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Assume I have a symmetric, positive-definite matrix $S \in \mathbb{R}^{p \times p}$. Assume that there is some matrix $L \in \mathbb{R}^{n \times p}$ that has full row-rank, i.e., has rank $n$ and where $L L^T = I$.

Define $N \in \mathbb{R}^{p \times p - n}$ as an orthogonal basis for the nullspace of $L$. From some numerical experiments, it seems that the following is true.

$$\det \left( N^T S N \right) = \det(S)\det \left( L S^{-1} L^T \right)$$

I'm not sure how to prove it as I can't figure out how to replace $N$ with $L$ or vice versa. It seems related to the Sherman-Morrison-Woodbury result but it seems like it is missing a few terms.

shadow1234
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1 Answers1

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Yes, your formula is correct.

From the information given, we can deduce that the block-matrix $$ U = \pmatrix{L\\ N^T} $$ is orthogonal (i.e. square with orthonormal rows/columns). Let $A$ denote the matrix $A = USU^T$ (noting that $U^T = U^{-1}$). We have $$ A = USU^T = \pmatrix{ LSL^T & LSN\\ N^TSL^T & N^TSN } = \pmatrix{A_{11} & A_{12}\\ A_{21} & A_{22}} $$ Let $B = A^{-1}$ We similarly have $$ B = A^{-1} = (USU^T)^{-1} = U^TS^{-1}U = \\ \pmatrix{ LS^{-1}L^T & LS^{-1}N\\ N^TS^{-1}L^T & N^TS^{-1}N } = \pmatrix{B_{11} & B_{12}\\ B_{21} &B_{22}}. $$ Now, following the logic of my previous answer here, we have $$ \det(A_{22}) = \det(A)\det(B_{11}). $$ On the other hand, we have $$ \det(A) = \det(USU^T) = \det(U)\det(S)\det(U^{-1}) = \det(S). $$ Putting everything together, we come to the conclusion that $$ \det(N^TSN) = \det(S) \det(LS^{-1}L^T), $$ which is what we wanted.

Ben Grossmann
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