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The matrix determinant lemma states that for any $n\times n$ invertible matrix $A$, and $n \times 1$ vectors $u$ and $v$:

$\text{det}(A + uv^T) = \text{det}(A)(1 + v^TA^{-1}u)$

Is there some generalization of this lemma for two collections of $I>1$ vectors, $(u_i)_{i\leq I}$ and $(v_i)_{i\leq I}$, that has something to say about the determinant of:

$A + \sum_{i=1}^I u_i v_i^T$

Help is much appreciated.

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  • $\begingroup$ Use \det instead. $\endgroup$ Apr 9, 2020 at 15:09

1 Answer 1

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Note that $\sum_{i=1}^I u_i v_i^T = UV^T$, where $U$ is the matrix whose columns are $u_i$ and $V$ is the matrix whose columns are $v_i$. With that, we can apply the generalization given here. $$ \det\left(A + \sum_{i=1}^I u_i v_i^T\right) = \det(A + UV^T) = \det(\mathbf 1_I + V^TA^{-1}U)\det(A), $$ where $\mathbf 1_I$ is the $I \times I$ identity matrix. Note that $[V^TA^{-1}U]_{i,j} = v_i^TA^{-1}u_j$.

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  • $\begingroup$ Thanks so much! $\endgroup$
    – FGS
    Apr 9, 2020 at 15:04

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