Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if dim 𝑉 = 1 and 𝑇 ∈ ℒ(𝑉), then there exists 𝜆 ∈ 𝐅 such that 𝑇𝑣 = 𝜆𝑣 for all 𝑣 ∈ 𝑉. solution Suppose dim 𝑉 = 1 and 𝑇 ∈ ℒ(𝑉). Let 𝑢 be a nonzero vector in 𝑉. Then every vector in 𝑉 is a scalar multiple of 𝑢. In particular, 𝑇𝑢 = 𝜆𝑢 for some 𝜆 ∈ 𝐅. Now consider a typical vector 𝑣 ∈ 𝑉. There exists 𝑏 ∈ 𝐅 such that 𝑣 = 𝑏𝑢 . Thus 𝑇𝑣 = 𝑇(𝑏𝑢) = 𝑏𝑇(𝑢) = 𝑏(𝜆𝑢) = 𝜆(𝑏𝑢) = 𝜆𝑣. Edward Frenkel --- Commentary: $\textbf{Exercise 7.}$ Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\dim V = 1$ and $T \in \mathcal{L}(V)$, then there exists $\lambda \in \mathbb{F}$ such that $Tv = \lambda v$ for all $v \in V$. $\textbf{Solution 7.}$ Suppose $\dim V = 1$ and $T \in \mathcal{L}(V)$. Let $u$ be a nonzero vector in $V$. Then every vector in $V$ is a scalar multiple of $u$. In particular, $Tu = \lambda u$ for some $\lambda \in \mathbb{F}$. Now consider a typical vector $v \in V$. There exists $b \in \mathbb{F}$ such that $v = bu$. Thus$ \begin{align*} Tv &= T(bu) \\ &= bT(u) \\ &= b(\lambda u) \\ &= \lambda(bu) \\ &= \lambda v. \end{align*}$ $\textit{Commentary:}$ This exercise shows that linear maps on a one-dimensional vector space are very simple: they are just scalar multiplication by some fixed scalar. The proof relies on the fact that in a one-dimensional space, every vector is a scalar multiple of any nonzero vector. So we choose a nonzero vector $u$, and observe that $Tu$ must be a scalar multiple of $u$, say $Tu = \lambda u$. Then, for any other vector $v$, we write $v$ as a scalar multiple of $u$, say $v = bu$, and use the linearity of $T$ to compute: $Tv = T(bu) = bT(u) = b(\lambda u) = \lambda(bu) = \lambda v.$ This shows that $T$ acts as multiplication by $\lambda$ on every vector $v$. This result is a special case of the more general fact that linear maps on finite-dimensional vector spaces can be represented by matrices. In the one-dimensional case, the matrix is just a $1 \times 1$ matrix, i.e., a scalar. $\textit{Examples:}$ 1. Let $V$ be the space of polynomials of degree at most 0 over $\mathbb{R}$, i.e., the space of constant real polynomials. This is a one-dimensional space. Every linear map $T : V \to V$ is of the form $Tp = \lambda p$ for some fixed $\lambda \in \mathbb{R}$. 2. Let $V$ be the space of complex numbers $\mathbb{C}$, viewed as a one-dimensional vector space over $\mathbb{C}$. Every linear map $T : \mathbb{C} \to \mathbb{C}$ is of the form $Tz = \lambda z$ for some fixed $\lambda \in \mathbb{C}$. 3. Let $V$ be any one-dimensional vector space over a field $\mathbb{F}$, and let $u$ be a nonzero vector in $V$. Every vector in $V$ can be uniquely written as $au$ for some $a \in \mathbb{F}$. Every linear map $T : V \to V$ is of the form $T(au) = \lambda(au)$ for some fixed $\lambda \in \mathbb{F}$. These examples illustrate the principle in various one-dimensional spaces, including polynomial spaces, the complex numbers, and abstract one-dimensional spaces. In each case, once we fix a nonzero vector $u$, every linear map is determined by where it sends $u$, which must be a scalar multiple of $u$. This scalar is the $\lambda$ that characterizes the map.