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Let $A = (a_{ij})$ be an $n \times n$ matrix with entries in the nonnegative real numbers $\mathbb{R}_+$. Suppose that, for each $i = 1,\ldots, n$, the sum $b_i := a_{i1} + \cdots + a_{in}$ of the entries in the $i$th row is positive. Let $$\kappa := \frac{1}{b_1 b_2 \cdots b_n} \cdot \det A.$$

Is $A - \kappa \cdot \mathrm{diag}(b_1,\ldots, b_n)$ a conical combination of rank $1$ matrices in $M_n(\mathbb{R}_+)$ (i.e. $n\times n$ matrices of rank $1$ with nonnegative real entries?)

For example, when $n = 2$, writing $A = \left(\matrix{ a & b \\ c & d }\right)$, a direct computation shows that $A - \kappa \cdot \mathrm{diag}(a + b, c+d)$ is equal to the rank $1$ matrix $\left(\matrix{ \frac{(a + b) c}{c + d} & b \\ c & \frac{( c + d) b}{(a + b)} }\right) \in M_2(\mathbb{R}_+)$.

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    $\begingroup$ Each matrix with nonnegative entries is obviously a sum of rank 1 matrices with nonnegative entries (put each tow into a separate matrix). So you are asking just whether the entries of your matrix are nonnegative? $\endgroup$
    – Ilya Bogdanov
    Commented Nov 24, 2016 at 16:17
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    $\begingroup$ If Ilya interpreted your question correctly, the answer appears to be no. Try {{1/16, 1, 1}, {1, 1/2, 1}, {3/2, 1, 1}}. $\endgroup$
    – Pace Nielsen
    Commented Nov 25, 2016 at 5:14
  • $\begingroup$ Yes, your reduction shows that I was asking whether the entries of the matrix $A - \kappa diag(b_1,...,b_n)$. Actually the original intention was for the rank 1 matrices to have no zero rows. $\endgroup$
    – user94803
    Commented Nov 25, 2016 at 7:02

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