Gauss Markov Theorem for linear combinations I know that the Gauss Markov theorem implies that under some conditions, OLS estimates have the smallest variance of all unbiased linear estimators.
In particular if I have a model like $ y = \alpha + \beta_1.x_1 + \beta_2.x_2$ then I know that my OLS estimate $\hat{\beta}_1$ has smallest variance and my estimate $\hat{\beta}_2$ has also smallest variance.
But is it true that my estimate of $\hat{\beta}_2 - \hat{\beta}_1$ has the smallest variance?

$\DeclareMathOperator{\Var}{Var}$
Suppose that $a'y$ is an unbiased linear estimator of $\beta_2 - \beta_1 =: c'\beta$ where $c = (0, -1, 1)'$ and $\beta = (\alpha, \beta_1, \beta_2)'$. Note that, the following proof clearly generalizes to all linear functionals of $\beta$ and any design matrix $X \in \mathbb{R}^{n \times p}$ with $\operatorname{rank}(X) = p$.
By $E[a'y] = c'\beta$, it follows that $a'X\beta = c'\beta$, where $X$ is the design matrix.  Because this holds for all $\beta \in \mathbb{R}^p$, this requires $c = X'a$.  It then follows that
\begin{align}
 & \Var(a'y) - \Var(c'\hat{\beta}) \\
=& \sigma^2(a'a - c'(X'X)^{-1}c) \\
=& \sigma^2(a'a - a'X(X'X)^{-1}X'a) \\
=& \sigma^2a'(I - H)a \\
=& \sigma^2 ((I - H)a)'((I - H)a) \geq 0.
\end{align}
In the above derivation, $H = X(X'X)^{-1}X'$ is the hat matrix, which is idempotent and symmetric.  Consequently, $I - H$ is idempotent, which entails the desired result.
