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Show that there does not exist a isomorphism $\phi:E\rightarrow E^*$ that it takes every basis of $E$ to its dual basis. ($E$ is a vector space over field K and $\text{dim}E=n$ .)

My attempt: There exist some basis $(e_1,\cdots,e_n)$ that $(\phi(e_1),\cdots,\phi(e_n))$ is its dual basis since there exist the isomorphism. Now I probably have to make a basis that under $\phi$ is not its dual basis.

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    $\begingroup$ check the quantity $\phi(e).e$ $\endgroup$ – hHhh Jul 1 '15 at 20:20
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You should seem $(e_1+e_2,\cdots,e_n)$ and $(\phi(e_1+e_2),\cdots,\phi(e_n))$ .

since $\phi(e_1+e_2)\in E^{**}$ then $1=\phi(e_1+e_2)(e_1+e_2)=\phi(e_1)(e_1+e_2)+\phi(e_2)(e_1+e_2)=\cdots=1+1$. this is a contradiction.

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