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Suppose $A$ is a positive-definite matrix and $b$ is a vector

which satisfies $ b\leq \mbox{diag}(A)$ for all entries of $b$, i.e. $b_i= b^T e_i\leq e_i^T A e_i $.

The linear equations holds: $Ax=b$ where $x$ is a vector.

The question is to prove that the sum of the entries of $x$ is bounded between $0$ and $1$ $$0 \leq \sum_{i=1}^n{x_i} \leq 1.$$

Thank you very much.

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i tried using carmer's rule to solve the sum of x. – tomer Jan 22 '13 at 16:48
then the quastion can be formulate as an inequality of determinants of the matrix A – tomer Jan 22 '13 at 16:50
Can you edit your post and show how far you've come til now? – Bob Jan 22 '13 at 16:56
And more important, how can you explain this? $$b \leq diag(A)$$ Because $b$ is a vector and $diag(A)$ a number – Bob Jan 22 '13 at 17:04
trace(A) is the sum of the diag. – tomer Jan 22 '13 at 17:18

We cannot prove that because the assertion is false. Just consider the scalar case where $A=1$ and $b=-1$. For a less trivial case, consider $$ \begin{pmatrix}2&-1\\-1&1\end{pmatrix}\begin{pmatrix}2\\3\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}. $$

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