Suppose we have an $m\times n$ matrix $A$ of rank $n$, (where $m>n$,) and an $n\times n$ projection matrix $P$.
Is it always possible to partition $A$ as $A' = \begin{bmatrix}X\\Y\end{bmatrix},$ so that $X$ is unaffected by P and the Frobenius norm of $A'$ remains the same as $A$?
Maybe one can assume that $X$ is $p \times n$, where $p$ is the number of non-zero eigenvalues of $P$. I feel like the answer should be yes, but I am not sure how to get there.
It seems like one has to start from finding the eigenvectors of $P = Q\begin{bmatrix}I_p & 0 \\ 0 & 0\end{bmatrix}Q^T$. $Q$ can then be partitioned into $Q=\begin{bmatrix}Q_1&Q_2\end{bmatrix}$, where $Q_1$ are the $p$ non-zero eigenvectors.
If it is now possible to express $X$ and $Y$ in terms of $Q_1$ and $Q_2$, then the statement would be proved, but I am not certain on how this can be done.
Edit 1: Giving this more thought, if $A'$ can be expressed as $$A' = \begin{bmatrix}X'&0\\0&Y'\end{bmatrix}Q^T = \begin{bmatrix}X'Q_1^T\\Y'Q_2^T\end{bmatrix},$$ then this matrix has the desired properties. $X'$ should be $p\times p$ and $Y'$ - $(m-p)\times (n-p)$.
First, the top block is unaffected by the projection $P$: $$\begin{multline}A'P=\begin{bmatrix}X'&0\\0&Y'\end{bmatrix}Q^TP=\begin{bmatrix}X'&0\\0&Y'\end{bmatrix}Q^TQ\begin{bmatrix}I_p & 0 \\ 0 & 0\end{bmatrix}Q^T=\\=\begin{bmatrix}X'&0\\0&Y'\end{bmatrix}\begin{bmatrix}I_p & 0 \\ 0 & 0\end{bmatrix}Q^T=\begin{bmatrix}X' & 0 \\ 0 & 0\end{bmatrix}Q=\begin{bmatrix}X' & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix}Q_1^T\\Q_2^T\end{bmatrix}= \begin{bmatrix}X'Q_1^T \\ 0\end{bmatrix}.\end{multline}$$
Now the question is if the transformation $T$, defined as $$AT=\begin{bmatrix}X'&0\\0&Y'\end{bmatrix},$$ does not change the norm? Or, alternatively, can it be shown or reasoned that $TT^T=I_n$?
Edit 2: The structure of $A'$ above is probably unobtainable with a norm preserving transformation, the one here $$A' = \begin{bmatrix}X'&0\\Y_2'&Y_1'\end{bmatrix}Q^T = \begin{bmatrix}X'Q_1^T\\Y_1'Q_1^T+Y_2'Q_2^T\end{bmatrix}$$ can be obtained with QL factorization. I think from here it is possible to show what is needed.
The partitioning can be accomplished by first factoring $A$ into a product of a lower triangular matrix $L$ and an orthogonal matrix $W$. If in the first matrix the $p\times p$ block at the top can be denoted as $X'$ and the two lower non-zero blocks as $Y_1'$ and $Y_2'$, i.e. $$A=LW=\begin{bmatrix}X' & 0 \\ Y_1' & Y_2'\end{bmatrix}W.$$
The matrix $A'$ can then be defined as $$A'=AW^TQ^T=\begin{bmatrix}X' & 0 \\ Y_1' & Y_2'\end{bmatrix}Q^T=\begin{bmatrix}X'Q_1^T \\ Y_1'Q_1^T + Y_2'Q_2^T\end{bmatrix}.$$ As $W$ and $Q$ are orthogonal, then the norms $A$ and $A'$ are will be the same.
Also, the top block will be unchanged by the projection $P$: $$\begin{multline}A'P=\begin{bmatrix}X'Q_1^T \\ Y_1'Q_1^T + Y_2'Q_2^T\end{bmatrix}\begin{bmatrix}Q_1 & Q_2\end{bmatrix}\begin{bmatrix}I_p & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix}Q_1^T \\ Q_2^T\end{bmatrix}=\\=\begin{bmatrix}X'Q_1^T \\ Y_1'Q_1^T + Y_2'Q_2^T\end{bmatrix} Q_1Q_1^T=\begin{bmatrix}X'Q_1^TQ_1Q_1^T \\ (Y_1'Q_1^T + Y_2'Q_2^T)Q_1Q_1^T\end{bmatrix}=\begin{bmatrix}X'Q_1^T \\ Y_1'Q_1^T\end{bmatrix}.\end{multline}$$