1
$\begingroup$

While trying to define an inner product, I realized that what I ended up with was a special case of the following. It seems like it should be a standard way to define an inner product, but I'm not sure what it's called.

Let $V$ be a vector space over $\mathbb{C}$ and let $\varphi: V \rightarrow \mathbb{C}$ be a linear functional. Define $\langle \cdot, \cdot \rangle_{\varphi} : V \times V \rightarrow \mathbb{C}$ as follows for $\alpha$, $\beta$ $\in V$:

$$\langle \alpha, \beta \rangle_{\varphi}=\varphi(\alpha)\overline{\varphi(\beta)}$$

$\endgroup$
  • 5
    $\begingroup$ If $dim V\geq 2$ then it fails to be positive definite. $\endgroup$ – azarel Jun 2 '12 at 2:37
  • 1
    $\begingroup$ Somewhat related, as you may know, is that if $V$ is finite dimensional and given any inner product, then there exists $\beta\in V$ such that for all $\alpha\in V$, $\varphi(\alpha)=\langle \alpha,\beta\rangle$. (This also applies if $V$ is a complete inner product space and $\varphi$ is continuous.) $\endgroup$ – Jonas Meyer Jun 2 '12 at 2:42
  • $\begingroup$ Thanks @azarel. I forgot that not only does $\langle \alpha, \alpha \rangle_{\varphi} \ge 0$, but equality only occurs when $\alpha=0$. $\endgroup$ – Jackson Jun 2 '12 at 2:56
  • 2
    $\begingroup$ This "semi"-inner product is composition of $\phi$ with an inner product on the range of $\phi$, which is $1$ dimensional. More generally if $T:V\to W$ is linear and $W$ is an inner product space, then $\langle \alpha,\beta\rangle_T=\langle T(\alpha),T(\beta)\rangle$ defines a "semi"-inner product on $V$, which is definite if and only if $T$ is injective. (I don't know of any name. I also put "semi" in quotes because semi-inner product sometimes means something other than this.) $\endgroup$ – Jonas Meyer Jun 2 '12 at 3:01

Your Answer

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

Browse other questions tagged or ask your own question.