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

[fbelotti]

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.

 

[kevinferreira]

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

 

[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.

 

[Mute]

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

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?

 

[joeblow]

Oops. Only now noticed the determinant.

 

[fbelotti]

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...