[proofplan] The proof is just the quotient definition of [projective space](/page/Projective%20Space) written out in coordinates. We denote the two nonzero vectors by $a$ and $b$, recall that $\mathbb{P}^n_k$ is the quotient of $k^{n+1}\setminus\{0\}$ by multiplication by elements of $k^\times$, and translate equality of quotient classes into the defining [equivalence relation](/page/Equivalence%20Relation). The forward direction unpacks equality in the quotient, and the reverse direction repacks a nonzero scalar multiple relation as equality of homogeneous coordinate points. [/proofplan] [step:Unpack the quotient relation defining projective space] Define the vectors $a \in k^{n+1}\setminus\{0\}$ and $b \in k^{n+1}\setminus\{0\}$ by $a=(a_0,\ldots,a_n)$ and $b=(b_0,\ldots,b_n)$. By definition, $\mathbb{P}^n_k$ is the quotient of $k^{n+1}\setminus\{0\}$ by the equivalence relation $\sim$ given by \begin{align*} x \sim y \quad \Longleftrightarrow \quad \text{there exists } \mu \in k^\times \text{ such that } y=\mu x. \end{align*} The homogeneous coordinate symbols $[a_0:\cdots:a_n]$ and $[b_0:\cdots:b_n]$ denote the equivalence classes of $a$ and $b$ under this relation. [guided] Let us name the two representatives: $a=(a_0,\ldots,a_n)$ and $b=(b_0,\ldots,b_n)$. Both vectors lie in $k^{n+1}\setminus\{0\}$ by hypothesis, so they are valid representatives of projective points. The definition of projective space is the quotient \begin{align*} \mathbb{P}^n_k=(k^{n+1}\setminus\{0\})/{\sim}, \end{align*} where the relation $\sim$ identifies two nonzero vectors exactly when one is obtained from the other by multiplying by a nonzero scalar. More explicitly, for $x,y \in k^{n+1}\setminus\{0\}$, \begin{align*} x \sim y \quad \Longleftrightarrow \quad \text{there exists } \mu \in k^\times \text{ such that } y=\mu x. \end{align*} The restriction $\mu \in k^\times$ matters: multiplying by $0$ would send every vector to $0$, which is not an allowed representative in projective space. Under this quotient notation, the homogeneous coordinate point $[a_0:\cdots:a_n]$ is the equivalence class of $a$, and $[b_0:\cdots:b_n]$ is the equivalence class of $b$. [/guided] [/step] [step:Translate equality of classes into a nonzero scalar multiple] Assume $[a_0:\cdots:a_n]=[b_0:\cdots:b_n]$. By the quotient definition just recalled, the representatives $a$ and $b$ lie in the same equivalence class, so $a \sim b$. Hence there exists $\lambda \in k^\times$ such that $b=\lambda a$. Equality of vectors in $k^{n+1}$ is coordinatewise equality, so for every integer $i$ with $0 \leq i \leq n$, \begin{align*} b_i=\lambda a_i. \end{align*} [/step] [step:Convert a nonzero scalar multiple back into equality of projective points] Conversely, assume there exists $\lambda \in k^\times$ such that $b_i=\lambda a_i$ for every integer $i$ with $0 \leq i \leq n$. Then, by coordinatewise equality in $k^{n+1}$, $b=\lambda a$. Since $\lambda \in k^\times$, this means $a \sim b$ under the scalar-multiplication equivalence relation. Therefore $a$ and $b$ determine the same equivalence class in $\mathbb{P}^n_k$, which is precisely \begin{align*} [a_0:\cdots:a_n]=[b_0:\cdots:b_n]. \end{align*} This proves both implications. [/step]