Discussion Overview
The discussion revolves around proving a property of ranks for an n x m matrix A where n < m, specifically that there exists some λ such that rank(A^T A + λ I) = m. Participants explore various aspects of matrix properties, eigenvalues, and conditions for invertibility.
Discussion Character
- Exploratory
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant suggests that if A is an n x m matrix with n < m, there exists some λ such that rank(A^T A + λ I) = m, but they struggle to prove it for more than two columns being independent.
- Another participant questions how many eigenvalues transpose(A)*A has, noting that A is any n x m matrix.
- A later reply indicates that the rank will be full if the determinant is non-zero, stating that the determinant of A^T A + λ I is a polynomial in λ and will be non-zero for all but finitely many λ.
- One participant provides hints, stating that A^T A + λ I is full rank if it is invertible and questions the diagonal elements of A^T A + λ I.
- Another participant mentions that A^T A is symmetric and positive semi-definite, and by taking any λ > 0, the matrix A^T A + λ I becomes positive definite, hence non-singular and invertible.
- There is a query about whether -B is positive semi-definite if B is symmetric, while also noting that it is possible to choose λ such that the matrix is positive definite.
- One participant clarifies that they meant A^T A is symmetric in their previous post.
Areas of Agreement / Disagreement
Participants express differing views on the conditions under which the rank can be proven to be full, and there is no consensus on the implications of eigenvalues or the conditions for positive definiteness.
Contextual Notes
Some assumptions regarding the properties of matrices and eigenvalues are not fully explored, and the discussion includes unresolved mathematical steps related to the rank and invertibility of the matrix.