Difference between revisions of "Linear map/Definition"
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* <math>T(\lambda x)=\lambda T(x)</math> | * <math>T(\lambda x)=\lambda T(x)</math> | ||
Or indeed: | Or indeed: | ||
| − | * <math>T(x+\lambda y)=T(x)+\lambda T(y)</math> | + | * <math>T(x+\lambda y)=T(x)+\lambda T(y)</math><ref>Linear Algebra via Exterior Products - Sergei Winitzki</ref><noinclude> |
| + | ==References== | ||
| + | <references/> | ||
| + | </noinclude> | ||
Latest revision as of 10:34, 12 June 2015
Given two vector spaces [ilmath](U,F)[/ilmath] and [ilmath](V,F)[/ilmath] (it is important that they are over the same field) we say that a map, [math]T:(U,F)\rightarrow(V,F)[/math] or simply [math]T:U\rightarrow V[/math] (because mathematicians are lazy), is a linear map if:
- [math]\forall \lambda,\mu\in F[/math] and [math]\forall x,y\in U[/math] we have [math]T(\lambda x+\mu y) = \lambda T(x) + \mu T(y)[/math]
Which is eqivalent to the following:
- [math]T(x+y)=T(x)+T(y)[/math]
- [math]T(\lambda x)=\lambda T(x)[/math]
Or indeed:
- [math]T(x+\lambda y)=T(x)+\lambda T(y)[/math][1]
References
- ↑ Linear Algebra via Exterior Products - Sergei Winitzki
