Mathematical Immaturity

2.4 Exercises

28. $\quad$ Let $V$ be the linear space of all real convergent sequences $\{x_n\}.$ Define a transformation $T:V\to V$ as follows: If $x=\{x_n\}$ is a convergent sequence with limit $a,$ let $T(x)=\{y_n\},$ where $y_n = a - x_n$ for $n\ge1.$ Prove that $T$ is linear and describe the null space and range of $T.$

Proof. $\quad$ 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 $x = \{x_n\}$ and $y = \{y_n\}$ be two elements of $V$ which converge to limits $a$ and $b,$ respectively. Since $x$ and $y$ are elements of $V,$ we know that $x + y = \{x_n + y_n\}$ and $cx = \{cx_n\}$ for any scalar $c,$ giving us: \begin{align*} T(x + y) &= \{a - x_n + b - y_n\} \\ &= \{a - x_n\} + \{b - y_n\} \\ &= T(x) + T(y) \\ \\ T(cx) &= \{ca - cx_n\} \\ &= c\{a - x_n\} \\ &= cT(x) \end{align*} We can see from this that $T$ is a linear transformation. Its null space is the set of all sequences $x = \{x_n\}$ in $V$ such that $x_n = a$ for $n = 1, 2, \dots,$ where $a$ is the limit of $x.$ In other words, the null space of $T$ is the set of all constant sequences in $V.$ The range of $T$ is the set of $x$ in $V$ such that for some $m, n \geq 1,$ $x_m \neq x_n.$ In other words, $T(V)$ is the set of non-constant convergent real sequences $\{y_n\} = \{a - x_n\}.$ But if $\{x_n\}$ has limit $a,$ then it follows that $\{y_n\} = \{a - x_n\}$ has limit $0.$ Thus, $T(V)$ is the set of nonzero sequences $y = \{y_n\}$ with limit $0. \quad \blacksquare$