Transformation to remove column in regression proof I'm trying to show that the $F$ statistic where the smaller model is just the removal of the $j$th predictor is equivalent to the square of the $t$ statistic for question $3.1$ in Elements of Statistical Learning.
$$F = \frac{RSS_0 - RSS_1}{RSS_0/(N-p-1)}$$
$$z_j = \frac{\hat{\beta_j}}{\hat{\sigma}\sqrt{v_{jj}}}$$
where $v_{jj} = (X^TX)^{-1}_{jj}$, and $RSS_0$ is the residual sum of squares of the bigger model, and $RSS_1$ is the residual sum of squares of the model with predictor $j$ removed.
This boils down to showing that $$RSS_0 - RSS_1 = \frac{\hat{\beta_j^2}}{v_{jj}}$$
I try and compute $RSS_0 - RSS_1$ (I will refer to $H = X(X^TX)^{-1}X^Ty$):
\begin{align}
RSS_0 - RSS_1 &= ||y-\hat{y}_{full}||^2 - ||y-\hat{y}_{no \beta_j}||^2
\\&= ||(I-H_{full})y||^2 - ||(I-H_{no\beta_j})y||^2
\\&= y^T(I-H_{full})y - y^T(I-H_{no\beta_j})y = y^T(H_{no\beta_j} - H_{full})y
\end{align}
It is over here where I am stuck - I don't know how to get an expression for $H_{no\beta_j}$ -
If I knew some transformation $T: \mathbb{R}^{n\times p} \rightarrow \mathbb{R}^{n\times (p-1)}$ then maybe I could proceed, but other ideas are appreciated.
 A: Consider the symbols:
$\textbf{x}_{(j)}$: $j$th column of matrix $\textbf{X}$.
$\textbf{X}_{(j)}$: $\textbf{X}$ matrix without $j$th column.
$\textbf{H}_{(j)}=\textbf{X}_{(j)}'(\textbf{X}_{(j)}\textbf{X}_{(j)})^{-1}\textbf{X}_{(j)}$, hat matrix construct without $j$th column.
$\textbf{H}=\textbf{X}'(\textbf{X}\textbf{X})^{-1}\textbf{X}$, full hat matrix.
Let us consider $m = [\textbf{x}_{(j)}'(\textbf{I}_n-\textbf{H}_{j})\textbf{x}_{(j)}]^{-1} = (\textbf{X}'\textbf{X})_{jj}^{-1} ....(1)$
Now consider, $$\textbf{H} = [\textbf{X}_{(j)}, \textbf{x}_{(j)}]\left[\begin{matrix}
\textbf{X}_{(j)}'\textbf{X}_{(j)} & \textbf{X}_{(j)}'\textbf{x}_{(j)} \\
\textbf{x}_{(j)}'\textbf{X}_{(j)} & \textbf{x}_{(j)}'\textbf{x}_{(j)} \\ 
\end{matrix}\right]^{-1} \left[\begin{matrix} \textbf{X}_{(j)}'\\ \textbf{x}_{(j)}'\end{matrix}\right].......(2) \\
= \textbf{H}_j + m[\textbf{H}_j\textbf{x}_{(j)}\textbf{x}_{(j)}'\textbf{H}_{j}-\textbf{x}_{(j)}\textbf{x}_{(j)}'\textbf{H}_{j}-\textbf{H}_{j}\textbf{x}_{(j)}\textbf{x}_{(j)}' + \textbf{x}_{(j)}\textbf{x}_{(j)}']\\
= \textbf{H}_{j} + m[(\textbf{I}_{n}-\textbf{H}_{j})\textbf{x}_{(j)}\textbf{x}_{(j)}'(\textbf{I}_{n}-\textbf{H}_{j})]$$

I used this identity on inverse of block matrix,
$\left[\begin{matrix}
\textbf{A}_{11} & \textbf{A}_{12} \\
\textbf{A}_{21} & \textbf{A}_{22} \\ 
\end{matrix}\right]^{-1} =  \left[\begin{matrix}
\textbf{A}_{11}^{-1}+\textbf{B}_{12}\textbf{B}_{22}^{-1}\textbf{B}_{21} & -\textbf{B}_{12}\textbf{B}_{22}^{-1} \\
-\textbf{B}_{22}^{-1}\textbf{B}_{21} & \textbf{B}_{22}^{-1} \\ 
\end{matrix}\right]........(*)$
where $\textbf{B}_{22}=\textbf{A}_{22}-\textbf{A}_{21}\textbf{A}_{11}^{-1}\textbf{A}_{12}, \textbf{B}_{12}=\textbf{A}_{11}^{-1}\textbf{A}_{12}, \textbf{B}_{21}=\textbf{A}_{21}\textbf{A}_{11}^{-1}$.

For expression (1),
$(\textbf{X}'\textbf{X})^{-1}=\left[\begin{matrix}
\textbf{X}_{(n)}'\textbf{X}_{(n)} & \textbf{X}_{(n)}'\textbf{x}_{(n)} \\
\textbf{x}_{(n)}'\textbf{X}_{(n)} & \textbf{x}_{(n)}'\textbf{x}_{(n)} \\ 
\end{matrix}\right]^{-1}$
Therefore $(\textbf{X}'\textbf{X})_{nn}^{-1} = [\textbf{x}_{(n)}'(\textbf{I}_n-\textbf{H}_{n})\textbf{x}_{(n)}]^{-1}$, from expression (*)
Let $\textbf{P}$ be the permutation matrix $\textbf{I}_n$ with $j$th and $n$th columns interchanged. then $\textbf{P}^2=\textbf{I}_n$. So, $\textbf{P}$ is symmetric orthogonal matrix, and its own inverse.
Therefore, $(\textbf{X}'\textbf{X})^{-1}=\textbf{P}(\textbf{P}\textbf{X}'\textbf{X}\textbf{P})^{-1}\textbf{P}=\textbf{P}\left[\begin{matrix}
\textbf{F} & \textbf{g} \\
\textbf{g}' & m \\ 
\end{matrix}\right]\textbf{P}$
Where $m=(\textbf{X}'\textbf{X})_{jj}^{-1}$, we interchanged $\textbf{x}_{(n)}$ and $\textbf{x}_{(j)}$ and $\textbf{X}_{(n)}$ and $\textbf{X}_{(j)}$. Hence the result.
and we take (*) in expression of (2) also.
