# Showing a mapping is bijective if and only if a matrix is invertible

Let $$\mathbf{A}$$ be an $$n\times n$$ matrix and let $$\mathbf{c}$$ and $$x_{\star}$$ be point in $$\mathbb{R}^{n}$$. Define the affine mapping $$\mathbf{G} : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$$ by

$$\mathbf{G(x)} = \mathbf{c + A(x - x_{\star})}$$

for $$\mathbf{x}$$ in $$\mathbb{R}^{n}$$. Show that the mapping $$\mathbf{G} : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$$ is one-to-one and onto if and only if $$\mathbf{A}$$ is invertible.

I am not too sure about how to approach this problem. I've also got the following theorem that I think might help:

Let $$\mathcal{O}$$ be an open subset of $$\mathbb{R}^{n}$$ and suppose $$\mathbf{F} : \mathcal{O} \rightarrow \mathbb{R}^{n}$$ is continuously differentiable. Let $$x_{\star}$$ be a point in $$\mathcal{O}$$ at which the derivative matrix $$\mathbf{DF(x_{\star})}$$ is invertible. Then there is a neighborhood $$U$$ of $$x_{\star}$$ and a neighborhood $$V$$ of its image $$\mathbf{F(x_{\star})}$$ such that $$\mathbf{F} : U \rightarrow V$$ is one-to-one and onto.

I've tried taking the derivative of both sides of the equation, etc, but didn't get anywhere.

Any help is appreciated.

## 2 Answers

No need to take differential approaches.

If $$A$$ is invertible, you can explicitly write up the inverse of $$G$$, as $$x-x_*=A^{-1}(G(x)-c)$$ Conversely, if $$G$$ is injective, so must be $$x\mapsto Ax$$, too, which implies in finite dimension that $$A$$ is invertible.

You have overthinking this.

If $$A$$ is one-to-one, then if $$G(x)=G(y)$$ you have $$c+A(x-x_*)=c+A(y-x_*).$$ After subtracting $$c$$ and $$-Ax_*$$ from both sides, you get $$Ax=Ay$$; as $$A$$ is one-to-one, $$x=y$$ and then $$G$$ is one-to-one. Conversely, if $$G$$ is one-to-one, you can write $$Ax=G(x)-c+A(x_*),$$ and used the same idea as above to conclude that $$A$$ is one-to-one.

Similarly, assume that $$A$$ is onto. Given $$z$$, there exists $$x$$ such that $$Ax=z-c+A(x_*)$$; then $$G(x)=z$$, so $$G$$ is into. And, conversely, if $$G$$ is onto given $$z$$ you can get $$x$$ such that $$G(x)=c+z-Ax_*$$, so $$A(x)=z$$.

Thus $$G$$ is bijective if and only if $$A$$ is bijective, i.e., invertible.