Let \(V\) and \(W\) be subspaces of \(\RNrSpc{k}\). Given a basis \(\EuScript{A}=\Set{ \Vect{a}_1,\dots ,\Vect{a}_n}\) of \(V\) and arbitrary vectors \(\Vect{z}_1\), ... , \(\Vect{z}_n\) in \(W\), there is exactly one linear transformation \(L\from V\to W\) with
\[L(\Vect{a}_1)=\Vect{z}_1\ ,\quad \dots \ ,\quad L(\Vect{a}_n)=\Vect{z}_n\]We begin by constructing a function \(L\from V\to W\) satisfying
\[L(\Vect{a}_1)=\Vect{z}_1,\quad \dots ,\quad L(\Vect{a}_n)=\Vect{z}_n\]Indeed, we know that every vector \(\Vect{x}\in V\) has a unique expression as a linear combination
| \(\Vect{x}\) | \(=\) | \(x_1 \Vect{a}_1 + \cdots + x_n \Vect{a}_n\) |
Therefore, setting \(L(\Vect{x}) \DefEq x_1 L(\Vect{a}_1) + \cdots + x_n L(\Vect{a}_n)\) is a function from \(V\) to \(W\). We need to verify that \(L\) is linear:
Verification of the additivity property: With \(\Vect{x}=x_1 \Vect{a}_1+\cdots + x_n \Vect{a}_n\) and \(\Vect{y}=y_1 \Vect{a}_1+\cdots + y_n \Vect{a}_n\), we find
| \(L(\Vect{x}+\Vect{y})\) | \(=\) | \(L\left( (x_1 \Vect{a}_1 +\cdots x_n \Vect{a}_n) + (y_1 \Vect{a}_1+\cdots + y_n \Vect{a}_n) \right)\) |
| \(\) | \(=\) | \(L\left( (x_1+y_1)\Vect{a}_1 + \cdots + (x_n+y_n) \Vect{a}_n \right)\) |
| \(\) | \(=\) | \((x_1+y_1)L(\Vect{a}_1) + \cdots (x_n+y_n) L(\Vect{a}_n)\) |
| \(\) | \(=\) | \(\left( x_1L(\Vect{a}_1) +\cdots + x_n L(\Vect{a}_n)\right)\ +\ \left(y_1L(\Vect{a}_1) + \cdots + y_nL(\Vect{a}_n)\right)\) |
| \(\) | \(=\) | \(L(\Vect{x})\ +\ L(\Vect{y})\) |
Verification that \(L\) commutes with scalar multiplication. With \(\Vect{x} = x_1 \Vect{a}_1 + \cdots x_n \Vect{a}_n\in V\) and \(t\in \RNr\) we find
| \(L(t \Vect{x})\) | \(=\) | \(L\left( t(x_1 \Vect{a}_1 +\cdots + x_n \Vect{a}_n)\right)\) |
| \(\) | \(=\) | \(L\left( (tx_1)\Vect{a}_1 + \cdots + (tx_n)\Vect{a}_n\right)\) |
| \(\) | \(=\) | \((tx_1)L(\Vect{a}_1) + \cdots + (tx_n)L(\Vect{a}_n)\) |
| \(\) | \(=\) | \(t\left( x_1L(\Vect{a}_1) +\cdots + x_nL(\Vect{a}_n)\right)\) |
| \(\) | \(=\) | \(t L(\Vect{x})\) |
It remains to show that the function \(L\) we just constructed is the only linear function \(V\to W\) with
\[L(\Vect{a}_1)=\Vect{z}_1,\quad \dots ,\quad L(\Vect{a}_n)=\Vect{z}_n\]So suppose \(M\from V\to W\) is linear and satisfies
\[M(\Vect{a}_1)=\Vect{z}_1,\quad \dots ,\quad M(\Vect{a}_n)=\Vect{z}_n\]We need to show that \(L=M\). For an arbitrary vector \(\Vect{x}=x_1 \Vect{a}_1 + \cdots + x_n \Vect{a}_n\) we find
| \(L(\Vect{x})\) | \(=\) | \(x_1 L(\Vect{a}_1) + \cdots + x_n L(\Vect{a}_n)\) |
| \(\) | \(=\) | \(x_1 \Vect{z}_1 + \cdots + x_n \Vect{z}_n\) |
| \(\) | \(=\) | \(x_1 M(\Vect{a}_1) + \cdots + x_n M(\Vect{a}_n)\) |
| \(\) | \(=\) | \(M(x_1 \Vect{a}_1 + \cdots + x_n \Vect{a}_n)\) |
| \(\) | \(=\) | \(M(\Vect{x})\) |
This means that \(L=M\), and the proof is complete.