[proofplan] We isolate the part of the predictor $x_j$ that is orthogonal to the intercept and the remaining predictors. This residual vector $z_j$ is the only direction in the design that can identify the coefficient $\beta_j$ after the other columns have been accounted for. We then express $\hat{\beta}_j-\beta_j$ as the scalar projection of the noise vector $\varepsilon$ onto $z_j$, compute its conditional variance using homoskedasticity, and finally use the auxiliary regression identity $z_j^\top z_j=S_{jj}(1-R_j^2)$. [/proofplan] [step:Residualize $x_j$ against the remaining columns of the design] Let \begin{align*} A := X_{-j} \in \mathbb{R}^{n \times p}. \end{align*} Since $X$ has full column rank, the columns of $A$ are linearly independent, so $A^\top A \in \mathbb{R}^{p \times p}$ is invertible. Define the [linear map](/page/Linear%20Map) \begin{align*} P_A: \mathbb{R}^n &\to \mathbb{R}^n \\ v &\mapsto A(A^\top A)^{-1}A^\top v. \end{align*} This is the [orthogonal projection](/theorems/437) onto $\operatorname{Range}(A)$ with respect to the Euclidean inner product $u^\top v$ on $\mathbb{R}^n$. Define the residual vector \begin{align*} z_j := (I_n-P_A)x_j \in \mathbb{R}^n. \end{align*} Then $A^\top z_j=0$, because \begin{align*} A^\top z_j &= A^\top x_j-A^\top A(A^\top A)^{-1}A^\top x_j \\ &= A^\top x_j-A^\top x_j \\ &=0. \end{align*} Also $z_j \ne 0$. If $z_j=0$, then $x_j=P_Ax_j \in \operatorname{Range}(A)$, so the column $x_j$ would be a linear combination of the remaining columns of $X$, contradicting the full column rank of $X$. [guided] We want to separate the contribution of $x_j$ from the contribution of the other regressors. Let \begin{align*} A := X_{-j} \in \mathbb{R}^{n \times p}. \end{align*} The matrix $A$ contains the intercept column and every predictor column except $x_j$. Since the full design matrix $X$ has full column rank, deleting one column leaves a matrix $A$ whose columns are still linearly independent. Therefore $A^\top A$ is invertible. Define the linear map \begin{align*} P_A: \mathbb{R}^n &\to \mathbb{R}^n \\ v &\mapsto A(A^\top A)^{-1}A^\top v. \end{align*} This map is the Euclidean orthogonal projection onto $\operatorname{Range}(A)$. The residual part of $x_j$ after regressing on the other columns is \begin{align*} z_j := (I_n-P_A)x_j \in \mathbb{R}^n. \end{align*} The key property of $z_j$ is orthogonality to every column of $A$. Indeed, \begin{align*} A^\top z_j &= A^\top (I_n-P_A)x_j \\ &= A^\top x_j-A^\top A(A^\top A)^{-1}A^\top x_j \\ &= A^\top x_j-A^\top x_j \\ &=0. \end{align*} The vector $z_j$ cannot vanish. If $z_j=0$, then $x_j=P_Ax_j$, so $x_j$ lies in the span of the columns of $A$. That would make the columns of $X$ linearly dependent, contradicting the assumed full column rank of $X$. [/guided] [/step] [step:Express the coefficient error as a projection of the noise onto $z_j$] Let $\hat{\beta}_j$ denote the coordinate of $\hat{\beta}$ corresponding to the predictor column $x_j$. Since $\hat{\beta}$ satisfies the normal equations, \begin{align*} X^\top(Y-X\hat{\beta})=0. \end{align*} In particular, \begin{align*} z_j^\top(Y-X\hat{\beta})=0, \end{align*} because $z_j$ is orthogonal to every column of $A=X_{-j}$ and hence only tests the $x_j$ direction. Write the true coefficient vector as \begin{align*} \beta=(\beta_0,\beta_1,\dots,\beta_p)^\top \in \mathbb{R}^{p+1}. \end{align*} Using $Y=X\beta+\varepsilon$ and $z_j^\top A=0$, we obtain \begin{align*} 0 &= z_j^\top(Y-X\hat{\beta}) \\ &= z_j^\top(X\beta+\varepsilon-X\hat{\beta}) \\ &= z_j^\top\varepsilon+z_j^\top X(\beta-\hat{\beta}) \\ &= z_j^\top\varepsilon+z_j^\top x_j(\beta_j-\hat{\beta}_j). \end{align*} Since $z_j=(I_n-P_A)x_j$ and $P_A$ is an orthogonal projection, $z_j^\top x_j=z_j^\top z_j$. Therefore \begin{align*} \hat{\beta}_j-\beta_j=\frac{z_j^\top\varepsilon}{z_j^\top z_j}. \end{align*} [guided] The normal equations for ordinary least squares say that the fitted residual vector is orthogonal to every column of the design: \begin{align*} X^\top(Y-X\hat{\beta})=0. \end{align*} The vector $z_j$ is not itself a column of $X$, but it lies in the span of the columns of $X$: it is $x_j$ minus its projection onto the span of the remaining columns. Therefore the same orthogonality gives \begin{align*} z_j^\top(Y-X\hat{\beta})=0. \end{align*} Write \begin{align*} \beta=(\beta_0,\beta_1,\dots,\beta_p)^\top \in \mathbb{R}^{p+1}. \end{align*} Substitute the model equation $Y=X\beta+\varepsilon$ into the preceding identity: \begin{align*} 0 &= z_j^\top(Y-X\hat{\beta}) \\ &= z_j^\top(X\beta+\varepsilon-X\hat{\beta}) \\ &= z_j^\top\varepsilon+z_j^\top X(\beta-\hat{\beta}). \end{align*} Now $X$ consists of the columns in $A=X_{-j}$ together with $x_j$. Since $z_j$ is orthogonal to every column of $A$, all terms in $z_j^\top X(\beta-\hat{\beta})$ vanish except the one involving $x_j$. Hence \begin{align*} 0=z_j^\top\varepsilon+z_j^\top x_j(\beta_j-\hat{\beta}_j). \end{align*} Because $z_j=(I_n-P_A)x_j$ and $P_Ax_j$ is orthogonal to $z_j$, we have \begin{align*} z_j^\top x_j &= z_j^\top(P_Ax_j+z_j) \\ &= z_j^\top P_Ax_j+z_j^\top z_j \\ &= z_j^\top z_j. \end{align*} Since $z_j\ne 0$, the scalar $z_j^\top z_j$ is positive, so division is valid. We get \begin{align*} \hat{\beta}_j-\beta_j=\frac{z_j^\top\varepsilon}{z_j^\top z_j}. \end{align*} This identity is the Frisch-Waugh-Lovell residualization mechanism in this setting: the uncertainty in $\hat{\beta}_j$ is exactly the uncertainty in the noise projected onto the part of $x_j$ not explained by the other regressors. [/guided] [/step] [step:Compute the conditional variance of the residualized coefficient] Since $X$ is fixed and $z_j$ is determined by $X$, the vector $z_j$ is non-random conditional on $X$. Using $\operatorname{Var}(\varepsilon\mid X)=\sigma^2 I_n$, we compute \begin{align*} \operatorname{Var}(\hat{\beta}_j\mid X) &= \operatorname{Var}\left(\frac{z_j^\top\varepsilon}{z_j^\top z_j}\,\middle|\,X\right) \\ &= \frac{1}{(z_j^\top z_j)^2}\operatorname{Var}(z_j^\top\varepsilon\mid X) \\ &= \frac{1}{(z_j^\top z_j)^2}z_j^\top\operatorname{Var}(\varepsilon\mid X)z_j \\ &= \frac{1}{(z_j^\top z_j)^2}z_j^\top(\sigma^2 I_n)z_j \\ &= \frac{\sigma^2}{z_j^\top z_j}. \end{align*} [/step] [step:Identify the residual sum of squares with $S_{jj}(1-R_j^2)$] By definition, $R_j^2$ is the coefficient of determination from the least-squares regression of $x_j$ on the columns of $A=X_{-j}$. The fitted vector in that auxiliary regression is $P_Ax_j$, and the residual vector is $z_j=(I_n-P_A)x_j$. Since the predictors are centred and the intercept column belongs to $A$, the total sum of squares in the auxiliary regression is \begin{align*} S_{jj}=x_j^\top x_j. \end{align*} The residual sum of squares is \begin{align*} z_j^\top z_j. \end{align*} Therefore \begin{align*} R_j^2 &=1-\frac{z_j^\top z_j}{S_{jj}}, \end{align*} and hence \begin{align*} z_j^\top z_j=S_{jj}(1-R_j^2). \end{align*} [guided] The auxiliary regression is the least-squares regression with response vector $x_j\in\mathbb{R}^n$ and design matrix $A=X_{-j}$. Its fitted vector is the projection of $x_j$ onto the column space of $A$: \begin{align*} P_Ax_j. \end{align*} Its residual vector is therefore \begin{align*} x_j-P_Ax_j=(I_n-P_A)x_j=z_j. \end{align*} Thus the residual sum of squares in the auxiliary regression is \begin{align*} z_j^\top z_j. \end{align*} The total sum of squares is computed around the sample mean of the auxiliary response $x_j$. The predictor $x_j$ is centred by hypothesis, so its sample mean is $\bar{x}_j=0$. Hence \begin{align*} S_{jj} = \sum_{i=1}^n (x_{ij}-\bar{x}_j)^2 = \sum_{i=1}^n x_{ij}^2 = x_j^\top x_j. \end{align*} The coefficient of determination is \begin{align*} R_j^2 =1-\frac{\text{residual sum of squares}}{\text{total sum of squares}} =1-\frac{z_j^\top z_j}{S_{jj}}. \end{align*} Rearranging gives \begin{align*} z_j^\top z_j=S_{jj}(1-R_j^2). \end{align*} This identity is exactly where collinearity enters the variance calculation: if $x_j$ is nearly explained by the other columns, then $z_j^\top z_j$ is small and $R_j^2$ is close to $1$. [/guided] [/step] [step:Substitute the auxiliary regression identity into the variance formula] Combining the variance computation with the auxiliary regression identity gives \begin{align*} \operatorname{Var}(\hat{\beta}_j\mid X) &= \frac{\sigma^2}{z_j^\top z_j} \\ &= \frac{\sigma^2}{S_{jj}(1-R_j^2)}. \end{align*} Since $z_j\ne 0$, the denominator is positive. This proves the stated variance inflation factor formula. [/step]