Question

Let \( \{ x_1, x_2, \dots, x_n \} \) be a set of linearly independent vectors in \( \mathbb{R}^n \). Let the \( (i,j) \)-th element of matrix \( A \in \mathbb{R}^{n \times n} \) be given by \( A_{ij} = x_i^T x_j \), where \( 1 \leq i, j \leq n \). 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. \( z^T A z = 0 \) for some non-zero \( z \in \mathbb{R}^n \)

Hint:

For a Gram matrix formed by linearly independent vectors, the matrix is always positive definite, and therefore it is invertible. A positive definite matrix has all positive eigenvalues and a non-zero determinant.

Answer 1

The Correct Option is A

The matrix \( A \) is a Gram matrix, where each element \( A_{ij} = x_i^T x_j \) is the inner product between vectors \( x_i \) and \( x_j \). Since the vectors \( x_1, x_2, \dots, x_n \) are linearly independent, the matrix \( A \) is invertible. This is because the rank of the Gram matrix is \( n \), and a matrix with full rank is invertible.

Option (A): \( A \) is invertible. This is correct because the matrix is positive definite.
Option (B): 0 is a singular value of \( A \). This is incorrect, as the matrix is invertible and does not have 0 as a singular value.
Option (C): The determinant of \( A \) is 0. This is incorrect because \( A \) is invertible, meaning its determinant is non-zero.
Option (D): \( z^T A z = 0 \) for some non-zero \( z \in \mathbb{R}^n \). This is incorrect because for a positive definite matrix like \( A \), \( z^T A z \) is always positive for any non-zero vector \( z \).

Thus, the correct answer is Option (A).