Equality in the Cauchy-Bunyakowsky-Schwarz inequality for a semi-inner product 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 $\left|u(x,y)\right|^2=u(x,x)u(y,y)$?
 A: 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).
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, $\det G=0$.
A: 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 $
