How to show $||Y-\hat{Y}_H||^2 = ||Y-\hat{Y}||^2 + ||\hat{Y} - \hat{Y}_H||^2$

Question

Problem Setting:
If we have a testable hypothesis with the form $A\beta = 0$ where $A$ is an $q \times p$ matrix and $\beta$ is an $p \times 1$ vector in a linear model, and we define that $\hat{Y} = X \hat{\beta}$ are the fitted values under the unrestriced model and $\hat{Y}_H = X \hat{\beta}_H$ are the fitted values under the restricted model, we want to show
$$ ||Y-\hat{Y}_H||^2 = ||Y-\hat{Y}||^2 + ||\hat{Y} - \hat{Y}_H||^2. $$ Attempt: We know that $$ Y-\hat{Y}_H = Y-\hat{Y} + \hat{Y} - \hat{Y}_H, $$ so we can decomposite $$ ||Y-\hat{Y}_H||^2 = ||(Y -\hat{Y}) + (\hat{Y} - \hat{Y}_H)||^2 $$ $$ ||Y-\hat{Y}_H||^2 = ||Y-\hat{Y}||^2 + ||\hat{Y} - \hat{Y}_H||^2 + 2(\hat{Y}^T\hat{Y}_H - Y^T\hat{Y}_H), $$ since $Y^T\hat{Y} - \hat{Y}^T\hat{Y} = 0$. But I don't know how to show $$ \hat{Y}^T\hat{Y}_H - Y^T\hat{Y}_H = 0. $$ One hint mentions that $$ Y^T(I-P)(\hat{Y} - \hat{Y}_H) = 0, $$ how this hint is related to the equation?

Answer 1

The cross term is $(Y-\hat{Y})^\top (\hat{Y} - \hat{Y}_H)$. You haven't given the definition of $P$ but I suspect it is such that $\hat{Y} = PY$. So substituting $Y-\hat{Y} = (I-P)Y$ (and noting that $I-P$ is symmetric) into the cross term shows why the hint is relevant.