# A theorem concerning unique linear mapping between vector spaces: What does it say?

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!

-
A possible interpretation is that finite dimensional vector spaces resemble finite sets when it comes to homomorphisms (="linear mappings"). Indeed, to construct a homomorphism between finite dimensional vector spaces requires exactly the same work it takes to construct a mapping from a finite set into another. – Giuseppe Negro Jan 29 '13 at 0:42

## 2 Answers

The theorem says that the values of a linear transformation $T:V\to U$ are completely and freely determined by the values $T$ attains on a basis of $V$. In more detail, if $T,S:V\to U$ are two linear transformations and $\{v_1 ,\cdots v_n\}$ is a basis for $V$ and $T(v_i)=S(v_i)$ holds for all $v_i$ then $S=T$ (which means that for any $v\in V$ holds that $T(v)=S(v)$). This is the precise meaning of $T$ is completely determined by its values on a basis.

Further, given any (free) choice of vectors $u_1 ,\cdots, u_n$ in $U$ there is a linear transformation $T:V\to U$ such that $T(v_i)=u_i$. This is the meaning of $T$ is freely determined by its values on a basis.

These properties are extremely useful when studying and constructing linear transformations since they reduce the study of an arbitrary linear transformations $T:V\to U$ to understanding what $T$ does on a basis. If $V$ has a finite basis then this process reduces considering the value of $T$ on infinitely many values to just finitely many.

For example, taking the indefinite integral of a function is a linear operator (since it respects addition of functions and scalar multiplication). If you consider the linear space $P$ of all polynomials then you obtain a linear transformation $\int :P\to P$. It is useful to know that the values of $\int$ are determined uniquely by its values on a basis, for instance the basis $\{1,x,x^2,x^3,\cdots\}$.

Another example considers the space $\mathbb R^2$. Given any two points $v_1,v_2$ it is convenient to know that there exists a linear transformation $T:\mathbb R^2\to \mathbb R^2$ with $T(e_1)=v_1$ and $T(e_2)=v_2$ since then you can study properties of these two values by studying properties of the linear transformation.

-

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.

-