# A question on matrices from Golan's “Linear Algebra”.

The question is given below:

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

• columns of $A$ are images of basis vectors – J. W. Tanner Sep 20 at 12:27

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.

• I do not think your answer is correct ..... I have a counter example for this. – Mathstupid Sep 23 at 15:49
• @Smart: feel free to post your counterexample, so it could be scrutinized – J. W. Tanner Sep 23 at 15:51
• 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? – Mathstupid Sep 23 at 16:08
• to satisfy the condition that if $v$ is a vector means it's true for any $v$ – J. W. Tanner Sep 23 at 16:18
• 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 – J. W. Tanner Sep 23 at 16:20

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 $$i=p$$.

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.