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Let $M(i)$ denote the matrix $M$ with its $i$th row and column deleted. Then for the matrix $M$ in the question, the determinant after deletion of the $ mk $th (last) row is

Let $M(i)$ denote the matrix $M$ with its $i$th row and column deleted. Then for the matrix $M$ in the question, the determinant after deletion of the $ mk $th (last) row and column is

We carry out a similarity with $I_{k}\otimes U_{m}$ (inverse $I_{k}\otimes U_{m}^{-1}= I_{k}\otimes U_{m}^{T}$) to give $N_{2}$ \begin{align} N_{2} &=I_{mk}+t\left( \left( I_{k}\otimes U_{m}^{-1}\right) \left( D_{a}^{1/2}J_{k}D_{a}^{1/2}\otimes I_{m}\right) \left( I_{k}\otimes U_{m}^{-1}\right) \right. \\ &+ \left. I_{k}\otimes U_{m}^{-1}\left( I_{k}\otimes D_{b}^{1/2}J_{m}D_{b}^{1/2}\right) \left( I_{k}\otimes U_{m}\right) \right) \\ &=I_{mk}+t\left( D_{a}^{1/2}J_{k}D_{a}^{1/2}\otimes I_{m}+I_{k}\otimes \Lambda _{m}\right) \end{align}

We carry out a similarity with $I_{k}\otimes U_{m}$ (inverse $I_{k}\otimes U_{m}^{-1}= I_{k}\otimes U_{m}^{T}$) to give $N_{2}$ \begin{align} N_{2} &=I_{mk}+t\left( \left( I_{k}\otimes U_{m}^{-1}\right) \left( D_{a}^{1/2}J_{k}D_{a}^{1/2}\otimes I_{m}\right) \left( I_{k}\otimes U_{m}\right) \right. \\ &+ \left. I_{k}\otimes U_{m}^{-1}\left( I_{k}\otimes D_{b}^{1/2}J_{m}D_{b}^{1/2}\right) \left( I_{k}\otimes U_{m}\right) \right) \\ &=I_{mk}+t\left( D_{a}^{1/2}J_{k}D_{a}^{1/2}\otimes I_{m}+I_{k}\otimes \Lambda _{m}\right) \end{align}

The approach will be to first find the eigenvectors of a matrix congruent to $M$, and then convert them to eigenvectors of the submatrix. We will make frequent use of the mixed-product property of the Kronecker product, $\left( A\otimes B\right) \left( C\otimes D\right) =AC\otimes BD$. Let $J_{m}$ be the $m\times m$ matrix of $1$s. We will require the eigenvalues and eigenvectors of $ D_{b}^{1/2}J_{m}D_{b}^{1/2}=(D_{b}^{1/2}1_{m})(D_{b}^{1/2}1_{m})^{T}$. \ This is a rank 1 matrix with $m-1$ zero eigenvalues, and non-zero eigenvalue $ (D_{b}^{1/2}1_{m})^{T}(D_{b}^{1/2}1_{m})=b^{T}1_{m}$ with corresponding eigenvector $D_{b}^{1/2}1_{m}$.

The approach will be to first find the eigenvectors of a matrix congruent to $M$, and then convert them to eigenvectors of the submatrix. We will make frequent use of the mixed-product property of the Kronecker product, $\left( A\otimes B\right) \left( C\otimes D\right) =AC\otimes BD$. Let $J_{m}$ be the $m\times m$ matrix of $1$s. We will require the eigenvalues and eigenvectors of $ D_{b}^{1/2}J_{m}D_{b}^{1/2}=(D_{b}^{1/2}1_{m})(D_{b}^{1/2}1_{m})^{T}$. This is a rank 1 matrix with $m-1$ zero eigenvalues, and non-zero eigenvalue $ (D_{b}^{1/2}1_{m})^{T}(D_{b}^{1/2}1_{m})=b^{T}1_{m}$ with corresponding eigenvector $D_{b}^{1/2}1_{m}$.