[proofplan] The standard representation remembers the action of an orthogonal matrix on every vector of $\mathbb{R}^n$. To prove injectivity, we take two orthogonal matrices whose images under $\rho$ are equal as linear maps and show that their columns agree. Since the columns of a matrix are exactly the images of the standard basis vectors, equality of the represented linear maps forces equality of the original matrices. [/proofplan] [step:Unpack equality of the standard representations] Let $A,B\in O(n)$ and suppose that $\rho(A)=\rho(B)$ in $GL(\mathbb{R}^n)$. By the definition of the standard representation, $\rho(A)$ and $\rho(B)$ are the linear maps \begin{align*} \rho(A):\mathbb{R}^n\to\mathbb{R}^n,\quad x\mapsto Ax \end{align*} and \begin{align*} \rho(B):\mathbb{R}^n\to\mathbb{R}^n,\quad x\mapsto Bx. \end{align*} Equality in $GL(\mathbb{R}^n)$ is equality as functions $\mathbb{R}^n\to\mathbb{R}^n$. Hence, for every $x\in\mathbb{R}^n$, \begin{align*} Ax=Bx. \end{align*} [guided] Let $A,B\in O(n)$ and assume $\rho(A)=\rho(B)$. The point is to interpret this equality in the codomain $GL(\mathbb{R}^n)$. An element of $GL(\mathbb{R}^n)$ is an invertible [linear map](/page/Linear%20Map) from $\mathbb{R}^n$ to itself, and equality in $GL(\mathbb{R}^n)$ means equality as maps. For the standard representation, the image of $A$ is the linear map \begin{align*} \rho(A):\mathbb{R}^n\to\mathbb{R}^n,\quad x\mapsto Ax. \end{align*} Similarly, the image of $B$ is the linear map \begin{align*} \rho(B):\mathbb{R}^n\to\mathbb{R}^n,\quad x\mapsto Bx. \end{align*} Thus the assumption $\rho(A)=\rho(B)$ says that these two functions have the same value at every vector $x\in\mathbb{R}^n$. Therefore, for every $x\in\mathbb{R}^n$, \begin{align*} Ax=Bx. \end{align*} This is the only property of the standard representation needed: it records the actual action of the matrix on each vector. [/guided] [/step] [step:Evaluate on the standard basis to recover the matrices] For each $i\in\{1,\dots,n\}$, let $e_i\in\mathbb{R}^n$ denote the $i$-th standard basis vector. Substituting $x=e_i$ into $Ax=Bx$ gives \begin{align*} Ae_i=Be_i. \end{align*} The vector $Ae_i$ is the $i$-th column of $A$, and the vector $Be_i$ is the $i$-th column of $B$. Hence the $i$-th columns of $A$ and $B$ are equal for every $i\in\{1,\dots,n\}$. Therefore all columns of $A$ and $B$ agree, so $A=B$ as matrices. [/step] [step:Conclude that the representation is injective] We have shown that whenever $A,B\in O(n)$ satisfy $\rho(A)=\rho(B)$, one has $A=B$. This is exactly the definition of injectivity of $\rho:O(n)\to GL(\mathbb{R}^n)$. Hence the standard representation is injective. [/step]