From both this paper and Wikipedia, it is mentioned that for an orthogonal projection matrix $(I – A^+A)$ its pseudo inverse is itself, i.e., $$(I – A^+A)^+ = I – A^+A$$ Why is this the case? Can someone please help me understand how this can be proved?
Why is the pseudoinverse of an orthogonal projection matrix itself
linear algebramatricesprojection-matricespseudoinverse
Best Answer
After a few months I think I am able to answer my own question, as @greg said the key is to use the four Penrose conditions:
Then if we see $A$ as $(I - A^\dagger A)$ and $M$ also as $(I - A^\dagger A)$ we can check whether they satisfy the above four conditions.
First, we need show that $(I - \mathbf{A}^\dagger\mathbf{A})(I - \mathbf{A}^\dagger\mathbf{A}) = (I - \mathbf{A}^\dagger\mathbf{A})$, this can be shown via direct calculation
\begin{align*} (I - \mathbf{A}^\dagger\mathbf{A})(I - \mathbf{A}^\dagger\mathbf{A}) &= I - \mathbf{A}^\dagger\mathbf{A} - \mathbf{A}^\dagger\mathbf{A} + \mathbf{A}^\dagger\mathbf{A}\mathbf{A}^\dagger\mathbf{A}\\ &= I - \mathbf{A}^\dagger\mathbf{A} - \mathbf{A}^\dagger\mathbf{A} + \mathbf{A}^\dagger\mathbf{A} & \mathbf{A}\mathbf{A}^\dagger\mathbf{A} = \mathbf{A}\\ &= I - \mathbf{A}^\dagger\mathbf{A}. \end{align*}
Then we can have \begin{align} (I - \mathbf{A}^\dagger\mathbf{A})(I - \mathbf{A}^\dagger\mathbf{A}))^T &= (I - \mathbf{A}^\dagger\mathbf{A})^T = I^T - (\mathbf{A}^\dagger\mathbf{A})^T = I - \mathbf{A}^\dagger\mathbf{A}\\ ((I - \mathbf{A}^\dagger\mathbf{A})(I - \mathbf{A}^\dagger\mathbf{A}))^T &= I - \mathbf{A}^\dagger\mathbf{A}\\ (I - \mathbf{A}^\dagger\mathbf{A})(I - \mathbf{A}^\dagger\mathbf{A})(I - \mathbf{A}^\dagger\mathbf{A}) &= I - \mathbf{A}^\dagger\mathbf{A}\\ (I - \mathbf{A}^\dagger\mathbf{A})(I - \mathbf{A}^\dagger\mathbf{A})(I - \mathbf{A}^\dagger\mathbf{A}) &= I - \mathbf{A}^\dagger\mathbf{A} \end{align}
Thus, we show that $(I - \mathbf{A}^\dagger\mathbf{A})^\dagger = (I - \mathbf{A}^\dagger\mathbf{A})$.