Equality in the Cauchy-Bunyakowsky-Schwarz inequality for a semi-inner product

Question

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 $\left|u(x,y)\right|^2=u(x,x)u(y,y)$?

Answer 1

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, $\det G=0$.

edit 2 (2024-09-23)

$G$ is positive semidefinite $\iff$ $\forall z\in \Bbb C^2\quad z^*Gz \ge 0$.

We can rewrite the last expression as $\iff\forall (\alpha,\beta)\in \Bbb R^2\quad (\bar\alpha,\bar \beta)G(\alpha,\beta)^T \ge 0$.

Now use the definition of $G$:

$$(\bar\alpha,\bar \beta)G(\alpha,\beta)^T = u(\bar\alpha x + \beta y, \bar\alpha x + \beta y) \ge 0$$ by the definition of $u$ (second bullet point).

Answer 2

We will show the following. Note that the above definition of a semi-inner product makes sense for both a real and a complex vector space, so we state the result for both fields below.

Let $\mathscr{X}$ be a vector space over $\mathbb{F}$, where $\mathbb{F}$ is fixed as one of $\mathbb{R}$ or $\mathbb{C}$. Let $u$ be a semi-inner product on $\mathscr{X}$. Let $x, y\in \mathscr{X}$. The following are equivalent.

$({\rm i})$ We have $|u(x, y)|^{2} = u(x, x) u(y, y)$.

$({\rm ii})$ We have $u(y, y) = 0$ or otherwise have $u(x + \lambda y, x + \lambda y) = 0$ for some $\lambda\in\mathbb{F}$.

$({\rm ii})$ There are $\alpha , \beta \in \mathbb{F}$ with at least one of $\alpha$ or $\beta$ not equal to $0$ such that $u(\alpha x + \beta y) = 0$.

We remind the reader that the Cauchy-Schwarz inequality holds in the setting of semi-inner product spaces (see here). More specifically, for any $z_{1}, z_{2}\in \mathscr{X}$ we have \begin{equation} |u(z_{1}, z_{2})|^{2} \leq u(z_{1}, z_{1}) u(z_{2}, z_{2}) . \tag{1} \end{equation}

$({\rm ii}) \Rightarrow ({\rm iii})$. This is trivial.

$({\rm iii}) \Rightarrow ({\rm ii})$. Assume there are $\alpha , \beta \in \mathbb{F}$ with at least one of $\alpha$ or $\beta$ not equal to $0$ such that $u(\alpha x + \beta y) = 0$. If $\alpha = 0$ then $\beta \neq 0$ and \begin{equation} u(y, y) = |\beta |^{-2} u(\beta y, \beta y) = |\beta |^{2} u(\alpha x + \beta y, \alpha x + \beta y) = 0 . \end{equation} Otherwise, if $\alpha \neq 0$ then \begin{equation} u(x + (\tfrac{\beta}{\alpha})y, x + (\tfrac{\beta}{\alpha})y) = |\alpha |^{-2} u(\alpha x + \beta y, \alpha x + \beta y) = 0 . \end{equation} Regardless of the value of $\alpha$ we see that $({\rm ii})$ holds.

$({\rm ii}) \Rightarrow ({\rm i})$. If $u(y, y) = 0$ then $(1)$ implies $|u(x, y)|^{2} \leq 0$ and the conclusion in $({\rm i})$ readily follows. Otherwise, there is $\lambda\in\mathbb{F}$ such that $u(x + \lambda y, x + \lambda y) = 0$. We claim that for each $z\in\mathscr{X}$ we have \begin{equation} u(x, z) = -\lambda u(y, z) . \tag{2} \end{equation} Let $z\in\mathscr{X}$. By the Cauchy-Schwarz inequality $(1)$ we have \begin{equation} |u(x, z) - (-\lambda u(y, z)| = |u(x + \lambda y, z)| \leq u(x + \lambda y)^{\tfrac{1}{2}} u(z, z)^{\tfrac{1}{2}} = 0 . \end{equation} Hence $(2)$ follows. Using $(2)$, we deduce \begin{equation} u(x, y) = -\lambda u(y, y) \tag{3} \end{equation} and \begin{equation} u(x, x) = -\lambda u(y, x) = (-\lambda ) \overline{u(x, y)} = (-\lambda ) \overline{ (-\lambda ) u(y, y) } = |\lambda |^{2} u(y, y) . \tag{4} \end{equation} From $(3)$ and $(4)$ we deduce \begin{equation} u(x, x) u(y, y) - |u(x, y)|^{2} = {} \big( |\lambda |^{2} u(y, y) \big) u(y, y) - |\lambda |^{2} u(y, y)^{2} = 0 \end{equation} and $({\rm i})$ follows.

$({\rm i}) \Rightarrow ({\rm ii})$. Assume $|u(x, y)|^{2} = u(x, x) u(y, y)$. If $u(y, y) = 0$ then the conclusion of $({\rm ii})$ is immediate. Otherwise, we have $u(y, y) \neq 0$. For each $\lambda\in\mathbb{F}$ we have \begin{equation} u(x + \lambda y, x + \lambda y) = u(x, x) + 2 \, {\rm Re} (\overline{\lambda} u(x, y) ) + |\lambda |^{2} u(y, y) . \tag{5} \end{equation} By taking $\lambda = -\frac{u(x, y) }{u(y, y) }$ in $(5)$ and using the assumption from $({\rm i})$, we get \begin{align*} u(x + \lambda y, x + \lambda y) &= u(x, x) - 2\frac{|u(x, y)|^{2} }{u(y, y) } + \frac{|u(x, y)|^{2} }{u(y, y) } \\ &= \frac{u(x, x) u(y, y) - |u(x, y)|^{2} }{u(y, y) } = 0 . \end{align*} Hence we deduce that $({\rm ii})$ holds. This completes the proof.

Answer 3

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 $