How can I represent R squared in matrix form? This question is a follow-up to a prior question.

Basically, I wanted to study under what conditions when we regress the residuals to $x_1$, we will get $\small R^2$ of 20%. 
As a first step to attack this problem, my question is, how do I express $\small R^2$ in matrix form?
Then I will try to express "$\small R^2$ of regressing residuals to $x_1$" using matrix form.
Also, how can I add regression weights into the expression?
 A: You can write the coefficient-of-determination as a simple quadratic form of the correlation values between the individual variables (see this answer for details).  Consider a 
multiple linear regression with $m$ explanatory vectors and an intercept term.  Let $r_i = \mathbb{Corr}(\mathbf{y},\mathbf{x}_i)$ and $r_{i,j} = \mathbb{Corr}(\mathbf{x}_i,\mathbf{x}_j)$ and define:
$$\boldsymbol{r}_{\mathbf{y},\mathbf{x}} = \begin{bmatrix} r_1 \\ r_2 \\ \vdots \\ r_m \end{bmatrix} \quad \quad \quad \boldsymbol{r}_{\mathbf{x},\mathbf{x}} = \begin{bmatrix} 
r_{1,1} & r_{1,2} & \cdots & r_{1,m} \\
r_{2,1} & r_{2,2} & \cdots & r_{2,m} \\
\vdots  & \vdots  & \ddots & \vdots  \\
r_{m,1} & r_{m,2} & \cdots & r_{m,m} \\ \end{bmatrix}.$$
With a bit of linear algebra it can be shown that:
$$R^2 = \boldsymbol{r}_{\mathbf{y},\mathbf{x}}^\text{T} \boldsymbol{r}_{\mathbf{x},\mathbf{x}}^{-1} \boldsymbol{r}_{\mathbf{y},\mathbf{x}}.$$
The square-root of the coefficient-of-determination gives the multiple correlation coefficient, which is a multivariate extension of the absolute correlation.
A: We have
$$\begin{align*} R^2 = 1 - \frac{\sum{e_i^2}}{\sum{(y_i - \bar{y})^2}} = 1 - \frac{e^\prime e}{\tilde{y}^\prime\tilde{y}}, \end{align*}$$
where $\tilde{y}$ is a vector $y$ demeaned. 
Recall that $\hat{\beta} = (X^\prime X)^{-1} X^\prime y$, implying that $e= y - X\hat{\beta} = y - X(X^\prime X)^{-1}X^\prime y$. Regression on a vector of 1s, written as $l$, gives the mean of $y$ as the predicted value and residuals from that model produce demeaned $y$ values; $\tilde{y} = y - \bar{y} = y - l(l^\prime l)^{-1}l^\prime y$.
Let $H = X(X^\prime X)^{-1}X^\prime$ and let $M = l(l^\prime l)^{-1}l^\prime$, where $l$ is a vector of 1's. Also, let $I$ be an identity matrix of the requisite size. Then we have
$$\begin{align*} R^2 &= 1- \frac{e^\prime e}{\tilde{y}^\prime\tilde{y}} \\
&= 1 - \frac{y^\prime(I - H)^\prime(I-H)y}{y^\prime (I - M)^\prime(I-M)y} \\
&= 1 - \frac{y^\prime(I-H)y}{y^\prime (I-M)y},
\end{align*}$$
where the second line comes from the fact that $H$ and $M$ (and $I$) are idempotent.
In the weighted case, let $\Omega$ be the weighting matrix used in the OLS objective function, $e^\prime \Omega e$. Additionally, let $H_w = X \Omega^{1/2} (X^\prime \Omega X)^{-1} \Omega^{1/2} X^\prime$ and $M_w = l \Omega^{1/2}(l^\prime \Omega l)^{-1} \Omega^{1/2} l^\prime$. Then,
$$\begin{align*} R^2 &= 1 - \frac{y^\prime \Omega^{1/2} (I-H_w) \Omega^{1/2} y}{y^\prime \Omega^{1/2}(I-M_w) \Omega^{1/2}y},
\end{align*}$$
