- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.9 Linear transformations with prescribed values
$\quad$ If $V$ is finite-dimensional, we can always construct a linear transformation $T: V \to W$ with prescribed values at the basis elements of $V.$
$\quad$ Theorem 2.12. $\quad$ Let $e_1, \dots, e_n$ be a basis for an $n$-dimensional linear space $V.$ Let $u_1, \dots, u_n$ be $n$ arbitrary elements in a linear space $W.$ Then, there is one and only one linear transformation $T: V \to W$ such that \begin{align*} (2.7) \qquad T(e_k) &= u_k \qquad for \quad k = 1, 2, \dots, n. \end{align*} This $T$ maps an arbitrary element $x$ in $V$ as follows: \begin{align*} (2.8) \qquad If \quad x &= \sum_{k=1}^n x_ke_k, \quad then \quad T(x) = \sum_{k=1}^n x_ku_k \end{align*}
$\quad$ Proof. $\quad$ Let $T': V \to W$ be a linear transformation. Then, if $T'(e_k) = u_k,$ for each $x$ in $V,$ we have \begin{align*} T'(x) &= T'\left[\sum_{k=0}^n x_ke_k\right] \\ &= \sum_{k=0}^n T'(x_ke_k) \\ &= \sum_{k=0}^n x_ku_k \\ &= T(x) \end{align*} But if this is true for all $x$ in $V,$ then $T' = T. \quad \blacksquare$