On a reversed Cauchy-Schwarz inequality for an indefinite quadratic form

Question

Let $f$ be a quadratic form over $\mathbb{R}$ and write $$f(x)=xAx^t,$$ where $A\in M_n(\mathbb{R_{+}})$ is symmetric with positive entries and $x=(x_1,x_2,\ldots,x_n)$. Suppose that $A$ has exactly one positive eigenvalue and $n-1$ negative eigenvalues.

IS IT TRUE that for any $0\neq u\in \mathbb{R}_{\geq 0}^n$ and $v\in \mathbb{R}^n$ that is not parallel to $u$, $$(u^tAv)^2>(u^tAu)(v^tAv).$$

I tried to find proof of this result but failed. Anyone can help me or provide some references? Thanks a lot!

Answer 1

Denote the eigenvalues of $A$ in decreasing order as $\lambda_1(>0)>\lambda_2\geq...\geq\lambda_n$ and their corresponding orthonormal eigenvectors as $z_1, z_2, ..., z_n$.

So $z_1$ with $\lVert z_1 \rVert=1$ is the eigenvector to the positive eigenvalue $\lambda_1$, let $U_{\perp}$ be the subspace spanned by the other eigenvectors to the negative eigenvalues. If $v\in U\setminus{0}$, then $v^TAv\leq0$ and the right-hand side of the inequality is zero or negative. Also $u$ cannot be in $U$, otherwise $u^TAu\leq0$, a contradiction to $u>0$

Thus since we know that $u^TAu>0$, going ahead we will only concern ourselves with $v^TAv>0$ because otherwise the inequality is obvious. So indeed, we are concerned with this case: while neither of these is in $U_\perp$, since they are non-parallel, some combination of the eigenvectors corresponding to negative eigenvalues, unique upto scaling, is in $U_\perp$ (or else we have $n+1$ linearly independent vectors in $\mathbb{R}^n$). In this combination, let $z^*$ be one of these eigenvectors with a non-zero weight. If we drop $z^*$, then we must have a linearly independent set again.

Define an $n\times n$ matrix $S$ with the first row as $u^T$ and second row as $v^T$. and the other rows as $z_j^T$s with $z^j$s other than $z_1$ and $z^*$ in decreasing order of corresponding eigenvalues. Thus: $$S=\begin{bmatrix} \cdots & u^T & \cdots \\ \cdots & v^T & \cdots \\ \hline \vdots & z_j^T\;|\; z_j\notin \{z_1,z^*\} & \vdots \\ \end{bmatrix}$$

Clearly $S$ is non-singular because all rows are linearly independent as we argued before. By Sylvester's law of inertia, $SAS^T$ also has one positive and $n-1$ negative eigenvalues. This is how $SAS^T$ looks:

$$SAS^T=\left[\begin{array}{cc|cc} u^TAu & u^TAv & \cdots \\ v^TAu & v^TAv & \cdots \\ \hline \cdots & \cdots & \mathscr{D} \\ \end{array}\right]$$ where $\mathscr{D}$ is the diagonal matrix with diagonal entries being (in decreasing order), the $\lambda_j$s corresponding to the $z_j$s in $S$. Denote the principal submatrix with the first two rows and columns to be $X$. Thus: $$X=\left[\begin{array}{cc|cc} u^TAu & u^TAv\\ v^TAu & v^TAv\\ \end{array}\right]$$

• First we use the eigenvalue interlacing theorem: which in our case guarantees that the lowest (second) eigenvalue of $X$ is less than the second highest eigenvalue of $SAS^T$ which is negative (as we noted from Sylvester's only, one is positive, others are negative). Thus the second eigenvalue of $X$ is negative.
• Next we note that the trace of $X$ is positive but this means that the sum of its eigenvalues is positive. But that means that the first eigenvalue of $X$ is positive.

But this means that the product of eigenvalues of $X$ is negative. But this equals the determinant. Thus: $$u^TAu\cdot v^TAv -u^TAv\cdot v^TAu < 0 \iff (u^TAv)^2 > u^TAu\cdot v^TAv$$$$\tag*{$\blacksquare$}$$

Answer 2

The way this has been constructed means $\mathbf u^T A \mathbf u \gt 0$ and to avoid trivialities we only need to consider $\mathbf v$ such that $\mathbf v^T A \mathbf v \geq 0$.

More generally the question becomes:
for any real symmetric $A$ with signature $(1,n-1)$ with
$[\mathbf v, \mathbf u] := \mathbf v^T A \mathbf u$
if we restrict ourselves to the non-negative cases of this form, i.e. $\mathbf x\in \mathbb R^n$ such that $[\mathbf x, \mathbf x]\geq 0$
is it true that we have 'reversed Cauchy-Schwarz' where
$[ \mathbf v ,\mathbf v]^\frac{1}{2}[ \mathbf u ,\mathbf u]^\frac{1}{2} \leq \Big \vert [ \mathbf v, \mathbf u]\Big \vert$
with equality iff $\mathbf v$ and $\mathbf u$ are linearly dependent?

To standardize the question, first consider that by Sylvester's Law of Intertia we have an invertible $M$ such that
$M^TAM = \begin{bmatrix} 1 & 0 \\ 0 & -I_{n-1} \end{bmatrix}$
so the equivalent question is whether 'reversed Cauchy-Schwarz' holds with
$[\mathbf y, \mathbf z]_M := \mathbf y^TM^TAM\mathbf z= [\mathbf u, \mathbf v]$
where $\mathbf y = M^{-1}\mathbf u$ and $\mathbf z = M^{-1}\mathbf v$

(this means $\mathbf y$ may be any vector in $\mathbb R^n$ such that $[\mathbf y, \mathbf y]_M \geq 0$ and any $\mathbf z$ such that $[\mathbf z, \mathbf z]_M \geq 0$. Since $M$ is invertible this implies the result for $[\mathbf v, \mathbf u]$.)

The answer is yes; this is Minkowski's Light Cone Inequality. There are many proofs of this; a simple one follows.
$\mathbf y = \begin{bmatrix} y_0 \\ \mathbf y' \end{bmatrix}$, $\mathbf z = \begin{bmatrix} z_0 \\ \mathbf z' \end{bmatrix}$

for reasons of positive re-scaling, we can assume WLOG that $ y_0 =1= z_0 $. Then we have
$[\mathbf y, \mathbf y]_M^\frac{1}{2}[\mathbf z, \mathbf z]_M^\frac{1}{2}$
$\leq \frac{1}{2}\Big([\mathbf y, \mathbf y]_M+ [\mathbf z, \mathbf z]_M\Big)$
$=1 - \frac{1}{2}\Big(\big\Vert \mathbf y'\big \Vert_2^2 + \big\Vert \mathbf z'\big \Vert_2^2\Big)$
$\leq 1 - (\mathbf y')^T\mathbf z'$
$=[\mathbf y, \mathbf z]_M$

the first inequality is $\text{GM}\leq \text{AM}$ and the second one is the additive form of Cauchy Schwarz or equivalently the fact that $\big \Vert \mathbf y'- \mathbf z'\big \Vert_2^2 \geq 0$, which tacitly deals with the strictness of the inequality (and linear dependence).

Answer 3

As $A$ is entrywise positive but $u$ is nonnegative and nonzero, $u^TAu$ must be positive. Now consider the restriction of $A$ on $P=\operatorname{span}\{u,v\}$. Let $\{w_1,w_2,\ldots,w_n\}$ be an eigenbasis of $A$ where $w_i$ corresponds to a negative eigenvalue for each $i\ge2$. Since $\dim P+\dim\operatorname{span}\{w_2,\ldots,w_n\}>n$, $P\cap\operatorname{span}\{w_2,\ldots,w_n\}$ is a nonzero subspace. Therefore there exists some $w\in P$ such that $w^TAw<0$. But we also have $u^TAu>0$. Hence $A$ is indefinite on $P$.

Let $\{y,z\}$ be an orthonormal eigenbasis of $A|_P$ with $y^TAy=\lambda_1$ and $z^TAz=\lambda_2$. Then $X=\{x\in P: x^TAx=0\}$ is the union of the two lines extended from the vectors $\sqrt{|\lambda_2|}y+\sqrt{|\lambda_1|}z$ and $\sqrt{|\lambda_2|}y-\sqrt{|\lambda_1|}z$. Since the line $L=\{tu+v:t\in\mathbb R\}$ is not parallel to any line in $X$ (because $u^TAu>0$) and does not pass through the origin (because $u$ and $v$ are linearly independent), $L$ must intersect $X$ at two distinct points. Hence the quadratic equation $(tu+v)^TA(tu+v)=0$ has two distinct real roots and the discriminant of this equation must be positive, meaning that $(v^TAu)^2>(u^TAu)(v^TAv)$.