[proofplan] Choose a nonzero vector $v\in L$, which is possible because $L$ is one-dimensional. If $L$ is invariant, then $T(v)\in L$, so one-dimensionality writes $T(v)$ uniquely as a scalar multiple of $v$; linearity then shows that the same scalar works for every nonzero vector in $L$. Conversely, if all nonzero vectors in $L$ are multiplied by the same scalar $\lambda$, then $T$ sends each nonzero vector of $L$ back into $L$, and linearity sends $0$ to $0$, so $T(L)\subset L$. [/proofplan] [step:Extract the scalar action from invariance of the line] Assume first that $L$ is invariant under $T$, so $T(L)\subset L$. Since $L$ is one-dimensional, choose a nonzero vector $v\in L$. The singleton $\{v\}$ is a basis of $L$, so there exists a unique scalar $\lambda\in k$ such that \begin{align*}T(v)=\lambda v.\end{align*} Now let $w\in L$ be nonzero. Since $\{v\}$ is a basis of $L$, there exists a scalar $a\in k$ such that \begin{align*} w=av. \end{align*} Because $w\ne 0$ and $v\ne 0$, this scalar satisfies $a\ne 0$. By linearity of $T$, \begin{align*} T(w)=T(av)=aT(v)=a\lambda v=\lambda av=\lambda w. \end{align*} Thus every nonzero vector in $L$ is an eigenvector of $T$ with the same eigenvalue $\lambda$. [guided] Assume that $L$ is invariant under $T$, meaning precisely that $T(L)\subset L$. Since $L$ is one-dimensional, it contains a nonzero vector; choose one and call it $v\in L$. A one-dimensional subspace is spanned by any of its nonzero vectors, so $\{v\}$ is a basis of $L$. The invariance hypothesis now gives the essential point: because $v\in L$, we have $T(v)\in T(L)\subset L$. Since $\{v\}$ is a basis of $L$, the vector $T(v)$ has a unique coordinate in this basis. Therefore there exists a unique scalar $\lambda\in k$ such that \begin{align*}T(v)=\lambda v.\end{align*} We must prove that this same scalar $\lambda$ works for every nonzero vector in $L$, not only for the chosen vector $v$. Let $w\in L$ be nonzero. Since $\{v\}$ is a basis of $L$, there exists a scalar $a\in k$ such that \begin{align*} w=av. \end{align*} The scalar $a$ cannot be zero, because $a=0$ would give $w=0v=0$, contradicting the choice of $w$ as nonzero. Using linearity of $T$ and the already established identity $T(v)=\lambda v$, we compute \begin{align*}T(w)=T(av)=aT(v)=a\lambda v=\lambda av=\lambda w.\end{align*} Thus every nonzero vector $w\in L$ satisfies $T(w)=\lambda w$. The scalar $\lambda$ is independent of $w$, so all nonzero vectors in $L$ are eigenvectors of $T$ with the same eigenvalue. [/guided] [/step] [step:Use a common eigenvalue to prove invariance of the line] Conversely, assume that there exists a scalar $\lambda\in k$ such that every nonzero vector $w\in L$ satisfies \begin{align*}T(w)=\lambda w.\end{align*} Let $x\in L$. If $x\ne 0$, then $T(x)=\lambda x\in L$ because $L$ is a subspace and is closed under scalar multiplication. If $x=0$, then linearity gives \begin{align*}T(x)=T(0)=0\in L.\end{align*} Therefore $T(x)\in L$ for every $x\in L$, which is exactly $T(L)\subset L$. Hence $L$ is invariant under $T$. [/step]