# Inverse of a Spectral Matrix

• Oct 26th 2010, 11:23 PM
Antem
Inverse of a Spectral Matrix
I'm working on a problem that involves multiplying the inverse of a spectral matrix by a matrix containing a subset of that matrices spectral components. Specifically:

$
{\bf S}_N = \sum\limits_{i = 1}^N {\lambda _i {\bf v}_i {\bf v}_i ^H }
$

$
{\bf S}_{N - 1} = \sum\limits_{i = 1}^{N - 1} {\lambda _i {\bf v}_i {\bf v}_i ^H }
$

How can I determine
$
\left[ {{\bf S}_N } \right]^{ - 1} S_{N - 1} = ?
$

S is a covariance matrix.
${\lambda _i }$ is the ith eigenvalue.
${{\bf v}_i }$ is the ith eigenvector.

Also, can the inverse of a specral matrix be writen in terms of the inverse of each spectral component ${\lambda _i {\bf v}_i {\bf v}_i ^H }$ somehow?

Thank you
• Oct 27th 2010, 03:07 AM
Ackbeet
Is $\{\mathbf{v}_{i}\}$ an orthonormal set?
• Oct 27th 2010, 07:19 AM
Antem
Yes $\left\{ {{\bf v}_{\bf i} } \right\}$ is an orthonormal set.

Also, in genral I'm interested in figuring out how to solve

$\left[ {S_N } \right]^{ - 1} S_{N - k} = ?$

for an arbitrary 0 < k < N

Thanks
• Oct 27th 2010, 07:34 AM
Ackbeet
Well then, it sounds like you've got yourself a diagonalized matrix. You can write

$S_{N}=P^{-1}DP,$ where

$P=[\mathbf{v}_{1}\;\mathbf{v}_{2}\;\dots\;\mathbf{v}_ {N}],$ and

$\displaystyle D_{ij}=\delta_{ij}\lambda_{i}.$

In that case, you have $S_{N}^{-1}=(P^{-1}DP)^{-1}=P^{-1}D^{-1}P.$

Now $D^{-1}$ is easy to calculate, since

$(\delta_{ij}\dfrac{1}{\lambda_{i}})(\delta_{ij}\la mbda_{i})=\delta_{ij}^{2}=\delta_{ij}.$

Therefore,

$D_{ij}^{-1}=\delta_{ij}\dfrac{1}{\lambda_{i}}.$

Make sense?