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If column vectors of ${\bf B}$ are independent and normalized, can we conclude the row summation of $({{\bf B}^T{\bf B}})^{-1}$ nonnegative? By row summation, I mean $({{\bf B}^T{\bf B}})^{-1}{\bf 1}$. ${\bf 1}$ is a vector with components all one's.

Sorry for this question. Seems what I need is only ${\bf 1}^T({\bf B}^T{\bf B}){\bf 1}>0$. But this is trivial.

Thank you all so much!

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    What's the background, motivation for this question?2012-04-16

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No.

This is rather crude, but let $ B = \left( \begin{array}{ccc} 1 & 0.85 & 0.52669 \\ 0 & 0.52678 & -0.83086 \\ 0 & 0 & 0.17963 \end{array} \right).$ We have $||B e_i|| = 1$, so the columns are normalized, the eigenvalues of $B^T B$ are $0.05, 0.99, 2.00$, so the columns of $B$ are independent, but $(B^T B)^{-1} \left( \begin{array}{c} 1\\ 1\\ 1 \end{array} \right)= \left( \begin{array}{c} -40.544 \\ 35.243 \\ 22.002 \end{array} \right) .$

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    @mining: I $f$ixed by example, I thin$k$ it satis$f$ies all the conditions, but the row sum in negative.2012-04-16