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I am tasked with the following:

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I am thus tasked with proving:

$1)$ $\phi(T)$ is linear, so that it respects closure under scalar multiplication and addition.

$2)$ $\phi(T)$ is a bijection.

I only need justification as to whether or not my explanation as to whether $ \phi(T)$ is a bijection is indeed valid or not. First, I will note that linear functionals are isomorphisms, so are bijective themselves and respect closure under addition and scalar multiplication. I work to first prove that $\phi$ is injective.

Injectivity

Suppose $\phi$ is injective.

$$\implies \phi(T_n) = \phi(T_m) \iff T_n = T_m$$

$$\implies T^*_n = T^*_m \iff T_n = T_m$$ $$\implies f \ (T_n) = f \ (T_m) \iff T_n = T_m$$ $$\implies f(T_n - T_m) = 0 \iff T_n = T_m$$

Let's suppose $T_n \ne T_m$.

$$\implies f(T_n-T_m) = 0 \implies T_n - T_m \in \text{Ker} \ f$$

Which is a contradiction, as $f$ is surjective.

Surjectivity

Assume $\exists \ T \in \mathcal L (V,W)$ such that $\phi(T) = 0$.

$$\implies T^* = 0$$. $$\implies f(T) = 0$$ $$\implies T \in \text{Ker} \ f$$

Which is a contradiction. Hence, $\phi$ is surjective, making it a bijection and $\phi$ an isomorphism.

Did I make any mistakes here?

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  • $\begingroup$ You prove injectivity of $\;\phi\;$ ...by assuming $\;\phi\;$ is injective?! $\endgroup$ – DonAntonio Nov 16 '18 at 20:25
  • $\begingroup$ Ack! I hope not. I thought I proved by contradiction eventually with suppose $T_n \ne T_m$ $\endgroup$ – sangstar Nov 16 '18 at 20:26
  • $\begingroup$ Also observe that in (1)-(2) you're supposed, I guess, to prove $\;\phi\;$ is linear and injective, not $\;\phi(T)\;$ , which is merely a linear map $\;W^*\to V^*\;$ ... $\endgroup$ – DonAntonio Nov 16 '18 at 20:28
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Fill in details:

By definition, if $\;T:V\to W\;$ is a linear map, then $\;T^*:W^*\to V^*\;$ is defined as

$\;T^*(g)v:=g(Tv)\;,\;\;v\in V\;,\;\;g\in W^*\;$ , thus

$\;\phi\;$ is linear, because

$$\phi(T+S):=(T+S)^*=T^*+S^*$$

and the last equality follows from

$$(T+S)^*(g)v:=g(T+S)v=g(Tv+Sv)=g(Tv)+g(Sv)=:T^*(g)v+S^*(g)v$$

And also

$$T\in\ker \phi\implies \phi(T):=T^*=0^*\implies\forall\,g\in W^*\,,\,\,T^*(g)v=g(Tv)=0\,,\,\,\forall v\in V\implies$$

$$Tv\in\ker g\,,\,\,\forall g\in W^*\implies Tv=0\,\,\forall\,v\in V\implies T\equiv0.$$

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DonAntonio already touched linearity and injectivity questions. For surjectivity, you seem to be proving injectivity instead. However, to correctly prove surjectivity, you are going to need to use a dimension-counting argument. This is because, if $W$ is infinite-dimensional and $V\neq 0$, then $$\dim_\mathbb{K}\big(\mathcal{L}(V,W)\big)<\dim_\mathbb{K}\big(\mathcal{L}(W^*,V^*)\big)\,,$$ where $\mathbb{K}$ is the ground field. However, the dual map $\phi:\mathcal{L}(V,W)\to\mathcal{L}(W^*,V^*)$ is still an injective linear map, regardless of the dimensions of $V$ and $W$. The proofs of linearity and injectivity are essentially unchanged.

Since $V$ and $W$ in the problem statement are both finite-dimensional, $$\dim_\mathbb{K}\big(\mathcal{L}(V,W)\big)=\dim_\mathbb{K}(V)\,\dim_\mathbb{K}(W)=\dim_\mathbb{K}(W^*)\,\dim_\mathbb{K}(V^*)=\dim_\mathbb{K}\big(\mathcal{L}(W^*,V^*)\big)\,.$$ Thus, any injective linear map from $\mathcal{L}(V,W)$ to $\mathcal{L}(W^*,V^*)$ is automatically surjective, whence bijective.

Interestingly, if $W$ is finite-dimensional and $V$ is infinite-dimensional, the map $\phi$ is still an isomorphism. We are left to show that $\phi$ is surjective. To show this, let $S:W^*\to V^*$ be a linear map. Let $n:=\dim_\mathbb{K}(W)$. Pick a basis $\{w_1,w_2,\ldots,w_n\}$ of $W$, along with the dual basis $\{f_1,f_2,\ldots,f_n\}$ of $W^*$ (i.e., $f_i(w_j)=\delta_{i,j}$ for $i,j=1,2,\ldots,n$, where $\delta$ is the Kronecker delta). For each $w\in W$, write $w^{**}\in W^{**}$ for its double dual. Ergo, we see that $S$ takes the form $$S=\sum_{i=1}^n\,e_i\otimes w_i^{**}$$ for some $e_1,e_2,\ldots,e_n\in V^*$ (namely, $e_i:=S(f_i)$ for $i=1,2,\ldots,n$). Define $$T:=\sum_{i=1}^n\,w_i\otimes e_i\,.$$ Then, for all $j=1,2,\ldots,n$ and $v\in V$, we have $$\big(T^*(f_j)\big)(v)=f_j\big(T(v)\big)=f_j\left(\sum_{i=1}^n\,e_i(v)\,w_i\right)=\sum_{i=1}^n\,e_i(v)\,f_j(w_i)=\sum_{i=1}^n\,e_i(v)\,\delta_{i,j}=e_j(v)\,.$$ However, as $e_j=S(f_j)$, we get $$\big(S(f_j)\big)(v)=e_j(v)$$ for all $j=1,2,\ldots,n$ and $v\in V$. This proves that $S(f_j)=T^*(f_j)$ for $j=1,2,\ldots,n$. Because $f_1,f_2,\ldots,f_n$ span $W^*$, we get $S=T^*=\phi(T)$. Therefore, $\phi$ is surjective whenever $W$ is finite-dimensional. (Consequently, the dual map $\phi:\mathcal{L}(V,W)\to\mathcal{L}(W^*,V^*)$ is an isomorphism if and only if $W$ is finite-dimensional or $V=0$.)

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    $\begingroup$ Surjectivity is automatic once we have injectivity since all the involved linear spaces are finite dimensional...and of the same (finite) dimension, of course. $\endgroup$ – DonAntonio Nov 16 '18 at 21:49
  • $\begingroup$ That is what I wrote. I merely wanted to discuss what happens when $V$ or $W$ is not finite-dimensional. $\endgroup$ – Batominovski Nov 16 '18 at 21:50
  • $\begingroup$ Why does involved linear spaces being finite dimensional imply subjectivity when injective? $\endgroup$ – sangstar Nov 17 '18 at 16:51
  • $\begingroup$ Because the dimension of the image of $\phi$ equals the dimension of the domain, but the dimension of the domain also equals the dimension of the codomain. Hence, the image is a subspace of the codomain with the same finite dimension. $\endgroup$ – Batominovski Nov 17 '18 at 17:09

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