Derivative of Log Determinant of a Matrix w.r.t a scalar parameterby fbelotti Tags: derivative, determinant, matrix, parameter, scalar 

#1
Jan2913, 12:28 PM

P: 2

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
Jan2913, 12:42 PM

P: 123

Have you thought about what the logarithm of a matrix means?




#3
Jan2913, 12:52 PM

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.




#4
Jan2913, 01:59 PM

HW Helper
P: 1,391

Derivative of Log Determinant of a Matrix w.r.t a scalar parameterfbelotti, 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? 



#5
Jan2913, 03:22 PM

P: 71

Oops. Only now noticed the determinant.




#6
Jan3013, 05:06 PM

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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