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$V$ and W are finitely dimensional linear spaces over the field $K$. Is that true that for every linear transformation $\phi : V^* \to W^*$ there is a linear transformation $\psi: W \to V$ such that $\psi^* = \phi$?

  • W* means dual space of W, that is a space of all functionals $W \to K$
  • V* means dual space of V, that is a space of all functionals $V \to K$

It is like an invert definition of dual mapping. It must be something simple, I just can't think of it anymore.

mymathc
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    do you know about the relationship between a finite dimensional vector space and its second dual? – peek-a-boo Jan 22 '21 at 20:48
  • One way to answer this question is to choose bases for $V$ and $W$, write $\phi$ as a matrix with respect to the dual bases for $V^$ and $W^$, and $\psi$ as a matrix with respect to the bases of $V$ and $W$, and figure out how the two matrices are related if you assume $\psi^* = \phi$. – Deane Jan 22 '21 at 21:09
  • Are you sure? I've just read about sceond dual but I don't know how to use it. – mymathc Jan 22 '21 at 21:13
  • @Deane, I think I can't choose basis for V and W because then I will just prove that equation for those basis in those dimensions. I could look for counterargument that way. – mymathc Jan 22 '21 at 21:17
  • @mymathc, if you can work this out without making any assumptions about the dimensions, bases, and the map $\phi$, then it will be a proof. – Deane Jan 22 '21 at 21:20

1 Answers1

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Yes, it is true. You can consider the transpose map $\phi^*\colon W^{**}\to V^{**}$ and the canonical isomorphisms $\omega_W\colon W\to W^{**}$, $\omega_V\colon V\to V^{**}$.

Set $\psi=\omega_V^{-1}\circ\phi^*\circ\omega_W$ and finish up.

egreg
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