With u = (1, 2, 2)^T and v = (2, 1, 1)^T, compute det(I_3 + u v^T).

With u = (1, 2, 2)^T and v = (2, 1, 1)^T, compute det(I_3 + u v^T).

Approach: Use the matrix determinant lemma for a rank one update of the identity, then check the answer against the eigenvalues of u v^T.

7. The matrix determinant lemma gives det(I + u v^T) = 1 + v^T u, and v^T u = 2*1 + 1*2 + 1*2 = 6, so the determinant is 7. The check is that the rank one matrix u v^T has eigenvalue v^T u = 6 on the direction u and eigenvalue 0 on the two dimensional space orthogonal to v, so I + u v^T has eigenvalues 7, 1, 1 and their product is 7. The same lemma is what makes a rank one covariance update cheap: no 3 by 3 determinant is ever formed.

Follow-up: What is det(A + u v^T) when A is invertible, and how much work does it cost once you already hold A^{-1}?

Key concepts: matrix determinant lemma, rank one matrix, eigenvalues, determinant.