Given the matrix
$$P(D)=I-A(A^HA+D)^{-1}A^H$$
that is parameterized on the diagonal matrix $D$ whose diagonal entries are not necessarily the same, and $A$ is full column rank, I want find the impact of the diagonal matrix $D$ on the ratio between the squared Euclidean norms
$$r=\frac{||P(0)\mathbf{a}||^2_2}{||P(D)\mathbf{a}||^2_2}$$
Obviously, when the entries of $D$ are small as compared to the corresponding diagonal entries of $A^HA$,
$$r\to 1$$
But what is the difference between the true value of $r$ and 1. Can the difference be writen in terms of $D$, $\mathbf{a}$ and $A$? Or can the difference be estimated, say, for example, when the trace of $D$ is 10% of the trace of $A^HA$, what is the difference between $r$ and 1?
The only way I can think of is to find the derivative of $P(D)$ with respect to each diagonal entry of $D$ and using the Taylor expansion, but the procedure is tedious and not intuitive.
Can you please help? Thank you in advance.
The singular value decomposition (SVD) scheme can help some further simplifications, which, I hope, is helpful enough.
Since $A$ is full column rank, its SVD can be written as $$ A=U\begin{bmatrix}D_1\\0\end{bmatrix}V, $$ where $U_{m\times m}$ and $V_{n\times n}$ are unitary, $D_1$ is an $n\times n$ diagonal and invertible matrix, and $0_{(m-n)\times n}$ is a rectangular zero matrix. Henceforth, we will omit the matrix orders, since they will be clear from the context. Through replacement, we achieve $$ I-A(A^HA+D)^{-1}A^H{ = I-U\begin{bmatrix}D_1\\0\end{bmatrix}V(V^H\begin{bmatrix}D_1&0\end{bmatrix}U^HU\begin{bmatrix}D_1\\0\end{bmatrix}V+D)^{-1}V^H\begin{bmatrix}D_1&0\end{bmatrix}U^H \\= I-U\begin{bmatrix}D_1\\0\end{bmatrix}V(V^HD_1^2V+D)^{-1}V^H\begin{bmatrix}D_1&0\end{bmatrix}U^H \\= I-U\begin{bmatrix}D_1\\0\end{bmatrix}(D_1^2+VDV^H)^{-1}\begin{bmatrix}D_1&0\end{bmatrix}U^H \\= I-U\begin{bmatrix}(I+D_1^{-1}VDV^HD_1^{-1})^{-1}&0\\0&0\end{bmatrix}U^H \\= U\begin{bmatrix}I-(I+D_1^{-1}VDV^HD_1^{-1})^{-1}&0\\0&I\end{bmatrix}U^H . } $$ Therefore, by defining $\mathbf{b}=U^H\mathbf{a}$ and $\mathbf{b}=\begin{bmatrix}b_1\\b_2\end{bmatrix}$ where $b_1$ is $n\times 1$ and $b_2$ is $(m-n)\times 1$, we obtain $$ ||P(D)\mathbf{a}||_2^2{= \left|\left|U\begin{bmatrix}I-(I+D_1^{-1}VDV^HD_1^{-1})^{-1}&0\\0&I\end{bmatrix}U^H\mathbf{a}\right|\right|_2^2 \\= \left|\left|\begin{bmatrix}I-(I+D_1^{-1}VDV^HD_1^{-1})^{-1}&0\\0&I\end{bmatrix}\mathbf{b}\right|\right|_2^2 \\= \left|\left|\begin{bmatrix}I-(I+D_1^{-1}VDV^HD_1^{-1})^{-1}&0\\0&I\end{bmatrix}\begin{bmatrix}b_1\\b_2\end{bmatrix}\right|\right|_2^2 \\= \left|\left|\begin{bmatrix}[I-(I+D_1^{-1}VDV^HD_1^{-1})^{-1}]b_1\\b_2\end{bmatrix}\right|\right|_2^2 \\= \left|\left|[I-(I+D_1^{-1}VDV^HD_1^{-1})^{-1}]b_1\right|\right|_2^2 + \left|\left|b_2\right|\right|_2^2, } $$ where without loss of generality we can assume $||b_2||_2=1$, since $\frac{||P(0)\mathbf{a}||_2^2}{||P(D)\mathbf{a}||_2^2}$ is invariant under scaling $\mathbf{a}$. Finally, $$ \frac{||P(0)\mathbf{a}||_2^2}{||P(D)\mathbf{a}||_2^2} = \frac{1}{\left|\left|[I-(I+D_1^{-1}VDV^HD_1^{-1})^{-1}]b_1\right|\right|_2^2+1} . $$ The latter equation cannot be simplified generally, but under some special cases.
Special Case: $D=\alpha I$
When $D$ is a multiple of the identity matrix, we can proceed further and write $$ \frac{||P(0)\mathbf{a}||_2^2}{||P(D)\mathbf{a}||_2^2} {= \frac{1}{\left|\left|[I-(I+\alpha D_1^{-2})^{-1}]b_1\right|\right|_2^2+1} \\= \frac{1}{1+\alpha^2\sum_{i=1}^n\left[\frac{b_i}{\lambda_i^2+\alpha}\right]^2}, } $$ where $\lambda_i$ is the $i$-th eigenvalue of $A$ and $b_i$ is the $i$-th component of $\mathbf{b}$.
The following figure, illustrates its behavior for $n=10$, $b_i\sim\text{Uniform}(0,1)$ and $\lambda_i\sim\text{Uniform}(0,2)$: