1.Let $V$ be a finite dimensional vector space over the field $K$, with a non-degenerate scalar product. Let $W$ be a subspace. Show that $W^{\perp\perp}=W$.

2.Show that the same conclusion as in the preceding exercise is valid if by $W^{\perp}$ we mean the orthogonal complement of $W$ in the dual space $V^*$.

  1. I have already proven the first result. W is on the same vector space as $W^{\perp}$

2.Apparently $W^\perp$ is on the dual space and $W$ on $V$. My problem is to understand how can we prove that $W^{\perp\perp}=W$, despite the fact they may lie on different spaces. That is what I think the question meant. Considering the feedback I had on the comments it seems I have to find an isomorphism between $V$ and $V^{**}$. However I do not see how I prove the isomorphic relation nor I can understand how the isomorphism is going to lead me to prove the question claim.

I am self-studying. And I want to see more examples to create more intuitions about the workings of functionals.


1) Can someone give me a complete proof of the question claim?

Thanks in advance!

  • $\begingroup$ It helps to visualize everything with vectors: If $V$ is written as column vectors, the elements of $V^*$ can be written as row vectors. You can use the dual basis to see this fact, or just start off with the standard space and standard product as an example. Taking it from there, it gets easier to understand the general case. $\endgroup$ – Dirk Jul 12 '17 at 13:19
  • $\begingroup$ First you are saying that you have proved $1$, but then asking "how can we prove $(W^\perp)^\perp = W$". I didn't get it . $\endgroup$ – onurcanbektas Jul 12 '17 at 15:43

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