Vector inequality $(\langle Ax,y \rangle^2) \le (\langle Ax,x\rangle * \langle Ay,y\rangle)$

Question

Let $A = (a_{ij})_{1\le i \le j \le n} \in M_{n \times n}^{\Bbb R}$

such that $$(a_{ij})=\begin{cases} 2, i = j \\ 1, i\ne j \end{cases}$$

Prove that $(\langle Ax,y \rangle^2) \le (\langle Ax,x\rangle * \langle Ay,y\rangle)$ for every $x,y \in \Bbb R^n$

Where $\langle\cdot,\cdot\rangle$ stands for the standard inner product in the space.

I think I'm supposed to solve this with Cauchy-Schwarz inequality.

Any ideas?

Thanks!

Answer 1

$A$ is the matrix associated with the quadratic form:

$$x_1^2+x_2^2+\cdots+x_n^2+(x_1+x_2+\cdots+x_n)^2$$

This form is $>0$ except in the case where $x_1=x_2=...=x_n=0$. Thus it is a positive definite quadratic form, and $A$ is symmetric positive definite.

It is well known that, with such a matrix, $X^TAY$ is an inner product associated with the norm $\|X\|_A=\sqrt{X^TAX}$ said "$A$-norm".

Thus the target property is a form of the Cauchy-Schwarz inegality.

see for example http://www-users.cs.umn.edu/~saad/csci5304/FILES/LecN7_R.pdf

A direct proof is possible, generalizing the classical proof of Cauchy-Schwarz inequality. Here it is.

Let us expand:

$$(\lambda X+Y)^TA(\lambda X+Y)=\lambda^2(X^TAX)+\lambda(X^TAY+Y^TAX)+Y^{T}AY \ \ (1)$$

As $A^T=A$, $Y^TAX=Y^TA^TX \in \mathbb{R}$ ; being a number, it is the same as its transpose $X^TAY$. Therefore, relationship (1) can be written:

$$(\lambda X+Y)^TA(\lambda X+Y)=\lambda^2(X^TAX)+2\lambda(X^TAY)+(Y^{T}AY) \ \ (2)$$

This positive expression (being of the form $U^TAU$ with a symmetric p.d.m.) considered as a quadratic expression in $\lambda$ cannot have 2 distinct roots (were it the case, the expression could be $<0$ for certain values). Thus its discriminant must be $\leq 0$. Computing this discriminant gives, up to a factor $4$:

$$(X^TAY)^2-(X^TAX)Y^{T}AY \leq 0$$

ending the proof.

Answer 2

What is below is now obsolete, because the OP has made a correction in his text:

I understand now why we (@Michael Burr, me and probably others) had difficulties to prove this property, simply because it is FALSE.

Here is a very elementary counter-example:

$$A=\begin{bmatrix}1&2\\2&1\end{bmatrix}, \ \ X=\begin{bmatrix}1\\0\end{bmatrix}, \ \ Y=\begin{bmatrix}0\\1 \end{bmatrix}$$

$2^2$ cannot be less than $1 \times 1$...