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
[hide]Definition
Suppose U and V are normed vector spaces with the norm ∥⋅∥U and ∥⋅∥V respectively, a linear isometry preserves norms
It is a linear map L:U→V where ∀x∈U we have ∥L(x)∥V=∥x∥U
Notes on definition
This definition implies L is injective.
Proof
Suppose it were not injective but a linear isometry, then we may have have L(a)=L(b) and a≠b, then ∥L(a−b)∥V=∥L(a)−L(b)∥V=0 by definition, but as a≠b we must have ∥a−b∥U>0, contradicting that is an isometry.
Thus we can say L:U→L(U) is bijective - but as it may not be onto we cannot say more than L is injective. Thus L 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
