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Determine if a matrix if positive definite

  1. Mar 6, 2012 #1
    Given a symmetric matrix
    [tex]A=\left(\begin{array}{ccccc}
    \sum a_{1s} & & & & \\
    & \ddots & & a_{ij} \\
    & & \ddots & & \\
    &a_{ij} & & \ddots & \\
    & & & & \sum w_{as}
    \end{array}\right) \in\mathbb{R}^{n\times n},
    [/tex]
    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,
    [tex]
    x^TAx>0
    [/tex]
    for some non-zero vector x.
     
  2. jcsd
  3. Mar 6, 2012 #2

    AlephZero

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    The http://en.wikipedia.org/wiki/Gershgorin_circle_theorem 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. Mar 7, 2012 #3
    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. Mar 7, 2012 #4

    morphism

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    Actually, unless I'm mistaken, the matrix is invertible (hence positive definite) if n>3.
     
  6. Mar 7, 2012 #5
    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 http://planetmath.org/?op=getobj&from=objects&id=7483
     
  7. Mar 7, 2012 #6
    It can be shown that the inequality holds for n>3, but not in general case, as is observed above.
     
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