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Give an example of non-degenerate scalar product $\langle,\rangle$ on a vector space $V$ such that the application $$\varphi : V\to V^* : v\mapsto L_v$$defined by $L_v(w)=\langle v,w\rangle$ is not an isomorphism.

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    $\begingroup$ Sorry, it is not clear for me what you are asking. Could you be more precise? $\endgroup$ – Janik Dec 28 '16 at 17:05
  • $\begingroup$ I hope the question clear $\endgroup$ – Shabeeb Al-alawi Dec 28 '16 at 17:11
  • $\begingroup$ @ZevChonoles and I edited your question almost simultaneously. Feel free to roll back to his version if you prefer it (this applies to everyone). $\endgroup$ – Arnaud D. Dec 28 '16 at 17:27
  • $\begingroup$ In fact, a vector space is isomorphic to its dual if and only if it is finite-dimensional. Thus you just need to find an infinite-dimensional one. $\endgroup$ – Hanul Jeon Dec 28 '16 at 18:36
  • $\begingroup$ Can you give an example ? $\endgroup$ – Shabeeb Al-alawi Dec 28 '16 at 18:57
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Consider the vector space $V$ of all infinite sequences $(x_i)_{i\geq0}$ of real numbers with finitely many non-zero terms, and endow it with the inner product such that $$\langle(x_i)_{i\geq0},(y_i)_{i\geq0}\rangle=\sum_{i\geq0}x_iy_i.$$ There is a linear function $\phi:V\to\mathbb R$ such that $$\phi((x_i)_{i\geq0})=\sum_{i\geq0}x_i$$ for all sequences $(x_i)_{i\geq0}\in V$, and you can easily check that it is not in the image of the canonical map $V\to V^*$. Indeed, no vector $(\xi_i)_{i\geq0}$ in $V$ exists such that $$\sum_{i\geq0}x_i\xi_i=\sum_{i\geq0}x_i$$ for all $(x_i)_{i\geq0}\in V$.

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  • $\begingroup$ I did not know how check that it is not in the image of the canonical map V→V∗ ? $\endgroup$ – Shabeeb Al-alawi Dec 28 '16 at 19:22
  • $\begingroup$ Well, the last sentence in the answer gives you an almost complete hint. $\endgroup$ – Mariano Suárez-Álvarez Dec 28 '16 at 19:30

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