How covariance-variance inequality implies Kantorovich inequality

Question

Hi, I read this article OPERATOR INEQUALITIES RELATED TO CAUCHY-SCHWARZ AND H\"OLDER-McCARTHY INEQUALITIES, Nihonkia Math. J. 1997,117-122. In the introduction part, it states covariance-variance inequality implies Kantorovich inequality, but I don't know how (can anyone give me some hint?). Is this statement a well known result?

The covariance-variance inequality can be found here http://en.wikipedia.org/wiki/Covariance

Let $A$ be a real symmertric positive definite matrix and $0< m\le A\le M$, the Kantorovich inequality is $$(x^TAx)(x^TA^{-1}x)\le \frac{(M+m)^2}{4Mm},$$ where $x^T$ means the transpose of $x$ and $\|x\|=1$.

Answer 1

Here is a longish, but simple proof (some ideas taken from Prof. Bhatia's book on positive definite matrices, but adapted to match the setting of the paper you cite).

To simplify notation, assume without loss of generality that all matrices are real and symmetric. Then, the cited paper defines the covariance for an arbitrary unit vector $x$ as

$$C(A,B) = x^TBAx - (x^TAx)(x^TBx),$$

using which it further defines $V(A) := C(A,A)$.

The standard covariance-variance inequality is

$$(*)\hskip 10pt |C(A,B)| \le \sqrt{V(A)}\sqrt{V(B)}.$$

Using the above inequality, here is one way in which we can prove the Kantorivich inequality.

First, define $\delta=(x^TAx)(x^TA^{-1}x)$; then, observe that $C(A,A^{-1}) = 1 - \delta$.

Inequality (*) says that

$$|1-\delta| \le \sqrt{V(A)}\sqrt{V(A^{-1})}.$$

So let us first bound $V(A)$ and $V(A^{-1})$. Assume therefore $mI \preceq A \preceq MI$ (here $\preceq$ denotes the L\"owner ordering).

Notice that $(MI-A)(A-mI) \succeq 0$, or in other words \begin{eqnarray*} A^2 + mMI &\preceq& (m+M)A\\\\ x^TA^2x-(x^TAx)^2 + mM &\le& x^TAx[(m+M) - x^TAx]\\\\ x^TA^2x-(x^TAx)^2 + mM &\le& \frac{1}{4}(m+M)^2\\\\ x^TA^2x-(x^TAx)^2 &\le& \frac{1}{4}(M-m)^2. \end{eqnarray*} Similarly, we obtain $V(A^{-1}) \le \frac{1}{4}(1/m-1/M)^2$.

Finally, since $\delta \ge 1$, we have that $|1-\delta| = \delta - 1$. Putting the pieces together we obtain from (*) $$\delta - 1 \le \frac{1}{4}(M-m)(1/m-1/M),$$ which you can simplify to finish the proof.

Answer 2

Since $A$ is diagonalizable in an orthonormal basis, for every vector $x$ such that $\|x\|=1$, $$ x^TAx=\sum_ia_iy_i^2\quad\mbox{and}\quad x^TA^{-1}x=\sum_ia_i^{-1}y_i^2, $$ where the $a_i$ are the eigenvalues of $A$ and $y$ is a vector such that $\|y\|=1$. Hence $$ x^TAx=E(\xi)\quad\mbox{and}\quad x^TA^{-1}x=E(\xi^{-1}), $$ where $\xi$ is a random variable such that $P(\xi=a_i)=y_i^2$ for every $i$, in the simple case when the $a_i$ are distinct. In the general case, for every $a$, $$ P(\xi=a)=\sum_iy_i^2\ \mathbf{1}_{a_i=a}. $$ One is left with the task to prove that $E(\xi)E(\xi^{-1})\le(M+m)^2/(4Mm)$ for every random variable $\xi$ such that $m\le\xi\le M$ almost surely... This is a true inequality but I am not sure that the easiest way to prove it is to write it as a variance-covariance inequality.

To prove Kantorovich inequality, note that, since every eigenvalue of $A$ is between $m$ and $M$, $$ A+MmA^{-1}\le(M+m)I. $$ Thus $a+b\le c$ with $$ a=x^TAx,\qquad b=Mm(x^TA^{-1}x),\qquad c=M+m. $$ But $a+b\le c$ implies that $ab\le\frac14c^2$ and this is Kantorovich inequality written as $$ Mm(x^TAx)(x^TA^{-1}x)\le\frac14(M+m)^2. $$