- 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.
12. $\quad$ $T$ maps each point onto its reflection with respect to a fixed line through the origin.
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 $N$ be the vector normal to the line and let $X$ and $Y$ be points in $V_2.$ Then, for some points $P$ and $Q$ on the line, we can express $X$ and $Y$ as the orthogonal decompositions: \begin{align*} X &= P + aN, \quad Y = Q + bN \end{align*} where $a$ and $b$ are scalars (It is easy to verify that all lines through the origin satisfy the closure axioms and are thus subspaces of $V_2$ with inner product $P \cdot Q$ for all $P$ and $Q$ on the line.)
Using the orthogonal decomposition of $X$ and $Y,$ we can express the reflection of $X$ and $Y$ across the line as: \begin{align*} T(X) &= P - a N, \quad T(Y) = Q - b N \end{align*} In other words, this transformation makes the orthogonal components of $X$ and $Y$ negative. Thus, with addition and scalar multiplication of elements in $V_2$ defined in the typical component-wise way, we have: \begin{align*} T(X + Y) &= T([P + Q + (a + b)N]) \\ &= (P + Q) - (a + b)N \\ &= P - a N + Q - b N \\ &= T(X) + T(Y) \\ \\ T(cY) &= T(cQ + cb N) \\ &= cQ - cb N \\ &= c(Q - b N) \\ &= cT(Y) \end{align*} As we can see, the transformation is linear. To find the null space of $T,$ we find $X = P + a N$ such that $T(X) = P - a N = O$ or in other words, a point $P$ on the line such that $P = a N.$ But the only scalar $a$ that satisfies this equation is $0,$ making $N(T) = \{O\}$ and $T(V_2) = V_2,$ giving $T$ nullity $0$ and rank $2. \quad \blacksquare$