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Determinant and symmetric positive definite matrix

  1. May 26, 2015 #1
    As a step in a solution to another question our lecture notes claim that the matrix (a,b,c,d are real scalars).

    \begin{bmatrix}
    2a & b(1+d) \\
    b(1+d)& 2dc \\
    \end{bmatrix}

    Is positive definite if the determinant is positive. Why? Since the matrix is symmetric it's positive definite if the it got positive (real) eigenvalues and ##det (A) = \prod \lambda_i## but two negative eigenvalues would give a positive determinant too.

    Earlier in the text its given that the matrix
    ##
    S =
    \begin{bmatrix}
    a & b \\
    b& c \\
    \end{bmatrix}
    ##
    is positive definitive while ##d## is without any restrictions if that is somehow relevant.
     
  2. jcsd
  3. May 26, 2015 #2
    Edit: nevermind I solved it

    If i calculate the eigenvalues I get
    ##0 = (2a-\lambda )(2dc-\lambda) - b^2(1+d)^2 = 4adc +\lambda^2 - (2a+2dc)\lambda - b^2(1+d)^2 = det(A) + \lambda^2 -(2a+2dc)\lambda##
    then
    ## \lambda^2-(2a+2dc)\lambda < 0## since ##det(A) > 0##
    equal when
    ##(\lambda-(a+dc))^2-(a+dc)^2 = 0 \Longleftrightarrow \lambda = (a+dc) \pm (a+dc)##
    so the smallest eigenvalue is always ##\lambda> 0 ##.
     
    Last edited: May 26, 2015
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