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Suppose I have two inner products $\langle \cdot ,\cdot \rangle_{1}$ and $\langle \cdot ,\cdot \rangle _{2}$ in a finite-dimensional $\Bbb C$ vector space. Does there exists $\phi: V \to V$ invertible linear operator s.t $\langle \phi(v),\phi(w)\rangle _{1}=\langle v,w\rangle_{2}$? If so then how to construct it.

My guess is to define $H_1=(h_{1ij})=\langle e_i,e_j\rangle _{1}$ and $H_2=(h_{2ij})=\langle e_i,e_j\rangle _{2}$.

Then do I have to transform $H_1$ to $H_2$?

Please give me some precise answer.

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    $\begingroup$ Use $\langle\rangle$ for $\langle\rangle$. $\endgroup$
    – Shaun
    Apr 3, 2018 at 20:22
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    $\begingroup$ @Shaun The \langle \rangle fairy smiles upon you. $\endgroup$ Apr 3, 2018 at 20:25

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Let $\{e_i\}_{i=1}^n$ be an orthonormal basis of $(V, \langle\cdot, \cdot \rangle_1)$ and $\{f_i\}_{i=1}^n$ an orthonormal basis of $(V, \langle\cdot, \cdot \rangle_2)$ Define the map: $$T :(V, \langle\cdot, \cdot \rangle_1) \to (V, \langle\cdot, \cdot \rangle_2)$$ By $T(e_i) =f_i$ for each $i$. This is linear and preserves the norm as you would like, because if: $$ x = \sum_{j=1}^n \alpha_j e_j, y = \sum_{\ell=1}^n \beta_\ell e_\ell \in (V, \langle\cdot, \cdot \rangle_1)$$ Then: $$\langle T(x), T(y) \rangle_2 =\left\langle \sum_{j=1}^n \alpha_j f_j , \sum_{\ell=1}^n \beta_\ell f_\ell\right\rangle_2 = \sum_{j=1}^n \sum_{\ell=1}^n \alpha _j \beta_ \ell \underbrace{\langle f_j ,f_\ell\rangle_2}_{=\delta_{j\ell}} $$ Where $\langle f_j ,f_\ell\rangle_2 = \langle e_j ,e_\ell\rangle_1 = \delta_{j\ell}$ by orthonormality, so plugging this in: $$\langle T(x), T(y) \rangle_2 = \sum_{j=1}^n \sum_{\ell=1}^n \alpha _j \beta_ \ell \langle e_j ,e_\ell\rangle_1 = \langle x,y\rangle_1 $$ Also, this is invertible, as $T$ is basically a change of basis matrix which has full rank.

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  • $\begingroup$ Everything is okay. I think you have to put $<,>_1$ and $<,>_2$ accordingly. $\endgroup$
    – user548580
    Apr 3, 2018 at 20:52

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