Let $V$ be an $n$-dimensional Euclidean space with inner product $\langle\cdot,\cdot\rangle$, with basis $e_1,\dots,e_n$. Then the Gram matrix is $A=(a_{ij})$ with $a_{ij}=\langle e_i,e_j\rangle$. It is well-known that $A$ is positive definite.

Suppose now that if $$\langle e_i,e_j\rangle\leq 0,\ i\neq j,$$ then could we have the following statement:

if $Ax\geq 0$, then $x\geq 0$. Here, $x,0$ are column vectors, and a vector $y\geq 0$ means that each coordinate is $\geq 0$.


1 Answer 1


Your statement is correct. The reason is that the inverse matrix $A^{-1}$ has all its entries $\ge 0$.

To check the last statement: reduce to a matrix with $1$ on the diagonal, that is of form $I - L$, where $L$ has $0$ on the diagonal and all entries $\ge 0$. We know that the eigenvalues of $I - L$ are $>0$ and therefore all the eigenvalues of $L$ are $< 1$. From Perron-Frobenius it follows that the eigenvalues of $L$ are in $(-1,1)$ and therefore the series $\sum_{n \ge 0} L^n$ is convergent and gives the inverse $(I-L)^{-1}$ with positive entries.

  • $\begingroup$ If all entries of $A^{-1}$ is great than $0$, then how can I prove that $x\geq 0$ if $AX\geq 0$..... $\endgroup$
    – xldd
    Oct 17, 2014 at 12:25
  • 3
    $\begingroup$ $x = A^{-1}( A x)$ $\endgroup$
    – orangeskid
    Oct 17, 2014 at 12:27
  • $\begingroup$ Thanks. And do you have some hints on the fact that the eigenvalues of $I-L$ are $>0$... $\endgroup$
    – xldd
    Oct 17, 2014 at 23:23
  • $\begingroup$ @xldd: The matrix $I-L$ is the Gram matrix of a system of linearly independent vectors of length $1$ and so positive definite and so all the eigenvalues are $>0$. $\endgroup$
    – orangeskid
    Oct 18, 2014 at 2:08
  • $\begingroup$ Thank you very much...I have got it. $\endgroup$
    – xldd
    Oct 18, 2014 at 2:51

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