Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\|x\|=1$.

share|improve this question
    
Any chance you want to state the actual inequalities? –  Deane Yang May 25 '11 at 2:14
    
I think you forgot a factor $|x|^4$ on the RHS, to get the same homogeneity of the LHS (or you may require $|x|=1$) –  Pietro Majer May 25 '11 at 13:18
    
Yes, thanks. I should have require $|x|=1$. Corrected. –  Sunni May 25 '11 at 13:22

2 Answers 2

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.

share|improve this answer
    
+1 for your proof. From your proof, I have sense that it is not appropriate to say Kantorovich inequality follows from Covariance-variance inequality. What do you think? Perhaps the first assertion of this implication is not through your method. –  Sunni May 27 '11 at 0:04
    
I agree, from the above proof, all the work was done while bounding the variance; after that just the cov-var inequality was used to conclude. maybe the authors of the paper you cited had something else in mind. –  Suvrit May 27 '11 at 7:14

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 $$ Mm(x^TAx)(x^TA^{-1}x)\le\frac14(M+m)^2. $$

share|improve this answer
    
I thought the OP knew how to prove Kantorovich's inequality, but not how to get it from the cited Covariance inequality..... –  Suvrit May 25 '11 at 19:17
    
@Suvrit: You can omit the second part of the post (starting with To prove Kantorovich inequality) if this part bothers you... –  Did May 26 '11 at 0:12
    
@ Didier: I am asking how Kantorovich inequality follows from Covariance inequality. Your proof did not even mention what is covariance and how that leads to Kantorovich inequality.... I know how to proof Kantorovich inequality. Please clarify it. –  Sunni May 26 '11 at 0:56
    
@Didier, no, no, it does not bother me at all! I was just trying to point out the part that yet remains to be shown :-) thanks for your comment. –  Suvrit May 26 '11 at 7:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.