Let $G:\mathbb{R}^n \to\mathbb{R}^n$ be transformation such that $G(x):=Ax+b$ where $A\in\mathcal{M}_{nxn}(\mathbb{R})$ and $b\in\mathbb{R}^n$ such that $det(A-I)\neq0$ .
How would you prove G has a unique fixed point $p\in\mathbb{R}^{n}$ ?
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Hint:
Rewrite: $Ax+b=x$ to $(A-I)x=-b$ and use that $det(A-I)\neq 0$.
What does $det(B)\neq 0$ says about the associated linear map?
Marm
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Having a fix point means $x=Ax+b$, i.e. $0=(A-I)x+b$. But now $A-I$ is invertible, in particular there exists one and only one element which hits $-b$.
Daniel Valenzuela
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1Which hits $-b$ :) – Marm Mar 17 '15 at 17:41
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1It's equivalent :) – Daniel Valenzuela Mar 17 '15 at 17:42
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1But I edited it. – Daniel Valenzuela Mar 17 '15 at 17:42
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1Right, but its more clear now for the OP :) – Marm Mar 17 '15 at 17:42