Let A = I_n + z*z^T, where I_n is the n x n identity matrix and z in R^n, z^T*z = 1. Which of the following options is/are correct? A. Rank of A is n B. A is invertible C. 0 is an eigenvalue of A D. A^(-1) has a negative eigenvalue

GATE 2025 · Engineering Mathematics · Eigen Value · medium

Answer: Options A and B are correct: A has rank n and is invertible. Answer: A, B

  1. Find eigenvalues of A: Case 1: v = z (parallel to z) A*z = z + z*(z^T*z) = z + z*1 = 2z So lambda = 2 is an eigenvalue with eigenvector z. Case 2: v is orthogonal to z (z^T*v = 0) A*v = v + z*(z^T*v) = v + 0 = v So lambda = 1 is an eigenvalue with multiplicity n-1.
  2. Check option A: Rank of A is n: All eigenvalues of A are 1 or 2, none are zero. Therefore A has full rank n. Option A is TRUE.
  3. Check option B: A is invertible: det(A) = product of eigenvalues = 2 * 1^(n-1) = 2 != 0. So A is invertible. Option B is TRUE.
  4. Check options C and D: Option C: 0 is NOT an eigenvalue of A (eigenvalues are 1 and 2). Option C is FALSE. Option D: eigenvalues of A^(-1) are 1/2 and 1, both positive. A^(-1) has NO negative eigenvalue. Option D is FALSE.