Let \(\{x_1, x_2, \ldots, x_n\}\) 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 \(A_{ij} = x_i^{\top} x_j\), \(1 \le i, j \le n\). Which one of the following statements is correct?
Answer
(A)
Explanation
Put the vectors in the columns of \(X = [x_1\ x_2\ \cdots\ x_n]\). Then \(A_{ij} = x_i^{\top} x_j\) says \(A = X^{\top} X\), a Gram matrix.
For any non-zero \(z\), \(z^{\top} A z = (Xz)^{\top}(Xz) = \|Xz\|^{2}\). The columns of \(X\) are linearly independent, so \(Xz \ne 0\) and \(z^{\top} A z \gt 0\). Hence \(A\) is positive definite, which rules out (D).
A positive definite matrix has only positive eigenvalues, so \(\det A = (\det X)^{2} \gt 0\) and \(A\) is invertible. Being symmetric, its singular values are these same positive eigenvalues, so 0 is not one of them. Options B and C fail.
Answer: A