2
$\begingroup$

Suppose there is a $3$ dimensional space and transformation that

  1. transforms zero into zero
  2. preserves distance between any two points

Prove:

  1. transformation is linear
  2. transformation is invertible
  1. According to definition:

$$\begin{array}{l}f(x + y) = f(x) + f(y)\\\alpha f(x) = f(\alpha x)\end{array} $$

How I can try to prove it in this context?

  1. There is the proof from external resource:

    • Introduce inner product:

    $$(x,y) = \frac{{{{\left\| {x + y} \right\|}^2} - {{\left\| {x - y} \right\|}^2}}}{4} = xy $$

    • it's defined with distance only(preserved from our second axiom) $\Rightarrow $ inner product is preserved $\Rightarrow $ angles preserved

    • construct matrix A: columns - basis vectors, they are orthogonal, they form an orthonormal system

    • ${A^T}A = A{A^T} = E $ because multiply orthogonal vectors (1 only if multiply by itself)

    • we explicitly find ${A^{-1}} \Rightarrow$ transformation is invertible

Questions:

  1. Why inner product has this form? Why not simple $(x,y) = $$\left\| {x + y} \right\| $

  2. Why columns of matrix $A$ form orthonormal system?

$\endgroup$
3
  • $\begingroup$ See isometric operators in Google. $\endgroup$
    – Piquito
    Jan 7, 2016 at 17:47
  • $\begingroup$ What is a "3N space"? $\endgroup$ Jan 7, 2016 at 18:00
  • $\begingroup$ three dimensional space $\endgroup$ Jan 7, 2016 at 18:26

1 Answer 1

3
$\begingroup$
  1. Why inner product has this form? Why not simple $(x,y) = \| x + y \|$

The inner product would not have the properties it is supposed to have if you defined it as $\| x + y \|$. For example, $(x,-x)=-(x,x)$ is supposed to be nonzero iff $x$ is nonzero. But in your definition, $(x,-x)=\| x + -x \|=\|0\|=0$.

  1. Why columns of matrix A form orthonormal system?

That is literally what $A^\top A=I_n$ says. The $i$'th row of $A^\top$ times the $j$th column of $A$ is $1$ iff $i=j$ says that all the vectors are normalized, and the product is $0$ iff $i\neq j$ says that the vectors are pairwise orthogonal.

As for the underlying problem in the post, you ought to be able to find more versions of solutions here on the site. Here are some:

Showing that an Isometry on the Euclidean Plane fixing the origin is Linear

Are isometries always linear?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .