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Given an affine function $f:\mathbb{R}^n \to \mathbb{R}^n$ defined for all $x\in \mathbb{R}^n$ by $$f(x)=T(x)+a$$ such that $T$ is an invertible Linear map and $a\in \mathbb{R}^n$, is $f$ a diffeomorphism?

h3110w0r1d
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It is a diffeomorphism iff $T$ is invertible.
It is easy to see that $f$ is invertible iff $T$ is invertible with inverse $$f^{-1}(x) = T^{-1}(x - a).$$

As $T^{-1}$ is a linear transformation, it is differentiable everywhere.

Aryaman Maithani
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  • Hello and thanks! I forgot to mention that $T$ is invertible, added to the oroginal post. – h3110w0r1d Jun 02 '20 at 16:12
  • No worries, I belive my answer still answers your question regardless? – Aryaman Maithani Jun 02 '20 at 16:14
  • A function $g$ is diffeomorphism if it is a bijection, differentiable and $g^{-1}$ is also differentiable. So by your answer the function $f$ is a diffeomorphism, right? – h3110w0r1d Jun 02 '20 at 16:20
  • Yep! That is what I've written as well. – Aryaman Maithani Jun 02 '20 at 16:22
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    Thank you very much! – h3110w0r1d Jun 02 '20 at 16:29
  • Another thing I want to ask - Let $T:\mathbb{R}^n \to \mathbb{R}^n$ be a linear map and $Q\subset \mathbb{R}^n$ a Rectangle. The image $T(Q)$ is also a Rectangle? – h3110w0r1d Jun 02 '20 at 16:29
  • You mean $Q \subset \Bbb R^n$. And no, even in the case that $T$ is invertible, you can have something skewed. Consider $n = 2$ and the map $(x, y) \mapsto (x, x + y)$. The rectangle $[0, 1] \times [0, 1]$ gets mapped to a parallelogram. (I'm taking the definition of a rectangle to be a set of the form $[a_1, b_1] \times \cdots \times [a_n, b_n]$). – Aryaman Maithani Jun 02 '20 at 16:31
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    Aha! I should have thought about it, thank again – h3110w0r1d Jun 02 '20 at 16:46