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 $G_{ij} = \beta(v_i,v_j)$

  • $\begingroup$ 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

Take $\;v,w\in V\;$ and express them as linear combinations of the given orthonormal basis:


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Oh I see now. Thanks a lot! :) $\endgroup$ – AlonAlon Jan 11 '15 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.