# Proof regarding transpose mapping

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1. May 20, 2017

1. The problem statement, all variables and given/known data
Suppose T:V→U is linear and u ∈ U. Prove that u ∈ Im T or that there exists $\phi$ ∈ V* such that TT($\phi$) = 0 and $\phi$(u)=1.

2. Relevant equations
N/A

3. The attempt at a solution
Let $\phi$ ∈ Ker Tt, then Tt($\phi$)(v)=$\phi$(T(v))=0 ∀T(v) ∈ Im T. So obviously if u ∈ Im T than $\phi$(u)=0. I now need to prove that if u ∉ Im T, than there exists a linear functional which answers the above criteria, and this is where I'm stuck. I don't know which mapping I define that would answer the criteria.

Any help would be appreciated,

2. May 20, 2017

### andrewkirk

There seems to be something wrong with how this problem is written. If $\phi\in V^*$ then the domain of $\phi$ is $V$ which , from the problem specification, does not necessarily have any intersection with $U$, so $\phi(u)$ is undefined.

There must be some missing information, or some implicit assumptions, which need to be brought out into the open.

3. May 20, 2017

Ah, yes I forgot to mention that. I just assumed it meant $\phi$∈U* and not V*. Damn book's full of mistakes.

4. May 20, 2017

### andrewkirk

Some more info would be helpful. What are V and U? Vector spaces, or modules? If vector spaces, are they finite dimensional? Do they both have inner products? What about norms?

Also, what, exactly, is $T^T$ (which sometimes is written above as $T^t$)? It looks like it's supposed to be some sort of inverse pushforward.

5. May 21, 2017

V and U are vector spaces, dimension is not specified (I quoted the question word for word) which naturally means I cannot use arguments of dimension as the vector spaces can be of both finite and infinite dimension. Inner products are not mentioned and are probably irrelevant here as the chapter of this question focuses mostly on linear functionals, same with norms.

Tt:U*→V* is the transpose mapping for T. If $\phi$ is a linear functional in U*, then Tt($\phi$)=$\phi \circ T$, thus Tt($\phi$)(v)=$\phi$(T(v)).

6. May 21, 2017

### zwierz

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7. May 21, 2017

### zwierz

This is the same as to prove that if $u\notin \mathrm{Im}\,T$ then...
Prove that the space $U$ is decomposed as follows $U=\mathrm{Im}\,T\oplus\mathrm{span}\,\{u\}\oplus L.$

Last edited: May 21, 2017