Let $D \in \mathbb{R}^{n \times n}$ defined as \begin{equation} D := \begin{pmatrix} 1 & 0 & \cdots & \cdots & 0 \\ -1 & 1 & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & -1 & 1 \end{pmatrix} \end{equation} or in another form as \begin{equation} D_{i,j} := \begin{cases} 1 & \textrm{ if } i=j , \\ -1 & \textrm{ if } i=j+1 , \\ 0 & \textrm{ otherwise }. \end{cases} \end{equation} Then consider the matrix $L \in \mathbb{R}^{2n^{2} \times 2n^{2}}$ defined as \begin{equation} L := \left( d+1 \right) \begin{pmatrix} \mathbb{I}_{n} \otimes D \\ D \otimes \mathbb{I}_{n} \end{pmatrix} . \end{equation}
I would like to compute (or at least estimate) the smallest eigenvalue of $LL^{T}$.
The exact form of $LL^{T}$ can be computed. Indeed, \begin{equation} L^{T} := \left( d+1 \right) \begin{pmatrix} \left( \mathbb{I}_{n} \otimes D \right) ^{T} , \left( D \otimes \mathbb{I}_{n} \right) ^{T} \end{pmatrix} = \left( d+1 \right) \begin{pmatrix} \mathbb{I}_{n} \otimes D^{T} , D^{T} \otimes \mathbb{I}_{n} \end{pmatrix} . \end{equation} Thus \begin{align} LL^{T} & = \left( d+1 \right) ^{2} \begin{pmatrix} \mathbb{I}_{n} \otimes D \\ D \otimes \mathbb{I}_{n} \end{pmatrix} \begin{pmatrix} \mathbb{I}_{n} \otimes D^{T} , D^{T} \otimes \mathbb{I}_{n} \end{pmatrix} \\ & = \left( d+1 \right) ^{2} \begin{pmatrix} \mathbb{I}_{n} \otimes DD^{T} & D^{T} \otimes D \\ D \otimes D^{T} & DD^{T} \otimes \mathbb{I}_{n} \end{pmatrix} . \end{align} Probably we can use the Schur Complement to compute but it still too complicated. Since each of the above 4 blocks, the eigenvalues can be compute, I wonder is there any way to connect these eigenvalues with the one of $LL^{T}$.
Updated according to the comment As I tried with small $d$ in matlab, the eigenvalue is small but strictly positive (around $10^{-6}$) and tends to increase as $d$ increase. Since in my problem $d$ is $512$ or even $1024$, I think that this value is not that small
