---
tag: NM2636CNW5
kind: proof
name: Proof of the matrix determinant lemma from Sylvester's determinant theorem
msc: [15A15]
proves: 0BC3Q6K92H
uses: [0TWTS88TWN, 0RXD6EZYBK, 073E4XC016, 04CPJ6AF00, 0G24VQ1088, 016PE4VZ8Q, 06DW9BAWB1]
sources:
  - url: https://en.wikipedia.org/wiki/Matrix_determinant_lemma
    locator: Proof
---

## Proof

We factor out $A$ and reduce the rest to a $1 \times 1$ determinant.

1. Let $A$ be an invertible $n \times n$ matrix over $K$, and let $u, v \in K^n$ be column vectors. Let $w = A^{-1} u$.
2. By [[distributivity|0RXD6EZYBK]] and [[associativity|073E4XC016]], $A (I_n + w v^T) = A + (A A^{-1}) u v^T$.
3. By the definition of the [[inverse|0TWTS88TWN]], $A A^{-1} = I_n$. So $A (I_n + w v^T) = A + u v^T$, with the [[identity matrix|06DW9BAWB1]] $I_n$.
4. By the [[product rule|04CPJ6AF00]], $\det(A + u v^T) = \det A \cdot \det(I_n + w v^T)$.
5. The matrix $w$ is $n \times 1$ and $v^T$ is $1 \times n$. By [[Sylvester's determinant theorem|0G24VQ1088]], $\det(I_n + w v^T) = \det(I_1 + v^T w)$.
6. The matrix $I_1 + v^T w$ is the $1 \times 1$ matrix with entry $1 + v^T A^{-1} u$.
7. By the [[Leibniz formula|016PE4VZ8Q]] for $n = 1$, which has one term, a $1 \times 1$ matrix has its entry as determinant. So $\det(I_1 + v^T w) = 1 + v^T A^{-1} u$.
8. By steps 4 to 7, $\det(A + u v^T) = (1 + v^T A^{-1} u) \det A$.

## Idea

Write $A + u v^T = A (I + A^{-1} u v^T)$. The second factor is the identity plus a column times a row. Sylvester's theorem swaps the order of the column and the row, which turns the $n \times n$ determinant into the number $1 + v^T A^{-1} u$.