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 $\left(\matrix{ \frac{(a + b) c}{c + d} & b \\ c & \frac{( c + d) b}{(a + b)} }\right) \in M_2(\mathbb{R}_+)$.

  • 1
    $\begingroup$ 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 Nov 24 '16 at 16:17
  • 1
    $\begingroup$ 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 Nov 25 '16 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 Nov 25 '16 at 7:02

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.