Error in Sadri Hassani's Mathematical Physics

[cleaf]

I am reading his excellent book "Mathematical Physics Part 1, Second Edition", which has benefited me a lot in many ways.

However, I have a doubt about the correctness of the theorem 2.3.23, which states that for any subspace U of V, the map T ' : V/U -> T(V) defined by T ' ([a]) = T|a> is well defined isomorphism.

I can not understand this theorem. If the theorem is correct, for any b and a in the same class [a], T(|a>-|b>) must be zero in T(V). That doesn't seem right for any subspace but Ker T.

What do the author really mean？

 

[Office_Shredder]

I'm also confused. It might help if we can see the entire preceding example to get a bit more context.

Is the thing that follows intended to be a proof (if so post the whole proof would also help)? It doesn't look like it is.

 

[martinbn]

As written seems wrong.

 

[xaos]

I think its safe to assume the book is implicitly stating distinct cosets correspond to distinct coset representatives. So if b is anything in [a] then necessarily "[ b] = [a] " and defines an equivalence relation.

T' receives the set U identified to [0] and sends it to the value T'([0])=T|0> = 0 since T is a linear map. It can be seen that the kerT'=[0]. If there is any nonzero [a] in kerT' must map to 0 therefore [a]=[0] by the above remarks. Notice we're looking at kerT' not kerT.

 

[cleaf]

I think its safe to assume the book is implicitly stating distinct cosets correspond to distinct coset representatives. So if b is anything in [a] then necessarily "[ b] = [a] " and defines an equivalence relation.

T' receives the set U identified to [0] and sends it to the value T'([0])=T|0> = 0 since T is a linear map. It can be seen that the kerT'=[0]. If there is any nonzero [a] in kerT' must map to 0 therefore [a]=[0] by the above remarks. Notice we're looking at kerT' not kerT.

Thanks very much!

I think the key point is T' is not a well defined map.

As defined, for any equivalent class [a]=[c]={a, c, ...}, T'([a]) = T(a), T'([c]) = T(c), However, a-c is not guaranteed to be in Ker T (it is in U, but U is not Ker T). As a result, T'([a]) - T'([c]) = T(a-c) may not be zero in T(V). That is [a] = [c] , But T'([a]) != T'([c]), which violates the requirement of a well defined mapping.

 

[xaos]

If c is in [a] then we have [a-c] = (a-c) + U = (a + U) - (c +U) = [a] - [c] = [a] - [a] = [0].
So it [a]=[c] then T'([a-c])=T'([0]) = T(0) = 0

 

[cleaf]

If c is in [a] then we have [a-c] = (a-c) + U = (a + U) - (c +U) = [a] - [c] = [a] - [a] = [0].
So it [a]=[c] then T'([a-c])=T'([0]) = T(0) = 0

Wonderful exposition！

Thanks a lot!

By definition, I can not understand why T(a) = T(b) if a and b are both in U.

That is T'([0]) = T(0), T'([a∈U, a != 0]) = T(a), why T(0) - T(a) = T(0)?

If we assume any element in U must be sent to zero in W, then U must be a subspace of Ker T.

 

[xaos]

We're sending U to the equivalence class [0] in V/U. So by equivalence if a is in [0] then [a]=[0].
This means a + U = [a] = [0] = 0 + U = U, so a and 0 are both in U. T'([a]) = T'([0]) = T(0) = 0 so [a] is in the KerT'=[0] by 1-1 ( T'([a]) != T'([ b]) => T(a) != T(b) )

 

[cleaf]

We're sending U to the equivalence class [0] in V/U. So by equivalence if a is in [0] then [a]=[0].
This means a + U = [a] = [0] = 0 + U = U, so a and 0 are both in U. T'([a]) = T'([0]) = T(0) = 0 so [a] is in the KerT'=[0] by 1-1 ( T'([a]) != T'([ b]) => T(a) != T(b) )

So you mean the U=[0] can only be represented by 0?

By definition, T'([0]) = T(0), T'([a]) = T(a). If we always translate [a] to [0], it is correct. However, there is no rule against the direct sending, where T(a) != T(0) and ambiguity emerges.

Please see the attached example 2.3.22 in the same book.

 

[martinbn]

@xaos The text syas that $V/U$ is isomorphic to $T(V)$, where $U$ is a subspace of $V$. How can that be true for any subspace!

 

[xaos]

The cosets {a + U | a in V} partition V into distinct equivalence classes under the equivalence relation [a] = [ b] iff a + U= b + U iff a-b is in U. If 'a' is in the coset 0 + U = U is another representation of [0], then [a]=[0]. If 'a' is not in U then there is a unique coset a+U with 'a' representative so [a] is its equivalence class. Representatives in the same coset/class intersect so are equivalent. Representatives of different cosets/classes are disjoint so are not equivalent. We can then choose 0 to be the preferred representative of U, and we can choose 'a' to be the preferred representative of a+U when 'a' is not in U.

By identifying each symbol T'([a]) with its equivalent symbol T(a) we can prove:
i. Onto. For all T(a) in T(V) there is a [a] in V/U such that T'([a])=T(a)
ii. One-To-One. T'([a]) != T'([ b]) => T(a) != T(b)
iii. Homomorphism. T'([a]+[ b]) = T'([a+b]) = T(a+b) = T(a) + T(b) = T'([a]) + T'([ b])
iv. Well Defined. If [a]=[c] then T'([a]) - T'([c]) = T'([a-c])=T'([0]) = T(0) = 0 (see previous)

My guess is we can do this because the homomorphism property of the maps is a very strong constraint. Effectively we appear to be identifying the two homomorphisms, which is an even stronger constraint.

 

[xaos]

[edit] v. T' is linear. T'(a[x]+b[y]) = T'([ax + by]) = T(ax+by) = aT(x) + bT(y) = aT'([x])+bT'([y])
then I can correct ii.One-to-one. 0=T(a)-T(b) = T(a-b) = T'([a-b])=T'([a]-[ b])=T'([a])-T'([ b])

Wait. Are you rejecting the proofs or the theorems in the book? I wasn't looking at the proofs. Give me some time to look at them.

 

[Infrared]

iv. Well Defined. If [a]=[c] then T'([a]) - T'([c]) = T'([a-c])=T'([0]) = T(0) = 0 (see previous)

This isn't true. Consider $T(x,y)=(y,x)$ and $U=\text{Span}(1,0).$ Then $[(1,1)]=[(0,1)]$ but $[T(1,1)]\neq [T(0,1)].$

 

[xaos]

T((1,1)) = T'([(1,1)] = T'([(0,1)]) = T((0,1)) So they have the same image under T. What does [T(x,y)] mean? The equivalence classes are not in W but in V/U.

 

[Infrared]

The problem is that $(1,1)$ is in the same equivalence class as $(0,1)$, but $T(1,1)\neq T(0,1)$. This means that $T'([v])=T(v)$ is not well-defined. Yes, the brackets in $[T(x,y)]$ were a typo.

 

[xaos]

I think the proof#1 in the book is assuming a priori this defining. Once we start with a defining property, we're showing it must be an isomorphism. Proof#2 is using a different defining property. So Proof#1 != Proof#2.

 

[Infrared]

Forget about a proof for a second. What is the $\textit{statement}$ of what you think is true?

 

[xaos]

If you allow for the moment that T'([a])=T(a)... That there maps that do not have this property then T isn't one of them. True in general, maybe not. But this is where we are starting with by saying "let us define..." By doing this we are automatically making strong constraints on what T can be.

 

[Infrared]

Even if that $T'$ is well-defined (and this is not mentioned as a hypothesis in the pdf), it doesn't have to be an isomorphism. Let $T:V\to V$ be the zero map and let $U=\{0\}.$ Then $T':V/U\to T(V)$ is well-defined (it's still the zero map), but it's not an isomorphism unless $V=\{0\}.$

 

[xaos]

Proof#1. Let T':V/U -> T(V) s.t. T'([x]) = T(x). Full stop. No further properties are assumed. This is a very strong constraint. Everything has to be amicable to this assumption for it to work. This is a hypothetical definition, not a '≡' identity definition.
Proof#2. Let T': V/KerT -> T(V) s.t. T'([x])=T(x). Again this will be a defining characteristic of T' and T in the proof. This doesn't mean all T and T' must have this property, but we are choosing the ones that do.

 

[Infrared]

I can't tell what you're trying to prove. Can you clearly state the proposition that you think is true?

 

[xaos]

Let T:V->W be a linear map. Let p:V->V/U be the canonical projection. Then the linear map T' induced by the given identication T'(p(x))=T(x) is an isomorphism onto the image T(V). I think we need T' to be linear for this to work.

 

[Infrared]

No, even if that map $T'$ is well-defined, it does not have to be injective (see my post 19). You do not need to assume that $T'$ is linear; that will always be true (as long as $T'$ is well-defined).

 

[xaos]

Yes, I concede. I've been missing a step.
[edit] Try: [a-b] is in the KerT' then [a]+KerT' = [ b]+KerT' -> [a]= [ b] as coset representatives ?

 

[xaos]

Refering to #13. For Well-defined. Maybe this works: [a]= [ b] then a=b as coset representatives then T(a)=T(b) since linear maps are well defined. Then use the identification T'([c])=T(c) to say T'([a])=T'([ b])

 

[Infrared]

You can't prove a false statement by writing an argument a different way. I gave a counterexample. The condition $[x]=[y]$ does not imply $T(x)=T(y).$ This implication is true if and only if $U$ is a subspace of the kernel of $T$.

 

[xaos]

Let T'(p(U)) = T(U) s.t. p(U)=[0]. Then T(U)=T'([0])=T(0)=0 so U is in the KerT.
I'm not getting the coset idea to work unless a+U =b + U -> a+U - U = b+U-U -> a=b. which might not work.

 

[Infrared]

You're assuming that $T'$ is well-defined there. Given an arbitrary subspace $U\subset V$ and an arbitrary linear map $T:V\to W$, there is no reason why $U\subseteq\ker(T)$ should be true.

 

[xaos]

I'm just using what was given in the hypothesis. T and T' are not arbitrary, they're related by the identification T'([ b])=T(b). This is where I'm starting.

 

[Infrared]

I can't tell what you're using as a hypothesis and what you're trying to prove. It seems like you're trying to prove that there is a well-defined linear map $T':V/U\to W$ defined by the formula $T'([x])=T(x)$. What hypothesis are you assuming to try to show this?