Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
What have you tried? –  Bob Jan 22 '13 at 16:39
    
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
1  
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
show 3 more comments

1 Answer 1

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 $$ \begin{pmatrix}2&-1\\-1&1\end{pmatrix}\begin{pmatrix}2\\3\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}. $$

share|improve this answer
add comment

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.