Let \(A = I_{n} + xx^{\top}\), where \(I_{n}\) is the \(n \times n\) identity matrix and \(x \in \mathbb{R}^{n}\), \(x^{\top}x = 1\). Which of the following options is/are correct?
Answer
(A), (B)
Explanation
Find the eigenvalues of \(A\) using \(x^{\top}x = 1\):
- \(Ax = x + x(x^{\top}x) = 2x\), so 2 is an eigenvalue.
- Any \(v\) orthogonal to \(x\) has \(x^{\top}v = 0\), so \(Av = v\). These vectors fill an \((n - 1)\)-dimensional subspace, so 1 is an eigenvalue of multiplicity \(n - 1\).
No eigenvalue is 0, so \(A\) is invertible with rank \(n\): A and B are true and C is false. The eigenvalues of \(A^{-1}\) are the reciprocals \(\frac{1}{2}\) and 1, all positive, so D is false.
Answer: A, B