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I have some questions about advanced linear algebra. Let $V$ be a vector space and $V^*$ be the dual space.

  1. Why is $V=V^*$ called non-natural, and $V=V^{**}$ called natural?

  2. $V$ is a vector space with dimension $\dim V=\infty$. Give an example where $V$ is not equal to $V*$ and $V$ is not equal to $V**$ ?

  3. If $\langle .,.\rangle$ is a non-degenerate scalar product on $V$, and $$\varphi: V \to V^{*}: v \mapsto L_v(w) = \langle w,v \rangle$$ is not an isomorphism. Give an example.

sTertooy
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Shabeeb Al-alawi
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    I prefer canonic rather than natural. Canonic or natural in linear algebra means, that we can find an isomorphism between two spaces which does not depend on a choice of a basis in the two spaces. It is well known, that the double dual space is isomorphic to the underlying space independently on a choice of an explicit basis. – TheGeekGreek Dec 25 '16 at 12:59
  • Relevant: http://math.stackexchange.com/questions/1519457/basis-for-dual-in-infinite-dimensional-vector-space, http://mathoverflow.net/questions/13322/slick-proof-a-vector-space-has-the-same-dimension-as-its-dual-if-and-only-if-i. – Martín-Blas Pérez Pinilla Dec 25 '16 at 14:07
  • In this answer, I provide an explanation of why the isomorphism $V \to V^{**}$ is natural. – Ben Grossmann Dec 25 '16 at 14:37

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First, $V\ne V^*$ and $V\ne V^{**}$ always, but in finite dimension $V\cong V^*$ and $V\cong V^{**}$ trivially because all have the same dimension. But in the second case there is a natural (definable independently of basis) isomorphism, namely: $$v\longmapsto e_v,\qquad e_v(f) = f(v),\qquad v\in V, f\in V^*.$$ Now, check what happens when $\dim V = \infty$. For example, $V = \Bbb R\oplus\Bbb R\oplus\cdots =$ space of eventually zero real sequences.

Martín-Blas Pérez Pinilla
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