Is it true that any inner product $\langle \cdot, \cdot \rangle : V \times V \to \mathbb{C}$ is given by $$\langle \boldsymbol{x}, \boldsymbol{y} \rangle = \boldsymbol{x}^T \boldsymbol{M} \boldsymbol{\bar{y}}$$ where $\boldsymbol{M}$ is a positive definite, hermitian matrix, $\boldsymbol{x}, \boldsymbol{y} \in V$?

Is it true that any inner product is also given by $$\langle \boldsymbol{x}, \boldsymbol{y} \rangle = \boldsymbol{y}^* \boldsymbol{M} \boldsymbol{x}$$ where $\boldsymbol{x}^*$ denotes the conjugate transpose of $\boldsymbol{x}$?

  • 2
    $\begingroup$ What are your assumptions on $V$? (i.e. finite dimensional, base field $\Bbb{C}$, etc) $\endgroup$ – icurays1 Nov 25 '12 at 21:09
  • $\begingroup$ Say $V= \mathbb{C}^2$ $\endgroup$ – Sam Nov 25 '12 at 21:29
  • 1
    $\begingroup$ Hint: Put elements of the standard basis in place of $\bf x$ and $\bf y$. $\endgroup$ – Berci Nov 25 '12 at 21:55
  • $\begingroup$ Since we multiply matrices by vectors, $V$ must be a set of column vectors. $\endgroup$ – GEdgar Nov 26 '12 at 14:56

The answer to the first question is yes.

Let $V$ be a finite dimensional complex vector space, and let $\langle\cdot,\cdot\rangle:V\times V\rightarrow\Bbb{C}$ be an inner product. Fix a basis $\{e_1,\ldots,e_n\}$, so that for any two vectors $u,v\in V$, we have

$$ \langle u,v\rangle=\left\langle\sum_{j=1}^n\alpha_je_j,\sum_{k=1}^n\beta_ke_k\right\rangle=\sum_{j=1}^n\sum_{k=1}^n\alpha_j\bar{\beta}_j\langle e_j,e_k\rangle $$

So if we define $M=(M_{ij})_{n\times n}=(\langle e_i,e_j\rangle)_{n\times n}$, we have exactly the expression

$$ \langle u,v\rangle = u^\intercal M \bar{v} $$

as desired. It is easy to check that $M$ is Hermitian - this follows from conjugate symmetry of the inner product: $\overline{\langle x,y\rangle}=\langle y,x\rangle$. Positive definiteness follows from the positivity of the inner product - $\langle x,x\rangle>0$ for all $x\in V\backslash\{0\}$.

For the second claim, it suffices to show the identity $x^\intercal M \bar{y}=y^*\overline{M} x$. This is straightforward:

$$ x^\intercal M\bar{y}=\sum_j\alpha_j\sum_kM_{jk}\bar{\beta}_k=\sum_k\bar{\beta}_k\sum_jM_{jk}\alpha_j=y^*\overline{M}x $$

Notice of course that we must then use the matrix $\overline{M}$, i.e. the same matrix won't work. (Thanks to @user1551 for pointing this out)

  • $\begingroup$ Sorry, for the second one I meant $$\langle \boldsymbol{x} , \boldsymbol{y} \rangle = \boldsymbol{y}^* M \boldsymbol{x}$$ $\endgroup$ – Sam Nov 25 '12 at 22:15
  • $\begingroup$ Correct - sorry, its fixed now. $\endgroup$ – icurays1 Nov 26 '12 at 15:08

It gets even better...imo, much better: if you have an orthonormal basis $\,\{u_1,...,u_n\}\,$ , then the inner product is given by the extremely simple formula

$$x=\sum_{k=1}^nx_iu_i\,\,,\,\,y=\sum_{k=1}^ny_iu_i\,\,\,,\,\,x_i,y_i\in\Bbb C\Longrightarrow\langle\,x\,,\,y\,\rangle=\sum_{i=1}^nx_i\overline y_i$$

  • 1
    $\begingroup$ The orthonormal basis will be different for each distinct inner product, of course. $\endgroup$ – icurays1 Nov 25 '12 at 23:24
  • $\begingroup$ I'm not sure what you're asking: there are many different orthonormal basis for the same inner product, but of course the representation of each vector wrt each of those will be different. $\endgroup$ – DonAntonio Nov 26 '12 at 0:30

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