- 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
2.8 Exercises
26.$\quad$ Let $S$ and $T$ be the linear transformations of $V_3$ into $V_3$ defined by the formulas $S(x, y, z) = (z, y, x)$ and $T(x, y, z) = (x, x + y, x + y + z),$ where $(x, y, z)$ is an arbitrary point of $V_3.$
(a) $\quad$ Determine the image of $(x, y, z)$ under each of the following transformations: $ST,$ $TS,$ $ST - TS,$ $S^2,$ $T^2,$ $(ST)^2,$ $(TS)^2,$ $(ST - TS)^2.$
(b) $\quad$ Prove that $S$ and $T$ are one-to-one on $V_3$ and find the image of $(u, v, w)$ under each of the following transformations: $S^{-1},$ $T^{-1},$ $(ST)^{-1},$ $(TS)^{-1}.$
(c) $\quad$ Find the image of $(x, y, z)$ under $(T^{-1})^n$ for each $n \geq 1.$
Solution.
(a) $\quad$ Determine the image of $(x, y, z)$ under each of the following transformations: $ST,$ $TS,$ $ST - TS,$ $S^2,$ $T^2,$ $(ST)^2,$ $(TS)^2,$ $(ST - TS)^2.$
\begin{align*}
ST(x, y, z) &= S[T(x, y, z)]
\\
&= S(x, x + y, x + y + z)
\\
&= (x + y + z, x + y, x)
\\
\\
TS(x, y, z) &= T[S(x, y, z)]
\\
&= T(z, y, x)
\\
&= (z, z + y, z + y + x)
\\
\\
(ST - TS)(x, y, z) &= ST(x, y, z) - TS(x, y, z)
\\
&= (x + y + z, x + y, x) - (z, z + y, z + y + x)
\\
&= (x + y, x - z, - y - z)
\\
\\
S^2(x, y, z) &= S[S(x, y, z)]
\\
&= S(z, y, x)
\\
&= (x, y, z)
\\
\\
T^2(x, y, z) &= T[T(x, y, z)]
\\
&= T(x, x + y, x + y + z)
\\
&= (x, 2x + y, 3x + 2y + z)
\\
\\
(ST)^2(x, y, z) &= ST[ST(x, y, z)]
\\
&= ST(x + y + z, x + y, x)
\\
&= S(x + y + z, 2x + 2y + z, 3x + 2y + z)
\\
&= (3x + 2y + z, 2x + 2y + z, x + y + z)
\\
\\
(TS)^2(x, y, z) &= TS[TS(x, y, z)]
\\
&= TS(z, z + y, z + y + x)
\\
&= T(z + y + x, z + y, z)
\\
&= (x + y + z, x + 2y + 2z, x + 2y + 3z)
\\
\\
(ST - TS)^2(x, y, z) &= (ST - TS)[(ST - TS)(x, y, z)]
\\
&= (ST - TS)(x + y, x - z, -y - z)
\\
&= (2x + y - z, x + 2y + z, 2z + y - x)
\quad \blacksquare
\end{align*}
(b) $\quad$ Prove that $S$ and $T$ are one-to-one on $V_3$ and find the image of $(u, v, w)$ under each of the following transformations: $S^{-1},$ $T^{-1},$ $(ST)^{-1},$ $(TS)^{-1}.$
$\quad$ Proof. $\quad$ Referring to Theorem 2.9, a function $T: V \to W$ is said to be one-to-one if for all $x$ and $y$ in $V,$ $T(x) = T(y)$ implies $x = y.$ Now, let $(x, y, z)$ and $(p, q, r)$ be elements of $V_3$ with transformations
\begin{align*}
S(x, y, z) &= (z, y, x)
\\
S(p, q, r) &= (r, p, q)
\\
\\
T(x, y, z) &= (x, x + y, x + y + z)
\\
T(p, q, r) &= (p, p + q, p + q + r)
\end{align*}
If $S(x, y, z) = S(p, q, r),$ this implies that $x = p,$ $y = q,$ and $z = r,$ or $(x, y, z) = (p, q, r).$ Hence, $S$ is one-to-one on $V_3.$ If $T(x, y, z) = T(p, q, r),$ we have $x = p,$ $x + y = p + q,$ and $x + y + z = p + q + r.$ In other words, we have $x = p,$ $y = q,$ and $z = r,$ making $T$ one-to-one on $V_3.$
$\quad$ Now, let $(u, v, w)$ be an element of $V_3$ given by $(u, v, w) = S(x, y, z).$ We then have
\begin{align*}
(u, v, w) &= (z, y, x)
\\
S^{-1}(u, v, w) &= (w, v, u)
\end{align*}
If $(u, v, w) = T(x, y, z),$ we have
\begin{align*}
(u, v, w) &= (x, x + y, x + y + z)
\\
T^{-1}(u, v, w) &= (u, v - u, w - v)
\end{align*}
If $(u, v, w) = ST(x, y, z),$ we have
\begin{align*}
(u, v, w) &= S[T(x, y, z)]
\\
&= S(x, x + y, x + y + z)
\\
&= (x + y + z, x + y, x)
\\
(ST)^{-1}(u, v, w) &= T^{-1}S^{-1}(u, v, w)
\\
&= T^{-1}(w, v, u)
\\
&= (w, v - w, u - v)
\end{align*}
If $(u, v, w) = (TS)(x, y, z),$ we have
\begin{align*}
(u, v, w) &= T[S(x, y, z)]
\\
&= T(z, y, x)
\\
&= (z, z + y, z + y + x)
\\
(TS)^{-1}(u, v, w) &= S^{-1}T^{-1}(u, v, w)
\\
&= S^{-1}(u, v - u, w - v)
\\
&= (w - v, v - u, u) \quad \blacksquare
\end{align*}
(c) $\quad$ Find the image of $(x, y, z)$ under $(T - I)^n$ for each $n \geq 1.$ \begin{align*} (T - I)(x, y, z) &= T(x, y, z) - I(x, y, z) \\ &= (x, x + y, x + y + z) - (x, y, z) \\ &= (0, x, x + y) \\ \\ (T - I)^2(x, y, z) &= (T-I)(T-I)(x, y, z) \\ &=(T - I)(0, x, x + y) \\ &= (0, x, 2x + y) - (0, x, x + y) \\ &= (0, 0, x) \\ \\ (T - I)^3(x, y, z) &= (T - I)(T - I)^2(x, y, z) \\ &= (T - I)(0, 0, x) \\ &= (0, 0, x) - (0, 0, x) \\ &= (0, 0, 0) \\ \\ (T - I)^n(x, y, z)&= (T - I)(0, 0, 0) \\ &= (0, 0, 0) \quad \text{for $n \gt 3$} \quad \blacksquare \end{align*}