Mathematics - 2018-06-21 (page 3 of 4)

@mercio:
Let $H \subset G$ (be finite groups (is that really required?)).

\subsection*{Conjecture}

If $G$ does not contain any irreducible representation
that branches into more than two under restriction to any of its subgroups
and $V$ is an irreducible representation of $G$, such that
when restricted to $H$ branches into exactly two irreps $V = \V_1 \oplus V_2$,
then there exists one irreducible representation $W$ of $G$ such that
$ W = [V \oplus V]^- = \subseteq V_1 \oplus V_2 $ which is one and

Drew Brady

In particular, I'm struggling to show that P(B) = P(\cup_k G_k) >= a.

The other inequality is straightforward, since P(B) =P(\cup_k G_k) = P(\cup_k H_k) = lim_k P(H_k) <= a since P(H_k) <=a for all k.

in the penultimate equality, I use the continuity of measure from below.

Mike Miller

@Rudi_Birnbaum What does branching mean? When a restriction of an irrep is no longer irreducible?

Rudi_Birnbaum

@MikeMiller: the decomposition of an irrep of a group restricted to a subgroup into the irreps of the subgroup is the branching.

Mike Miller

7:44 PM

Sure. I am a little confused. There are not many irreps of cyclic groups (they are all 1-dimensional over C) so your irreps which don't branch more than twice must be at most 2-dimensional?

Tobias Kildetoft

@Rudi_Birnbaum So by branching into two, you mean having at most two distinct irreducible summands?

(but possibly having those summands multiple times each)

MatheinBoulomenos

Hi @TobiasKildetoft

Tobias Kildetoft

@MatheinBoulomenos Hi

Rudi_Birnbaum

@MikeMiller: I stopped that dimension stuff, since that lead to confusion (the one talks over $\Bbb C$ and the other $\Bbb R$ (me)). Since only when you talk over $\Bbb R$ you can say that branching in two at most is also 2-dim. But then people didn't like thinking over R.

Drew Brady

I'm probably missing some inequality, but I don't understand how to show the lower bound, and a hint would be great.

Rudi_Birnbaum

7:48 PM

Hi @MatheinBoulomenos; @TobiasKildetoft!

MatheinBoulomenos

Hi @Rudi_Birnbaum

Mike Miller

I see, so this is like saying that there are at most two weights of the cyclic group

mercio

"under restriction to any of its subgroups " that's very restricting given that the trivial subgroup only has trivial irreducibles

so it implies that G's irreducible representations have dimension at most 2

Rudi_Birnbaum

It depends if we talk over $\Bbb R$, or over $\Bbb C$.

MatheinBoulomenos

@Tobias I think I solved the positive basis problem in the semisimple case, but I need to make sure I understand the results in your paper correctly

Tobias Kildetoft

7:50 PM

@MatheinBoulomenos Cool

mercio

wait where did i copy that from

when yo usay "any" do you mean "for all" or "there exists" ?

Rudi_Birnbaum

@mercio: But "at most two" is correct anyway

MatheinBoulomenos

@TobiasKildetoft so if we suppose that $A$ is a positively based algebra, then we can look at the cell module and this will have a Perron-Frobenius element, right? And that gives us a subquotient of dimension 1?

Rudi_Birnbaum

@mercio: "for all" but I wanted emphasise that its always the same one $W$ for all $H$.

Tobias Kildetoft

@MatheinBoulomenos No, the Perron-Frobenius element is an element of the algebra, and it is not one of the basis elements (so it is not an element of any cell)

MatheinBoulomenos

7:52 PM

oh, nevermind then

Tobias Kildetoft

Woops, you said cell-module, not cell

The cell module is a quotient of an ideal of the algebra

MatheinBoulomenos

yeah

but the only thing that really matters for me is: do we get a one-dimensional module?

Tobias Kildetoft

with a basis corresponding to the given cell

Not in general, no

mercio

so are you saying that if there is a representation $V$ that can branch into 2 for some subgroup $H$, then there is an irrep in the deocmposition of $(V \otimes V)^-$ that never branches no matter the subgroup ?

MatheinBoulomenos

ah, damn, I interpreted a remark like that

Rudi_Birnbaum

7:54 PM

@mercio: yep! thats the special one I need!!

Tobias Kildetoft

@MatheinBoulomenos The cell module will have the same dimension as the number of elements in the cell, and it sometimes happens that the cell module is simple

MatheinBoulomenos

@TobiasKildetoft I thought that corollary 3 said that you have one, my bad

Tobias Kildetoft

(without the cell having just one element that is)

mercio

so you are saying that if there is a representation $V$ that can branch into 2 for some subgroup $H$, then there is an irrep in the decomposition of $(V \otimes V)^−$ that has dimension 1 ?

Rudi_Birnbaum

@mercio: I once compared it to "-1" but some thought its a bad comparison

Tobias Kildetoft

7:55 PM

@MatheinBoulomenos Ahh, the $[\cdot:\cdot]$ there refers to composition multiplicity

Rudi_Birnbaum

@mercio: yes

MatheinBoulomenos

@TobiasKildetoft ah, okay that's what confused me

Tobias Kildetoft

Note that $L$ might be neither a submodule nor a quotient (at least a priori, no idea of an example where it really is neither)

MatheinBoulomenos

okay hmm

at least I managed to show that having a one-dimensional module is sufficient for semisimple algebras (that's really elementary)

mercio

that's a pretty strange statement

Rudi_Birnbaum

7:58 PM

would an example help?

MatheinBoulomenos

@TobiasKildetoft do you know an example of a semisimple positively-based algebra with no one-dimensional module?

Rudi_Birnbaum

@mercio: wait

Tobias Kildetoft

@MatheinBoulomenos Hmm

MatheinBoulomenos

I'd conjecture that this is necessary and sufficient for semisimple modules to have a one-dimensional module

Rudi_Birnbaum

@mercio: no fine its correct: "If there is a representation $V$ that can branch into 2 for some subgroup $H$, then there is an irrep in the decomposition of $(V \otimes V)^−$ that has dimension 1"

MatheinBoulomenos

8:01 PM

all group algebras and I think quiver algebras satisfy this

Tobias Kildetoft

interesting

true

mercio

do you have examples of all 3 cases of the truth table ?

Rudi_Birnbaum

@mercio: me?

mercio

yeah

Rudi_Birnbaum

i don't get the question

what is a truth table?

MatheinBoulomenos

8:04 PM

@TobiasKildetoft in case you're interested, the proof for sufficiency (it works over any field contained in $\Bbb C$) define a non-unital positive basis in the obvious way. Then if $A$ and $B$ have non-unital positive basis, so does $A \times B$ and $k \times A$ has a positive basis containing $1$ (where $k$ is the ground field.) The matrix algebra has a non-unital positive basis given by standard matrices, i.e. one entry 1, all other 0, so it follows from Artin-Wedderburn

mercio

one case where both sides are false, one case where both sides are true, and one case where the hypothesis is false and the conclusion is true

Tobias Kildetoft

@MatheinBoulomenos Ahh, nice

Rudi_Birnbaum

@mercio: oh I see. OK. Do we need true/true as well?

@mercio: Anyway:

mercio

it's good to have examples for all cases

if you don't then it would be evidence that something even stronger is true

Tobias Kildetoft

@MatheinBoulomenos So showing that it is necessary is the same as showing that there will be some two-sided cell containing just a single element. Probably this will tend to either be the one containing $1$ or a maximal one with respect to the two-sided order

Sorry, a two-sided cell which is also both a left and a right cell

Rudi_Birnbaum

8:13 PM

@mercio: D3 = group of order six has 3 irreps A1 (trivial 1dim) A2 (one dimensional) and E (=complex whatever dimension). Under distortion to C3 (this is "direction" A2) they branch to A1->A, A2->A, E->E. Now in D3 $[ExE]^- = A2$ and in C3 $E x E = A + [A] + E$, and A2->A so it holds. OK?

Tobias Kildetoft

So question: If we have a basis element $x$ such that $1$ occurs in $yxz$ for some $y$ and $z$. Does it hold that $1$ occurs in $wx$ for some $w$?

mercio

um

E branches into 2 parts if you restrict to the cyclic 3-subgroup

(C3 is cyclic right ?)

Rudi_Birnbaum

Yes and one A is in the symmetric part and the other A is the anti-symmetric part.

and Yes cyclic order three. E branches to E

mercio

but you wrote E -> E

Rudi_Birnbaum

@mercio: E in D3 restricts to E in C3

mercio

8:16 PM

no

Rudi_Birnbaum

yes

mercio

C3 is abelian all its irreducible representations have dimension 1

Rudi_Birnbaum

@mercio: My tables say "The single “E” representation is reducible but almost behaves like a true irreducible representation.
Its norm, however, is twice the group order. Therefore, is has been marked with an asterisk in the table.
This is essential when trying to decompose a reducible representation into “irreducible” ones using the familiar projection formula."

mercio

õ..o

Rudi_Birnbaum

@mercio: So I need to cover these cases in this way. Maybe there is a more stringent way for that ...?

mercio

8:19 PM

so is E reducible or not ????????????????

Rudi_Birnbaum

@mercio: Or maybe we get the same result when we reduce it.

mercio

what result

Rudi_Birnbaum

Conjecture 1

mercio

well then the hypothesis is true for E

Rudi_Birnbaum

so to what does E then branch in C3

?

mercio

8:20 PM

since is splits for most subgroups

it branches into two complex 1-dimensional representations

C3 has 3 irreducible representations, basically (1), (ζ3), (ζ3²)

the three cubic roots of unity

Rudi_Birnbaum

Yes and then A will be of course contained as well in the product of these, so: no panic!

Tobias Kildetoft

@MatheinBoulomenos I need to go now, but the above question is what I think needs to be shown to show that the condition is necessary.

MatheinBoulomenos

@TobiasKildetoft okay, I'll think about it. See you!

mercio

anyway, any irreducible representation of dimension 2 will have both sides to true

Rudi_Birnbaum

yes

mercio

8:23 PM

so now you should give irreducible representations of higher dimension

Rudi_Birnbaum

SO true true is done?

yes

mercio

well yeah

Rudi_Birnbaum

simplest is "T" 12 elements a "chiral" tetrahedron

mercio

o..o

Rudi_Birnbaum

lol

mercio

8:24 PM

orientation-preserving isometries of the tetrahedron ?

A4 ?

Rudi_Birnbaum

sounds right

there are three tetrahedral groups, thats the smallest one

The other two have 24 elements

each

we call them T_d and T_h

In T there is again that complex E ... but I hope we can get over that ...

mercio

I would rant but A4 isn't much better if you don't know it's an alternating group

so A4 has 3 irreducible representations, right ?

Rudi_Birnbaum

Yes: A,E,T, they are 1,2,3 dimesnional

mercio

oh wait

are you suggesting that 1²+2²+3² = 12

Rudi_Birnbaum

but here its again the strange 2-dimensions which are actually two complex one dimensional

mercio

8:28 PM

okay it has 4 then

Rudi_Birnbaum

it you miss weight 1/2

õ..o

1^2 + 1/2 2^2 + 3^2

1²+1²+1²+3²

Lets take Th

8:30 PM

we aren't done with T

does that 3-dimensional representation branch into 2 for some of the many subgroups of A4 ?

Rudi_Birnbaum

But we keep on confusing us about that "E" that is for me 2-

and for you 2 x 1 one dimesnsional ...

OK

My rotation is now contained in the irrep $T$

mercio

the theorem is going to be 4 times as complicated to state if you work over $\Bbb R$ instead of $\Bbb C$ like any respectable mathematician would do

what is T

I thought it was A4

Rudi_Birnbaum

A = 1.dim, E=2-dim and T=3 dim.

mercio

"simplest is "T" 12 elements a "chiral" tetrahedron"

Rudi_Birnbaum

sry ...

:-)

mercio

8:32 PM

did your mom never tell you not to give 2 people the same name

Rudi_Birnbaum

OK so ExE contains A ⊕ [A] ⊕ E

mercio

so next you have to check the branching of T over all the subgroups of T

what does that bracket mean

Rudi_Birnbaum

T always branches since no smaller group has Ts

antisymmetric part

mercio

yeah but you said "exactly 2" in your statement

it could branch into 3 things

but T has dimension 3 so the antisymmetric part should have dimension 3

Rudi_Birnbaum

yes but thats for those cases where the conjecture holds

now we look at a case where it doesn't hold

mercio

8:35 PM

are you saying your conjecture is false ?

then we can pack up

Rudi_Birnbaum

Stop, I don't get it

mercio

the ocnjecture is of the form "if A is true then B is true"

you meant to say that this is a case where A is false

Rudi_Birnbaum

yes

T is a counter example

mercio

in that case the conjecture would be true

sigh

Rudi_Birnbaum

Yes its always true

mercio

8:36 PM

a counter example to what ?

so don't say that the conjecture doesn't hold

Drew Brady

it can't be a counterexample if the conjecture doesn't apply!

Rudi_Birnbaum

sry forget it

it hold always

mercio

so in order to show that the assumption (A) is false, you need to check that the branching always branch into 3 things

for every subgroup of A4

Rudi_Birnbaum

What is the assumption (A)?

mercio

it is that E branches into exactly 2 irreducible parts for some subgroup H of A4

and the conclusion (B) is that the antisymmetric part of E² has an irreducible representation (for G) of dimension 1

Rudi_Birnbaum

8:39 PM

But T is no example for that. Its an counterexample. its non-(A)

mercio

but you haven't showed me that it was non A

Drew Brady

its a little strange that you're doing this algebra stuff but can't get first order logic down, I'll admit!

mercio

maybe it is A

you have to check all the subgroups

(as you have written your conjecture)

Rudi_Birnbaum

It has ony four subgrpoups (simplest): D2, C3, C2, C1

mercio

that D2 is cracking me up

they are all commutative, right ?

then yeah E always branch into 3 1-dimensional representations

Rudi_Birnbaum

8:43 PM

E branches in "A+A", "E", "A+A" ... hmmm maybe better to replace "exactly" by "at most" ..

mercio

it's amazing how we keep contradicting each other

Rudi_Birnbaum

Yes that "exactly two" is non-sense

mercio

the fact that you work over $\Bbb R$ is also a pain

Rudi_Birnbaum

I thought the maximum number of branchings would be a nice way to circumvent that

Alessandro Codenotti

@mercio The more math I learn the more common this sentence becomes

mercio

8:46 PM

of course

$\Bbb R$ is ugly

Drew Brady

not analytically!

Rudi_Birnbaum

So the conjecture actually does not use R or C. Problem is I have all my intuition from $\Bbb R$

mercio

well yo ucould add "over $\Bbb R$" in the statement

Rudi_Birnbaum

I admitt its ugly

mercio

but then I would have to rethink all the examples

Rudi_Birnbaum

8:47 PM

But I don't want it ugly

And I have still no idea how to say "E" in complex

mercio

E $\otimes \Bbb C$

geocalc33

Hey

Rudi_Birnbaum

The tensor product of "any over reals two dimensional irrep with the complex numbers"?

hi @geocalc33

mercio

well I should have written E $\otimes_{\Bbb R} \Bbb C$ to be precise

are you picturing a rotation of order 3 in the plane when you think about E ?

Rudi_Birnbaum

for example

but wait, this is so to say the very core of the problem..

Can one picture an irrep for a group operation? That is what I do no get at all.

mercio

8:54 PM

well a representation over the reals of dimension 2

is an action of G on the plane

the real plane

Rudi_Birnbaum

"real plane"?

mercio

$\Bbb R^2$

as opposed to $\Bbb C^2$, his prettier brother

Rudi_Birnbaum

yes in the representation space

but the $C_3$ operation lives as a group element

mercio

actually E is a representation of all of G, not just C3

should be*

Rudi_Birnbaum

exactly

and thats where I hang, beacuse the following

:

lets assume I could assign the E to any rotation operations in the group

mercio

8:57 PM

?w??

Rudi_Birnbaum

then the thing would be trivial

mercio

??w?

Rudi_Birnbaum

in the two Es there is that rotation and by multiplying these two Es, I get it back in the antisymmetric part.

thats kind of how it feels

mercio

Im not following

Rudi_Birnbaum

That "peculiar irrep" we have called "W" in the conjecture above, remember?

mercio

8:59 PM

is it the one that I said is reducible ?

that for you it is of dimension 2 over the reals ?

Rudi_Birnbaum

no the one which is contained

mercio

because the anitsymmetric part was about the product of the representation of dimension 3 with itself

the one which is contained ?

Rudi_Birnbaum

no of dim2 with itself

three does not work, its not in the conjectire

mercio

no we are looking at T with itself

well 1 does not work either

and even over the reals, you will again get a decomposition of the tensor square into 3+1 so yes there is an irrep of dimension 1 in the antisymmetric part

Rudi_Birnbaum

in irrep TxT it is actually contained. but not in ExE

the antisymmetric part of TxT only contains T

mercio

9:02 PM

maybe

MatheinBoulomenos

Hi @Ted

Ted Shifrin

hi @Mathein

MatheinBoulomenos

my talk on Bezout went really well :)

Ted Shifrin

oh great :)

were there some good questions?

MatheinBoulomenos

they asked for counterexamples with less assumptions for some stuff

Ted Shifrin

9:04 PM

ah, well, you had thought about that one

Rudi_Birnbaum

What I mean is that the groups without any higher dimensional irreps than those E ones bet it two complex 1-d or one real 2-d. In those the conjecture holds.

MatheinBoulomenos

I had a lot of counterexamples: projective points that you can't see in the affine part, multiplicities, non-real points, weird stuff in finite fields where a curve is the whole projective plane, quaternions where the zero set of a one-variable polynomial is topologically a 2-sphere (so not quite finite) etc

mercio

yeah cuz the one of dimension 1 can't branch and the one of dimension 2 have an antisymmetric square of dimension 1

Rudi_Birnbaum

Yes

Ted Shifrin

cool @Mathein :)

Rudi_Birnbaum

9:08 PM

But also in all those cases the product of the branches gives that irrep. In other cases not.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

hi @Alessandro

Rudi_Birnbaum

So the branching lifts that degeneracy from these higher order axis.

geocalc33

What is the limit as s approaches 1 of zeta(s)^1/log(x)?

Rudi_Birnbaum

But when we trace the irreps into which that branches we get the right irrep back in the product

MatheinBoulomenos

9:11 PM

I managed to simplify some proofs (and actually complete some proofs, some steps were unjustified in my opinion) over the the source material by establishing some easy properties of the operator $k[X_1, \dots, X_N] \to \Bbb{A}^n(k), f \mapsto V(f)$. (So only for principal ideals) The weak Nullstellensatz for principal ideals is for example a straightforward consequence of the definition of algebraically closed.
I used homogenization to get the statements in the projective case

Rudi_Birnbaum

And the reason for that must be somehow that these E irreps come from the higher order axis. And multiplying them gives you in the antisymmetric part the rotation around the main axis. When I now distort the symmetry that connection still holds as long as I know which are the two irreps that came from the E.

MatheinBoulomenos

And in some way I proved the first theorem on resultants before even defining them: The resultant is zero iff the polynomials have a common factor. Because that proof is the motivation for the definition in the first place. Just writing down the Sylvester matrix is very unmotivated

Rudi_Birnbaum

But that argument suffers from that it connects some single group element with some single irrep. A strange correspondence ..

geocalc33

@TedShifrin what is the limit as s approaches 1 of zeta(s)^1/lnx?

MatheinBoulomenos

@TedShifrin one thing the author used without proof was that finite unions of curves and points never cover the projective plane. It's not hard to see, but I don't think it's completely obvious if you don't have any notion of dimension etc. (and you need at least that your field is infinite)

Ted Shifrin

9:20 PM

I don't know, @geocalc.

Rudi_Birnbaum

@mercio: I have to go to bed now.

geocalc33

How would I analytically go about solving such a limit

mercio

good night

Rudi_Birnbaum

good night!

Ted Shifrin

Right, @Mathein: Although maybe there's a way (in the case of infinite fields, of course) to reduce to some single-variable facts about polynomials only having finitely many roots.

(Intersect the picture with a generic line.)

MatheinBoulomenos

9:23 PM

@TedShifrin that's exactly how I did it. Just do it inductively, view the polynomial as a one-variable polynomial in $k[x_1, \dots, x_{n-1}] [x_n]$, then by the inductive hypothesis, you can find $a_1, \dots, a_{n-1}$ such that the leading coefficient doesn't vanish, then $f(a_1, \dots, a_{n-1},x)$ is not the zero polynomial

Ted Shifrin

Right ... not a big deal.

MatheinBoulomenos

yeah, I didn't say it's difficult, but I think you have to say something about it

Ted Shifrin

When I wrote my algebra book and included a chapter on projective geometry and my favorite result (how many lines meet 4 lines in general position in $\Bbb P^3$?) ... I had to write a few exercises to do very elementary arguments like that.

(Especially since we had really only studied polynomials in one variable carefully.)

MatheinBoulomenos

sometimes Gauss lemma arguments can reduce it to that, that's what I did in my talk at some places

Ted Shifrin

Hmm ... Artin actually states a generalized Gauss lemma for more variables.

MatheinBoulomenos

9:28 PM

it's just the usual Gauss lemma applied a bunch of times, it gives you that $k[x_1, \dots, x_{n-1}]$ is a UFD, then you can use the relation of irreducible polynomials in $k[x_1, \dots, x_{n-1}][x_n]$ and $k(x_1, \dots, x_{n-1})[x_n]$

it's a bit confusing, there are some related statement that are all called Gauss lemma

Drew Brady

Hey, I'd asked this question a couple of hours with no luck so I'll shameless bump it. The problem is here: imgur.com/a/p9Sb4FW. It is part (b) (I solved (a) above it). Following the hint, it is rather straightforward to see that P(B) <= a, since P(B) = P(\cup_k H_k) = lim_k P(H_k) <= a, since 0 < P(H_k) <= a for all k.

But I am having difficulty getting the lower bound P(B) >= a.

MatheinBoulomenos

(and some unrelated statements, too, but I don't want to talk about normal coordinates)

@TedShifrin what I did in my talk was applying a projective transformation such that $f(0,0,1) \neq 0 \neq g(0,0,1)$ for the polynomials I had and then this implies that there is some $Z^k$ term with no $X$ or $Y$, thus the content as an element in $k[X,Y] [Z]$ is $1$ and then Gauss lemma applies

Ted Shifrin

OK.

MatheinBoulomenos

I thought that's clever when I saw it in the reference

Jake S

10:06 PM

Hi all, I have a question

Drew Brady

ask it!

Jake S

If $A = \{u_1, u_2\}$ and $B = \{v_1, v_2\}$ are subsets of a vector space $V$ so that the union of $A$ and $B$ is linearly independent, and I need to find z $\in V$ , knowing that z is a linear combination of the elements of $A$ and $B$ simultaneously

Is it true that z is the 0 vector? Or is it just $av_1 + bv_2 + cu_1 + du_2$ ?

Drew Brady

what does it mean that "z is a linear combination of the elements of A and B simultaneously?"

Maximus

you need to find z such that z=av_1+bv_2+cu_1+du_2 like you said

user441848

Hi 😉, Does someone know what is the dual problem of the linear problem minimize $c^tx$ subject to Ax=b, $x\ge 0$?

Jake S

10:10 PM

Yes

Drew Brady

what are the restrictions on (a, b, c, d)?

can it be (1, 0, 0, 1) for example?

or does it need to have all nonzero entries.

user441848, what is the Lagrangian

Jake S

I was under the impression that a set $A$ is only LI if the coefficients in the span are all 0

rather

Drew Brady

ugh. Jake S you're not understanding my question.

when you say "z is a linear combination of the elemtns of A and B simultaneously" do you mean that it is a nontrivial linear combination of every element in A and B or a nontrivial linear combination of at least 1 element of A and B?

Jake S

Well, the exercise is written exactly how I stated. I think it would be the former.

user441848

@DrewBrady $L(x,u,v)=f(x)+\sum ug_i(x)+\sum vh_i(x)$, in this problem would be $L(x,v)=c^tx+\sum v(Ax-b)$ ?

Drew Brady

10:14 PM

user441848, the way you figure it out is you write the lagrangian L(x, u, v) = c^T x + u^T(b - Ax) - v^Tx

Jake S

My reasoning was:

$A \cup B$ is LI. $u \in A = (\alpha u_1 + \alpha u_2)$ and likewise for $v \in B$

user441848

@DrewBrady where does -v^Tx come from?

Drew Brady

because x >= 0 is the inequalities -x_i <= 0 for i = 1, ..., n

Jake S

Then $z \in V$, linear combination of $A$ and $B$, is $z = \alpha_1 u_1 + \alpha_2 u_2 + \beta_1 v_1 + \beta_2 v_2$

Drew Brady

now you need to infimize L(x, u, v) in the primal variable x.

Jake S

10:17 PM

sorry, forgot to add subindices to distinguish alphas and betas

But assuming what I said is true, I'm not sure if z is what I wrote and to leave it there, or if i can furthermore conclude that z must be the 0 vector.

Drew Brady

it will be helpful to rewrite the lagrangian as L(x, u, v) = (c -v - A^T u)^T x + u^Tb

Jake S

What do you think, @DrewBrady?

Drew Brady

this immediately implies that one dual problem is: maximize u^T b subject to c - v - A^T u = 0.

oops with the constraint also that v>= 0

Jake S

Sorry Drew, I can see that the other problem you're helping user441848 with is probably more engaging, I would just like to know if my reasoning is faulty and I should express z as I already have or if I can conclude more about z.

Drew Brady

obviously, this is equivalent to maximize u^T b subject to c - A^T u >= 0, equivalently, maximize u^T b subject to A^Tu <= c

Jake S I don't understand what you're asking.

You know that A cup B is a set of linearly independent vectors.

so now what? what are you trying to do

Jake S

10:28 PM

I know that A cup B is a set of LI vectors

Drew Brady

okay...

so what is the question then

Jake S

Previously, I used that to simply conclude that $z \in V = \alpha_1u_1 + \alpha_2u_2 + \beta_1v_1 + \beta_2v_2$

Drew Brady

where V = A cup B?

user441848

would you mind to explain the immediately implies that one dual problem is: maximize u^T b subject to c - v - A^T u = 0.
with the constraint v>= 0 ? I don't see it

Drew Brady

or rather where V is the span of A cup B?

user441848 what is the Lagrangian

Jake S

10:30 PM

I assume so, but I'm not given that data, I'm merely told that A and B are subsets of V and that their union is linearly independent

Is that enough to conclude that V is the span of A cup B?

Drew Brady

I don't know! what is V and what is z? I'm unclear about the question!!!

user441848

@DrewBrady this L(x, u, v) = (c -v - A^T u)^T x + u^Tb

Drew Brady

Right. Suppose that c - v- A^T u is non-zero. What does that mean.

Jake S

Haha, but I'm not given any more information! If it's lacking data I should assume the simplest answer, in all likelihood.

Because it was meant to be solvable as presented.

Drew Brady

no. Jake tell me the entire question again verbatim, you're obviously missing details.

Jake S

10:33 PM

Let $A = \{u_1, u_2\}$ and $B = \{v_1, v_2\}$ be two subsets of the vector space $V$, such that $A \cup B$ is linearly independent. Find $z \in V$, such that $z$ is a linear combination of the elements of $A$ and the elements of $B$ simultaneously.

user441848

hmm

Jake S

That is the entire question, verbatim from the worksheet I'm staring at.

user441848

@DrewBrady I don't know what could I say about it

geocalc33

@MatheinBoulomenos can you take a limit along a line such as y=x instead of from the left or right?

Drew Brady

not that if c is a vector, then c is zero iff c^T x = 0 for all x in R^n.

geocalc33

10:34 PM

Or along a geodesic

user441848

@DrewBrady that's right

Drew Brady

Jake S. I don't understand. why wouldn't you just say u_1 + u_2 + v_1 + v_2? that is obviously a linear combination of the elements of A and B. It doesn't have anything to do with the fac that A cup B is linearly indepednent..

okay. so now apply that to this problem....

^ user441848

Jake S

I guess that's a sufficient answer, maybe

user441848

@DrewBrady then we just have (-v-A^T u)^T x+u^ T b

Drew Brady

lol try again.

note that just because I used c in the example above does not mean I'm talking about the same c in your problem statement!

user441848

10:42 PM

@DrewBrady :( have to go now hope to find you here later :) And thnk you so much for your help

MatheinBoulomenos

@geocalc33 if your function is defined on $\Bbb R^2$, then you can take the limit along such a a line, yeah

Drew Brady

Hey, I'd asked this question a more than a couple of hours with no luck so I'll shameless bump it. The problem is here: imgur.com/a/p9Sb4FW. It is part (b) (I solved (a) above it). Following the hint, it is rather straightforward to see that P(B) <= a, since P(B) = P(\cup_k H_k) = lim_k P(H_k) <= a, since 0 < P(H_k) <= a for all k.
But I am having difficulty getting the lower bound P(B) >= a.

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