- 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 $V_3$.
17. $\quad$ $T(x, y, z) = (x, y, 0).$
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, with addition and scalar multiplication defined component-wise, we have: \begin{align*} T(X + Y) &= T(x_1 + y_1, x_2 + y_2, x_3 + y_3) \\ &= (x_1 + y_1, x_2 + y_2, 0) \\ &= (x_1, x_2, 0) + (y_1, y_2, 0) \\ &= T(X) + T(Y) \\ \\ T(cX) &= T(cx_1, cx_2, cx_3) \\ &= (cx_1, cx_2, 0) \\ &= cT(X) \end{align*} As we can see, $T(x, y, z) = (x, y, 0)$ is a linear transformation. The null space is the set of all elements in $V_3$ whose first and second components are zero. Thus, $N(T)$ is spanned by $\{(0, 0, 1)\},$ giving $T$ nullity $1.$ The range of $T$ is the set of all elements in $V_3$ whose third component is zero, which means $T(V)$ is spanned by the independent set $\{(1, 0, 0), (0, 1, 0)\},$ giving $T$ rank $2. \quad \blacksquare$