from Claude --- $\textbf{Exercise 3.}$ Suppose $V$ is finite-dimensional and $T \in \mathcal{L}(V)$. Prove that the following are equivalent: $ \begin{align*} \text{(a) } & T \text{ is invertible.} \\ \text{(b) } & Tv_1, \ldots, Tv_n \text{ is a basis of } V \text{ for every basis } v_1, \ldots, v_n \text{ of } V. \\ \text{(c) } & Tv_1, \ldots, Tv_n \text{ is a basis of } V \text{ for some basis } v_1, \ldots, v_n \text{ of } V. \end{align*}$ $\textbf{Solution 3.}$ First suppose (a) holds, so $T$ is invertible. Suppose $v_1, \ldots, v_n$ is a basis of $V$. To show that $Tv_1, \ldots, Tv_n$ is linearly independent, suppose $c_1, \ldots, c_n \in \mathbb{F}$ are such that $\begin{equation*} c_1Tv_1 + \cdots + c_nTv_n = 0. \end{equation*}$ Then $\begin{equation*} T(c_1v_1 + \cdots + c_nv_n) = 0. \end{equation*}$ Because $T$ is invertible, the equation above implies that $\begin{equation*}c_1v_1 + \cdots + c_nv_n = 0.\end{equation*}$ Because $v_1, \ldots, v_n$ is a basis of $V$, the equation above implies that $c_1 = \cdots = c_n = 0$. Thus the list $Tv_1, \ldots, Tv_n$ is linearly independent and hence is a basis of $V$, completing the proof that (a) implies (b). Clearly (b) implies (c). Now suppose (c) holds. Thus there exists a basis $v_1, \ldots, v_n$ of $V$ such that $Tv_1, \ldots, Tv_n$ is a basis of $V$. Suppose $v \in V$ is such that $Tv = 0$. There exist $c_1, \ldots, c_n \in \mathbb{F}$ such that $\begin{equation*} v = c_1v_1 + \cdots + c_nv_n. \end{equation*}$ Because $Tv = 0$, the equation above implies that $\begin{equation*} 0 = c_1Tv_1 + \cdots + c_nTv_n. \end{equation*}$ Because $Tv_1, \ldots, Tv_n$ is a basis of $V$, the equation above implies that $c_1 = \cdots = c_n = 0$. Hence $v = 0$. This shows that $T$ is injective and thus is invertible, completing the proof that (c) implies (a). --- End of version 3 $\textbf{Exercise 3.}$ Suppose $V$ is finite-dimensional and $T \in \mathcal{L}(V)$. Prove that the following are equivalent: $\begin{align*} \text{(a) } & T \text{ is invertible.} \\ \text{(b) } & Tv_1, \ldots, Tv_n \text{ is a basis of } V \text{ for every basis } v_1, \ldots, v_n \text{ of } V. \\ \text{(c) } & Tv_1, \ldots, Tv_n \text{ is a basis of } V \text{ for some basis } v_1, \ldots, v_n \text{ of } V. \end{align*}$ $\textbf{Solution 3.}$ First suppose (a) holds, so $T$ is invertible. Suppose $v_1, \ldots, v_n$ is a basis of $V$. To show that $Tv_1, \ldots, Tv_n$ is linearly independent, suppose $c_1, \ldots, c_n \in \mathbb{F}$ are such that $\begin{equation*} c_1Tv_1 + \cdots + c_nTv_n = 0. \end{equation*}$ Then $\begin{equation*} T(c_1v_1 + \cdots + c_nv_n) = 0. \end{equation*}$ Because $T$ is invertible, the equation above implies that $\begin{equation*} c_1v_1 + \cdots + c_nv_n = 0. \end{equation*}$ Because $v_1, \ldots, v_n$ is a basis of $V$, the equation above implies that $c_1 = \cdots = c_n = 0$. Thus the list $Tv_1, \ldots, Tv_n$ is linearly independent and hence is a basis of $V$, completing the proof that (a) implies (b). Clearly (b) implies (c). Now suppose (c) holds. Thus there exists a basis $v_1, \ldots, v_n$ of $V$ such that $Tv_1, \ldots, Tv_n$ is a basis of $V$. Suppose $v \in V$ is such that $Tv = 0$. There exist $c_1, \ldots, c_n \in \mathbb{F}$ such that $\begin{equation*} v = c_1v_1 + \cdots + c_nv_n. \end{equation*}$ Because $Tv = 0$, the equation above implies that $\begin{equation*} 0 = c_1Tv_1 + \cdots + c_nTv_n. \end{equation*}$ Because $Tv_1, \ldots, Tv_n$ is a basis of $V$, the equation above implies that $c_1 = \cdots = c_n = 0$. Hence $v = 0$. This shows that $T$ is injective and thus is invertible, completing the proof that (c) implies (a). --- ![[sol-5.pdf#page=3]]