[proofplan] The affine relation $X = a1_n + bX^*$ lets every linear predictor written with $X$ be rewritten uniquely as a linear predictor written with $X^*$, with the same fitted vector. We apply this change of parameters to the original least squares minimizer and show that it achieves the minimum value for the transformed problem. Full column rank of the transformed design matrix gives uniqueness of the transformed coefficient vector, so the transformed coefficients must be exactly the rewritten coefficients. [/proofplan] [step:Rewrite the original fitted predictor using the rescaled covariate] Since $X^* = (X-a1_n)/b$ and $b \neq 0$, we have \begin{align*} X = a1_n + bX^*. \end{align*} Define the transformed coefficient triple $(\tilde{\alpha}^*,\tilde{\beta}^*,\tilde{\gamma}^*) \in \mathbb{R} \times \mathbb{R} \times \mathbb{R}^m$ by \begin{align*} \tilde{\alpha}^* := \hat{\alpha} + a\hat{\beta}_X, \qquad \tilde{\beta}^* := b\hat{\beta}_X, \qquad \tilde{\gamma}^* := \hat{\gamma}. \end{align*} Substituting $X = a1_n + bX^*$ into the original fitted predictor gives \begin{align*} \hat{\alpha}1_n + \hat{\beta}_X X + Z\hat{\gamma} &= \hat{\alpha}1_n + \hat{\beta}_X(a1_n + bX^*) + Z\hat{\gamma} \\ &= (\hat{\alpha} + a\hat{\beta}_X)1_n + (b\hat{\beta}_X)X^* + Z\hat{\gamma} \\ &= \tilde{\alpha}^*1_n + \tilde{\beta}^*X^* + Z\tilde{\gamma}^*. \end{align*} Thus the original fitted vector is attainable in the transformed regression. [/step] [step:Transfer minimality from the original problem to the transformed problem] Let $(\alpha^*,\beta^*,\gamma^*) \in \mathbb{R} \times \mathbb{R} \times \mathbb{R}^m$ be arbitrary. Define $(\alpha,\beta,\gamma) \in \mathbb{R} \times \mathbb{R} \times \mathbb{R}^m$ by \begin{align*} \alpha := \alpha^* - \frac{a}{b}\beta^*, \qquad \beta := \frac{\beta^*}{b}, \qquad \gamma := \gamma^*. \end{align*} Using $X^* = (X-a1_n)/b$, we compute \begin{align*} \alpha^*1_n + \beta^*X^* + Z\gamma^* &= \alpha^*1_n + \beta^*\frac{X-a1_n}{b} + Z\gamma^* \\ &= \left(\alpha^* - \frac{a}{b}\beta^*\right)1_n + \frac{\beta^*}{b}X + Z\gamma^* \\ &= \alpha1_n + \beta X + Z\gamma. \end{align*} Since $(\hat{\alpha},\hat{\beta}_X,\hat{\gamma})$ minimizes the original residual sum of squares, this gives \begin{align*} |Y - \hat{\alpha}1_n - \hat{\beta}_X X - Z\hat{\gamma}|^2 \leq |Y - \alpha1_n - \beta X - Z\gamma|^2 = |Y - \alpha^*1_n - \beta^*X^* - Z\gamma^*|^2. \end{align*} By the equality from the previous step, \begin{align*} |Y - \tilde{\alpha}^*1_n - \tilde{\beta}^*X^* - Z\tilde{\gamma}^*|^2 = |Y - \hat{\alpha}1_n - \hat{\beta}_X X - Z\hat{\gamma}|^2. \end{align*} Therefore $(\tilde{\alpha}^*,\tilde{\beta}^*,\tilde{\gamma}^*)$ minimizes the transformed least squares objective. [/step] [step:Use full column rank to identify the transformed coefficients] Because $A^* = [\,1_n \ \ X^* \ \ Z\,]$ has full column rank, the [linear map](/page/Linear%20Map) \begin{align*} T^*: \mathbb{R}^{m+2} &\to \mathbb{R}^n \\ (\alpha^*,\beta^*,\gamma^*) &\mapsto \alpha^*1_n + \beta^*X^* + Z\gamma^* \end{align*} is injective. Hence the transformed least squares objective has at most one minimizer as a coefficient vector, and by hypothesis this minimizer is $(\hat{\alpha}^*,\hat{\beta}_{X^*},\hat{\gamma}^*)$. The previous step showed that $(\tilde{\alpha}^*,\tilde{\beta}^*,\tilde{\gamma}^*)$ is also a transformed minimizer. Therefore \begin{align*} (\hat{\alpha}^*,\hat{\beta}_{X^*},\hat{\gamma}^*) = (\tilde{\alpha}^*,\tilde{\beta}^*,\tilde{\gamma}^*) = (\hat{\alpha}+a\hat{\beta}_X,\ b\hat{\beta}_X,\ \hat{\gamma}). \end{align*} Consequently \begin{align*} \hat{\beta}_{X^*} = b\hat{\beta}_X, \qquad \hat{\gamma}^* = \hat{\gamma}. \end{align*} Using the equality of predictors from the first step with these identified coefficients yields \begin{align*} \hat{\alpha}^*1_n + \hat{\beta}_{X^*}X^* + Z\hat{\gamma}^* = \hat{\alpha}1_n + \hat{\beta}_X X + Z\hat{\gamma}. \end{align*} This proves that the coefficient on the rescaled covariate is multiplied by $b$, the coefficients on $Z$ are unchanged, and the fitted values are unchanged. [/step]