How to show $A\mathbf x=B\mathbf x\implies A=B$ (if it holds)? This question is from Mathematics for Economists by Simon and Blume.

Theorem: $f:\mathbb{R^k}\to\mathbb{R^m}$ is a linear function if and only if there exists an $m\times k$ matrix $A$ such that $f(\mathbf{x})=A\mathbf{x}$ for all $\mathbf{x}\in\mathbb R^k$.
  Theorem underlines the one-to-one correspondence between linear function from $\mathbb{R^k}$ to $\mathbb{R^m}$ and $m\times k$ matrices. Each linear $f$ is an $f_A$ for a unique $m\times k$ matrix $A$.

For a unique $m\times k$ matrix $A$ and that means for some matrix $B$, $$\text{ For all $\mathbf x$, }A\mathbf x=B\mathbf x\iff A=B.\text{ ( Right?)}$$ How to show $\text{ For all $\mathbf x$, }A\mathbf x=B\mathbf x\implies A=B$?
 A: Hint: 


*

*$A\mathbf{x}=B\mathbf{x}$ for all $\mathbf{x}$ if and only if $(A-B)\mathbf{x}=\mathbf{0}$ for all $\mathbf{x}$

*$C\mathbf{x}=\mathbf{0}$ for all $\mathbf{x}$ if and only if $C=\dots$

I usually do it in a different way, first showing uniqueness.
If $A$ exists, then we have
$$
A\mathbf{e}_i=f(\mathbf{e}_i)
$$
for $i=1,2,\dots,k$, where $\{\mathbf{e}_1,\dots,\mathbf{e}_k\}$ is the canonical basis of $\mathbb{R}^k$. But, if we write 
$A=\begin{bmatrix}\mathbf{a}_1&\mathbf{a}_2&\dots&\mathbf{a}_k\end{bmatrix}$, we clearly have
$$
A\mathbf{e}_i=\mathbf{a}_i
$$
and therefore
$$
A=\begin{bmatrix}
f(\mathbf{e}_1) & f(\mathbf{e}_2) & \dots & f(\mathbf{e}_k)
\end{bmatrix}
$$
so uniqueness follows, because the matrix we got depends only on $f$.
Then we can prove that this matrix has the requested property.
A: The statement you want is $[(\forall \mathbf{x} \in \mathbb{R}^k)(A\mathbf{x} = B\mathbf{x})] \Rightarrow A=B$.  To show this, consider what you get when computing $A\mathbf{e_i}$ (for $1\leq i\leq k$), where $\mathbf{e_i}$ has a $1$ in the $i$-th coordinate and $0$s elsewhere.
A: The columns of a matrix are just instances of $A\mathbf x$, namely for the standard basix vectors $\mathbf x=\mathbf e_j$ (in which case $A\mathbf x$ gives you column $j$ of the matrix$~A$). So saying that for all $\mathbf x$ one has $A\mathbf x=B\mathbf x$ implies in particular that every column of $A$ is equal to the corresponding column of $B$. Given that, it is hard for $A$ and $B$ to differ.
