Linear isometry

From Maths
Jump to: navigation, search


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.


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