Let $A$ be an $m\times n$ matrix. If $\forall \vec{x}\in\mathbb{R}^{n},\ A\vec{x}=\vec{0}$, show that $A$ is the zero matrix. I've thought about this a bit and it would seem the only way I can successfully do it is by an indirect proof; that is, assume that $A$ is not the zero matrix, and show that the null space of any such matrix must be a proper subset of $\mathbb{R}^n$. But I cannot think of how to get started.
The most I can think of is to say to let $a_{ij}=0$ except for some $(i,j)$ that is not zero. But then I'm stuck. So I don't know if I'm on the right track, or if I'm way off base and need to try something else.

Addition: Here's my answer as included in my homework assignment. I feel that it is not worded as best as it could be. Any tips?

Suppose that $A\vec{x}=\vec{0}$ for every $\vec{x}\in\mathbb{R}^{n}$ and for some non-zero matrix $A$. Therefore there must be a row vector $\vec{w}$ in $A$ that is not the zero vector. Since $\vec{w}\in\mathbb{R}_{n}$ and $\vec{x}\in\mathbb{R}^{n}$, let $\vec{x}=\vec{w}^{T}$.
  \begin{align*}
\vec{x}\cdot\vec{w}^{T}&=\vec{x}\cdot\vec{x}\\
&=\lVert\vec{x}\rVert^{2}\\
&>0
\end{align*}
  We have shown that there exists a vector $\vec{x}\in\mathbb{R}^{n}$ whereby $A\vec{x}\neq\vec{0}$, contradicting our assertion that $A\vec{x}=\vec{0}$ for every $\vec{x}\in\mathbb{R}^{n}$ and for some non-zero matrix $A$. Therefore, $A$ must be the zero matrix.

 A: I think the easiest way is to go a little more general, and assume some row in $A$ in not all zeros.  Then you need to show there's some vector whose dot product with that row is nonzero.
What happens when you dot a nonzero real vector with itself?
A: If some entry of $A$ is $\ne 0$, then some column of $A$ is $\ne \vec 0$, then you can find a nice vector $\vec x$ with $A\vec x\ne\vec 0$.
A: There is a particular set of vectors useful for this. They make up all other vectors. In particular, you can prove that $A\neq 0\implies \exists v:Av\neq 0$.
Sooner or later you will learn or read this, so I might as well say it. Let's agree to write out all vectors in column form. If you consider the function $$\eta(v)=Av$$ then you can check that $$\eta(v+w)=\eta(v)+\eta(w)$$ $$\eta(\lambda v)=\lambda \eta(v)$$
Now, pick any vector $v$. We can write this out in canonical coordinates as $$v=v_1e_1+\cdots+v_ne_n$$ where $e_i=(0,0,\ldots,0,\underbrace{1}_j,0,\ldots,0)$. Then by the above $$\eta(v)=v=v_1\eta(e_1)+\cdots v_n\eta(e_n)$$ This means that if you know what $\eta(e_i)=Ae_i$ are, you know essentially what $A$ is: it suffices you write $v$ in its canonical coordinates and take the dot product $$(Ae_1,\cdots,Ae_n) (v_1,\ldots,v_n)^t$$  You might have realized $Ae_j$ is just the $j$-th column of $A$. In particular if $Av=0$ for each $v$, then $Ae_j=(0,\ldots,0)$ for each canonical vector, so by the above the columns of $A$ must be all zero, or alternatively $$Av=v_1\cdot \vec 0+\cdots+v_n\cdot \vec 0=\vec 0$$ for each $v$, that is $A$ is the zero map.
A: Suppose that $A$ had a non-zero entry, which we will take to be in the $(1, 1)$-position. Let $\vec x = (1, 0, ..., 0)^T$. Then $A\vec x$ has a non-zero first coordinate.
More generally, if $A$ has a non-zero entry in the $(i, j)$-position, take a vector of all zeros and a $1$ in the $j$th coordinate.
A: For this question it suffices to realise that the columns of $A$ are instances of $A\vec x$, namely for $\vec x$ one of the standard basis vectors. If all such images are equal to $\vec0$, then all columns of $A$ are zero.
A: You can just use the fact that a linear transformation is completely determined by its action on basis vectors. In other words, since $Ax = 0$ for all $x$, you know that $Ae_{i} = 0$ for each basis vector $e_{i}$. Therefore $A$ must be the zero.
