- 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
2.4 Exercises
Do the same as above in each of Exercises 16 through 23 if a transformation $T: V_3 \rightarrow V_3$ is defined by the formula given for $T(x, y, z)$, where $(x, y, z)$ is an arbitrary point of $V_3$.
20. $\quad$ $T(x, y, z) = (x + 1, y + 1, z - 1).$
Solution. $\quad$ Recall from Section 2.1 that if $V$ and $W$ are linear spaces, then a function $T: V \to W$ is called a linear transformation from $V$ into $W$ if it has the following two properties:
$\quad$ (a) $\quad T(x + y) = T(x) + T(y) \quad$ for all $x$ and $y$ in $V,$
$\quad$ (b) $\quad T(cx) = cT(x) \quad$ for all $x$ in $V$ and all scalars $c.$
$\quad$ Let $X = (x_1, x_2, x_3)$ and $Y = (y_1, y_2, y_3)$ be elements of $V_3.$ Then, if addition and scalar multiplication of elements is defined in the usual way, we have: \begin{align*} T(X) &= (x_1 + 1, x_2 + 1, x_3 - 1) \\ T(Y) &= (y_1 + 1, y_2 + 1, y_3 - 1) \\ T(X) + T(Y) &= (x_1 + y_1 + 2, x_2 + y_2 + 2, x_3 + y_3 - 2) \\ \\ T(X + Y) &= T(x_1 + y_1, x_2 + y_2, x_3 + y_3) \\ &= (x_1 + y_1 + 1, x_2 + y_2 + 1, x_3 + y_3 - 1) \\ &\neq T(X) + T(Y) \end{align*} Thus, we can see that $T$ is not a linear transformation. $\quad \blacksquare$