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

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fbelotti
#1
Jan29-13, 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.
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kevinferreira
#2
Jan29-13, 12:42 PM
P: 123
Have you thought about what the logarithm of a matrix means?
joeblow
#3
Jan29-13, 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.

Mute
#4
Jan29-13, 01:59 PM
HW Helper
P: 1,391
Derivative of Log Determinant of a Matrix w.r.t a scalar parameter

Quote Quote by kevinferreira View Post
Have you thought about what the logarithm of a matrix means?
Quote Quote by joeblow View Post
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
#5
Jan29-13, 03:22 PM
P: 71
Oops. Only now noticed the determinant.
fbelotti
#6
Jan30-13, 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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