Determine if a matrix if positive definite

  1. Given a symmetric matrix
    \sum a_{1s} & & & & \\
    & \ddots & & a_{ij} \\
    & & \ddots & & \\
    &a_{ij} & & \ddots & \\
    & & & & \sum w_{as}
    \end{array}\right) \in\mathbb{R}^{n\times n},
    with strictly positive entries a_{ij}, and with the diagonal entries being sum of off-diagonal entries residing
    in the corresponding row/column, how to proceed with the proof for A being positive definite,
    for some non-zero vector x.
  2. jcsd
  3. AlephZero

    AlephZero 7,248
    Science Advisor
    Homework Helper

    The shows there are no negative eigenvalues, but it doesn't exclude the possibiltiy of zero eigenvalues (i.e. a singular matrix).

    In fact the matrix
    $$\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}$$
    is singular, and therefore not positive definite.
  4. Thanks for providing the example.
    I guess the author of the book stating the above positive-definiteness on the given matrix type
    somehow misinterpreted it.
  5. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    Actually, unless I'm mistaken, the matrix is invertible (hence positive definite) if n>3.
  6. This is true because the matrix is diagonally dominant. There is a theorem that says a Hermitian diagonally dominant matrix with real nonnegative diagonal entries is positive semidefinite. A proof of this is found here
  7. It can be shown that the inequality holds for n>3, but not in general case, as is observed above.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?