Premise: I have an $n × q$ matrix $X$ and a $q × a$ matrix $C$ with $n > q > a$.
I'm interested in the structure of the matrix $$ M = X X^+ - X_0 X_0^+ $$ where the superscript $^+$ indicates the Moore–Penrose pseudoinverse and $$ X_0 = X (I_q - C C^+). $$
I assume that $X$ is of full column rank and therefore $X^+ = (X' X)^{-1} X$ (where $'$ indicates the transpose).
Background: $X$ is the design matrix of a linear model, $C$ is a contrast, $X_0$ is a reduced design matrix, and $M$ occurs in the definition of standard test statistics.
$M$ is the difference of two orthogonal projection matrices, where the second projects into a subspace of the subspace the first projects into. This makes the difference an orthogonal projection matrix itself (symmetric and idempotent), which means it has a representation $$ M = X_\Delta X_\Delta^+. $$
Question: How do I obtain $X_\Delta$?
user1551 has correctly pointed out in an answer that $X_\Delta = M$ itself fulfills the equation. However, I'm looking for a "version" of $X$, meaning an $n \times q$ matrix of rank $a$.
My approach: I am guessing that $$ X_\Delta = X - X_0 X_0^+ X, $$ and this seems to be confirmed by numerical tests. But I am unable to come up with a proof, i.e. to show that $$ (X - X_0 X_0^+ X) (X - X_0 X_0^+ X)^+ = X X^+ - X_0 X_0^+. $$
The problem is how to deal with the pseudoinverse of a difference. One can write $$ X_\Delta = (I_n - X_0 X_0^+) X, $$ and according to Wikipedia, in the pseudoinverse of a product where one factor is an orthogonal projection, the orthogonal projection can be redundantly multiplied to the opposite side, meaning here $$ X_\Delta^+ = [(I_n - X_0 X_0^+) X]^+ = [(I_n - X_0 X_0^+) X]^+ (I_n - X_0 X_0^+) = X_\Delta^+ (I_n - X_0 X_0^+), $$ but that doesn't seem to help.
I can prove that $M$ is symmetric and idempotent, using the relations $$ X X^+ X_0 = X_0 \quad \text{and} \quad X_0 X_0^+ X X^+ = X_0 X_0^+, $$ which derive from the definition of $X_0$ and the properties of the pseudoinverse. I can also show that $$ X X_0^+ = X_0 X_0^+ $$ using the property of the pseudoinverse of a product involving an orthogonal projection (see above). But none of that helps either.