I want to simplify the following expression between matrices. Let $\mathbf{A}$ a $n\times n$ symmetric PSD matrix, $\mathbf{I}$ the identity matrix of order $n$ and $\lambda>0$ a positive constant.
I want to simplify the following expression:
$$\mathbf{A} -\mathbf{A} \left[\mathbf{A} + \lambda \mathbf{I}\right]^{-1} \mathbf{A}$$
My attempt: I use the SDV of $\mathbf{A}=\mathbf{PDP^T}$. Then as suggested in this answer, $$\left[\mathbf{A} + \lambda \mathbf{I}\right]^{-1}=\mathbf{P} \left(\mathbf{D} +\lambda \mathbf{I}\right)^{-1}\mathbf{P^T}.$$
Therefore:
$$\mathbf{A} -\mathbf{A} \left[\mathbf{A} + \lambda \mathbf{I}\right]^{-1} \mathbf{A} =\\ = \mathbf{PDP^T} - \mathbf{PDP^T} \left[\mathbf{P} \left(\mathbf{D} +\lambda \mathbf{I}\right)^{-1} \mathbf{P^T}\right] \mathbf{PDP^T}=\\ =\mathbf{PD}\left[\mathbf{I} -\left( \mathbf{D} + \lambda \mathbf{I} \right)^{-1} \mathbf{D}\right] \mathbf{P^T}=\\ =\mathbf{PD}\left[\mathbf{D}^{-1} -\left( \mathbf{D} + \lambda \mathbf{I} \right)^{-1}\right] \left(\mathbf{PD}\right)^{T}.$$
My questions are:
- Is it correct?
- Can the result be simplified further in terms of computational cost? I need to use it in an algorithm where the only thing that changes at each step is the value of $\lambda$.
Thanks a lot!
I didn't read your proof, because it looks overly complicated. Anyway, by a change of orthonormal basis, you may assume that $A$ is diagonal and hence the problem reduces to the scalar case. Since $x-\frac{x^2}{x+\lambda}=\frac{\lambda x}{x+\lambda}$, the expression becomes $\lambda A(A+\lambda I)^{-1}$, which in your notations is equal to $P\left[\lambda D(D+\lambda I)^{-1}\right]P^T$.