TITLE: A theorem concerning unique linear mapping between vector spaces: What does it say?
QUESTION [6 upvotes]: 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, it is not so hard. But I don't understand what this theorem means or what does it imply, how one can use it to deduce more results. Can anybody briefly explain it? Thanks in advance!

REPLY [0 votes]: It means what it says.
To detour a little, consider a quadratic function $g:\mathbb{R}\to\mathbb{R}$ of the form $g(x)=ax^2+bx+c$. Are there $a,b,c$ such that $g(0)=0,\ g(-1)=g(1)=1$ and $g(2)=5$? The answer is no, because $g(0)=0$ implies that $c=0$, then $g(-1)=g(1)=1$ implies that $a=1$ and $b=0$, i.e. $g(x)=x^2$. Yet this is not consistent with the requirement that $g(2)=5$. Hence, it is not always possible to find a quadratic function $g$ that take four prespecified values at four given points.
But with the right number of points, say, three points $x_1,x_2,x_3$ and the right condition --- say, $x_1,x_2,x_3$ are distinct --- you can always find a quadratic function $g$ (or strictly speaking, a polynomial function of degree at most $2$ because $a$ may turn out to be zero) such that $g(x_i)=y_i$ for any prespecified values $y_1,y_2,y_3$.
Now the analogous holds for a linear function. Given the right number of points ($n$) and the right condition ($v_1,\ldots,v_n$ are nonzero and distinct, which is automatically true if they form a basis), you can always find a linear function that maps these $n$ points to the prespecified values ($u_1,\ldots,u_n$). This is what the existence of $F$ in the theorem means.
Let us return to the quadratic function example. Under some circumstances, it possible to find more than one functions $g$ that satisfy our requirements. For instance, if $x_1=x_2=-1,\,x_3=1$ and $y_1=y_2=y_3=0$, there are infinitely many quadratic functions $g$ such that $g(x_i)=y_i$. In fact, for any real number $a$, the function $g(x)=ax^2-a$ will do. However, given the right condition ($x_1,x_2,x_3$ are distinct), it can be shown that the function $g$ that maps $x_1,x_2,x_3$ to any prespecified values $y_1,y_2,y_3$ is unique. Again, the analogous holds for $F$. Given the right condition ($v_1,\ldots,v_n$ form a basis of $V$), there is only one possible linear function $F$ that maps $v_1,\ldots,v_n$ to $u_1,\ldots,u_n$. This is what the uniqueness of $F$ is about.
{"set_name": "stack_exchange", "score": 6, "question_id": 289397}