TA)^-1 A^T)Ac=((A^TA)^-1 A^T)w
(A^TA)^-1(A^T A)c=Uw
Ic=Uw
c=Uw

[spiral star]

yea lol i was doubting myself so much i constructed a false sanity check

[hollow finch]

happens to us all

glad you pinged me; I might not have seen it. thats a much better solution than elimination

[spiral star]

yea but it only works because A is injective

otherwise the square matrix wouldnt be invertible

[hollow finch]

that is true

so basically just dont be a dummy and make sure your set of vectors are linearly independent

[spiral star]

there is a similar formula to calculate the right sided inverse for a surjective mapping

[old flame]

@Flow ohhhhhhhh, I see. I think my assumption of range S=range T was stopping me from developing, so I guess the main point was to see that STv_1 results in a vector of span v_1

[spiral star]

yea

since since we chose v_1 arbitrarily, it applies to all non zero vectors

but for the 0 vector its trivial

T(0) = 0 so you can pick any scalar you want

next step is showing that all vectors get scaled by the same amount

you can take a basis again and know that each basis vector gets scaled by some factor

then see if you can show that all their factors were the same

the key here is that a basis is linearly independent

[old flame]

Could I just ask how do I stop myself from getting stuck from assumptions I've made that's wrong ?

I guess asking is one but is there other methods ?

[spiral star]

i guess you should always be skeptical about what you claim

assume everything is false and make sure you can follow the line of reasoning for why something is right from start to end

after a while you will get the hang of it and you wont have to break down arguments into their basic pieces anymore

but if you are new to some topic you should be able to relate every claim you make back to an already proved theorem or an axiom

for example, you didnt give any proof for why "range S=range T" would be true

you just said it would be

[native rampart]

Also,Be aware of how your assumptions might help you, supposing they were correct. For example,Your assumption range S being same as range T didn't seem to be useful

Then,check whether your assumptions are correct,as mentioned above

[old flame]

Ah okay thank you

[old flame]

From the above, for $v \in V$, $Tv_i=\lambda_i v_i$, so $Tv=a_1\lambda_1v_1+...+a_n\lambda_nv_n$. As $(v_1,...,v_n)$ are linearly independent $a_1\lambda_1v_1+...+a_n\lambda_nv_n=0$ when $a_1\lambda_1=...=a_n\lambda_n=0$

[stoic python]

Otoro:

[old flame]

We know that the $a_i=0$ since the the basis is linearly independent

[stoic python]

Otoro:

[native rampart]

Yes, but what do we get to know about $\lambda_i$?

[stoic python]

DrunkenDrake:

[native rampart]

Solution
||Take S(v_1)=v_2,and S(v_i)=0,for other basis vectors||

[old flame]

So a_i(\lambda_m-\lambda_n)=0 ?

[native rampart]

Yes

For any a_i

[old flame]

So lambdam = lambda n right ? Is that it then ?

Since m and n is arbitrary ?

[native rampart]

Yes

Now,Condense your entire proof

[old flame]

Sure thing

I will send it later, not at home now

Thank you very much as well @native rampart

[jade mirage]

$30<x+y\18<x+z\16<y+z$

[stoic python]

madeira:

[jade mirage]

hi, can i solve this with linear algebra or is this a non-linear algebra?

do i solve this normally?

[wintry steppe]

It is not linear algebra

[jade mirage]

for this system, is that correct?
$\30+18+16<2x+2y+2z\64<2x+2y+2z\x+y+z>32$

[stoic python]

madeira:

[jade mirage]

or i cant do this because its an inequality

[native rampart]

You can do that

But,I don't know what you get from that

Subtraction is not valid

[half storm]

how would you solve this?

I think some geometric intuition might help

[dusky epoch]

i mean the answer is a region bounded by three planes

well

"answer"

the solution set

@jade mirage is there anything in specific that you're asked to do with these inequalities

[jade mirage]

nothing in particular, just a random question

thank you everybody ^^

[old flame]

Heres the entire proof. Take any $v_1 \in V \setminus {0}$. Extend to the basis $(v_1,...,v_n)$ of $V$ then define S by $S(v_1)=v_1$ and $S(v_i)=0$ for the other basis vectors. Since $ST=TS$, $STv_1=TSv_1=Tv_1$, but $range S = span v_1$ so this implies $range S = {\lambda v_1 \mid \lambda \in F$. For the zero vector $T(0)=0 \dot \lambda$.

[stoic python]

Otoro:

[old flame]

Then, take $v \in V$, from the basis $(v_1,...,v_n)$, $v=a_1v_1+ \dots + a_nv_n$, applying T, $Tv=a_1Tv_1+ \dots + a_nTv_n$. From the result above, $Tv_i=\lambda_iv_i$ for $i={1,...,n}, so $Tv=a_1\lambda_1v_1+ \dots + a_n\lambda_nv_n$. As $(v_1,...,v_n)$ are linearly independent, $a_1\lambda_1v_1+ \dots + a_n\lambda_nv_n=0$, when $a_1\lambda_1= \dots = a_n\lambda_n=0$. Since we know $a_1= \dots = a_n=0$, $a_i\lambda_i - a_j\lambda_j=0$, this implies $a_i(\lambda_i-\lambda_j)=0$, therefore, $\lambda_i=\lambda_j$. Hence all the constants $\lambda_i$ are equal. $Tv=\lambda(a_1v_1+ \dots + a_nv_n)=\lambda v$.

[spiral star]

you might wanna rework that 2nd part

[old flame]

where though ?

[spiral star]

the linear independence argument

[old flame]

whoops typo

[stoic python]

Otoro:
Compile Error! Click the reaction for details. (You may edit your message)

[spiral star]

uh okay lets see

[old flame]

I'm looking at how to conclude now

[spiral star]

why do you know that a_1 = ... = a_n = 0?

[native rampart]

Ok,I thought you saw my solution and concluded that

[old flame]

Because those are the coefficients for the basis and basis is linearly independent

oh sorry, may I ask why $S(v_1)=v_2$ ?

[stoic python]

Otoro:

[native rampart]

But,a1=0 so the 2 lambdas need not be equal

[spiral star]

^

[native rampart]

Try substituting it and see

[old flame]

oh okay sure'

[native rampart]

Always substitute, when in doubt

[old flame]

but then aren't we redefining S ?

[native rampart]

S is arbitrary

We are taking a different S, call it S`

S` will also satisfy the commutative condition

Only T is fixed

[old flame]

one question, for Tv_n, won't it become \lambda v_{n+1}

and that is not in the basis, so do we rewrite it in terms of basis vectors ?

[native rampart]

Why?

[old flame]

we have $\dim V=n$ at the start though

[stoic python]

Otoro:

[native rampart]

S(v1)=v2 and others are mapped to 0

Now,vary the output of v1 to be the other basis vectors

Similarly take S(v2) and vary it across different basis vectors,while mapping others to zero

I mean ,vary S

[old flame]

so we define n S's ?

[native rampart]

Yes

And all such S will satisfy

[old flame]

so we stop at v_{n-1} to v_n ?

[native rampart]

Yes

[old flame]

So formally, we would define maps $S_{i}$ where $S(v_i)=v_{i+1}, \forall i={1,...,n-1}$ ?

[stoic python]

Otoro:

[native rampart]

No,we define maps $S_{i}$ such that $S_{i}(v_1)=v_i $ and $S(v_j)=0,j not 1 ,\forall i$

And repeat with v2 instead of v1 and so on

[stoic python]

DrunkenDrake:

[old flame]

ah okay, let me try to continue then

[native rampart]

Ok

[old flame]

so similar to the first case, each implies that $Tvi=\lambda_i v_i$ right ?

[stoic python]

Otoro:

[native rampart]

Yes

[old flame]

so since the $Tv_i$'s are each the span of $v_i$, where $v_i$ are the basis vectors, $Tv_i$ are linearly independent am I right ?

[stoic python]

Otoro:

[native rampart]

Yes

[old flame]

and hence $a_1\lambda_1= \dots = a_n \lambda_n=0$ yeh ?

[stoic python]

Otoro:

[native rampart]

Again,what is the point

[old flame]

we need to find lambda ?

[native rampart]

Yes,but why does this help?

[old flame]

this shows that T is the map from $v \to \lambda v $?

[stoic python]

Otoro:

[native rampart]

Doesn't show $\lambda$ is same

[stoic python]

DrunkenDrake:

[native rampart]

And That's what your statement means

[old flame]

so theres one more thing required to show that right ?

[native rampart]

Yea, That's where you use those n maps

[old flame]

alright, let me finish thias

[prisma pier]

I have to prove that for a matrix C, there exists a basis ${e_1, e_2, ..., e_N}$ such that $(Ce_1=0$ and $Ce_{n+1}=e_n) \iff (C^N = 0$ and $C^{N-1}\neq{0})$

[stoic python]

smh:

[prisma pier]

I've done the forward proof

and I think I know how I would do the converse with Jordan normal form

but the textbook im using hasn't introduced jordan form yet

so there must be a way to show it without the use of jordan normal form, though I can't figure it out

[native rampart]

Ok, nvm

Take e,Ce, C^2e...

As a basis

For some e

Show they are linearly independent

And form a basis

[prisma pier]

ok ty for the hint I'll keep trying

Ah! I got it. Thanks again

[old flame]

@native rampart this ? range T = span v_1 = \dots = span v_n

[native rampart]

Well, span v1, span v2 are all different

Do not think of interms of range and span,think in terms of basis vectors and what the transform does to them

[old flame]

the transform scales each of them by a associated scalar

[native rampart]

What is TS(e1)?(considering S(e1)=e2,and S(ej)=0 otherwise)

And compute ST(e1)

Equate,and you are done(repeat with other S,varying S(e1))

[edgy orbit]

how do I prove that ||u||-||v||<=||u-v||

should I ask later, it seems like this channel is busy

[winged iris]

use the triangle inequality

[edgy orbit]

well yes, but I am clueless where to start

because it looks like the left and right are opposite in this

I think I got an idea

[native rampart]

Square both sides and expand

[old flame]

For each $i \in {1,..,n}$, $TS_{i}(v_1)=Tv_i$, whereas $S_{i}Tv_1=S_{i}(\lambda_i v_1)=\lambda_iS_{i}v_1=\lambda_iv_i$. So $\lambda_i v_i=Tv_i$ ?

[stoic python]

Otoro:

[native rampart]

Don't abstract too much

[edgy orbit]

oh, I was thinking of doing ||u|| = ||(u-v) + v||

[old flame]

take the first case one ?

[native rampart]

Yes

[old flame]

but how does it generalise though ?

[native rampart]

Just do it first,generalise later

You will get a feel of how it will generalise,if you do a specific case

For each $i \in {1,..,n}$, $TS_{i}(v_1)=Tv_i$, whereas $S_{i}Tv_1=S_{i}(\lambda_i v_1)=\lambda_iS_{i}v_1=\lambda_iv_i$. So $\lambda_i v_i=Tv_i$ ?
@old flame $Tv1$ will be $\lambda_1 v_1$ and $Tv_i$ is $\lambda_i v_i$

[stoic python]

DrunkenDrake:

[old flame]

omg I got it finally

lol

is this ok ? So from the above equality, $TSv_1=Tv_2=\lambda_2v_2$ and $STv_1 = S(\lambda_1v_1) = \lambda_1S(v_1)=\lambda_1v_2$. Implying that $\lambda_1=\lambda_2$. Applying this method to every pairs $v_i,v_j$, we obtain an equality of $\lambda_1=\dots=\lambda_n$.
Therefore, $Tv=a_1\lambda_1v_1+ \dots + a_n\lambda_nv_n=\lambda(a_1v_1+ \dots + a_nv_n)=\lambda v$.

[native rampart]

Well, That's how 90% of linear algebra works

Yes

[stoic python]

Otoro:

[native rampart]

You substitute some values

[old flame]

wow this took much longer than expected

[native rampart]

Yea,just plug in values

[old flame]

ok wait, so 90% of linear algebra is like this ?

[native rampart]

And it works

[old flame]

why though

[native rampart]

Well Not 90%,more like 40% the other time is finding a "basis like" thing you care about

Like canonical forms

[old flame]

but isn't guessing the required values difficult ?

like how do you guess in the first place

[native rampart]

Why do you think basis vectors exist?

[old flame]

and btw am I done with the question lol ?

[native rampart]

Yes

To get substituted

[old flame]

basis vectors exists, so that we could represent all the vectors with it

simplifies things ?

[native rampart]

Also, everything can be known,if we just take the basis vectors case

[old flame]

substitute ???? im sorry what are you referring to ?

[native rampart]

Here,we wanted to know about T

So,we plugged in S such that our process simplifies significantly

A choice would be the one where only 1 basis vector doesn't map to zero,but to a basis vector

(well,Such transforms form a basis of all linear transforms. Proof is homework)

That's our "basis vector" substitution

Also,one for the verification

[spiral star]

oh you finished the proof? 😄

[native rampart]

Yes

Finally

[spiral star]

lol

is this where i post my version as an alternative?

[native rampart]

Yes

Wait,ask him first

[spiral star]

waiting

well are you done or not

[native rampart]

The proof is done

But,ask him anyway

[spiral star]

alright

[old flame]

yeah I'm also very glad that I'm done, took way too long imo

[spiral star]

eh you will get faster once you figured out some general approaches

[old flame]

I mean, getting $Tv=\lambda v$ is the result right ?

[stoic python]

Otoro:

[native rampart]

Yes

[old flame]

when do you actually get to know those general approaches ?

[spiral star]

you just learned one

[old flame]

by doing more problems ?

[native rampart]

Yes

[old flame]

So I guess more struggling then , not saying that I dislike struggling, cause I enjoy the process to some extent though

[native rampart]

Linear algebra is probably the most straightforward

[old flame]

but it does take up so much time

[native rampart]

Also, Extremely important

[spiral star]

honestly, analysis was even worse for me lol

[old flame]

lol I tried doing analysis too, goddamn

[native rampart]

Have you tried group theory?

[old flame]

but could I just ask, is this normal ? or am I just taking too long or not understanding stuff

[spiral star]

xd

[native rampart]

Well,It's your first time

[old flame]

nope, haven't touched abstract algebra

[native rampart]

So, This is normal

[spiral star]

you just have to figure out how it works and then it will get much easier

once you understand the structures a bit better

[old flame]

Cause I'm struggling almost in all subjects I try to learn

[spiral star]

struggling with math is completely normal

[old flame]

i.e analysis 1, probability theory and this

[native rampart]

Well,establish a basis like thing. Solve .

[old flame]

"how it works" hmmm what does this mean though

[native rampart]

You will understand once you do more problems

[old flame]

I guess a technique I've learnt from this, is to define some objects in the question ?

[native rampart]

Yes

[old flame]

cause when it said TS=ST for every S, I don't know that you could specify

[spiral star]

your toolbox for linear algebra isnt that big. you just have basis, linear independence and maybe invariant subspaces

[old flame]

oh, so generally try to use these as an approach to questions yeah ?

[native rampart]

That's the gist,the particulars aren't obvious

[spiral star]

yea there isnt much you can do :p

[old flame]

but if were like this, then those wouldn't work I guess ? right ?

[spiral star]

and the more abstract your setting is, the fewer tools you can use, so that makes the number of possible approaches really small

[old flame]

but doesn't less approaches seems to make starting easier ?

[native rampart]

No

Those approaches are pretty unobvious.

[spiral star]

to me abstraction makes things easier

because im left with the only obvious choice

[native rampart]

Well,I like to start with an example,and then abstract

[spiral star]

anyway, otoro do you want to see my proof

i forgot to ask

[old flame]

ok.... so I guess different preferences of approach for different people ? I will have mines once I do more problems ?

[native rampart]

Yes

[old flame]

yeah of course !

please do

but the only source of problems is from textbook, not a ton, so where would you have more ?

[spiral star]

alright, i sprinkled in some "(why?)" annotations where i want you to convince yourself of the fact that i can make those claims.

[old flame]

sure thing

[native rampart]

but the only source of problems is from textbook, not a ton, so where would you have more ?
@old flame really,not a ton?

Maybe,You should change your book,in that case.

[spiral star]

isnt it LA done right?

[native rampart]

Math books have plenty of problems

[old flame]

@old flame really,not a ton?
@native rampart for example for this chapter axler has 26

[half storm]

just do problems from other books as well

[old flame]

make a collection of textbooks then ?

and one thing, could I have access to those advanced forums ? I'm starting my first yr maths next week :3

[spiral star]

advanced forums?

[native rampart]

,i am adv ones

[old flame]

type here ?

[spiral star]

oh those channels

[old flame]

yeah those......

[spiral star]

yea then just do what drake said

[old flame]

,i am adv ones

[spiral star]

,i am adv

usually you would do this in the bot channel

the channels also tell you how to get access

i think its in the description

[old flame]

ah alright thanks

[spiral star]

welp i think i got the command wrong

¯_(ツ)_/¯

[old flame]

its ok, I just wanna say thank you so much guys

took me long enough to complete the chapter

[spiral star]

when you have time to read through my proof, then you will see that i only used span and linear independence of a basis for this proof

[old flame]

I'm more than excited to start my degree very soon

sure thing, its 2am here and Im tired lol, I will read it tmr and find you if I have any problems, much appreciated !

[spiral star]

lol

but hey you completed the proof 🙂

[old flame]

yes, another 4 hours today, but I got it

overjoyed

It could be helpful if you could share some experience lol, since sometimes taking up so much time makes me feel incapable

[spiral star]

oh there is a thing i wanted to mention about my alternative argument for (1)

i originally thought it was a proof by contradiction

but it isnt

it is negation

so all is good

[old flame]

oh

I have heard that generally want to avoid contradiction proofs right ?

thanks for mentioning though 🙂

[spiral star]

well its kinda nice to use a different argument if you can

i guess for now you could say it's a more general proof if it doesnt use contradiction

usual suspects to avoid if you can, is contradiction and choice. but it's not like you will have to use them for linear algebra in finite dimensions anyway

[old flame]

alright

what is choices ? haven't heard of this

[spiral star]

axiom of choice

[old flame]

ohhh this one, I don't really understand how this works as a start for a proof though

[spiral star]

its used when you have to choose infinitely many elements at the same time

(roughly speaking)

but you wont need that for finite dimensional LA

well "at the same time" probably wasnt the best analogy

it's a powerful addition to the ZF axioms

but when you add it, then it implies some other things that might be questionable, so mathematicians arent really sure whether they should use it or not. classical math has widely adopted it but you usually say explicitly when you are going to use it

and its avoided if possible

depending on your math foundation you might not be able to use it for proofs at all

and if your approach is intuitionistic, then you can say goodbye to the full axiom of choice and also to contradiction

[old flame]

I see, so in general attempt a direct proof, if it couldn't be done then try these then

[spiral star]

proof by negation is also fine

like i did

[old flame]

gotcha !

[spiral star]

anyway, if you wanna know more you can probably ask in the foundations channel

[half storm]

if we were to reject AOC basis - which would be equivalent to rejecting infinite-dimensional - wouldn't we lose alot of math that we enjoy using?

LIke how a function under certain conditions can be expressed as a fourier series?

I think I remember this from my PDE's class but I could be wrong.

[old flame]

no worries, thanks for explaining

[spiral star]

yea we would lose some stuff lol

The Axiom of Choice is obviously true, the Well-ordering theorem obviously false, and who can tell about Zorn’s lemma?

[half storm]

lol I've heard this meme quote

Like alot of imporant shit like fourier analysis depends on it I think.

Because with it we get to assume that there exists an infinite dimensional basis for smooth functions, and fourier series comes out of that.

[spiral star]

i havent done fourier analysis yet

for the bit of analysis on infinite dimensional spaces i learned i didnt need choice

[half storm]

Really, you need a logical equivalent to choice - which would mean needing choice yea?

Like Zorn's Lemma/ Hausdorff Maximal Principle?

[spiral star]

i didnt need zorns lemma for any arguments so far.

except for the proof that every vector space has a basis

but i didnt need that argument for anything so far

i needed choice in universal algebra once i think

[half storm]

Oh I see what you're saying

just for that part on showing that every vector space has a basis

[spiral star]

i had some horrible proofs in universal algebra and category theory that made use of choice

i would expect it to turn up in analysis

maybe i will learn a few proofs needing it soon-ish

my math background is kinda weird

[half storm]

lol mine too

went to a liberal arts school where not many people went to grad school or really cared to take the higher level classes

we were bounded in what we could learn there

anythign else you had to learn on your own.

and I was a bad student so I have gaps in my knowledge compared to most people on here.

I know enough to help some people, but I make some mistakes sometimes.

I'm going back and trying to fill in those gaps now.

Taking a gap year to do it.

[spiral star]

makes sense

[half storm]

and then apply maybe for a graduate program in the fall of next year

[spiral star]

i studied computer science before

and i specialized in that area where theoretical comp sci and algebra meet

so i took a lot of more advanced algebra classes

and universal algebra + category theory

[half storm]

I was thinking about doing this actually lol.

[spiral star]

but i never studied much analysis

[half storm]

I think I want to do probabilistic logic or applicatoins of probablility to computer science.

[spiral star]

that's cool

now that im studying actual math i'm choking on analysis

[half storm]

You'll do fine though I'm sure. So you studied CS in undergrad and swiched to math in grad school?

[spiral star]

i finished my masters degree in cs and then started math from scratch

[half storm]

Got you.

[spiral star]

im now starting my 2nd year of math in undergrad

[half storm]

Oh wow

cool.

Maybe I can ask you some stuff about CS if you ever got time.

or are online.

[spiral star]

yea im probably always online lol

my next semester was delayed because of covid reasons

so i have nothing to do til end of october

except reading some books and lurking around here

[half storm]

What made you decide to switch to math?

[spiral star]

realized that i hate working as a programmer 😂

[half storm]

lol really?

You could have probably done more than that if you had a masters

software development?

[spiral star]

i actually had a decent job in research

i specialized in high performance computing

[half storm]

oh wow

[spiral star]

the problem solving aspect was nice, but writing the code was kinda awful

[floral thistle]

the problem solving aspect was nice, but writing the code was kinda awful
@spiral star Why?

[spiral star]

because nothing works in hpc

you rely on compilers like icc and nvcc and they are quite far behind gcc or clang when it comes to supporting c++ standards etc.

we constantly fought the compilers and reported compiler bugs

outdated libraries, proprietary compilers that dont support the features you need or have bugs

debugging GPU code is also not easy

it felt like making fire with sticks and stones

then we also kinda had to build a lot of libraries from scratch because they didnt exist before lol

and so we used some pre-alpha versions of libraries from like a different team of researchers

it's all open source of course, but it wasnt ready for usage and we were sort of alpha testing it

so then you add those bugs as well

in the end i was never working on my own project, but just attending meetings, and writing pull requests and bug reports for projects i wasnt actually involved in

and if the alternative to this is working a boring job writing ecommerce code in java where im actively losing iq points every day...

then i'd rather do something else

so i picked math

and it's much more fun already :))

[half storm]

That actually sounds really boring

lol

[spiral star]

i could meet a bunch of smart people and we tackled some pretty interesting problems

i learned a lot i would say

and you could make some business trips xd

i went to switzerland for a week

and france for 2 weeks

of course all work related 😉

[half storm]

lol

Did you ever try building your own compilers from scratch. That probably is a heavy-duty task though and possibly not feasible.

[spiral star]

only for toy languages but nothing serious

[half storm]

What do you want to do after getting your degree in math?

[wintry steppe]

how can a matrix be antisymmetrical?

def: M^T = -M

[limber sierra]

as an example

[wintry steppe]

ah I see, couldnt find an example in my head

thanks

[spiral star]

you can probably also take a 2x2 identity and negate one of the 1s

[limber sierra]

uh

[spiral star]

like $\mqty[0 & -1 \ 1 & 0]$

[stoic python]

Flow:

[limber sierra]

ah, i see

thats not an identity matrix though

[spiral star]

oh yea LOL

mb

[wintry steppe]

yeah i thoughr so, diagonals have to be zero others wouldnt work

[spiral star]

for 2x2 yea

or in general

actually

yea...

i cant think straight anymore

i'll see myself out

[wintry steppe]

uh example of hermitische and anti hermitsch matrix

hermitian is what it's called

ah right

all i becomes negative and then transpose the matrix right?

or all i become positive

you switch the sign on all the imaginary parts of the entries

right

that is the conjugate part

then you take the tranpose

gives you the conjugate transpose

maybe a silly example is any real symmetric matrix is automatically hermitian

and any real antisymmetric matrix is automatrically antihermitian

for a slightly less silly example of a hermitian matrix,
$$\begin{pmatrix}
0 & 1 - i \ 1 + i & 0\end{pmatrix}$$
is a nice one

[stoic python]

TTerra:

[wintry steppe]

(i called those examples silly because they don't even have complex entries, where you'd usually be working with matrices with non-zero imaginary part when discussing hermitian matrices)

[wintry steppe]

which of the following are symmetrical, antisymmetrical, hemirtain , anti hermitian

donmt know why we're not given a solution pdf...

do you know how to start?

but I find that
a is symmetrical

h is symmetrical

do it matrix-by-matrix

start with a and see which of the four conditions you're given it satisfies, if any

you got at least one of them down

b is nothing right?

none of those 4

hmmm

i think it satisfies at least one of those

well it's not symmetrical

also not anti-symmetrical

now what about (anti-)hermitian-ness

also not right

-3 doesnt become 3

when taking conjugate

i might have just had a brain fart

haha no worries, happens

i'm on a roll with those today

well i got you to check your work so it worked out

thanks for the help

i mean i didn't really do anything lol

just forgot how conjugates work

[wintry steppe]

is it correct to say A^2 = A*A?

for exercise 4

[rose coral]

Yeah A^2 is defined to be A * A

[wintry steppe]

you don't take the exponent inside the matrix, right. so it's not correct to say A^n = every number inside A to the nth exponent

[rose coral]

Yeah, its multiplication of matrices, not taking exponents of entries in A

[wintry steppe]

ok in excercize 6, what's the I refered to in (X- I) (x- 2I). is it the identity matrix?

and if so how to know what the size of the identity matrix?

[rose coral]

Yeah it is the identity. In this case it should be a 3x3 identity matrix

Usually they don't say what size it is and you have to figure it out based on context

So here, to be able to add/subtract like in X - I and X - 2I you need I to be 3x3

[wintry steppe]

right, ok thanks

[rose coral]

No problem

[lucid cedar]

Hello #linear-algebra,
I am back again with stupid questions. I am trying to make sense of statement d.

I think I have a proof of the statement. Namely:
b = (f + g)'(x) = f'(x) + g'(x) = b + b

The problem I am having is the consequence of this fact. Does this mean that the function y = 2x is not a subspace of R^2. That seems to go against my idea of a subspace as I understood it in my LinAlg + DiffyQ class. On top of this conclusion, I fail to see the importance of R^(0,3). Are these bounds important to the question in some way?

[limber sierra]

Does this mean that the function y = 2x is not a subspace of R^2.
I'd assume you mean the line that this function defines. If so, why would it?

also note that your proof only proves the "only if" direction

you still need to prove the "if" direction

[lucid cedar]

I dont know why i assumed 2x should be a subspace, bad intuition i suppose. Bad habit but sometimes leaning on intuition is the best way to get me through a final that i dont understand that well haha

im trying to see how i would prove the if direction; I have an idea but im not convinced it proves the "if" direction

[gray dust]

how do you show a set is a subspace?

[latent ledge]

Suppose $T \in\mathcal{L}(V)$ has a diagonal matrix $A$ with respect to some basis of $V$ and that $\lambda\in F$. Prove that $\lambda$ appears on the diagonal of $A$ precisely $\dim E(\lambda; T)$ times.

[stoic python]

Konoha:
Compile Error! Click the reaction for details. (You may edit your message)

[latent ledge]

so, I know that $V=null(T-\lambda I)\oplus range(T-\lambda I)$

[stoic python]

Konoha:

[latent ledge]

eigenvectors of lambda is a basis of null(T-\lambda I), the basis of range(T-\lambda I) are eigenvectors belong with other eigenvectors, that is what I have so far

[lucid cedar]

To clarify my confusion i forgot closed under multiplication was specifically scalar multiplication. I ended up answering most of my own questions with a quick review. Thanks again everyone.

[short fiber]

can someone tell me what the capital A means?

K(x) is a matrix that is obtained by some weird operation on a bunch of k_i

[wintry steppe]

context?

[short fiber]

k_i(x_i) is a 24 by 24 matrix

what is that capital letter A supposed to be doing on several k_i?

[wintry steppe]

ive never seen it before so i feel like this will be explained in whatever you're reading (or in what it references)

can someone give me an exmaple of adjunt matrix

like definition wise but with an example

what's the adj(A)

definition wise: Adj(A) = C^T with C = all cofactors

[half storm]

engineering notation christ

[wintry steppe]

guys I'm not even halfway and I've got exam in 6 hours, Don't even know how to calculate eigenvalues... keep going or just get some sleep?

calculating eigenvectors is just finding bases of subspaces @wintry steppe . are you comfortable with doing that?

recall that an eigenspace is just ker(A - lambda I)

wait

eigenvalues is just finding the roots of det(A - t I)

what about it confuses you?

I've never solved or even attempted solving it, but I can try with the formula you gave me. t is never given I assume?

det(A - tI) is a polynomial. the roots of this polynomial are the eigenvalues of the matrix A

sometimes you'll see det(tI - A), but the roots are the same so it doesn't matter

alright lets attempt one

i'll try a

[pulsar turret]

Like this?

[wintry steppe]

why did you change colors

lol

[pulsar turret]

I use 2 different accs

Lol

[wintry steppe]

ccr180

or something

dont know why its upside down, want me to take new pic?

a rough outline of what you should do is

1. calculate the roots of the polynomial det(A - t I). the roots are the eigenvalues of A
2. calculate bases of the eigenspaces ker(A - λ I), for each eigenvalue λ of A. these will be your basis eigenvectors (although any element of ker(A - λ I) will be an eigenvector, you want a basis because...)
3. if the multiplicity of the eigenvalue λ as a root of det(A - t I) equals the dimension of ker(A - λ I), for each eigenvalue λ of A, then A is diagonalizable. if not, then A is not diagonalizable

ok so I'm pretty sure a question will be about diagonlizing a matrix. doing the steps you showed does not diagonlize it, it's the steps needed to do to see if it is even able to diagonalize

soo let's say I want to diagonlize excercize a

how do you do step 2?

yeah i realize i kind of vomited a bunch of words at you without any explanation lol

I read about ker somwhere but it was very abstract that I could not make sense of it

it is the set of vectors that get sent to zero by the matrix

so ker(A - λ I) is the set of all vectors v with (A - λ I)v = 0, or if you do some rearranging, Av = λv; eigenvectors!

[gray dust]

in fact you should've started the lecture with Av=tv

[wintry steppe]

ree

i was preoccupied with a game

[gray dust]

excuses

[wintry steppe]

in fact i should've started with diagonalization, got eigenvectors from that

:^)

[gray dust]

should start with defining unsigned/oriented volume & det

[wintry steppe]

so ker(A - λ I) is the set of all vectors v with (A - λ I)v = 0, or if you do some rearranging, Av = λv; eigenvectors!
@wintry steppe I tried to make sense of it but, I think I need an example

I don't see how to compute Ker(B)

[gray dust]

ex, find ker(2x2 identity)

let $x=(x_1,x_2)^T\in\ker(I)$ so $Ix=\mathbf0$ or $$\m{1&0\0&1}\m{x_1\x_2}=\m{0\0}$$

[stoic python]

RokettoJanpu:

[wintry steppe]

so, x1 and x2 = 0?

[gray dust]

yes

try ker(2x2 with all entries 1)

[wintry steppe]

uh

x1+x2 =0

so x1 = x2

[gray dust]

bad math

[wintry steppe]

2 times

[x1 +x2 | 0
x1+x2 | 0]

[gray dust]

x1+x2=0 doesn't give x1=x2

[wintry steppe]

right lol

x1 = -x2

[gray dust]

what's the form of vector in ker?

[wintry steppe]

form?

no idea

what you mean by it

so x2 can be any number k, so x1 = -k

[gray dust]

let x=(x1,x2) be in the kernel. you actually should've said this, you never introduced x1,x2

you got x1=-x2 so x=(x1,x2)=(-x2,x2) where x2 is free to take any value

[wintry steppe]

ah I see

[gray dust]

if you're familiar with span you can write ker=span{(-1,1)}

[wintry steppe]

if we do matrix a

diagonlize it

[pulsar turret]

So this is how you’d do it? Notations are probably wrong, but overal going in the correct direction?

[wintry steppe]

so what exactly are my eigenvectors?

[gray dust]

pay attention to the computations you're doing. https://i.gyazo.com/25fbd855258e2989877580afe150ff04.png

in finding ker(A-5I) you find vectors v where (A-5I)v=0. what you did is let v=(x,y) and find restrictions on x,y

[pulsar turret]

Is that wrong?

Or maybe unpractical?

[gray dust]

i never said those, i'm summing up what you did

[floral thistle]

https://discordapp.com/channels/268882317391429632/540211747613704221/748720650751049858
Is that german?

[pulsar turret]

Ok so I found x and y coords of eigenvector v. But x and y can be any number

[gray dust]

general note, it's bad to introduce variables without saying what they are

[pulsar turret]

And thats logical because the eigenvector is actually like an infinite line. So every point on that line should be reached

[gray dust]

notice i say every time "let v=(x,y) be in ker... where x,y are any number"

[pulsar turret]

Ok I’ll remember it

[gray dust]

then you can stick x,y in there, do some computations to find restrictions on what x,y are

[pulsar turret]

Right I see

And what I said earlier thats correct right

That the eigenvector is actually infinite

And that’s why x,y can be any number

[gray dust]

not quite, x,y aren't free to take on ANY values. in getting ker(A-5I) you got 3y=4x, so you can write y in terms of x like y=4x/3, so v=(x,y)=(x,4x/3)=x(1,4/3). we say we have 1 free variable in defining the vectors in ker(A-5I)

it's true that A-5I is a singular matrix which is equivalent to A-5I having a nontrivial kernel which means the eigenvalue 5 has infinitely many corresponding eigenvectors

[pulsar turret]

I see, and for eigenvalue 2. I found both x and y were 0. That’s still considered an eigenvector, correct. v = (x,y) = (0,0)

Thank you for helping out, I appreciate the help

[gray dust]

@pulsar turret eigenvalue is -2, everything for that is wrong

you should also know

it's true that A-5I is a singular matrix which is equivalent to A-5I having a nontrivial kernel which means the eigenvalue 5 has infinitely many corresponding eigenvectors
is true for eigenvalues in general, they have infinitely many corresponding eigenvectors, so if you got a SINGLE vector, and especially if it's (0,0) then you messed up

[pulsar turret]

Eeh

Ok, Good to know

[gray dust]

everything wrong started here https://i.gyazo.com/5b8c2539b87fae12b498be8df2979589.png

[pulsar turret]

-(-) = +... I understand

[knotty raven]

<@&286206848099549185> How can I prove + understand whether or not I can find a 2x2 matrix with rows that are linearly dependent and columns that are linearly independent?

[native rampart]

There are no such matrices

The dimension of row space=dimension of column space

[native rampart]

Change of basis?

[stoic python]

n256:

[knotty raven]

@native rampart can it be proved directly from the definition of span and linear dependence? the 2x2 matrix thing from earlier?

[native rampart]

You have to know row rank=column rank

Don't think follows directly

[native rampart]

can someone help me to understand transition matrices? isn't the idea to find a matrix that takes in each vector in one basis and outputs a corresponding vector in another basis?
@wheat shoal Are you aware that
$[\vec{v}_1,\vec{v}_2]v$=$[\vec{u}_1,\vec{u}_2]u$

[stoic python]

DrunkenDrake:

[native rampart]

Because both represent the same vector

The v will be a pair of numbers (a,b), such that av1+bv2 is the vector

u will be pair (c,d) such that cu1+du2 is the vector

Multiply both sides by
$[\vec{v}_1,\vec{v}_2]^{-1}$

[stoic python]

DrunkenDrake:

[native rampart]

You get a map u to v

And that's your change of basis matrix

$[\vec{v}_1,\vec{v}_2]^{-1}[\vec{u}_1,\vec{u}_2]$

[stoic python]

DrunkenDrake:

[native rampart]

From u to v

[old flame]

@spiral star Just read your proof. For the first (why ?), is it because the zero vector $T(0)=0$, so $T(0)-0=0$. For $n=1$, the span is the vector itself, this just transforms the only vector into a scalar multiple of itself. For the second why, $n=1$ just got proved and $n=0$ is just $T(0)=0 \times \lambda=0$ right ?

[stoic python]

Otoro:

[old flame]

I am a bit confused on $\lambda \sum_{i=1}^{n}v_i = T(\sum_{i=1}^{n}v_i)$. Is this saying since T assigns a scalar to respective basis vectors, it assigns another one for the sum of it ?

[stoic python]

Otoro:

[spiral star]

just let $v := v_1 + \dots + v_n$. then from (1) we know that $T(v) = \lambda v$ for some $\lambda$.

[stoic python]

Flow:

[spiral star]

then you unwrap it

[native rampart]

I am a bit confused on $\lambda \sum_{i=1}^{n}v_i = T(\sum_{i=1}^{n}v_i)$. Is this saying since T assigns a scalar to respective basis vectors, it assigns another one for the sum of it ?
@old flame did you understand Tv,v dependent part?

[stoic python]

DrunkenDrake:

[native rampart]

For all v

[old flame]

just let $v := v_1 + \dots + v_n$. then from (1) we know that $T(v) = \lambda v$ for some $\lambda$.
@spiral star So here you're taking another arbitrary v in V and using what we know about T yeah ?

[stoic python]

Otoro:

[spiral star]

(1) applies to all vectors v

therefore it must also apply to the sum of the basis vectors

[old flame]

@old flame did you understand Tv,v dependent part?
@native rampart $Tv_i - \lambda_i v_i=0$, so each $Tv_i$ is linearly dependent to $v_i$. Since every v is made up of $v_i$, the sum is linearly dependent as well ?

[stoic python]

Otoro:

[native rampart]

He didn't quite do that

He formed a new basis and repeated the process

Since,You can form a basis which includes one particular vector v,This means v,Tv will be linearly dependent for all v

[old flame]

New basis being $(Tv_1,...,Tv_n)$ ?

[stoic python]

Otoro:

[native rampart]

No,new basis which starts with an arbitrary vector v,and is extended to span the entire space

Read the first part carefully

[old flame]

wait, in the first part it said linearly dependent, and you said its independent so is there a typo ?

[spiral star]

typo

writing "independent" is like a reflex lmao

[old flame]

so v,Tv is independent instead yeah ?

[spiral star]

(v, Tv) are dependent

[old flame]

oh LOL

[spiral star]

i wrote dependent in my proof

it is no typo there

[old flame]

yeah, but @native rampart say it is independent not ?

[spiral star]

that was a typo

[old flame]

oh I'm so sorry

[native rampart]

Reflex. Meant dependent

[old flame]

or is it because $Tv_i$ is a scalar multiple of $v_i$, so they can't be independent right

[stoic python]

Otoro:

[native rampart]

Yes

[old flame]

so this shows that T is dependent for all basis vectors, hence dependent for all vectors ?

[native rampart]

For any choice of basis vectors

I.e.,If you choose any basis,the basis vectors and their images will be linearly dependent

And you can always create a basis which contains a specific basis vector,we need

[spiral star]

(in-dependent)

[old flame]

we need ?

[native rampart]

I meant,for any vector we choose,we can find a basis,which contains the vector

[old flame]

so like just taking less basis vectors ?

[native rampart]

Different set

[old flame]

so like breaking down the vector into elementary components yeah ?

[native rampart]

Are you familiar with span and basis vectors?

[old flame]

so from what I know, the span is the set of linear combinations of the vector space, while basis spans V, it is also linear independent. The difference between spanning list and basis is that spanning list could be reduced down to a basis, since a spanning list could be linearly dependent

[native rampart]

And the spanning list may not cover the entire space

We are just choosing a completely different set of independent vectors,and forming a new basis

[old flame]

ohhhh alright

[charred nacelle]

https://i.imgur.com/SF0q0Bw.png

I dont really know what I should be observing

[marble lance]

The only way to make the second coordinate 0 is if you multiply x by 0

So then it would have to just be a multiple of y

[native rampart]

I don't think there is a simple answer to that

[prisma pier]

idk about a geometric interpretation for the adjugate matrix specifically, but there's definitely a geometric interpretation for cramer's rule which is connected to the adjugate

I think 3b1b has a vid for it

[lucid cedar]

Some of the proofs in Linear Algebra Done Right seem like over kill. Specifically around a vector times a scaler.
Here is the solution of "Prove that -(-v) = v for every v in V":

My proposed solution was:
-(-v) = (-1)(-1)(v) = (1)v = v

but my professor mentioned something in class about this that I didnt quite catch. Something about because were dealing with vectors the proofs cant be so direct.

Im wondering if there are some subtleties here that I am missing

[marble lance]

Tbh, I would say your proof is overkill. You do more work than the book does. The book literally just notices that v already satisfies the conditions to be -v's inverse.

But if you have shown that -v = (-1)v, your proof is technically fine

[lucid cedar]

how do i show that

that should be axiomatic right?

[marble lance]

I don't believe that's an axiom, no

[lucid cedar]

im being hyperbolic and not exact when i say axiom. I mean so obvious that it does not* need proving

[marble lance]

All of these things can be considered so obvious that they don't need proving

[lucid cedar]

okay fair enough

[marble lance]

But you should still be able to show them rigorously using only the axioms, and things you have shown already

[lucid cedar]

I think i have an idea of how to do that actually

which now that i think about it. whats in my head right now looks alot like axlers proof

[marble lance]

Anyway, the book is saying in the definition of -v, we see v also satisfies the conditions to be -v's inverse. That's it. It's literally doing as little as possible.

[lucid cedar]

I guess its just not what I would of thought to do so it seemed foreign.

thanks tho I gained alot from this

[marble lance]

Np

[limber sierra]

@lucid cedar have you already proven that -v = (-1)v?

since thats not necessarily self-evident

oh whoops

didnt read the full convo

sorry for pinging!

[lucid cedar]

its okay. Its actually a result proven in the chapter before the questions, so im just writing down "by blank result" and calling it a day

[limber sierra]

ah sure, thats fine then

[dreamy iron]

i don't see how :
$\left<0,y-x, 0, w-z \right> \in \mathbf{W}$

[stoic python]

ninnymonger:

[wintry steppe]

W is defined by the first and third coordinates of a point being zero. the first and third coordinates of that point are zero

[dreamy iron]

that's the problem.

this is how i tried to create W

did i do it wrong?

[wintry steppe]

idk i just read the first picture

in the last picture

is W a subspace?

[dreamy iron]

i hope my W is a subspace.

[wintry steppe]

is it?

[dreamy iron]

😟

[native rampart]

W is just {(0,z,0,w)}

[dreamy iron]

is it not?

[wintry steppe]

you tell me

[native rampart]

That is what your result boils down to

[wintry steppe]

the W in the very last subspace that you wrote isn't a subspace. it's not closed under scalar multiplication

so you should change what you wrote a little

drunkendrake wrote what you should have

[native rampart]

Also,you haven't defined x and y

[dreamy iron]

okay, i'll correct my $\mathbb{W}$.

[stoic python]

ninnymonger:

[dreamy iron]

can we go on some more about :

W is defined by the first and third coordinates of a point being zero. the first and third coordinates of that point are zero
@wintry steppe

[wintry steppe]

in the first picture you wrote

send

sent

fuck

i am illiterate

there's nothing special about that point being in W. look at the definition of W and tell me why (0, y-z, 0, w-z) is in W

in fact i already said why

but

[dreamy iron]

what i dont understand how is how 

$\left< 0,y-x,0,w-z\right> \in \bigg\{ \left< 0,z,0,w\right>\bigg\}$```

[wintry steppe]

W consists of the points with first and third coordinate zero.

[dreamy iron]

yes, i read that too.

[wintry steppe]

(0, y - z, 0, w - z) has first and third coordinates zero.

[native rampart]

a vector - an arbitary vector is an arbitary vector

[wintry steppe]

Therefore (0, y - z, 0, w - z) is in W.

[stoic python]

ninnymonger:

[wintry steppe]

idk

good luck

[dreamy iron]

y and x aren't defined in W

[wintry steppe]

drunkendrake i'm leaving this to you

[native rampart]

y-x will be an vector

And it will be arbitary

[half storm]

This problem kind of reminds me a problem that I did a while ago out of friedberg.

It was basically proving a general result that if you have a vector space V and a subspcae W, then there exists a vector space W' such that $V = W \bigoplus W'$

[stoic python]

JohntheDon:

[wintry steppe]

W consists of the points with first and third coordinate zero.
The point (0, y - z, 0, w - z) has first and third coordinates zero.
Therefore (0, y - z, 0, w - z) is in W, since (0, y - z, 0, w - z) satisfies the definition of W.

[half storm]

Yea I'm looking at this now and they basically made the subspcae that I would ahve made I think.

[dreamy iron]

Okay, maybe there's something I'm not getting about the set builder notation.

(BTW I can see that $\left<x,x,z,z\right> + \left< 0 , y-x,0,w-z\right> = \left<x,y,z,w \right>$ )

[stoic python]

ninnymonger:

[wintry steppe]

i don't understand the confusion. the point satisfies the condition required for being an element of W, so it's in W

[dreamy iron]

W consists of the points with first and third coordinate zero.
The point (0, y - z, 0, w - z) has first and third coordinates zero.
Therefore (0, y - z, 0, w - z) is in W, since (0, y - z, 0, w - z) satisfies the definition of W.
@wintry steppe

No disrespect, but this is the 3rd time you've provided this particular line or reasoning, and for the 3rd time I'm left confused. I get the right coordinates are zero in the right spot.

how does z= y-x , and w = w-z

[native rampart]

z and w on lhs are dummy variables

[wintry steppe]

it doesn't matter what the coordinates of the point are because W consists of every single point with first and third coordinate zero

[dreamy iron]

can we go to a,b,c,d?

$<0,a,0,d> \in \mathbb{W}$

[stoic python]

ninnymonger:

[wintry steppe]

if that makes it more clear, sure

[half storm]

I used to have that problem alot; getting hung up notation rather than thinking more about in general what the definition of the set is or what the notation is really saying.

[wintry steppe]

let's go to the definition of W

W = {(0, z, 0, w) : z, w in F}

z and w are placeholder elements of F here. they don't mean anything in particular; they are dummy variables used for us to be able to express the set W in set builder notation. the idea in this definition of W is that W consists of all points with 1st and 3rd coordinate zero; this is because we use dummy variables to express the other coordinates. later in the proof, z and w are fixed variables. however, since we know W consists of any point with first and third coordinate zero, we can say that that point is in W

[dreamy iron]

i mean, i'm talking my way through the requirements W must have.

i get that my definition of W is wrong here. (non scalar closure and what not.)

[half ice]

{0,a,0,b}
Is a subspace

[dreamy iron]

yes.

[half ice]

I feel like you get the nature of the solution, but it's an awkward solution haha

[dreamy iron]

W = {(0, z, 0, w) : z, w in F}
@wintry steppe

if i'm to say that
( 0 , y-x , 0 , w-z ) \in {(0, z, 0, w) : z, w in F}

i've not defined y and x

[wintry steppe]

yes, you did. you fixed a point (x,y,z,w) in F^4

[dreamy iron]

when did i do that?

[wintry steppe]

and right in the next line it says (0, y - x, 0, w - z) is in W

you know x,y,z,w here

in the next line you are allowed to say (0, y - x, 0, w - z) is in W, because W consists of all the points (every single one) whose first and third coordinates are zero. we might be using the same symbols here that we used to define W in the set builder notation at the start, but there they were dummy variables, standing for any element of F. here, they are specific elements of F, so the point will be in W

[dreamy iron]

imma need to digest this

tyty @wintry steppe

[pastel kettle]

how can you tell that $B^2=2B$

[stoic python]

im_molee:

[dusky epoch]

holy fuck the purple annotations are overwhelming

anyway, you can calculate B^2 and 2B explicitly

by cayley-hamilton you get that $A(A-I)(A-2I)(A-3I) = 0$ which you can use to reduce the expression for $B^2$

[stoic python]

Ann:

[pastel kettle]

@dusky epoch let me think! give me few minutes lol

i get $A(A-I)(A-2I)(A-3I) = 0$ because the eigenvalues are 0,1,2,3

[stoic python]

im_molee:

[dusky epoch]

yes

[pastel kettle]

also if i put those values in $B=A^2-3A+2I$

[stoic python]

im_molee:

[pastel kettle]

i get 0 and 2

[dusky epoch]

uh

well i guess that's one way of doing it

[pastel kettle]

like let B=x^2-3a+2

[dusky epoch]

no i'm not sure if that's airtight

,w expand (x^2 - 3x + 2)^2

[stoic python]

Results provided by WolframAlpha

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Requested by Ann#0413

[dusky epoch]

,w expand x(x-1)(x-2)(x-3)

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[dusky epoch]

$B^2 = A^4 - 6A^3 + 13A^2 - 12A + 4I$

[stoic python]

Ann:

[pastel kettle]

okay i get that $B^2 = A^4 - 6A^3 + 13A^2 - 12A + 4I$

[stoic python]

im_molee:

[dusky epoch]

$0 = A(A-I)(A-2I)(A-3I) = A^4 - 6A^3 + 11A^2 - 6A$

[stoic python]

Ann:

[dusky epoch]

so what you're left with is $B^2 = 2A^2 - 6A + 4I$

[stoic python]

Ann:

[pastel kettle]

imma write down

how did you reduce B^2 to that?

you subtract sth from B^2

$0 = A(A-I)(A-2I)(A-3I) = A^4 - 6A^3 + 11A^2 - 6A$

[stoic python]

im_molee:

[pastel kettle]

by squaring given $B=A^2-3A+2I$ i get $B^2 = A^4 - 6A^3 + 13A^2 - 12A + 4I$

[stoic python]

im_molee:

[dusky epoch]

yes

[pastel kettle]

but what do i do with $0 = A(A-I)(A-2I)(A-3I) = A^4 - 6A^3 + 11A^2 - 6A$

[stoic python]

im_molee:

[dusky epoch]

subtract A^4 - 6A^3 + 11A^2 - 6A from your expression for B^2

[pastel kettle]

Oh damn

Ur magicians

also to get the spectrum of B, do i put spectrum of A to the $B=x^2-3x+2$?

[stoic python]

im_molee:

[pastel kettle]

is that how we know the minimal polynomial of B is $x(x-2)$

[stoic python]

im_molee:

[wintry steppe]

So i have three points:

A = (1,2,3), B = (1,0,-2), C = (t,t,1), where t € R and together they form a triangle.

How do I find t such that the point C is closest to point A?

I've tried projecting C onto A, but then got 7 as t. However when i used elementary calculus method, where I tried to find the minimum distance for the vector CA by solving for when the derivative is 0, i got t=1.5, which is even less distance than when t=7. But using derivative doesn't seem like a linear algebra way of solving this.

My question is if there is a linear algebraic method to solve this?

[half storm]

Got any hints?

I need help proving the converse

[gray dust]

rank(A)=1 so there exists a nonzero col vector in A, say c, where A's i'th col is L_ic for some scalar L_i

so A=(L_1c,L_2c,L_3c)=c(L_1,L_2,L_3) factored as the product you seek @half storm

[spiral star]

@pastel kettle i guess it might defeat the purpose of the exercise if your goal is to only argue with polynomials, but you can read off all the answers from the matrices as well. since A has 4 distinct eigenvalues, it must be similar to a diagonal matrix with 0,1,2,3 on the diagonal. if you choose this eigenbasis, then B is simply a sum of diagonal matrices and therefore also diagonal with respect to this particular basis. B has {2, 0, 0, 2} on the diagonal, making the eigenvalues {0, 2} and the char. polynomial x²(2-x)². for the min polynomial you can use a consequence of fitting's lemma: the multiplicy in the minimal polynomial is the power at which the sequence of kernels stabilizes. B is diag., so dim Ker(B) must be the same as dim Ker(B²). B-2I is also diagonal, so dim Ker(B-2I) = dim Ker(B-2I)². that means the kernels stabilize at power 1, making the the minimal polynomial x(x-2). by cayley hamilton B(B-2) = B² - 2B = 0 and thus B² = 2B

[half storm]

Thats where I thought of but didn't know where to go from there

[pastel kettle]

@trim marlin

@pastel kettle i guess it might defeat the purpose of the exercise if your goal is to only argue with polynomials, but you can read off all the answers from the matrices as well. since A has 4 distinct eigenvalues, it must be similar to a diagonal matrix with 0,1,2,3 on the diagonal. if you choose this eigenbasis, then B is simply a sum of diagonal matrices and therefore also diagonal with respect to this particular basis. B has {2, 0, 0, 2} on the diagonal, making the eigenvalues {0, 2} and the char. polynomial x²(2-x)². for the min polynomial you can use a consequence of fitting's lemma: the multiplicy in the minimal polynomial is the power at which the sequence of kernels stabilizes. B is diag., so dim Ker(B) must be the same as dim Ker(B²). B-2I is also diagonal, so dim Ker(B-2I) = dim Ker(B-2I)². that means the kernels stabilize at power 1, making the the minimal polynomial x(x-2). by cayley hamilton B(B-2) = B² - 2B = 0 and thus B² = 2B
@spiral star oh my god this is actually wonderful

[half storm]

@gray dust I think that makes sense. I get that there exist a nonzero column and that because the rank of a matrix is equal to the dimension of the subspace spanned by its rows and columns, that it has to be that one of the rows is nonzero as well. But I'm having a hard time extrapolating exactly why you know that row is a scalar factor of the nonzero column.

[pastel kettle]

what is the reason to use the Polarization identities

this is th projection

[spiral star]

@pastel kettle i have to say i was a bit imprecise sometimes, for example B doesnt necessarily have 2 0 0 2 on the diagonal, but it is similar to a matrix that does. i hope i could give at least a broad idea for how to approach this. i like to think about such problems in terms of endomorphisms instead of matrices, because it shows that when you define B in terms of A, then you really just define a new endomorphism, so it doesnt matter very much what the matrices look like in particular

[pastel kettle]

@pastel kettle i have to say i was a bit imprecise sometimes, for example B doesnt necessarily have 2 0 0 2 on the diagonal, but it is similar to a matrix that does. i hope i could give at least a broad idea for how to approach this. i like to think about such problems in terms of endomorphisms instead of matrices, because it shows that when you define B in terms of A, then you really just define a new endomorphism, so it doesnt matter very much what the matrices look like in particular
@spiral star yes this helped a lot!!

I solved question but idk why we are using the polarisation

[gray dust]

@half storm typo, edited row to col

[half storm]

Oh I see.

[dusky epoch]

let $C = A+B$ (the matrix that the problem gives you)

[stoic python]

Ann:

[dusky epoch]

given that you know $A$ is symmetric and $B$ is antisymmetric, how can you write $C^T$ in terms of $A$ and $B$?

[stoic python]

Ann:

[pastel kettle]

what is the reason to use the Polarization identities
@pastel kettle can anyone answer this plsss

[dusky epoch]

@pastel kettle you're given a norm, polarization identities allow you to recover the inner product that gave rise to it

@young jewel what don't you get? do you know what it means for a matrix to be symmetric?

ok, so can you write down "A is symmetric" more symbolically?

[pastel kettle]

@pastel kettle you're given a norm, polarization identities allow you to recover the inner product that gave rise to it
@dusky epoch so whenever i get norm, it's recommended to use polarisation identites when the actual inner product is given

[dusky epoch]

@pastel kettle you're overthinking it

@young jewel can you write down "A is symmetric" as an equation?

but you said you knew what a symmetric matrix was!

...

can you write down "A is symmetric" as an equation?

so...

[wintry steppe]

[pastel kettle]

can you help me with (c)

[dusky epoch]

aight one of you two will have to move out i don't think i can keep up two convos here any longer

it looks like xclear either doesn't want to respond or is away but i don't know for sure

[pastel kettle]

i get x= 1,

[dusky epoch]

xclear, why is it taking you more than ten seconds to write it out? the equation i'm asking you for is very simple!

[pastel kettle]

okay i will wait

[dusky epoch]

why couldn't you just write $A^T = A$???

[stoic python]

Ann:

[pastel kettle]

ping me later pls!

[dusky epoch]

A^T = A and B^T = -B

you should have known these. and ideally you should've written them down when i asked you to.

but you didn't

whatever

point is, if $C = A + B$, then $C^T = A - B$

[stoic python]

Ann:

[dusky epoch]

and from that, you can get $2A = C + C^T$ and $2B = C - C^T$

[stoic python]

Ann:

[dusky epoch]

do you understand?

(yes/no)

does it say anything else?

A and B commute?

like AB = BA?

is that what you mean?

i don't speak turkish, so if you don't translate properly then i cannot help you at all

great

use the definitions of symmetric and antisymmetric again

$(AB)^T \overset?= -AB$

[stoic python]

Ann:

[dusky epoch]

this is what you want to show

$(AB)^T = (BA)^T = ... = ... = ... = -AB$

[stoic python]

Ann:

[dusky epoch]

i thought you would apply SOME effort here on your own and do it yourself

but i guess i'm wrong and i have to do everything for you since for you apparently "help" means "do everything for me so that i can free myself from the burden of doing any thinking at all"

[pastel kettle]

damn

[winged iris]

yea in maths u have to think for yourself sometimes(most of the time)

[wintry steppe]

did they just delete all of their messages

[pastel kettle]

Think so

[wintry steppe]

that's sad

[pastel kettle]

Like Ann wanted him to derive things

[pallid rampart]

They also left the server

[pastel kettle]

Nooooo he was supposed to help me

No way

Can u guys help me lol

[dusky epoch]

Nooooo he was supposed to help me
who

me?

i'm not a he, thank you very much

i'm gonna go to sleep

[pastel kettle]

i'm not a he, thank you very much
@dusky epoch my bad! i have a guy friend named Ann sorry! night!

[errant wyvern]

hi my book has a typing error can someone help me clear this up

3. point, in the proof instead of wrting that rankB=dimZ it uses rankB=dimW; which one of these are correct?

[pastel kettle]

isn't it because rank BA = rank A

so rank b = dim w?

[errant wyvern]

so the correct would be rankB=dimW ight

[pastel kettle]

i mean

B\inL(W->Z) on first line

B included

but 3 says rank b = dimZ implies that rank (BA) = rank A

but A included in L(V -> W)

so i think in the end of the proof, it would say rank(BA) = dim W

idk not sure

[spiral star]

yea, you wanna make B injective :)) hence rank B = dim W implies rank(BA) = rank A

[errant wyvern]

yeah it made sense to me this way, but having an error in my litterature made me skeptical that i made an oversight somewhere

[spiral star]

dim Z cant be true, you can construct a simple counter example where you just make the dimension of Z smaller than dim W

and then it wont hold

[pastel kettle]

flow can you help me with my question !

after Lonac's

[spiral star]

i dont think there is anything else to discuss lol

what was your question?

[pastel kettle]

(c)

[spiral star]

but the answer is in the purple bubble on the right

[pastel kettle]

i can't confirm myself because im having trouble with understanding 😦

like it says by using the lemma

[spiral star]

okay, well A³ = I is something you can see right?

[pastel kettle]

yeah

because -A^3 +I = 0

[spiral star]

and A != I is also quite simple. for example, two matrices cant be equal if they have different char. polynomials

and A has a different char. polynomial than I

[spiral star]

so i guess the only argument you really have to make is for A²

[pastel kettle]

how do i tell char.polynomial of A is different with A^3?

[spiral star]

can you try to formulate that again? doesnt make sense to me

[spiral star]

A³ = I follows by cayley hamilton, A != I follows from the fact that -x³ + 1 != (1-x)³

[pastel kettle]

also this is what i get from -x^3 +1 = 0

what is the "!"

[spiral star]

inequality

[pastel kettle]

oh

[spiral star]

sorry, i have a programming background