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The question is given below:

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I feel like the answer is yes but I do not know how to justify it, could anyone help me in this please?

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    $\begingroup$ columns of $A$ are images of basis vectors $\endgroup$ – J. W. Tanner Sep 20 at 12:27
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Yes. The standard basis vectors have non-negative components, so, given the condition,

their images, which are the columns of the matrix, have non-negative components.

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  • $\begingroup$ I do not think your answer is correct ..... I have a counter example for this. $\endgroup$ – Mathstupid Sep 23 at 15:49
  • $\begingroup$ @Smart: feel free to post your counterexample, so it could be scrutinized $\endgroup$ – J. W. Tanner Sep 23 at 15:51
  • $\begingroup$ it is the matrix $2 \times 2$ matrix A that has entries in order 1, -1, 2 ,0 and the vector $v$ that has entries 1,1 ..... what is your opinion ? is my example wrong? $\endgroup$ – Mathstupid Sep 23 at 16:08
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    $\begingroup$ to satisfy the condition that if $v$ is a vector means it's true for any $v$ $\endgroup$ – J. W. Tanner Sep 23 at 16:18
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    $\begingroup$ the question is asking if all the matrix entries must be non-negative if the condition holds for all vectors; it is not asking if all the matrix entries must be non-negative if the condition holds for some vector $\endgroup$ – J. W. Tanner Sep 23 at 16:20
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Let $A$ and $v\neq 0$ be as in the exercise. Components of $Av$ are non-negative if and only if $$ a_{i1}v_1 + a_{i2}v_2+...+a_{in}v_n \geq 0 \quad(i=1,...n) $$ Assume one if it's entries, say $a_{pq}$ is negative. Then $a_{pq}v_q$ is also negative and if $v_{k}$ ($k \neq q$) are small enough, then $$a_{p1}v_1 + a_{p2}v_2+...+a_{pn}v_n = a_{pq}v_q + \sum_{k \neq q} a_{pk}v_k < 0$$, which equals to the first sum when $i=p$.

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Yes $a_{ij}$ is nothing but the $i$-th coordinate of $Ae_j$ where $e_j$ has $1$ at the $j-$th place and $0$ elsewhere.

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