I wish to understand linear transformations of the type $M = \left(\alpha I_n+ D^TD\right)^{-1}$
where $\alpha \neq 0$ and $D$ is a (full-rank) $k$ by $n$ matrix. ($k<n$)
My understanding is that though $D^TD$ is singular and thus does not have an inverse, $\left(\alpha I_n+ D^TD\right)^{-1}$ for $\alpha \ll 1$ could be a "good" approximation for it. (or am I mistaken?)
Ideally I would like to relate the eigenvectors of $M$ to my original $D$ (probably dependent also on $\alpha$) (see footnote [1])
So my question is: Is it possible to compute the eigenvalues and eigenvectors of this matrix $M$? Is there any interesting fact that you know about these types of matrices (thinking them of linear transformations)
I tried to use Woodbury identity to arrive at an expression
$$M = \left(\alpha I_n+ D^TD\right)^{-1}\\ =\alpha^{-1}I_n - \alpha^{-1}D^T\left(I_k+\alpha^{-1}DD^T\right)^{-1}D\alpha^{-1}\\ =\alpha^{-1}I_n - \alpha^{-1}D^T\left(\alpha I_k+DD^T\right)^{-1}D\\ = \alpha^{-1}I_n - \alpha^{-1}D^T\left(\alpha^{-1}I_k - \alpha^{-1}D\left(\alpha I_n+D^TD\right)^{-1}D^T\right)D\\ = \alpha^{-1}I_n - \alpha^{-2}D^TD + \alpha^{-2}D^TD\left(\alpha I_n+D^TD\right)^{-1}D^TD\\ = \alpha^{-1}I_n - \alpha^{-2}D^TD + \alpha^{-2}D^TDMD^TD$$
and got to an implicit equation for $M$ but could not solve it.
Any help is appreciated either in finding a closed-form solution or geometrical interpretations of the effect of $M$
[1] Some initial experimentation with simple $D$ lead me to believe that the rows of $D$ are eigenvectors for $M$ is this possible?