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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy