- 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
2.12 Exercises
$\quad$ In all exercises involving the vector space $V_n,$ the usual basis of unit coordinate vectors is to be chosen, unless another basis is specifically mentioned. In exercises concerned with the matrix of a linear transformation $T: V \rightarrow W$ where $V = W,$ we take the same basis in both $V$ and $W$ unless another choice is indicated.
6. $\quad$ For the linear transformation in Exercise 5, choose both bases to be $(e_1, e_2, e_3),$ where $e_1 = (2, 3, 5),$ $e_2 = (1, 0, 0),$ $e_3 = (0, 1, -1),$ and determine the matrix of $T$ relative to the new bases.
Solution. $\quad$ We recall from Exercise 5 that the linear transformation $T: V_3 \to V_3$ is such that \begin{align*} T(\mathbf{k}) &= 2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}, \quad T(\mathbf{j} + \mathbf{k}) = \mathbf{i}, \quad T(\mathbf{i} + \mathbf{j} + \mathbf{k}) = \mathbf{j} - \mathbf{k}. \end{align*} Then, with the new bases given by $(e_1, e_2, e_3),$ we have \begin{align*} T(e_1) &= T(2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) \\ &= 2T(\mathbf{i} + \mathbf{j} + \mathbf{k}) + T(\mathbf{j} + \mathbf{k}) + 2T(\mathbf{k}) \\ &= 2(2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) + (\mathbf{i}) + 2(\mathbf{j} - \mathbf{k}) \\ &= 2e_1 + e_2 + 2e_3 \\ \\ T(e_2) &= T(\mathbf{i}) \\ &= T(\mathbf{i} + \mathbf{j} + \mathbf{k}) - T(\mathbf{j} + \mathbf{k}) \\ &= (\mathbf{j} - \mathbf{k}) - (\mathbf{i}) \\ &= 0e_1 - e_2 + e_3 \\ \\ T(e_3) &= T(\mathbf{j} - \mathbf{k}) \\ &= T(\mathbf{j} + \mathbf{k}) - 2T(\mathbf{k}) \\ &= (\mathbf{i}) - 2(2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) \\ &= -2e_1 + e_2 + 0e_3 \end{align*} Then, if we set $t_{ik}$ as the component multiplying the $i^{th}$ basis element of $T(e_k)$ and place this value into the $i^{th}$ row and $k^{th}$ column, we get the following matrix of $T:$ \begin{align*} (t_{ik}) &= \begin{bmatrix} 2 & 0 & -2 \\ 1 & -1 & 1 \\ 2 & 1 & 0 \end{bmatrix} \quad \blacksquare \end{align*}