 Matrix determinant lemma

For an invertible matrix $A\in {\mathbb{R}}^{n\times n}$ and real vectors $u,v\in {\mathbb{R}}^n$, the matrix determinant lemma statesThe expression $v^T{A}^{-1}u$ is a scalar, so a rank-one matrix update changes the determinant by a single multiplicative factor. One need not expand the updated determinant entry by entry.

For the preceding example with $A=I$, the dot product is $v^{T}u=3-2=1$. Hence the updated determinant is $1+1=2$. Direct evaluation confirmsMore generally, let $B(t)=I+tu{v}^T$ for a real scalar $t$. The same calculation gives $\det B(t)=1+t$. The determinant is one at $t=0$, two at $t=1$, and zero at $t=-1$.

A compact proof explains the scalar reduction. Put $w=A^{-1}u$ and factor $A+u{v}^T=A(I+w{v}^T)$. If $w\ne 0$, choose a basis starting with $w$. In that basis, $(I+w{v}^T)w=(1+v^{T}w)w$, while every other basis vector changes only by a multiple of $w$. Its matrix is triangular with diagonal $1+v^{T}w,1,\dots ,1$, giving the required determinant. If $w=0$, the correction vanishes. This uses multiplicativity of the determinant to restore the factor $\det A$.

The denominator in the Sherman–Morrison formula is exactly this factor. The rank-one update singularity criterion therefore identifies when the updated matrix inverse ceases to exist. Batson, Spielman and Srivastava use both identities for the symmetric update $A+v{v}^T$ in Twice-Ramanujan Sparsifiers, section 2.3, Lemmas 2.1–2.2.

The Matrix Determinant Lemma is a useful result in linear algebra that relates the determinant of a matrix that has been modified by adding an outer product to the determinant of the original matrix. Specifically, it provides a way to compute the determinant of a modified matrix in terms of the determinant of the original matrix.