Proving a mapping from Hom(V,V) to Hom(V*,V*) is isomorphic

[Adgorn]

Homework Statement

Let V be of finite dimension. Show that the mapping T→T^t is an isomorphism from Hom(V,V) onto Hom(V*,V*). (Here T is any linear operator on V).

Homework Equations

N/A

The Attempt at a Solution

Let us denote the mapping T→T^t with F(T). V if of finite dimension, say dim (V)=n. Than dimension of V*=n as well, and dim(Hom(V,V))=dim(Hom(V*,V*))=n². So in order to prove F(T) is an isomorphism I only need to prove it is linear and non-singular.

First, F(aT₁+bT₂) = (aT₁+bT₂)^t = $\phi$(aT₁+bT₂) = a$\phi$(T₁) + b$\phi$(T₂)= aT₁^t + bT₂^t = aF(T₁)+bF(T₂) for some a,b ∈ K and $\phi$ ∈ V*.

Thus, F is linear. Now the only thing left to proof is that Ker F={0}, and this is where I got stuck. I don't know how to prove that the function is non-singular, so I need some assistance.

 

[fresh_42]

Homework Statement

Let V be of finite dimension. Show that the mapping T→T^t is an isomorphism from Hom(V,V) onto Hom(V*,V*). (Here T is any linear operator on V).

Homework Equations

N/A

The Attempt at a Solution

Let us denote the mapping T→T^t with F(T). V if of finite dimension, say dim (V)=n. Than dimension of V*=n as well, and dim(Hom(V,V))=dim(Hom(V*,V*))=n². So in order to prove F(T) is an isomorphism I only need to prove it is linear and non-singular.

First, F(aT₁+bT₂) = (aT₁+bT₂)^t = $\phi$(aT₁+bT₂) = a$\phi$(T₁) + b$\phi$(T₂)= aT₁^t + bT₂^t = aF(T₁)+bF(T₂) for some a,b ∈ K and $\phi$ ∈ V*.

Thus, F is linear. Now the only thing left to proof is that Ker F={0}, and this is where I got stuck. I don't know how to prove that the function is non-singular, so I need some assistance.

You have an inverse, don't you? Linearity, regularity and with comparison of dimension you have all you need, but why is $F(\varphi) \in \operatorname{Hom}(V^*,V^*)$ for $\varphi \in \operatorname{Hom}(V,V)$? Where is the definition of a transposed mapping used?

 

[Adgorn]

Well, my book's definition of transpose mapping (which may not be the general one) is as follows:
Let T:V→U be a linear mapping from V to U, and let $\phi$ be a linear functional from U to U*. Then the transpose mapping T^t is a linear mapping from V* to U* is defined as T^t:V*→U*=$\phi \circ T$. Thus (T^t($\phi$))(v)=$\phi$(T(v)).

Therefore if T ∈ Hom(V,V) (since it's a linear operator on V), then F(T)=T^t ∈ Hom(V*,V*) (since it's a linear operator on V*).

Also I'm afraid I am not advanced enough in my knowledge of the field to know what solution to deduce from what you said about linearity, regularity, and comparison of dimension. So more specification is required for me.

 

[fresh_42]

Also I'm afraid I am not advanced enough in my knowledge of the field to know what solution to deduce from what you said about linearity, regularity, and comparison of dimension. So more specification is required for me.

You have shown that transposing $F$ is linear. You also have compared dimensions, which are equal for $\operatorname{Hom}(V,V)$ and $\operatorname{Hom}(V^*,V^*)$. Regular here is only another word for bijective. And as you already said, due to the dimensions, even injectivity is sufficient. Thus you asked about the kernel of transposing, but as $F^2(T)=T$ and thus $F^2=1$, it has to be zero. Or you can assume $F(T)=0$. Then $0=F(T)(\phi(V))=\phi(T(V))$ with the isomorphism $\phi : V \rightarrow V^*$. Now what does that mean for first $T(V)$ and then for $T$?