- 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
32. $\quad$ Refer to Exercise 28 of Section 2.4. Determine whether $T$ is one-to-one on $V.$ If it is, describe its inverse.
Solution. $\quad$ By definition, if a function $T$ is one-to-one on a set $V,$ then for any two elements $x$ and $y$ in $V,$ $x \neq y$ implies $T(x) \neq T(y).$
$\quad$ Referring to Exercise 28 of Section 2.4, we have $V$ as the linear space of all convergent sequences $\{x_n\}$ and the linear transformation $T: V \to V$ defined as follows:
$\quad$ If $x=\{x_n\}$ is a convergent sequence with limit $a,$ let $T(x)=\{y_n\},$ where $y_n = a - x_n$ for $n\ge1.$
$\quad$ We will now exhibit a counterexample to show that $T$ is not one-to-one on $V.$ Let $x = \{0\}$ and $y = \{1\}$ be two constant sequences in $V$ that converge to limits $a = 0$ and $b = 1,$ respectively. By definition, $x \neq y$ but we have:
\begin{align*}
T(x) &= \{a - x_n\}
\\
&= \{0 - 0\}
\\
&= \{0\}
\\
\\
T(y) &= \{b - y_n\}
\\
&= \{1 - 1\}
\\
&= \{0\}
\\
&= T(x)
\end{align*}
Thus, $x \neq y$ does not imply $T(x) \neq T(y)$ and $T$ is not one-to-one on $V. \quad \blacksquare$