Exercise 2.1.34

Question

Let $V$ and $W$ be vector spaces over a common field, and let $\beta$ be a basis for $V$. Then for any function $f: \beta 	o W$ there exists exactly one linear transformation $T: V 	o W$ such that $T(x) = f(x)$ for all $x \in \beta$.

Jephian C.-H. Lin · Accepted Answer

Since $\beta$ is a basis, any $x\in V$ can be written as $x=\sum_{v_i\in\beta}{a_iv_i}$ for some $a_i$'s.  Given a function $f:\beta\rightarrow W$, we may define the mapping $T$ as 
$$T(x)=T(\sum_{v_i\in\beta}{a_iv_i})=\sum_{v_i\in\beta}a_if(v_i),$$
where $a_i$'s depend on $x$.  One may check $T$ is a linear transformation with $T(x)=f(x)$ for $x\in\beta$, and this gives the existence.\par

Suppose $T'$ is another linear transformation that satisfies $T(x)=f(x)$ for $x\in\beta$.  Then by the definition of linear transformation we have $T(x)$ must be 
$$T(\sum_{v_i\in\beta}{a_iv_i})=\sum_{v_i\in\beta}{a_iT(v_i)}=\sum_{v_i\in\beta}{a_if(v_i)},$$
where $x=\sum_{v_i\in\beta}{a_iv_i}$ is the unique representation of $x$ with respect to the basis $\beta$. So $T'=T$, giving the uniqueness.

Exercise 2.1.34

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Exercise 2.1.34

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