[proofplan] We unpack equality in [projective space](/page/Projective%20Space) as equality of equivalence classes of nonzero vectors under multiplication by nonzero scalars. Thus $[v] = [v']$ gives a scalar $\lambda \in K^\times$ with $v' = \lambda v$. Linearity carries this scalar relation through $T$, while injectivity ensures that the images of nonzero representatives are still nonzero. Therefore $T(v)$ and $T(v')$ determine the same point of $\mathbb{P}(W)$. [/proofplan] [step:Unpack equality of projective classes in $\mathbb{P}(V)$] Let $v, v' \in V \setminus \{0\}$ and assume $[v] = [v']$ in $\mathbb{P}(V)$. By the definition of $\mathbb{P}(V)$ as the quotient of $V \setminus \{0\}$ by nonzero scalar multiplication, equality of the two projective classes means that there exists a scalar $\lambda \in K^\times$ such that \begin{align*} v' = \lambda v. \end{align*} [guided] Let $v, v' \in V \setminus \{0\}$ and suppose $[v] = [v']$ in $\mathbb{P}(V)$. The notation $[v]$ denotes the equivalence class of the nonzero vector $v$ under the relation identifying two nonzero vectors when they differ by multiplication by a nonzero scalar. By the definition of $\mathbb{P}(V)$ in the theorem statement, the equality $[v] = [v']$ is precisely the statement that $v$ and $v'$ lie in the same equivalence class. Therefore there is a scalar $\lambda \in K^\times$ satisfying \begin{align*} v' = \lambda v. \end{align*} The condition $\lambda \in K^\times$ is essential: projective equivalence uses only nonzero scalar multiples, so the scalar relating two representatives is invertible in $K$. [/guided] [/step] [step:Use injectivity to show the images are nonzero] Let $0_V$ denote the zero vector of $V$ and let $0_W$ denote the zero vector of $W$. Since $T$ is $K$-linear, $T(0_V) = 0_W$. If $T(v) = 0_W$, then $T(v) = T(0_V)$, and injectivity of $T$ gives $v = 0_V$, contradicting $v \in V \setminus \{0\}$. Hence $T(v) \neq 0_W$. The same argument applied to $v'$ gives $T(v') \neq 0_W$. Thus $T(v)$ and $T(v')$ are valid representatives of points in $\mathbb{P}(W)$. [/step] [step:Transport the scalar relation through the linear map] Since $T$ is $K$-linear and $v' = \lambda v$, we have \begin{align*} T(v') = T(\lambda v). \end{align*} By homogeneity of the $K$-[linear map](/page/Linear%20Map) $T$, \begin{align*} T(\lambda v) = \lambda T(v). \end{align*} Therefore \begin{align*} T(v') = \lambda T(v). \end{align*} Because $\lambda \in K^\times$, the nonzero vectors $T(v)$ and $T(v')$ differ by multiplication by a nonzero scalar. [/step] [step:Conclude equality in $\mathbb{P}(W)$] The defining [equivalence relation](/page/Equivalence%20Relation) on $W \setminus \{0\}$ identifies two nonzero vectors when one is a nonzero scalar multiple of the other. We have shown that $T(v), T(v') \in W \setminus \{0\}$ and that \begin{align*} T(v') = \lambda T(v) \end{align*} for some $\lambda \in K^\times$. Hence $T(v)$ and $T(v')$ represent the same projective point in $\mathbb{P}(W)$, so \begin{align*} [T(v)] = [T(v')]. \end{align*} This proves that the projective class of $T(v)$ depends only on the projective class of $v$. [/step]