Question

For a "positive definite" square matrix, the TRUE statement is

• A. the matrix is singular.
• B. all the eigen values of the matrix are greater than zero.
• C. all the eigen values of the matrix are zero.
• D. some of the eigen values can be less than zero.

Hint:

Positive definite matrices are always symmetric, always invertible, and always have strictly positive eigenvalues.

Answer 1

The Correct Option is B

A square matrix \(A\) is defined as positive definite if it satisfies the fundamental condition: \[ x^{T} A x > 0 \text{for all non-zero vectors } x. \] This is a strong condition that imposes strict requirements on the eigenvalues of the matrix. To understand why eigenvalues must be positive, recall that any symmetric matrix can be decomposed through spectral decomposition as: \[ A = Q \Lambda Q^{T}, \] where \(Q\) is an orthogonal matrix of eigenvectors and \(\Lambda\) is a diagonal matrix of eigenvalues \(\lambda_i\). Substituting this into the quadratic form: \[ x^T A x = x^{T} Q \Lambda Q^{T} x = y^{T} \Lambda y = \sum_{i} \lambda_i y_i^2, \] where \(y = Q^{T} x\) is a non-zero vector. Since each term \(y_i^2\) is non-negative, the sum can only be strictly positive for all non-zero vectors if each \(\lambda_i > 0\). If:
- any eigenvalue is zero → the matrix becomes singular and not positive definite.
- any eigenvalue is negative → the quadratic form becomes negative along the corresponding eigenvector direction.
Therefore, all eigenvalues of a positive definite matrix must be strictly positive. That is why option (B) is correct.