$\textbf{Exercise 33.}$ Suppose $T \in \mathcal{L}(V)$ and $m$ is a positive integer. (a) Prove that $T$ is injective if and only if $T^m$ is injective. (b) Prove that $T$ is surjective if and only if $T^m$ is surjective. $\textbf{Solution 33.}$ (a) First suppose $T$ is injective. We will prove that $T^m$ is injective by induction on $m$. The desired result holds if $m = 1$. Thus suppose that $m > 1$ and the desired result holds for $m - 1$. Suppose $v \in V$ and $T^mv = 0$. Thus $\begin{equation*} T(T^{m-1}v) = 0. \end{equation*}$ Because $T$ is injective, the equation above implies that $T^{m-1}v = 0$. Our induction hypothesis now implies that $v = 0$, which proves that $T^m$ is injective, as desired. To prove the implication in the other direction, now suppose $T^m$ is injective. Suppose $v \in V$ and $Tv = 0$. Then $T^mv = 0$. Because $T^m$ is injective, this implies that $v = 0$. Thus $T$ is injective, as desired. (b) First suppose $T$ is surjective. We will prove that $T^m$ is surjective by induction on $m$. The desired result holds if $m = 1$. Thus suppose that $m > 1$ and the desired result holds for $m - 1$. Suppose $v \in V$. Because $T$ is surjective, there exists $u \in V$ such that $v = Tu$. By our induction hypothesis, $T^{m-1}$ is surjective. Hence there exists $w \in V$ such that $u = T^{m-1}w$. Thus $\begin{align*} v &= Tu \\ &= T(T^{m-1}w) \\ &= T^mw. \end{align*}$ Hence $T^m$ is surjective, as desired. To prove the implication in the other direction, now suppose $T^m$ is surjective. Suppose $v \in V$. Thus there exists $u \in V$ such that $v = T^mu$. Now $\begin{equation*} v = T^mu = T(T^{m-1}u). \end{equation*}$ Thus $T$ is surjective, as desired. ---  --- Commentary: Problem 33 (Section 5A): $\textbf{Exercise 33.}$ Suppose $T \in \mathcal{L}(V)$ and $m$ is a positive integer. (a) Prove that $T$ is injective if and only if $T^m$ is injective. (b) Prove that $T$ is surjective if and only if $T^m$ is surjective. $\textbf{Solution 33.}$ (a) First suppose $T$ is injective. We will prove that $T^m$ is injective by induction on $m$. The desired result holds if $m = 1$. Thus suppose that $m > 1$ and the desired result holds for $m - 1$. Suppose $v \in V$ and $T^mv = 0$. Thus $\begin{equation*} T(T^{m-1}v) = 0. \end{equation*}$ Because $T$ is injective, the equation above implies that $T^{m-1}v = 0$. Our induction hypothesis now implies that $v = 0$, which proves that $T^m$ is injective, as desired. To prove the implication in the other direction, now suppose $T^m$ is injective. Suppose $v \in V$ and $Tv = 0$. Then $T^mv = 0$. Because $T^m$ is injective, this implies that $v = 0$. Thus $T$ is injective, as desired. (b) First suppose $T$ is surjective. We will prove that $T^m$ is surjective by induction on $m$. The desired result holds if $m = 1$. Thus suppose that $m > 1$ and the desired result holds for $m - 1$. Suppose $v \in V$. Because $T$ is surjective, there exists $u \in V$ such that $v = Tu$. By our induction hypothesis, $T^{m-1}$ is surjective. Hence there exists $w \in V$ such that $u = T^{m-1}w$. Thus $\begin{align*} v &= Tu \\ &= T(T^{m-1}w) \\ &= T^mw. \end{align*}$ Hence $T^m$ is surjective, as desired. To prove the implication in the other direction, now suppose $T^m$ is surjective. Suppose $v \in V$. Thus there exists $u \in V$ such that $v = T^mu$. Now $\begin{equation*} v = T^mu = T(T^{m-1}u). \end{equation*}$ Thus $T$ is surjective, as desired. $\textit{Commentary:}$ This exercise establishes that injectivity and surjectivity are preserved under taking powers of a linear operator. Part (a) shows that $T$ is injective if and only if $T^m$ is injective for any positive integer $m$. The proof is by induction. The base case $m = 1$ is trivial. For the inductive step, we assume $T^{m-1}$ is injective and prove that $T^m$ is injective. This follows by applying $T$ to the equation $T^{m-1}v = 0$ and using the injectivity of $T$. The converse is proved by taking $m$-th roots. Part (b) shows that $T$ is surjective if and only if $T^m$ is surjective for any positive integer $m$. The proof is again by induction. For the inductive step, we use the surjectivity of $T$ and $T^{m-1}$ to find a preimage under $T^m$. These results are useful in the study of linear dynamics and the long-term behavior of iterative processes. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and define $T: \mathbb{R}^2 \to \mathbb{R}^2$ by $T(x, y) = (2x, y/2)$. Then $T$ is injective because if $T(x, y) = (0, 0)$, then $2x = 0$ and $y/2 = 0$, so $x = y = 0$. However, $T$ is not surjective because there is no $(x, y)$ such that $T(x, y) = (1, 1)$ (because $y/2 = 1$ has no solution). For any positive integer $m$, the $m$-th power of $T$ is given by $T^m(x, y) = (2^mx, y/2^m)$. This is injective for the same reason as $T$, but not surjective (because there is no $(x, y)$ such that $T^m(x, y) = (1, 1)$). On the other hand, if we define $S: \mathbb{R}^2 \to \mathbb{R}^2$ by $S(x, y) = (2x, 3y)$, then $S$ is surjective because for any $(a, b) \in \mathbb{R}^2$, we can find $(x, y) = (a/2, b/3)$ such that $S(x, y) = (a, b)$. However, $S$ is not injective because $S(1, 0) = S(0, 0)$ but $(1, 0) \neq (0, 0)$. For any positive integer $m$, the $m$-th power of $S$ is given by $S^m(x, y) = (2^mx, 3^my)$. This is surjective for the same reason as $S$, but not injective. ---