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We want to show that orthogonal projection can be written as \begin{align} C\cdot C^+=\left(I_N-\frac{1}{N}\iota\iota'\right), \end{align} where $C=\left(\omega\iota'-I_N\right)$, $I_N$ is the identity matrix of dimension $N$, $\omega$ is an $N\times1$ vector, $\iota'\omega=1$ and $C^+$ denotes the Moore–Penrose pseudo-inverse of $C$. Numerically, we know that this holds for any vector $\omega$ fulfilling the summing-up-constraint.

Here, $C$ is a linear operator from $\mathbb{R}^{T\times N}$ onto $\mathbb{R}^{T\times N}$ and we want to show that despite of the choice of $\omega$, the orthogonal projector $CC^+$ is always of the same form.

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  • $\begingroup$ Orthogonal projection onto what subspace? What is the summing-up constraint? $\endgroup$ – user147263 May 15 '15 at 5:15
  • $\begingroup$ The summing-up-constraint is just $\iota'\omega=1$. $\endgroup$ – dsforecast May 15 '15 at 8:17
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Do you mean $C$ is a linear operator from $\mathbb{R}^{T\times N}$ onto $\mathbb{R}^{T\times N}$, mapping an element $x\in\mathbb{R}^{T\times N}\mapsto xC=x\omega\iota'-x$ and you want to show that despite of the choice of $\omega$ the orthogonal projector $CC^+$ is always of the same form?

Idempotence of $\left(I_N−\frac{1}{N}\iota_N\iota_N ′\right)$ is easy to show, therefore this term defines a projection on $\mathbb{R}^{T\times N}$. However, I do not know how to proove the orthogonality. First, as $CC^+$ is the orthogonal projection onto the range of $C$ it is necessary to show that range$(C)$ does not depend on $\omega$. In a second step it should be sufficient to show that $CC^+$ is a self-adjoint operator.

Another way would be to check if setting $C^+C=\left(I_N−\frac{1}{N}\iota_N\iota_N ′\right)$ fulfills the properties of the Moore-Penrose Pseudoinverse: \begin{align} CC^+C=&C\\ C^+CC^+=&C^+\\ (CC^+)'=&CC^+\\ (C^+C)'=&C^+C \end{align} The first and third equations hold, but I have trouble with the second and fourth: $C^+CC^+=C^+$ requires $C^+ \iota_N =0$ but as I do not have a closed form solution for $C^+$ I cannot confirm that this holds.

EDIT: The answer to this question gives you that $C^+$ is an element of the set: $$S=\{C+\omega y^T+\iota c^T:\;y,z\in\mathbb{R}^N\}.$$ Plug-in such an element and find $z$ such that $(CC^+)'=CC^+$ gives you $$z=\omega(\iota'z-1)+\frac{1}{N}\iota$$ The choice of $y$ is irrelevant for the representation of $CC^+$ as $\omega y'$ is in the nullspace of $C$. For one $z$ in the form given above you'll find $$C(C+\omega y'+z \iota')=I_N-\frac{1}{N}\iota \iota'.$$

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  • $\begingroup$ Yes, $C$ is a linear operator from $\mathbb{R}^{T\times N}$ onto $\mathbb{R}^{T\times N}$ and we want to show that despite of the choice of $\omega$, the orthogonal projector $CC^+$ is always of the same form. $\endgroup$ – dsforecast May 15 '15 at 16:33
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We find finally that $C^+$ has the following form: \begin{align} C^{+}= w \iota' - I \frac{1}{\omega'\omega} \omega \omega' - \left( \frac{1}{N} \frac{1}{\omega'\omega} + 1\right) \omega \iota' + \frac{1}{N} \iota \iota' \end{align}

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