Trying to find $T$ given $C=T^{-1}BT$
$B = A(A^TA)^{-1}A^T \in \mathbb{R}^{m \times m}$ given $A=\begin{bmatrix}A_1 \\ A_2\end{bmatrix}$ with $A_1 \in \mathbb{R}^{n \times n} $ is invertible and $A_2 \in \mathbb{R}^{k \times n}$ where $k = m-n$.
$C=\begin{bmatrix}I_{n \times n} & 0_{n \times k}\\ 0_{k \times n} & 0_{k \times k}\end{bmatrix}$
I got to this point: $$C=\begin{bmatrix}A_1A_1^{-1} & 0 \\ 0 & 0\end{bmatrix}=T^{-1}\begin{bmatrix}A_1 \\ A_2\end{bmatrix}(A_1A_1+A_2A_2)^{-1}\begin{bmatrix}A_1 & A_2\end{bmatrix}T$$
I somehow need to find a $T$ in terms of $A_1$ and $A_2$ that will cancel out all of the $A_2$s and leave me with $A_1A_1^{-1}=I$. How can I proceed?
Thanks!
$B$ is the orthogonal projection onto the column space of $A$. Its null space is the orthogonal complement to the column space of $A$. If you take $T=\pmatrix{A_1&-(A_2A_1^{-1})^T\\ A_2&I_k}$, its first $n$ columns are precisely the columns of $A$ and its last $k$ columns form a basis of the null space of $B$. Hence $T$ is invertible, $BT=TC$ and $C=T^{-1}BT$.