How to show the function $f(\mathbf{A},\mathbf{x})=\mathbf{Ax}$ injective? I have a map $f:\mathbb{R}^{n\times n}\times \mathbb{R}^n\to\Bbb R^n$ that takes a matrix $\mathbf{A}\in\mathbb{R}^{n\times n}$ that is full rank and a vector $\mathbf{x}\in\mathbb{R}^n$ that is non-zero and multiplies them i.e. $f(\mathbf{A},\mathbf{x})=\mathbf{Ax}$. I followed the standard route of showing $f(\mathbf{A}_1,\mathbf{x}_1)=f(\mathbf{A}_2,\mathbf{x}_2)$ only if $\mathbf{A}_1=\mathbf{A}_2$, $\mathbf{x}_1=\mathbf{x}_2$, however I get stuck with the manipulation as you can't fix one of the variables.
Any help would be appreciated.
 A: The comments already give a counter-example to injectivity and even more obvious is $f(\mathbf A,\mathbf 0)=\mathbf 0$ for any matrix $\mathbf A\in\Bbb R^{n\times n}$ and the zero-vector $\mathbf 0\in\Bbb R^n$. So for $\mathbf A\neq\mathbf B$ you have $f(\mathbf A,\mathbf 0)=f(\mathbf B,\mathbf 0)$.
A more theoretical approach to why this map can't be injective is the following:
The map $f\colon \Bbb R^{n\times n}\times \Bbb R^n\to\Bbb R^n, (\mathbf A,\mathbf x)\to\mathbf A\mathbf x$ is bilinear, where $\Bbb R^{n\times n}$ is equipped with the usual component-wise addition and scalar multiplication. Hence, it factors through the tensor product $\Bbb R^{n\times n}\otimes\Bbb R^n$ as a linear map $\overline f$:
$$
\begin{array}{ccccc}
\Bbb R^{n\times n} \times \Bbb R^n &\overset{\otimes}\longrightarrow& \Bbb R^{n\times n}\otimes \Bbb R^n &\overset{\overline f}\longrightarrow& \Bbb R^n, \\
(\mathbf A, \mathbf x) &\longmapsto& \mathbf A \otimes \mathbf x &\longmapsto& \mathbf A\mathbf x.
\end{array}
$$
If $f$ was injective, then the first map $\otimes$ has to be injective as well. However, for $\mathbf A\neq\mathbf B$ we still have $\mathbf A\otimes \mathbf 0=\mathbf B\otimes\mathbf 0=\mathbf 0_{n\times n}\otimes \mathbf 0$, which is the zero element of the tensor product.
This shows not just that $f$ can't be injective, but in fact there is no injective bilinear map at all, when one of the factors of the domain is at least of dimension $1$.
