- 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 11 through 15, a transformation $T: V_2 \rightarrow V_2$ is described as indicated. In each case determine whether $T$ is linear. If $T$ is linear, describe its null space and range, and compute its nullity and rank.
14. $\quad$ $T$ maps each point with polar coordinates $(r, \theta)$ onto the point with polar coordinates $(2r, \theta)$. Also, $T$ maps $O$ onto itself.
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 $X$ and $Y$ be two elements of $V_2$ with polar coordinates $(r_x, \theta_x)$ and $(r_y, \theta_y),$ respectively. $X$ and $Y$ can be expressed in terms of their rectangular coordinates as: \begin{align*} X &= r_x(\cos \theta_x, \sin \theta_x), \quad Y = r_y(\cos \theta_y, \sin \theta_y) \end{align*} Then, with addition of elements defined component-wise for elements in $V_2,$ we have: \begin{align*} \\ T(X + Y) &= T\left[(r_x \cos \theta_x + r_y \cos \theta_y, r_x \sin \theta_x + r_y \sin \theta_y) \right] \\ &= (2r_x \cos \theta_x + 2r_y \cos \theta_y, 2r_x \sin \theta_x + 2r_y \sin \theta_y) \\ &= 2r_x(\cos \theta_x, \sin \theta_x) + 2r_y(\cos \theta_y, \sin \theta_y) \\ &= T(X) + T(Y) \\ \\ T(cX) &= T\left[cr_x(\cos \theta_x, \sin \theta_x)\right] \\ &= 2cr_x(\cos \theta_x, \sin \theta_x) \\ &= cT(X) \end{align*} As we can see, the transformation is linear. To find the nullity of $T,$ we find the set of elements $X$ in $V_2$ such that $T(X) = O.$ But for $T(X) = O,$ it must the case that $r_x = 0,$ which implies that $X = O.$ And since $T$ maps $O$ to itself, we have $N(T) = \{O\}$ which gives $T$ nullity $0.$ And since the range of $T$ (that is, $V_2$) is spanned by the independent set $\{(1, 0), (0, 1)\},$ $T$ has rank $2. \quad \blacksquare$