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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 $\alpha$, $\beta$ $\in V$:

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

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If $dim V\geq 2$ then it fails to be positive definite. –  azarel Jun 2 '12 at 2:37
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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.) –  Jonas Meyer Jun 2 '12 at 2:42
    
Thanks @azarel. I forgot that not only does $\langle \alpha, \alpha \rangle_{\varphi} \ge 0$, but equality only occurs when $\alpha=0$. –  Jackson Walters Jun 2 '12 at 2:56
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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.) –  Jonas Meyer Jun 2 '12 at 3:01

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