In mathematics, in particular linear algebra, the matrix determinant lemma computes the determinant of the sum of an invertible matrix A and the dyadic product, u vT, of a column vector u and a row vector vT.
Statement Suppose A is an invertible square matrix and u, v are column vectors. Then the matrix determinant lemma states that
det ( A + u v T ) = ( 1 + v T A − 1 u ) det ( A ) . {\displaystyle \det(\mathbf {A} +\mathbf {uv} ^{\textsf {T}})=(1+\mathbf {v} ^{\textsf {T}}\mathbf {A} ^{-1}\mathbf {u} )\,\det(\mathbf {A} )\,.}
Here, uvT is the outer product of two vectors u and v. The theorem can also be stated in terms of the adjugate matrix of A:
det ( A + u v T ) = det ( A ) + v T a d j ( A ) u , {\displaystyle \det(\mathbf {A} +\mathbf {uv} ^{\textsf {T}})=\det(\mathbf {A} )+\mathbf {v} ^{\textsf {T}}\mathrm {adj} (\mathbf {A} )\mathbf {u} \,,}
in which case it applies whether or not the matrix A is invertible.
Proof First the proof of the special case A = I follows from the equality:
( I 0 v T 1 ) ( I + u v T u 0 1 ) ( I 0 − v T 1 ) = ( I u 0 1 + v T u ) . {\displaystyle {\begin{pmatrix}\mathbf {I} &0\\\mathbf {v} ^{\textsf {T}}&1\end{pmatrix}}{\begin{pmatrix}\mathbf {I} +\mathbf {uv} ^{\textsf {T}}&\mathbf {u} \\0&1\end{pmatrix}}{\begin{pmatrix}\mathbf {I} &0\\-\mathbf {v} ^{\textsf {T}}&1\end{pmatrix}}={\begin{pmatrix}\mathbf {I} &\mathbf {u} \\0&1+\mathbf {v} ^{\textsf {T}}\mathbf {u} \end{pmatrix}}.}
The determinant of the left hand side is the product of the determinants of the three matrices. Since the first and third matrix are triangular matrices with unit diagonal, their determinants are just 1. The determinant of the middle matrix is our desired value. The determinant of the right hand side is simply (1 + vTu). So we have the result:
det ( I + u v T ) = 1 + v T u . {\displaystyle \det(\mathbf {I} +\mathbf {uv} ^{\textsf {T}})=1+\mathbf {v} ^{\textsf {T}}\mathbf {u} .}
Then the general case can be found by setting u to be A−1u:
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