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Assume $A_{n \times n}$ is an invertible matrix, $B$ is a vector and $C$ is a row vector.

How can we prove that $\det(A+BC) = \det(A) (1+CA^{-1}B)$?

I tried proving it using the statement proved in this question $\det(M) = \det(A) \cdot \det(D-C A^{-1} B)$ (which was presented to me in the similar questions while posting this question), but had no success. I would be grateful if a proof can be made using the statement proved in the linked question.

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  • $\begingroup$ Note: this result is sometimes referred to as the matrix determinant lemma. $\endgroup$
    – Ben Grossmann
    Dec 3, 2020 at 15:59
  • $\begingroup$ @Jigsaw What you need is the following general result: let $m, n \in \mathbb{N}$ with $m \leqslant n$, $\mathbf{A}$ be a commutative ring and $M \in \mathscr{M}_{m, n}(\mathbf{A})$ respectively $N \in \mathscr{M}_{n, m}(\mathbf{A})$ two rectangular matrices. Then the following relation holds between the characteristic polynomials of $MN$ and $NM$: $\det(X\mathrm{I}_n-NM)=X^{n-m}\det(X\mathrm{I}_m-MN)$. $\endgroup$
    – ΑΘΩ
    Dec 3, 2020 at 16:01
  • $\begingroup$ Weinstein–Aronszajn $\endgroup$
    – Rodrigo de Azevedo
    Dec 3, 2020 at 16:26

2 Answers 2

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Note that we have $$ \det(A + BC) = \det(A[I + A^{-1}BC]) = \det(A)\det(I + [A^{-1}B]C). $$ with $U = A^{-1}B$ and $V = C$, it suffices to show that $$ \det(I + UV) = 1 + VU. $$ This can be seen as an instance of the W-A identity. One proof of this result is given in the linked page.

If you're still set on using the linked result, we can indeed prove this result as a consequence. Let $M$ denote the matrix $$ M = \pmatrix{A & -B\\ C & 1}. $$ If we apply the linked result directly to this matrix, then we have $$ \det(M) = \det(A)\det(D - CA^{-1}(-B)) = \det(A)(1 + CA^{-1}B). $$ On the other hand, we note that $$ \det \pmatrix{A & -B\\ C & 1} = (-1)^{n-1}\det\pmatrix{-B & A\\ 1 & C} = [(-1)^{n-1}]^2\det\pmatrix{1 & C\\ -B & A}. $$ That is, we have $$ \det(M) = \det\pmatrix{1 & C\\ -B & A}. $$ Applying the linked result to this matrix yields $$ \det\pmatrix{1 & C\\ B & A} = \det(1)\det(A - (-B)1^{-1}C) = \det(A + BC). $$ So, we have $\det(A)(1 + CA^{-1}B) = \det(A + BC)$ as desired.

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$A + B C = A(I + A^{-1} B C)$. If $S$ and $T$ are $n \times m$ and $m \times n$ matrices, $ST$ and $TS$ have the same nonzero eigenvalues.

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