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 [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.

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

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...