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Let $V$, a vector space over $\mathbb{C}$ from a finite dimension with an inner product $\beta \langle .,. \rangle$. Let $B=\{v_1,\ldots, v_n\}$, a basis for $V$. If $v_1,\ldots,v_n$ is an orthonormal basis, we have learn in class (apparently. I didn't) that:

For $v,w\in V$: $$ \beta(v,w) = \overline{([v]_B)^T} \cdot [w]_B $$

I need to prove the generalization (for any basis, not necessarily an orthonormal) of this, for:

$$ \beta(v,w) = \overline{([v]_B)^T} \cdot G \cdot [w]_B $$

Where $G\in M_n(\mathbb{C})$ such that $G_{ij} = \beta(v_i,v_j)$

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  • $\begingroup$ Yes...and thus $\;G=I=$ the identity matrix, and the formula you need to prove is completely trivial. Perhaps it was intended to be proved for any other basis, not precisely an orthonormal one? $\endgroup$ – Timbuc Jan 11 '15 at 10:42
  • $\begingroup$ Oh, it should make sense. $\endgroup$ – AlonAlon Jan 11 '15 at 10:43
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Take $\;v,w\in V\;$ and express them as linear combinations of the given orthonormal basis:

$$\begin{cases}v=\sum\limits_{k=1}^na_iv_i\\{}\\w=\sum\limits_{k=1}^nb_iv_i\end{cases}$$

so using the basic properties of a complex inner product:

$$\beta(v,w)=\beta\left(\sum\limits_{k=1}^na_iv_i\,,\,\,\sum\limits_{k=1}^nb_iv_i\right)=\sum_{i,j=1}^na_i\overline {b_j}\,\overbrace{\beta(v_i,v_j)}^{\delta_{ij}}=\sum_{i=1}^na_i\overline{b_i}=\left([v]_B\right)^t\overline{[w]_B}$$

where $\;[v]_B\;$ means column vector .

The above is what you say you apparently didn't learn, but now you have.

Now, if $\;B=\{v_1,...,v_n\}\;$ is not precisely an orthonormal basis, can you see how the matrix $\;G=(\beta(v_i,v_j))\;$ fits in the formula? Complete the argument.

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  • $\begingroup$ I'll have to ponder on it a little. Thank you in the meantime. $\endgroup$ – AlonAlon Jan 11 '15 at 10:58
  • $\begingroup$ Aren't you missing a bar in the RHS: $\sum_{i=1}^na_i\overline{b_i}=\left([v]_B\right)^t[w]_B$? $\endgroup$ – AlonAlon Jan 11 '15 at 11:15
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    $\begingroup$ Yes I am, thank you...and BTW: the bar must be over $\;w\;$ , not over $\;v\;$ . Editing now. $\endgroup$ – Timbuc Jan 11 '15 at 11:17
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    $\begingroup$ No @AlonAlon, as then what new thing would we be doing?! It is just to use your notation: assume you have another basis (call it this time $\;\{u_1,...,u_n\}\;$ if you want, to avoid confusion), and the matrix $\;G=\left(\beta(u_i,u_j)\right)\;$ and then do as shown above: write each vector as a l.c. of this basis, use the properties of inner product and etc. $\endgroup$ – Timbuc Jan 11 '15 at 12:08
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    $\begingroup$ Oh I see now. Thanks a lot! :) $\endgroup$ – AlonAlon Jan 11 '15 at 12:12

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