$SSR = \sum(\hat y_t - \bar y)^2 = \sum\hat y_t^2 - n \bar y^2$ From Statistical Methods for Forecasting by Abraham and Ledolter:

Let $\hat y = X \hat \beta$, where $\hat \beta$ is the least squares estimate $(X^TX)^{-1}X^Ty$.
Then $SSR = \sum(\hat y_t - \bar y)^2 = \sum\hat y_t^2 - n \bar y^2$.

I'm having trouble showing this equality.
I see that expanding and eliminating expressions similar to both sides we're left with showing:
$\sum(y_t - \hat y_t) = 0$, but I can't figure this either.
Anyone have any ideas?
 A: Let's decompose
\begin{equation}
 SSR = \sum\limits_{t=1}^n(\hat y_t - \bar y)^2 =
 \sum\limits_{t=1}^n \hat y_t^2 - 2\sum\limits_{t=1}^n\hat y_t \bar y + \sum\limits_{t=1}^n \bar y^2
\end{equation}
Notice that $\bar y$ does not depend on the summation index $t$ and hence could be extracted outside the sum as 
\begin{equation}
 SSR =
  \sum\limits_{t=1}^n \hat y_t^2 - 2\bar y\sum\limits_{t=1}^n\hat y_t  + \bar y^2\sum \limits_{t=1}^n 1
  =
 \sum\limits_{t=1}^n \hat y_t^2 - 2\bar y\sum\limits_{t=1}^n\hat y_t  + n\bar y^2
 \tag{1}
\end{equation}
It is true that (As you say) if $\sum(y_t - \hat y_t) = 0$, then we get the desired result. Notice that this happens when $X$ is a square and invertible matrix, because 
\begin{equation}
 \hat{\beta} = (X^TX)^{-1}X^Ty
 =X^{-1}X^{-T} X^{T}y = X^{-1}y
\end{equation}
where $X^{-T} = (X^{-1})^T$. Now, you can write $\hat{y}$ as 
\begin{equation}
 \hat{y} = X\hat{\beta} = XX^{-1}y = y
\end{equation}
This means that 
$\sum \hat{y}_t = \sum y = n\bar{y}$. Replacing this in $(1)$, we get
\begin{equation}
 SSR =
 \sum\limits_{t=1}^n \hat y_t^2 - 2n\bar y^2+ n\bar y^2
 =
\sum\limits_{t=1}^n \hat y_t^2 - n\bar y^2
\end{equation}
