Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following theorem:

Theorem (from Schaum's Linear Algebra) Let $V$ and $U$ be vector spaces and $\{v_1, \ldots, v_n\}$ be a basis on $V$. Let $\{u_1,\ldots, u_n\}$ be arbitrary vectors in $U$. Then there exists a unique linear mapping $F: V \to U$ such that $F(v_i) = u_i$.

I omit the proof of the statement, I focus on well definedness of $F$. In book, author says `since $a_i$'s are unique, the mapping $F$ is well defined'. What does it mean? I mean, is there a way to show well definedness rigorously?

Thanks in advance!

share|improve this question
    
Do you require $L$ to be linear? –  Michael Albanese Feb 21 '13 at 3:58

1 Answer 1

up vote 2 down vote accepted

It means that given a vector $v\in V$ then there is only one way to compute $F(v)$.

This follows from the fact that there is a unique way of writing $v=\sum\limits_{i=1}^n\alpha_i v_i$, so we must have $$F(v)=\sum\limits_{i=1}^n\alpha_i F(v_i)=\sum_{i=1}^n\alpha_i u_i.$$

Surely there are many ways to define a function between $V$ and $U$ which has the property $F(v_i)=u_i$, but if we require linearity then there are no other extensions. This is really just writing down the properties of a basis and a linear function.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.