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Lemma. If the matrices $\mathbf{X}$ ($n\times p$ design matrix of full column rank ) and $\mathbf{L}_{2}$ satisfy $\mathbf{L}_{2}^{\prime}\mathbf{X}=\mathbf{0}$ and $\Omega$ is positive definite then $$ \Omega - \Omega \mathbf{L}_{2}\left(\mathbf{L}_{2}^{\prime}\Omega^{-1}\mathbf{L}_{2}\right)^{-1}\mathbf{L}_{2}^{\prime}\Omega=\mathbf{X}\left(\mathbf{X}^{\prime}\Omega^{-1}\mathbf{X}\right)^{-1}\mathbf{X}^{\prime}. $$

Proof. If $\mathbf{X}^{*}=\Omega^{-\frac{1}{2}}\mathbf{X}$, a basis for the orthogonal complement of $\mathbb{R}\left(\mathbf{X}^{*}\right)$ is given by the column of $\mathbf{L}_{2}^{*}=\Omega^{\frac{1}{2}}\mathbf{L}_{2}$ as $\mathbf{L}_{2}^{\prime*}\mathbf{X}^{*}=\mathbf{L}_{2}^{\prime}\mathbf{X}=\mathbf{0}$. It is well known that the orthogonal projection onto the orthogonal complement of $\mathbb{R}\left(\mathbf{W}\right)$ is $\mathbf{I}-\mathbf{W}\left(\mathbf{W}^{\prime}\mathbf{W}\right)^{-1}\mathbf{W}^{\prime}$ for matrix $\mathbf{W}$, so that we have $$ \mathbf{L}_{2}^{*}\left(\mathbf{L}_{2}^{\prime*}\mathbf{L}_{2}^{*}\right)^{-1}\mathbf{L}_{2}^{\prime*}=\mathbf{I}_{n}-\mathbf{X^{*}}\left(\mathbf{X^{*}}^{\prime}\mathbf{X}^{*}\right)^{-1}\mathbf{X}^{*\prime}, $$

I've tough time to understand the left hand side of the last line in the proof. I would highly appreciate if someone explain this to me. Thanks in advance for your help and time.

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Hint: given a matrix $A$, what is the formula for orthogonal projection onto $\mathbb{R}(A)$? Now, note that if a vector space is a direct sum of two orthogonal subspaces, then the sum of the orthogonal projections onto each space is the identity map. – Aaron Feb 1 '12 at 20:58

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