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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!

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Do you require $L$ to be linear? – Michael Albanese Feb 21 '13 at 3:58
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.

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