Identity plus an outer product · Problem 117

s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.

Idea. $u v^{⊤}$ has rank one, so it has only one nonzero eigenvalue, and that eigenvalue is $v^{⊤} u$ . Adding the identity adds 1 to every eigenvalue.

Step 1: the eigenvalues of $u v^{⊤}$ . For any $w$ , $u v^{⊤} w = (v^{⊤} w) u$ , a multiple of $u$ . So the matrix has rank 1:

every $w$ orthogonal to $v$ is sent to 0, which gives the eigenvalue 0 twice;
$u v^{⊤} u = (v^{⊤} u) u$ , so $u$ is an eigenvector with eigenvalue $v^{⊤} u = 4 + 10 + 18 = 32$ .

Step 2: add the identity. $I + u v^{⊤}$ has eigenvalues $1 + 32 = 33$ , $1$ and $1$ .

Step 3. The determinant is the product of the eigenvalues: $det (I + u v^{⊤}) = 33 \cdot 1 \cdot 1 = 33.$

The general formula. $det (I + u v^{⊤}) = 1 + v^{⊤} u$ , the matrix determinant lemma. It is behind the Sherman-Morrison formula for updating an inverse after a rank-one change, used for example to update a covariance matrix with one new observation.