[proofplan] We translate identifiability into a statement about the kernel of the [linear map](/page/Linear%20Map) $\beta \mapsto X\beta$. A linear map is injective exactly when its kernel is $\{0\}$. The kernel consists precisely of the linear relations among the columns of $X$, so the kernel equals $\{0\}$ exactly when the columns are linearly independent. Since the rank of $X$ is the dimension of the span of its columns, this is equivalent to $\operatorname{rank}(X)=p$. [/proofplan] [step:Identify injectivity with trivial kernel] Let \begin{align*} T_X: \mathbb{R}^p &\to \operatorname{Range}(X) \\ \beta &\mapsto X\beta \end{align*} be the linear map in the statement. Its kernel is \begin{align*} \ker T_X = \{\beta \in \mathbb{R}^p : X\beta = 0\}. \end{align*} The map $T_X$ is injective if and only if $\ker T_X = \{0\}$. Indeed, if $T_X$ is injective and $\beta \in \ker T_X$, then $T_X(\beta)=0=T_X(0)$, hence $\beta=0$. Conversely, if $\ker T_X=\{0\}$ and $T_X(\beta_1)=T_X(\beta_2)$ for $\beta_1,\beta_2 \in \mathbb{R}^p$, then linearity gives \begin{align*} T_X(\beta_1-\beta_2)=T_X(\beta_1)-T_X(\beta_2)=0. \end{align*} Thus $\beta_1-\beta_2 \in \ker T_X=\{0\}$, so $\beta_1=\beta_2$. Hence $T_X$ is injective. [/step] [step:Relate the kernel to linear independence of the columns] Let $x_1,\dots,x_p \in \mathbb{R}^n$ denote the columns of $X$. For every $\beta=(\beta_1,\dots,\beta_p) \in \mathbb{R}^p$, matrix multiplication gives \begin{align*} X\beta=\sum_{j=1}^{p}\beta_j x_j. \end{align*} Therefore \begin{align*} \beta \in \ker T_X \iff \sum_{j=1}^{p}\beta_j x_j=0. \end{align*} Thus $\ker T_X=\{0\}$ if and only if the only linear relation among $x_1,\dots,x_p$ has all coefficients equal to zero. This is exactly the statement that the columns $x_1,\dots,x_p$ are linearly independent. [/step] [step:Convert column independence into full column rank] By definition, \begin{align*} \operatorname{rank}(X)=\dim \operatorname{span}\{x_1,\dots,x_p\}. \end{align*} The $p$ columns $x_1,\dots,x_p$ are linearly independent if and only if their span has dimension $p$. Hence \begin{align*} x_1,\dots,x_p \text{ are linearly independent} \iff \operatorname{rank}(X)=p. \end{align*} Combining this equivalence with the previous two steps gives \begin{align*} T_X \text{ is injective} \iff \ker T_X=\{0\} \iff x_1,\dots,x_p \text{ are linearly independent} \iff \operatorname{rank}(X)=p. \end{align*} Therefore $\beta \mapsto X\beta$ is injective from $\mathbb{R}^p$ to $\operatorname{Range}(X)$ if and only if $X$ has full column rank. [/step]