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I am stuck with an exercise that I found in a textbook by Conway. First, I would like to clarify what is meant by a semi-inner product.

Definition. Suppose that $\mathscr X$ is a vector space over the complex field $\mathbb C$. A semi-inner product on $\mathscr X$ is a function $u:\mathscr X\times\mathscr X\to\mathbb C$ such that for all $\alpha,\beta$ in $\mathbb C$, and $x,y,z$ in $\mathscr X$, the following are satisfied:

  • $u(\alpha x+\beta y,z)=\alpha u(x,z)+\beta u(y,z)$,
  • $u(x,x)\ge 0$,
  • $u(x,y)=\overline{u(y,x)}$,

where $\bar\alpha$ is the complex conjugate of $\alpha$.

The difference between an inner product and a semi-inner product is that an inner product also satisfies the following:

  • if $u(x,x)=0$, then $x=0$.

Now I formulate the exercise from the textbook.

Let $u(\cdot,\cdot)$ be a semi-inner product on $\mathscr X$. Then $\left|u(x,y)\right|^2=u(x,x)u(y,y)$ if and only if there are $\alpha$ and $\beta$ in $\mathbb C$, not both $0$, such that $u(\beta x+\alpha y,\beta x+\alpha y)=0$.

How can I show that if there are $\alpha$ and $\beta$ in $\mathbb C$, both not $0$, such that $u(\beta x+\alpha y,\beta x+\alpha y)=0$, then $\left|u(x,y)\right|^2=u(x,x)u(y,y)$?

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    That quantity *must* be $\pm2$, or else your equation cannot hold (because $\lvert u(x,y)\rvert\le u(x,x)u(y,y)$).2012-10-19

2 Answers 2

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Consider a matrix $G = \begin{pmatrix}u(x,x)&u(x,y)\\u(y,x)&u(y,y)\end{pmatrix}$ It allows to say that $$u(\alpha x + \beta y, \alpha x + \beta y) = \left(\alpha , \bar \beta\right )G\,\left(\bar \alpha, \beta\right)^T.$$

Now, since $\det G = u(x,x)u(y,y)-\left|u(x,y)\right|^2$, your exercise is equivalent to saying that $G$ is degenerate iff the system $\left(\alpha , \bar \beta\right )G\,\left(\bar \alpha, \beta\right)^T=0$ has nontrivial solutions, which is quite easy to show ($G$ is self-adjoint, it helps a lot).

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If $\det G =0 $, then by a well-known result from linear algebra ther exists a nontrivial pair $(\bar \alpha,\beta) \in \Bbb C^2$ such that $G (\bar \alpha,\beta)^T =0$, hence necessarily $( \alpha,\bar\beta)G(\bar \alpha,\beta)^T=0$.

In other direction, we know that $G$ is positive semidefinite. If $\det G \ne 0$, then all eigenvalues of $G$ are strictly positive ($G$ is symmetric, hence the structure of its eigenvalues is quite simple). Take any vector $w\in \Bbb C^2$ and its coordinates in the basis of eigenvectors of the matrix $G$: $w = a_1 v_1 + a_2v_2$. Then $w^*Gw = \sum_{i=1,2}\lambda_i|a_i|^2\|v_1\|^2 >0$ ($\lambda_i$ are eigenvalues and $v_i$ are eigenvectors), which yields a contradiction. Therefore, $\det G=0$.

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    Shouldn't the norm be raised to the power of $4$ (i.e. $\|v_i\|^4$)?2017-03-03
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Instead of $\mu \langle x, y\rangle $ I have used $ \langle x, y\rangle.$

Let $ \langle\mathcal{X}, .\rangle$ be semi inner product. Let $x $, $y $ be fixed vectors in $\mathcal{X}$ and $\gamma$ be scalar. Consider $ \langle x - \gamma y, x - \gamma y\rangle = \langle x, y\rangle - \gamma\langle y,x\rangle - \bar\gamma\langle x, y\rangle + |\gamma|^2\langle y, y\rangle $ Put $\langle y,x\rangle = b \mathrm{e}^{i\lambda} (b \ge 0) $ , $ \gamma = t\mathrm{e}^{-i\lambda}$ (t is real), $ a = \langle y, y\rangle, c = \langle x, x\rangle $ Note here that $\lambda, a, c $ are constants whereas $ t $ is real variable. With this, we have $ \langle x - \gamma y, x - \gamma y\rangle = c - 2bt + at^2 \tag{1} $ Now, $ |\langle x, y\rangle|^2 = \langle x, x\rangle\langle y, y\rangle \iff b^2 - ac = 0 \iff 4b^2 - 4ac = 0 $ $\iff c - 2bt + at^2 = 0 $ have equal roots. This is true if and only if $ c - 2bt + at^2 = 0$ has unique real root, say $ t_0 $. So by taking $\gamma_0 = t_0\mathrm{e}^{-i\lambda}$, from the equation $(1)$, we obtain that $ \langle x - \gamma_0 y, x - \gamma_0 y\rangle = c - 2bt_0 + at_0^2 = 0 $ Thus, the required scalars in your problem are $\beta = 1 $ and $\alpha = -\gamma_0 $

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    If $\langle y,x\rangle=be^{i\lambda}$, how do we know that $\gamma=t e^{-i\lambda}$? We only know that $\langle x-\gamma y,x-\gamma y\rangle=0$, where $\gamma\in\mathbb C$. We can express any complex number as $\gamma=te^{i\varphi}$, but how do we know that $\varphi=-\lambda$? This trick is used in the proof of the CBS inequality, but there we can choose $\gamma\in\mathbb C$ freely and here $\gamma\in\mathbb C$ is fixed.2015-02-03