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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.

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With your choice of $X_\Delta$, $M$ is indeed equal to $X_\Delta X_\Delta^+$.

Proof. Let $P=I-CC^+$. Note that the column space of $M=XX^+-(XP)(XP)^+$ is $\operatorname{col}(X)\cap\operatorname{col}(XP)^\perp$, while the column space of $X_\Delta X_\Delta^+$ is precisely the column space of $X_\Delta=\left[I-(XP)(XP)^+\right]X$.

Since $X_\Delta=X\left[I-P(XP)^+X\right]$, $\operatorname{col}(X_\Delta)\subseteq\operatorname{col}(X)$. Also, since \begin{aligned} (XP)^TX_\Delta &=(X_\Delta^TXP)^T\\ &=\left[X^T\left(I-(XP)(XP)^+\right)XP\right]^T\\ &=\left[X^T\left(XP-(XP)(XP)^+(XP)\right)\right]^T\\ &=\left[X^T\left(XP-XP\right)\right]^T=0, \end{aligned} we also have $\operatorname{col}(X_\Delta)\subseteq\operatorname{col}(XP)^\perp$. Thus $\operatorname{col}(X_\Delta)\subseteq\operatorname{col}(M)$.

We now show that the reverse inclusion is also true. Pick any $v\in\operatorname{col}(M)=\operatorname{col}(X)\cap\operatorname{col}(XP)^\perp$. Since $v\in\operatorname{col}(X)$, it can be written as $Xb$ for some vector $b$. Thus $$ X_\Delta b=\left[I-(XP)(XP)^+\right]Xb=v-(XP)(XP)^+v. $$ However, we also have $v\in\operatorname{col}(XP)^\perp$. Therefore $(XP)(XP)^+v=0$ and in turn $X_\Delta b=v$, meaning that $v\in\operatorname{col}(X_\Delta)$.

Thus $\operatorname{col}(X_\Delta X_\Delta^+)\equiv\operatorname{col}(X_\Delta)=\operatorname{col}(M)$. Hence $X_\Delta X_\Delta^+$ and $M$ must be equal, because they are orthogonal projections with identical column spaces.

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  • $\begingroup$ Now it's clear. Thanks again! $\endgroup$ – A. Donda Apr 7 '19 at 20:18
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    $\begingroup$ @A.Donda You are welcome. Note that if you need to verify the relation numerically, you need to use SVD and set thresholds to cut off the small singular values, otherwise the verification will likely fail because the calculation of pseudoinverse is not numerically stable. $\endgroup$ – user1551 Apr 7 '19 at 20:25

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