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

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 P: 2 Hi All, I'm trying to solve the following derivative with respect to the scalar parameter $\sigma$ $$\frac{\partial}{\partial \sigma} \ln|\Sigma|,$$ where $\Sigma = (\sigma^2 \Lambda_K)$ and $\Lambda_K$ is the following symmetric tridiagonal $K \times K$ 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.
 P: 123 Have you thought about what the logarithm of a matrix means?
 P: 71 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.
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P: 1,391
Derivative of Log Determinant of a Matrix w.r.t a scalar parameter

 Quote by kevinferreira Have you thought about what the logarithm of a matrix means?
 Quote by joeblow 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.
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?
 P: 71 Oops. Only now noticed the determinant.
 P: 2 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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