$c_1Av_1 + ... + c_nAv_n = 0 $ implies $c_i = 0$

so therefore $A(c_1v_1 + ... + c_nv_n) = 0$ implies $c_i = 0$

therefore the only solution to $A(c_1v_1 + ... + c_nv_n) = 0$ is the trivial solution

which is true iff A is nonsingular

you can put some of this on the same line, you know. :) my screen is pretty big but some people look at this on mobile devices.

oh sorry youre right

It was perfect up until the end. You have to argue that every vector $x$ in $\Bbb R^n$ can be written as a linear combination of the $v_i$.

presumably you have some list of characterizations of what it means for a matrix to be nonsingular. which characterization are you using here?

or ted's question.

The only solution of $Ax=0$ is the trivial solution.

exactly

Yes, but you must answer my question.

@TedShifrin ohh right

indeed, the characterization is not "the only solution to Ax = 0 is the trivial solution, when x happens to have the form sum c_j v_j, with A v_j linearly independent"

but you're close

because

otherwise you could have solutions for that that arent the trivial solution

conceivably

ex. if v_1=<1, 0> and v_2=<2,0> then theres also 2v_1 - v_2 = 0

it all ends up being the same thing, but you need to explain why in terms of something you've got

which isnt c_i = 0

instead of just pushing the exercise into a statement about solvability of a system

What word seems relevant to debate the question I asked?

umm

every?

Let's look for a linear algebra term.

are we tallking about the same line

Yes.

ummm

linear combination?

I'm losing more hair than I'm growing at this point. I'm trying to breach Variational Calculus and I'm trying to read up on it in a situation where I'm not doing Lagrangian mechanics, i.e. my functional is not Potential Energy minus Kinetic Energy. If I am looking to find a function $f(v)$ that minimizes some cost $C(f)$, do I have to restart the formulation from scratch? Or can I leverage the Euler–Lagrange equation and simply substitute the Largrangian for something else?

That was in my question.

The subtitles / closed captions on this are amazing.

im note sure

not sure*

Euler-Lagrange is totally general, @Axoren.

And of all the above, how do I turn this into a main-site question that can be answered because I don't even know if what I'm asking makes any sense.

What term is relevant to deciding whether every vector is a linear combination of a given set of vectors, @Allie?

@TedShifrin Hey Ted, that was my suspicion, but of all the articles and videos I'm seeing online, they are trying REALLY hard to convince me that this is the only one.

span?

Just look at a math book on calculus of variations.

allie: you have won linear algebra jeopardy.

OK, span is good. So do your $v_1,\dots,v_n$ have to span $\Bbb R^n$?

I'm lacking one at the moment. Do you recommend any specific ones?

Not off the top of my head. You can even see a derivation of Euler-Lagrange in the last section of my diff geo notes (freely downloadable).

I've been scouring lecture notes, wikis, math archives, and etc.

I'll see about grabbing those. I've seen the derivation quite a number of times, but I don't have a good sense for what the Lagrangian would be in a non-mechanics case.

Wiki should be fine for Euler-Lagrange.

Like, what functional should I be using when it's non-mechanics?

they have to in order to be nonsingular

You just want to minimize some function(al) $F$ given a constraint. No one mentions Lagrangian.

for A to be nonsingular

That sentence makes no sense, @Allie.

Let me see your derivation before I ask anymore questions. Maybe you derived it better.

Oh, but that's circular reasoning.

What in the given problem actually must guarantee that $v_1,\dots,v_n$ span $\Bbb R^n$? Can you bring in some allied notion?

The basic idea is that $"D_g C(f)" = \lim_{\varepsilon \to 0} (C(f + \varepsilon g) - C(f))/\varepsilon$ for functions $g, f$

@Axoren there are a couple of complete courses on NPTEL

maybe you can find something in the bibliography

@Allie your username possesses thw two starting letters that describe the words Ted is asking for.

Its a misconception that Lagrangian is a specific thing in physics (KE - PE). It's just a general function $TX \to \Bbb R$

But it is usually stated for Lagrangians

@Balarka Far more general even than that.

You can have as many derivatives involved as you want.

I'm going to need to disappear for a few hours.

Agree, but I do not know of an immediate use of such a setup

im so confused

The Frechet derivative on the other hand is totally general. I have used it to find minimizing geodesics

allie: why might v_1, .., v_n span R^n

When do $n$ vectors span $\Bbb R^n$, @Allie?

if they are linearly independent? if every vector in $R^n$ is expressible as a linear combination of the vectors?

same thing ig

Precisely. If and only if they are linearly independent. So why must they be here?

can you prove that v_1, ..., v_n are linearly independent?

@leslie is hereby appointed to replace me.

ted has to go do gatekeeping for the elitist institutions that are taking this country back to the stone age

That's the supreme court you're referring to. Nothing to do with me.

you know i don't think MIT shoulders much, if any, of the blame for our current predicament. so if that's what it is, you're OK.

probably some association with the military-industrial complex, but we're far beyond that now.

we just need more bullets

Anyhow, let poor Allie finish her proof.

allie, i'm here to continue/verify in place of ted, who is off arbitrarily deciding the leaders of tomorrow on the basis of his own whims. but you are almost there.

copper: i'd expect you to recommend something else. tick, tick, tick.

i just got an upvote & a downvote on a question i answered age ago.

Allie may or may not have done a preceding exercise that says precisely what we want, but it's easy enough to make the argument here.

@TedShifrin I think I see now. It's not that we start with $f(t, u(t), u'(t))$, it is that we start with the functional $F(u)$ (the cost) and the function $f(t, u(t), u'(t))$ naturally occurs within the integral formula for $F(u)$. Right?

Right.

Bye.

That was driving me so insane because everyone simply started with a given $L = P - K$

Take care.

@leslietownes ireland's rights are roughly similar to the us except for those that deal with the ira (Offences Against the State Acts) which give the state sweeping powers of detention, etc.

ohhh i did do the exercise ted was talking about

my first interview question, regardless of purpose or context, is, "what is the best madonna single, and why is it 'into the groove'?"

$T^{0}_{2}(U)$, the $(0,2)$ tensor bundle on a chart $(U,\phi)$ on the manifold $M$ is open in $T^{0}_{2}(M)$, right?

if you pass that, you win

so can i start from the beginning

we know $Av_i$ are all linearly independent

like a virgil

then so are the $v_i$.

allie, this is a minor nit but i have to do this because i got a phd, linear independence is a property of a collection

"all linearly independent" suggests linear indepenedence is an individual property enjoyed by particular vectors, which it isn't

i understand that

it was just ,a shorthand thing

yk

we know the list {Av_i, 1 <= i <= n} to be linearly independent

yes

today is your turn to be the resident pedant

i do apologize, i am contractually obligated to do this

so anyways

say pendant to the pink panther tune

@Allie presumably you are working with a square matrix of dimension $n \times n$?

indeed she is.

how did you guess?

i'm psychotic

@Allie can you show that the $v_i$ are li?

i'm barking up the wrong alley

sorry im atr work

yes copper hat

so $c_1Av_1 + ... + c_nAv_n = 0$ implies $c_i = 0$

there was a problem on main the other day. someone asked, you have linearly independent vectors in some R^k. you extend them to vectors in R^n (n > k) by adding entries arbitrarily to each vector. are they still linearly independent, and if so, why. this is a non-square-matrix version of the present theme, with A the map from R^n to R^k given by chopping off the last n-k coordinates.

so therefore $A(c_1v_1 + ... + c_nv_n) = 0$ implies $c_i = 0$

right

yes. it may help to fix a starting point and push in one direction only. one implication holds therefore another implication holds is more complicated than if a, then b.

suppose sum c_i v_i = 0. that's a. you have a goal in mind, z. can you get there.

anyone know the answer to my question?

i think im stuck

i feel so stupid

allie, suppose sum c_i v_i = 0, then A(sum c_i v_i) = A(0) = 0. can you conclude anything about the c's from your earlier reasoning and hypotheses

they must be 0?

yeah

im confused

im sorry

isnt that backwards

i'm not sure what you mean. in general, apart from this problem, to prove {v_i} linearly independent from the definition, one begins with scalars c_i and an assumption that sum c_i v_i = 0, and then proves that each of the scalars c_i must be 0

here, the reasoning in the middle happens to involve applying A and using the known linear independence of {Av_i}

other ways to do it, too, but this is the course that stays closest to the definitions

right

thats the definition in my book

so assume $\sum c_iv_i = 0$

thats how we start right

sure, if we're trying to prove {v_i} linearly independent

then $\sum Ac_iv_i = 0$

by distributivity

and since all the Av_i form a linearly independent set c_i = 0

sure.

note that this portion of the reasoning doesn't really need that you have n vectors in R^n being mapped into R^n by A. if you have k vectors in R^p being mapped into R^q, the proof is the same. {A v_i} linearly independent implies {v_i} linearly independent for exactly the same reasons.

once you want say that those v_i's to span something, of course, the values of these parameters become more relevant.

the intuition is that mapping by $A$ can only make a collection of vectors the same or 'less independent'.

yeah. shilov or someone said it, linear relations enjoyed by the vectors will also be enjoyed by A of those vectors. and then maybe more relations.

@copper.hat that was my intuition

is that only the case for square matrices?

notice that $A v_i$ span the whole space.

when you apply a linear operator to a subspace of dimension $d$ the resulting subspace must have dimension $\le d$.

regardless of square or not.

right

because u cant have dim > d since you only have d vectors

so what does that say about the $v_i$?

but you can have dim < d in the case of A having rank < d

right

@Allie not quite, but lets not go too far away from current problem for the moment.

i think

You guys are still here?!!!

no

@Allie: Here's a succinct way to think about that question. Take the contrapositive. If the $v_i$ were linearly dependent, could the $Av_i$ be linearly independent?

no

[as to me still being here]

Munchkin is your proxy?

pbhbhbhbht

Who is Emma?

Don't we need a prior?

it's a question i need help solving it

emma is just a name

I hope that Emma is okay. I'm really worried about her.

First step is to write down the definitions of "sensitivity" and "specificity"

(largely because I forgot)

I have no earthly idea what those words mean. I only know about false positives and false negatives.

So "predicted + given actual +" vs "predicted - given actual -"

So 100% of the negatives were predicted as negative? No false positives?

yeah FP is 0

So isn't the answer 0.825 ?

No they're asking for "actual +ve given predicted +ve"

0.825 is "predicted +ve given actual +ve"

("+ve" = "positive")

I'm confused. So .825 of the positive cases were predicted positive. That sounds like there had to be false negatives.

Yes

I leave this to DogAteMy to sort out. I am in the middle of something.

False negatives and no false positives

In any case, if there are no false positives, and she has a positive test result, surely it can't be false

Right.

The proper way to do this is with Bayes but I forget the formula

but I think the answer is 100%

P(pos|pos test)P(pos test) + P(pos|neg test)P(neg test)

I wrote exam questions like this. I just don't get the lingo here.

what will be the P(Actual +ve) ?

What I wrote down.

Yes, what he said :)

Cheba hut

Any logicians in the house ?

I have a question, which I rarely ever do here, but it's too soft for the main site

Yo @geocalc33

I am available for an hour break

If you want to get hands-on experience with C++ / D

What's with the silence in this room, ya'll acting like Jesus died or somethin...