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Let $A\in \mathbb{R}^{n\times n}$ be a positive definite symmetric matrix with eigenvalues $\lambda_1\ge \cdots\ge \lambda_n$, $X\in \mathbb{R}^{n\times k}$ such that $X'X=I_k$ ($X'$ means the transpose of $X$) and $n\ge 2k$, then $$\det(X'AXX'A^{-1}X)\le \prod\limits_{j=1}^k\frac{(\lambda_j+\lambda_{n-j+1})^2}{4\lambda_j\lambda_{n-j+1}}.$$

This result was first proved by Bloomfield and Watson(1975) and Knott(1975). I came across a note by H. Yang (A brief proof on the generalized variance bound of the relative efficiency in statistics, Communications in Statistics - Theory and Methods, 19(1990):12, 4587-4590), but the notation of his proof was rather confusing. Can any one make a clean proof or explain his proof?

(I don't have an e-version of that paper, sorry for inconvenience)

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1 Answer 1

Here is a crude sketch that could work. First recall Hadamard's determinant theorem:

\begin{eqnarray*} |X'AX| \le \prod_{i}(X'AX)_{ii} = \prod_i x_i^TAx_i. \end{eqnarray*}

Now recall Kantorovich's inequality (for $x^Tx=1$ ) \begin{eqnarray*} (x^TAx)(x^TA^{-1}x) \le \frac{(\lambda_1 + \lambda_n)^2}{4\lambda_1\lambda_n} \end{eqnarray*}

Now, inductively apply this inequality to the product

\begin{eqnarray*} \prod_i x_i^TAx_i \end{eqnarray*} by using $x_i^Tx_j = 0$, if $i\neq j$, so that for example, we have \begin{eqnarray*} (x_1^TAx_1)(x_1^TA^{-1}x_1) &\le& \frac{(\lambda_1 + \lambda_n)^2}{4\lambda_1\lambda_n}\\\\ (x_2^TAx_2)(x_2^TA^{-1}x_2) &\le& \frac{(\lambda_2 + \lambda_{n-1})^2}{4\lambda_2\lambda_{n-1}}, \end{eqnarray*} and so on uptil $x_k$.

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It is not clear how you get \begin{eqnarray*} (x_2^TAx_2)(x_2^TA^{-1}x_2) &\le& \frac{(\lambda_2 + \lambda_{n-1})^2}{4\lambda_2\lambda_{n-1}}. \end{eqnarray*} –  Sunni Sep 2 '11 at 0:18
    
Btw, did you read that article by Yang 1990? –  Sunni Sep 2 '11 at 0:19
    
I have not yet worked out the details, but "it could work"; as for Yang, nope, I have not read any of the papers that you have cited; no time. –  Suvrit Sep 2 '11 at 0:46

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