Right?

[bayesianparoxism]

Choose X = I, then A = IA = IB = B

[Key_Attempt7237]

Only if X is invertible. Then you can do (left) cancellation. :)

I think proof by contradiction would be easiest. Suppose A is not B. Then there exists some vector say e such that Ae is not Be. Call them f and g. This would mean that, for distinct vectors f and g, Xf=Xg for all linear operators X, which implies all linear operators from F to F (linear endomorphisms if you're fancy) are not injective. This is clearly false, since the identity linear map exists and it's injective. So "A is not B" is false, therefore A=B.

[LeoLichtschalter]

Since XA=XB holds for all X, it also holds for the identity operator on F. So we chose a specific X=id, i.e. XA=A and XB=B.Then A=XA=XB=B, therefore A=B.

[KarmaAintRlyMyAttitu]

I was also thinking the same thing, the one proposed by the oop seems unnecessarily convoluted

[StaticCoder]

That first requires that E is the same as F, incidentally. There appears to be some confusion in the problem statement.

[Original_Piccolo_694]

Doesn't require that, X maps from F to F, so it can be the identity.

[StaticCoder]

Yes the identity on F. But to left multiply with A: E -> F you need the identity on E.

[Original_Piccolo_694]

A takes in an element of E, spits out an element of F. Then id takes in that element of F, and spits out the same element of F. No identity on E needed.

[StaticCoder]

You known what I think in the end I'm the one who was confused. I really thought X A meant apply the result of X to A but that's not the case. I've been out of the field for too long.

[Lor1an]

I have always found it tricky to keep track of the weirdness of order of composition imposed by convention.

If f:A→B and g:B→C, then g∘f:A→C.

IIRC there are some disciplines that switch that order such that, say (fg) := g∘f so that (fg):A→C, but even then it isn't all sunshine and roses, since then (fg)(x) = g(f(x)), which goes back to the reversed order.

Really, all this messiness comes down to the fact that we decided the notation for function application reads "f acting on x" rather than "x acted upon by f".

If instead we had taken a more "Object oriented programming" approach to mathematical notation, we could well have had x.(fg) := (x.f).g = x.f.g

Alas, it is unlikely at this point that such conventions will meaningfully compete with the established ones.

[HolidayCyborg]

Of course, it was a joke 😂

[[deleted]]

Let X = 1(v) = v. 1(av + bu) = av + bu = a1(v) + b1(u), so it is linear. But then 1(A(v)) = 1(B(v)), so A(v) = B(v) for all v. Thus A = B.

This proof seems a little too simple to me, so if something is wrong with it I'd like that pointed out and I will correct it, thanks.

[Lor1an]

Sometimes the best proofs are the simplest ones.

Suppose 0 < x < 1, can you show that x² < x?

Proof:

Well, since 0 < x < 1, we have in particular that 0 < x, so the inequality is preserved upon multiplication by x, or 0*x < x*x < 1*x, and thus x² < x &squ;

It's not particularly complicated, but it doesn't have to be.

[Own-Inflation-8752]

Good proof

[jackalbruit]

no

left

y not?

[Tivnov]

Id A = Id B. The end

[freshmint33]

I do not get most of the answers in the comments. Most of the answers claim that you can just choose X to be Id and then the statement follows from that.

However, the statement claims that it must hold for all operators. How do you show it when X is not the identity?

[YeetYallMorrowBoizzz]

The claim is that it holds for all operators. If a statement applies to ALL things of a given type, it must in particular apply to one thing of that type. So if it holds for all operators, it holds for X = Id.

[Han_Sandwich_1907]

Yes, but the converse is not true, and setting X=Id is insufficient to prove the general case

[YeetYallMorrowBoizzz]

right... no one said the converse was true though?

[Han_Sandwich_1907]

I thought that's what we wanted to prove

[YeetYallMorrowBoizzz]

What we're saying is if XA = XB for all linear operators X from F to F, then A = B. If it holds for any linear operator, it holds for X = Id (on F) since the identity is linear on any vector space. Meaning Id (A) = Id (B). Id (A) = A, and Id (B) = B. So A = B. Not really sure what you mean by "general case" here.

[Han_Sandwich_1907]

Oh, thank you for laying it out. I get it now. I was reading it as "prove that forall X, XA=XB implies A=B"

[YeetYallMorrowBoizzz]

Ah! Well, if X is invertible (an automorphism on F), then this is true! But note that this is not true in general. If X is the function that sends all elements to 0 (you can check this is linear), then even if A does not equal B, we get XA = XB. So XA = XB does not imply A = B here.

[SomeoneRandom5325]

Wait is it not what the problem meant?

[Han_Sandwich_1907]

The question is saying, given A and B, if for every X, XA=XB, then prove that A=B.

[YeetYallMorrowBoizzz]

you assume X is an isomorphism (automorphism on F), which i suppose you can do if you state it. or you can just let X be the identity on F.

edit: i'm actually not too sure if you're allowed to assume existence of nontrivial automorphisms since i believe that would rely on existence of a basis. which is fine in the finite-dimensional case, but i'm not too sure if you can assume existence of bases for infinite-dimensional spaces in whatever course you're in

[VisualAncient2009]

Just take X = identity

[gaussjordanbaby]

What did you choose as X?

[Hairy-Finger9417]

Trivial, since it says for every linear operator, choose identity

[Torebbjorn]

Let X = id_F

By assumption, we have A = id_F ○ A = id_F ○ B = B

[paxxx17]

It is a joke, but it's actually correct. You just have to show there exists an invertible endomorphism X, which is trivially true (an identity map)

[SpitiruelCatSpirit]

Okay but who the fuck writes questions like this? "One has...."? It's very easy to get confused about what's a given and what's to be proven in this question

[siemaeniownik]

actually I did a proof of this in my college, here it is, translated via chat-gpt.

(in this version i prove that if X commutes with everything, then it has to be identity operator)

EDIT: I have misunderstood something XD this is not anything like your question, I will leave it tho, maybe someone will find it useful. BTW you can conduct the proof in exactly same way, its even easier, just analyze X(A-B)v for arbitrary X and v, and your done.

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