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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 $\mathbb C$, not both $0$, 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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    I rather strongly suspect that the requirement should say “not both $0$”, not “both not $0$”.2012-10-19
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    Have you tried expanding out $u(\beta x+\alpha y,\beta x+\alpha y)$ and look at the result? Note that the result must always be nonnegative for all choices of $\alpha$, $\beta$, and zero for the given pair. This should be useful.2012-10-19
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    The second condition in the definition should be $u(x,x) \geq 0$ not $u(x,y) \geq 0$. (If $u(x,y) \geq 0$ for all $x$ and $y$ then $u$ is identically equal to $0$).2012-10-19
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    Thank you, Yury, for correcting my typo.2012-10-19
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    Reply to Harald's remark: the requirement in the textbook says "both not $0$". I guess that otherwise the proposition is false. If $\alpha=0$ and $\beta\ne0$, then $u(x,x)=0$. Assume that $x\ne0$ (the equality is valid if $x=0$), but $u(x,x)=0$. Unless we show that $x(x,y)=0$ for all $y\in\mathscr X$ if $u(x,x)=0$, we get a contradiction. Is it possible to construct an example of a semi-inner product that for some $x\ne 0$ $u(x,x)=0$, but there exists $y\in\mathscr X$ such that $u(x,y)\ne0$?2012-10-19
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    @Harald: Yes, that is exactly what I tried to do. Let us denote $\gamma=\alpha/\beta$. Then we obtain the following: $$u(x+\gamma y,x+\gamma y)=u(x,x)+\bar\gamma u(x,y)+\gamma u(y,x)+\left|\gamma\right|^2u(y,y).$$ I am not sure what to do next... If we restrict our attention to the vector space $\mathscr X$ over the real field $\mathbb R$, then we can get the result in the following way. The quadratic polynomial $$q(t)=u(x,x)+2u(x,y)t+u(y,y)t^2$$ is non-negative and is equal to $0$ for some $t\in\mathbb R$. Hence the discriminant is equal to zero $$4\left|u(x,y)\right|^2-4u(x,x)u(y,y)=0.$$2012-10-19
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    Well, that was my point; if $u(x,x)=0$ then $u(x,y)=0$ for all $y$ by the C–B–S inequality.2012-10-19
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    Here is a point to consider: Note that $u(y,x)=\overline{u(x,y)}$ (this follows by polarization). So $\bar\gamma u(x,y)+\gamma u(y,x)$ is real. Now replace $\gamma$ by $\gamma t$ with $\gamma$ fixed and $t$ real and variable, and employ your discriminant argument. You're almost there.2012-10-19
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    @Harald: Thank you very much for your advice. It seems that I only need one final step. Let $\gamma=\left|\gamma\right|e^{i\theta}$ and $u(x,y)=\left|u(x,y)\right|e^{-i\varphi}$. We have that $$\left(\bar\gamma u(x,y)+\gamma u(y,x)\right)^2=4\left|\gamma\right|^2u(y,y)u(x,x)$$ and $$\left|u(x,y)\right|^2\left(e^{-i\theta}e^{i\varphi} +e^{i\theta}e^{-i\varphi}\right)^2=4u(y,y)u(x,x).$$ How can I show that $e^{-i\theta}e^{i\varphi}+e^{i\theta}e^{-i\varphi}=2$? Or am I making a mistake somewhere?2012-10-19
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    @Harald: I agree with your remark about the requirement. It should say "not both $0$". My previous comment is wrong.2012-10-19
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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 a 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).

edit

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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    Thanks for the answer (+1)! Should I try to diagonalise the matrix $G$ and use the fact that the determinant is a product of eigenvalues to show that the system has nontrivial solutions iff $\operatorname{det}G=0$? Is there a simpler way?..2017-03-03
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    @Cm7F7Bb you can do it thsi way, too. I shown a similar one in my edits.2017-03-03
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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