Orthogonal projection matrix of a matrix with one column sign switched Suppose that I have a $n \times k$ real matrix with full column rank. Say $k=3$ and I write
$$X = [\mathbf x_1:\mathbf x_2:\mathbf x_3]$$
where lower-case $\mathbf x$'s are $n \times 1$ vectors.
I go on and form the orthogonal projection matrix
$$P_X = X \left(X'X\right)^{-1}X'$$.
Consider now the matrix
$$W = [\mathbf x_1:-\mathbf x_2:\mathbf x_3]$$
Namely it is equal to $X$ matrix, but in (any) one column, the sign of the elements are switched.
Question: Can we express the projection matrix of $W$, $P_W=W \left(W'W\right)^{-1}W'$, in terms of the projection matrix of $X$, $P_X$, or at least state some relation between them?
I tried to explore this with what little matrix algebra I know, but could not come up with anything. In reality the $k$ dimension is bigger that $3$ but I guess this does not matter.
 A: We have $P_X = P_W$.
We can show this using matrix algebra by noting that $W = XQ$, where $Q$ is the orthogonal matrix
$$
Q = \pmatrix{1&0&0\\0&-1&0\\0&0&1}.
$$
With that, we note that
$$
\begin{align}
P_W &= W(W'W)^{-1}W' = 
[XQ]([XQ]'[XQ])^{-1}[XQ]'
\\ & = XQ[Q' (X'X) Q]^{-1}Q'X
\\ & = XQ[Q'(X'X)^{-1}Q]Q'X
\\ & = X[QQ'](X'X)^{-1}[QQ']X = X(X'X)^{-1}X = P_X,
\end{align}
$$
as was desired.
A: The reason why the matrices are the same is because you are projecting on the same subspace. There is no need for a proof here. Even if you replace your columns by some linear combinations that have the same span, you get the same matrix. They do not have to be independent either.
A: The projection is defined uniquely by the properties that it is the identity map on the column space of $X$ and zero on the orthogonal complement. The matrix that represents a linear transformation in a given basis (here, the standard basis) is unique, so if you use any other full-rank matrix $Y$ that has the same column space as $X$, you’ll end up with the same projection matrix.
