Talk:Linear Algebra/Definition of Homomorphism

The formulation of his theorem seems a little odd to me.

A homomorphism is determined by its action on a basis. That is, if $$ \langle \vec{\beta}_1,\dots,\vec{\beta}_n \rangle $$ is a basis of a vector space $$ V $$ and $$ \vec{w}_1,\dots,\vec{w}_n $$ are (perhaps not distinct) elements of a vector space $$ W $$ then there exists a homomorphism from $$ V $$ to $$ W $$ sending $$ \vec{\beta}_1 $$ to $$ \vec{w}_1 $$, ..., and $$ \vec{\beta}_n $$ to $$ \vec{w}_n $$, and that homomorphism is unique.
 * Theorem 1.9

I'd formulate it as:

A homomorphism $$h$$ is determined by its action on a basis. That is, if $$ \langle \vec{\beta}_1,\dots,\vec{\beta}_n \rangle $$  is a basis of a vector space $$ V $$ and $$ \vec{w}_1,\dots,\vec{w}_n $$ are the images of the base, i.e. $$\vec{w}_i= h(\vec{\beta}_i)$$, then the image of any $$\vec{v}\in V $$, with:
 * Theorem 1.9
 * $$\vec{v}= \sum_i \lambda_i \vec{\beta}_i$$

is uniqely determined by:


 * $$h(\vec{v})= h(\sum_i \lambda_i \vec{\beta}_i)=\sum_i \lambda_i h(\vec{\beta}_i)=\sum_i \lambda_i \vec{w}_i$$

This may be read as: the image of a certain linear combination of the vectors of a base, is the same combination of the images of these vectors. Nijdam (talk) 18:22, 15 March 2010 (UTC)