Inverse of a Spectral Matrix

**Antem** · October 26th 2010, 10:23 PM

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

**Ackbeet** · October 27th 2010, 02:07 AM

Is $\{\mathbf{v}_{i}\}$ an orthonormal set?

**Antem** · October 27th 2010, 06:19 AM

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

**Ackbeet** · October 27th 2010, 06:34 AM

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?

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