I have a problem with understanding of a rather simple concept in linear algebra. I have seen in a book, a following question:
Suppose $U,V$ are vector spaces over $K$ and $u_1,\dots,u_n$ is a basis of $U$. Let $v_1,\dots, v_n$ be any sequence of $n$ vectors in $V$. Prove that there is a unique map $T:U\rightarrow V$ with $T(u_i)=v_i$, for $1\leq i \leq n$.
The proof given in this book is:
Let $u \in U$. Then since $u_1,\dots, u_n$ is a basis of $U$, there exists uniquely determined $\alpha_1,\dots,\alpha_n \in K$ with $u=\alpha_1 u_1+\dots+\alpha_n u_n$. So that, if $T$ exists, then we must have:
$$T(u)=T(\alpha_1 u_1+\dots+\alpha_n u_n)=\alpha_1 v_1+\dots+\alpha_n v_n$$
and so $T$ is uniquely determined.
Honestly this proof makes little sense to me. How does it prove uniqueness? Is this proof complete? Could someone explain this to me? Thank you very much.
Best Answer
Let's say we have another mapping $f:U \to V$ such that $f(u_i)=v_i, \forall i$. Then, for each $u \in U:$ $T(u)=T(a_1u_1+\dots + a_nu_n)=a_1T(u_i)+\dots + a_nT(u_n)=a_1v_1+ \dots a_nv_n=a_1f(u_1)+ \dots + a_nf(u_n)$
$=f(a_1u_1+ \dots +a_nu_n)=f(u)$
Hope that answers your question.