Twos on the diagonal · Problem 23

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

Idea. Do not expand the determinant. The matrix is the identity plus a rank-one matrix, whose eigenvalues are known, and the determinant is the product of the eigenvalues.

Step 1: eigenvalues of $J$ . Let $1$ be the vector of ones. Then $J 1 = 4 \cdot 1$ , so $4$ is an eigenvalue. Any $v$ with $v_{1} + v_{2} + v_{3} + v_{4} = 0$ satisfies $J v = 0$ , and these vectors form a space of dimension $3$ . So the eigenvalues of $J$ are $4, 0, 0, 0$ .

Step 2: eigenvalues of $A$ . Adding the identity adds $1$ to every eigenvalue (same eigenvectors), so $A$ has eigenvalues $5, 1, 1, 1$ .

Step 3. $det A = 5 \cdot 1 \cdot 1 \cdot 1 = 5$ .

In general. For the $n \times n$ matrix $a I + b J$ , the eigenvalues are $a + nb$ (once) and $a$ ( $n - 1$ times), so $det (a I + b J) = a^{n - 1} (a + nb)$ . This structure appears often: it is the covariance matrix of assets that share one common factor.