Implication of Matrices in $AX=0$ I'm not sure how to phrase my question and this is the first time I'm posting on math exchange but basically I'm looking for some feedback on my answer to this question:
Let $A$ denote an $m×n$ matrix. If $AX=0$ for every $n×1$ matrix $X$, show that $A=0$.
I claim that since $X$ is every column vector in the set $R^n$, and $AX=0$, it is implied that $A=0$ by considering the constituents of the products of $AX$: $a_j×x_i$ where $a_j$ represents the $j$th column of $A$ and $x_i$ represents the $i$th row of $X$, and $i=j$.
If $a_j×x_i=0$ and vector $X$ is every possible column vector (where $x_i=n$), then it's implied that $A$ must be a zero matrix (where $a_j=0$) to satisfy the equation $AX=0$.
I feel like I'm running in circles with this argument...
 A: Observe that any matrix $A$ multiplied by the $i$th standard basis vector $e_i$ (whose $i$th coordinate is $1$ and the rest is $0$) gives the $i$th column of $A$.
Apply this for $X=e_1,\dots, e_n$ to show that every column of $A$ is the zero vector. 
A: With
$A = [a_{ij}] \tag 1$
and
$X = (x_1, x_2, \ldots, x_n)^T, \tag 2$
it is easy to see that
$AX = \left (\displaystyle \sum_1^n a_{1k}x_k , \sum_1^n a_{2k} x_k, \ldots, \sum_1^n a_{mk} x_k \right )^T; \tag 3$
since $X$ is arbitrary, we are free to choose
$x_k = \delta_{kj} \tag 4$
for any $j$; then (3) becomes
$AX = (a_{1j}, a_{2j}, \ldots, a_{mj}); \tag 5$
but we are given that this vanishes for any choice of $X$; thus
$a_{ij} = 0, \; 1 \le i \le m; \tag 6$
allowing $j$ to run from $1$ to $n$ shows that
$a_{ij} = 0, \; 1 \le i \le m, \; 1 \le j \le n, \tag 7$
whence
$A = 0. \tag 8$
A: Suppose that $A$ is not the $m\times n$ zero matrix. That is, for some index $j\in\{1,\ldots,n\}$, 
the $j$-th column $A_j$ of $A$ is not the $m\times 1$ zero matrix. Let $X$ be the $n\times 1$ matrix whose only nonzero entry is a $1$ on its $j$-th row. $AX=A_j$ is nonzero.
