Let {x_1, x_2, ..., x_n} be a set of linearly independent vectors in R^n. Let the (i,j)-th element of matrix A be given by A_ij = x_i^T * x_j. Which one of the following statements is correct? A. A is invertible. B. 0 is a singular value of A. C. Determinant of A is 0. D. A*z = 0 for some non-zero z.

GATE 2025 · Engineering Mathematics · Vector Space · medium

Answer: A. A is invertible.

1. Identify the Gram matrix as X^T X: Let X be the n x n matrix whose j-th column is x_j. Then (X^T X)_ij = (i-th row of X^T)(j-th col of X) = x_i^T x_j = A_ij. So A = X^T X.
2. X is invertible because {x_1,...,x_n} are linearly independent: An n x n matrix is invertible iff its columns are linearly independent. Given that {x_1,...,x_n} are linearly independent in R^n, X has rank n, so X is invertible.
3. A is positive definite: For any non-zero z in R^n: Xz != 0 (since X is invertible), so |Xz|^2 > 0. Thus z^T A z > 0 for all z != 0, confirming A is positive definite.
4. Conclude A is invertible; eliminate other options: Since A is positive definite, det(A) = product of eigenvalues > 0, so A is invertible. This rules out: B (0 is a singular value would mean A is singular — false), C (det=0 — false), D (Az=0 for non-zero z means z is in null space — null space is trivial since A is invertible — false).