- 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
In each of Exercises 1 through 10, a transformation $T: V_2 \rightarrow V_2$ is defined by the formula given for $T(x, y),$ where $(x, y)$ is an arbitrary point in $V_2$. In each case determine whether $T$ is linear. If $T$ is linear, describe its null space and range, and compute its nullity and rank.
9. $\quad$ $T(x, y) = (x - y, x + y).$
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 $c.$
Let $x = (x_0, x_1),$ $y = (y_0, y_1)$ be elements of $V_2,$ and let $c$ be any scalar. Then, $T(x) = (x_0 - x_1, x_0 + x_1)$ and $T(y) = (y_0 - y_1, y_0 + y_1).$ With $x$ and $y$ being elements of $V_2,$ we have component-wise addition and scalar multiplication of elements, giving us: \begin{align*} T(x + y) &= T(x_0 + y_0, x_1 + y_1) \\ &= (x_0 + y_0 - x_1 - y_1, x_0 + x_1 + y_0 + y_1) \\ &= (x_0 - x_1, x_0 + x_1) + (y_0 - y_1, y_0 + y_1) \\ &= T(x) + T(y) \\ \\ T(cx) &= T(cx_0, cx_1) \\ &= (cx_0 - cx_1, cx_0 + cx_1) \\ &= c(x_0 - x_1, x_0 + x_1) \\ &= cT(x) \end{align*} Thus, $T(x, y) = (x-y, x+y)$ is a linear transformation. Its null space is the set of elements $(x, y)$ such that $x - y = x + y = 0.$ In other words, $N(T)$ contains the pairs $(x, y)$ such that $x = -y = y.$ But, this is satisfied only by $(0, 0),$ making $N(T) = \{O\}.$ Thus, $T$ has nullity $0,$ and rank $2. \quad \blacksquare$