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So, I'm trying to show that $T^{**}\circ \varphi_1 = \varphi_2\circ T$ where $\varphi_1 : T\rightarrow V^{**}$ and $\varphi_2 : W\rightarrow W^{**}$. Also, $T^{**} : V^{**}\rightarrow W^{**}$, and yes $V$ and $W$ are finite dimensional vector spaces of a field $\mathbb{F}$.

It seems to follow given the diagram below $\require{AMScd}$ \begin{CD} V @>\varphi_1>>V^{**} \\ @VTVV @VVT^{**}V \\ W @>>\varphi_2> W^{**}, \end{CD}

but I must show via multiple equivalence definitions. I assume $T^{**}=(T^*)^*$, so does this mean that $T^{**}\circ \varphi_1=(T^*)^*\circ \varphi_1=\varphi_1\circ T^*$, but then $T^*$ takes elements $w^*\in W^*$ and maps them to $V^*$, which means $\varphi_1$ is taking as an input something from $V^*$, but $\varphi_1$ takes elements from $V$ not $V^*$. Any help is appreciated.

So, someone posted this as an answer to the above question:

I presume that $\phi_1 : V \to V^{\ast\ast}$ and $\phi_2 : W \to W^{\ast\ast}$ are the canonical isomorphisms given by $$ \phi_1(v) := \left(V^\ast \ni f \mapsto f(v)\right), \quad \phi_2(w) := \left(W^\ast \ni g \mapsto g(w) \right). $$ In that case then, for any $v \in V$, $g \in W^{\ast}$, $$ (\phi_2 \circ T)(v)(g) \overset{1}{=} \phi_2(Tv)(g) \overset{2}{=} g(Tv) \overset{3}{=} (T^\ast g)(v) \overset{4}{=} \phi_1(v)(T^\ast g) \overset{5}{=} T^{\ast\ast}(\phi_1(v))(g) \overset{6}{=} (T^{\ast\ast}\circ \phi_1)(v)(g), $$ simply by applying the relevant definitions repeatedly.

But, I would like a thorough explanation of each step $1-6$ if possible.

Martin
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Trancot
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  • How can a vector be mapped to a linear functional? – Trancot May 07 '13 at 19:38
  • The mapping $T^:W^\rightarrow V^$ is defined by $T^(g)=gT$ for all $g\in W^$, as is explained in Friedberg's Linear Algebra*. – Trancot May 07 '13 at 19:44
  • So I guess $\phi_1$ and $\phi_2$ are the canonical injections in the biduals? – Julien May 07 '13 at 19:45
  • I believe so. Rather, a distinguished (canonical, natural) isomorphism. – Trancot May 07 '13 at 19:48
  • @Julien I assume you're French. Perhaps the bidual goes by another name more familiar. – Trancot May 07 '13 at 19:52
  • Right, the dual of the dual. And I forgot this was finite-dimensional here. So the canonical injections are cononical isomorphisms. – Julien May 07 '13 at 19:55
  • @BarisaBarukh, I'd really like to know how somebody's being French has something to do with this. =) – Christopher A. Wong May 07 '13 at 20:07
  • Frenchmen know the double dual by the name bidual. Just checking, that's all. – Trancot May 07 '13 at 20:20
  • French women call it bidual too. But you're right, I should have said double dual. – Julien May 07 '13 at 20:23
  • Yes, that's right. I should have said humans who call themselves French often times call the double dual the bidual. – Trancot May 07 '13 at 20:34
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    If you are talking about Branimir Cacic's answer below, your edit is a strange and rude way to reward someone who took some time to help you for free. Also, you obviously failed to understand that accepting an answer means that one is happy about it. – Julien May 08 '13 at 02:57
  • @julien Not everyone accepts cognitive dissonance when its presented to them; that is, I am happy with the answer, but I am also not satisfied completely. My kind of people try to be clear when seriously answering something, at least . . . I represent myself to be that way--not necessarily here. – Trancot May 08 '13 at 03:12
  • @BarisaBarukh My apologies if things weren't clear: (1) is by definition of the composition $\phi_2 \circ T$; (2) is by definition of $\phi_2$; (3) is by definition of $T^\ast$; (4) is by definition of $\phi_1$; (5) is by definition of $T^{\ast\ast} := (T^\ast)^\ast$; (6) is by definition of the composition $T^{\ast\ast} \circ \phi_1$. There's nothing deep here at all, just unpacking and applying the same (contextually elementary) definitions again and again. If this is your first encounter with this material, it's a "yoga" you'll have to get used to. – Branimir Ćaćić May 08 '13 at 09:12

1 Answers1

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I presume that $\phi_1 : V \to V^{\ast\ast}$ and $\phi_2 : W \to W^{\ast\ast}$ are the canonical isomorphisms given by $$ \phi_1(v) := \left(V^\ast \ni f \mapsto f(v)\right), \quad \phi_2(w) := \left(W^\ast \ni g \mapsto g(w) \right). $$ In that case then, for any $v \in V$, $g \in W^{\ast}$, $$ (\phi_2 \circ T)(v)(g) = \phi_2(Tv)(g) = g(Tv) = (T^\ast g)(v) = \phi_1(v)(T^\ast g) = T^{\ast\ast}(\phi_1(v))(g) = (T^{\ast\ast}\circ \phi_1)(v)(g), $$ simply by applying the relevant definitions repeatedly.

Branimir Ćaćić
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