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Derivative of Log Determinant of a Matrix w.r.t a scalar parameter

  1. Jan 29, 2013 #1
    Hi All,

    I'm trying to solve the following derivative with respect to the scalar parameter [itex]\sigma[/itex]

    $$\frac{\partial}{\partial \sigma} \ln|\Sigma|,$$

    where [itex]\Sigma = (\sigma^2 \Lambda_K)[/itex] and [itex]\Lambda_K[/itex] is the following symmetric tridiagonal [itex]K \times K[/itex] matrix
    $$
    \Lambda_{K} =
    \left(
    \begin{array}{ccccc}
    2 & -1 & 0 & \cdots & 0 \\
    -1 & 2 & -1 & \cdots & 0 \\
    0 & -1 & \ddots & \ddots & \vdots \\
    \vdots & \ddots & \ddots & \ddots & -1 \\
    0 & 0 & \ldots & -1 & 2 \\
    \end{array}\right).
    $$

    Is there a rule for these case?

    Thanks in advance for your time.
     
  2. jcsd
  3. Jan 29, 2013 #2
    Have you thought about what the logarithm of a matrix means?
     
  4. Jan 29, 2013 #3
    Typically to define a function for matrices that is consistent with the usual elementary functions, you use Taylor's theorem in the indeterminate x and replace x with the matrix. The differentiation is straightforward, I think.
     
  5. Jan 29, 2013 #4

    Mute

    User Avatar
    Homework Helper

    Am I missing something here? fbelotti is taking the derivative of the determinant of a matrix. The matrix logarithm shouldn't need to come into this at all, no?

    fbelotti, if your matrix is just ##\Sigma = \sigma^2 \Lambda_K##, then by the property of determinants, ##|cB| = c^n |B|## for an nxn matrix B, are you not just taking the derivative of ##\log(|\sigma^2 \Lambda_K|) = \log(\sigma^{2K} |\Lambda_K|)##, where ##|\Lambda_K|## is just a constant?
     
  6. Jan 29, 2013 #5
    Oops. Only now noticed the determinant.
     
  7. Jan 30, 2013 #6
    Mute, you are perfectly right. Many thanks for pointing that out. It was too simple... maybe it was too late and I was too tired...
     
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