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Consider a system of linear equations $Ax = b$, where \begin{eqnarray} % A & = & \left[ \begin{array}{ccccc} 2 & 1 & & & \\ 1 & 4 & 1 & & \\ & \ddots & \ddots & \ddots & \\ & & 1 & 4 & 1 \\ & & & 1 & 2 \\ \end{array} \right] \in \mathbb{R}^{n \times n} \end{eqnarray} and $n\geq 3$.

I would like to prove the following two statements, where the inequalities are element-wise: \begin{eqnarray} b \geq 0 & ~~~~ \Longrightarrow ~~~~ & x \geq 0 \\ b > 0 & \Longrightarrow & x > 0 ~ . \end{eqnarray}

APPARENT FACTS: The matrix $A$ is

  • square
  • symmetric
  • tri-diagonal
  • non-negative (element-wise)
  • positive definite
  • invertible
  • strictly diagonally dominant.

I have tried proving the two statements based on the fact that each element of $A$ is non-negative, but my attempts at proofs are not water-tight. The matrix $A$ is positive-definite and symmetric, and so is its inverse, but I was not able to exploit this. Any assistance and insight is much appreciated. If you know of any further properties of the system of equations related to the specifics of $A$ I would also appreciate your insight.

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1 Answer 1

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You are essentially trying to prove that the inverse of $A$ is a positive matrix, which is not true as you can check by computing the inverse for, say, $n=3$: Wolfram Alpha

You can use the inverse to directly find a counterexample: when $$b=\left[\begin{array}{c}1\\0\\0\end{array}\right]$$ we have $$x = \left[\begin{array}{c}\frac{7}{12}\\\frac{-1}{6}\\\frac{1}{12}\end{array}\right] \not\geq 0.$$

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