Difference between revisions of "Linear isometry"
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==Definition== | ==Definition== | ||
| − | Suppose {{M|U}} and {{M|V}} are [[Norm|normed]] [[Vector space|vector spaces]] with the norm <math>\|\cdot\|_U</math> and < | + | Suppose {{M|U}} and {{M|V}} are [[Norm|normed]] [[Vector space|vector spaces]] with the norm <math>\|\cdot\|_U</math> and <math>\|\cdot\|_V</math> respectively, a linear isometry preserves [[Norm|norms]] |
It is a [[Linear map|linear map]] <math>L:U\rightarrow V</math> where <math>\forall x\in U</math> we have <math>\|L(x)\|_V=\|x\|_U</math> | It is a [[Linear map|linear map]] <math>L:U\rightarrow V</math> where <math>\forall x\in U</math> we have <math>\|L(x)\|_V=\|x\|_U</math> | ||
Latest revision as of 11:23, 12 May 2015
Contents
Definition
Suppose [ilmath]U[/ilmath] and [ilmath]V[/ilmath] are normed vector spaces with the norm [math]\|\cdot\|_U[/math] and [math]\|\cdot\|_V[/math] respectively, a linear isometry preserves norms
It is a linear map [math]L:U\rightarrow V[/math] where [math]\forall x\in U[/math] we have [math]\|L(x)\|_V=\|x\|_U[/math]
Notes on definition
This definition implies [math]L[/math] is injective.
Proof
Suppose it were not injective but a linear isometry, then we may have have [math]L(a)=L(b)[/math] and [math]a\ne b[/math], then [math]\|L(a-b)\|_V=\|L(a)-L(b)\|_V=0[/math] by definition, but as [math]a\ne b[/math] we must have [math]\|a-b\|_U>0[/math], contradicting that is an isometry.
Thus we can say [math]L:U\rightarrow L(U)[/math] is bijective - but as it may not be onto we cannot say more than [math]L[/math] is injective. Thus [math]L[/math] may not be invertible.
Isometric normed vector spaces
We say that two normed vector spaces are isometric if there is an invertible linear isometry between them.
Pullback norm
- See Pullback norm
