Let $A\in M_{n\times n} (\mathbb{R})$ in which each entry is positive.

Let $v$ be a nonzero vector in $\mathbb{C}^n$ where $v_k\neq 0$

Suppose $|\sum_{j=1}^n A_{kj} v_j|=\sum_{j=1}^n |A_{kj} v_j|$.

Then, how do i prove that "there exist nonnegative real numbers $c_1,...,c_n$ and a nonzero $z\in \mathbb{C}$ such that $A_{kj}v_j=c_j z$"?

If i take $d_j = \frac{v_j \overline{v_k}}{|v_k|^2}$, then $v_j=d_j v_k$ for all $j$, but i'm not sure if $d_j$'s are all nonnegative real numbers..

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    $\begingroup$ What does it mean for a complex number to be positive? Do you mean that $A\in M_{n\times n}(\mathbb{R})$? $\endgroup$ – vadim123 Apr 28 '13 at 3:55
  • $\begingroup$ @vadim123 Isn't it usual to say 'positive complex number' to mean positive real? If you don't like it, i'll edit it. I just wanted to emphasize that the field is $\mathbb{C}$ $\endgroup$ – Jj- Apr 28 '13 at 3:58
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    $\begingroup$ In some contexts we're interested in complex numbers with positive real component (i.e. in the right half-plane of $\mathbb{C}$). But honestly I've never heard anyone use the term "positive complex number" before in any context. $\endgroup$ – vadim123 Apr 28 '13 at 4:01
  • $\begingroup$ Note that $A$ and $v$ are relevant here only for the $n$ numbers $A_{k,j}v_j$. So you might simplify the question by giving just those values a name. Not only could you then remove the (matrices) tag, but you'd be a lot closer to a solution of the problem. $\endgroup$ – Marc van Leeuwen Apr 28 '13 at 4:30

You have an (extended) triangle inequality where actually equality holds. That means that all the elements ($v_j$) are nonnegative multiples of each other. You could prove this by induction on $n$, or using Cauchy-Schwarz.


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