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I have a matrix $Z$ of the form $Z = \left[Q^{-1}-Q^{-1}A^T\left(AQ^{-1}A^T\right)^{-1}AQ^{-1}\right]\Phi$

where,

  • $\Phi$ is a diagonal matrix of real non-negative values.
  • $\Theta$ (not explicitly in the equation above) is another diagonal matrix of real non-negative values.
  • $Q = \Phi + \Theta$ such that $Q$ is a diagonal matrix of real positive values.
  • $A$ is a rectangular matrix of the form $\left[J \quad -I\right]$ where the elements of $J$ are real values and $I$ is the identity matrix of appropriate size.

Matrix dimensions are as follows:

  • $\Phi, \Theta, Q$ are of dimension $(n+m)\times (n+m)$.
  • $A$ is of dimension $m\times (n+m)$.
  • $J$ is of dimension $m\times n$.
  • $I$ is of dimension $m\times m$.

$Z$ is the plant matrix of a discrete time linear time-invariant state space equation, and I have reason to believe that, given the properties above, $Z$ is always convergent. As the state space equation is in discrete time, the conditions for a convergent plant are that all eigenvalues must have magnitude $\leq 1$. Also, should there be an eigenvalue with magnitude $= 1$, its multiplicity must be 1 to be at least Lyapunov stable.

I would appreciate any help in either proving or disproving my claim.

So far, I have been able to prove that $Q^{-1}\Phi$ and $Q^{-1}A^T\left(AQ^{-1}A^T\right)^{-1}AQ^{-1}\Phi$ are both symmetric positive semi-definite. I have tried applying the Sherman–Morrison–Woodbury formula only to show that $Z$ is singular. I can also show that $Q^{-1}\Phi$ has a spectral radius $\leq 1$ but not the same for $Q^{-1}A^T\left(AQ^{-1}A^T\right)^{-1}AQ^{-1}\Phi$.

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Sorry, didn't see the dimensions. –  nonlinearism Aug 20 '13 at 18:40

1 Answer 1

up vote 2 down vote accepted

In general, for (possibly noninvertible) complex square matrices $A$ and $B$, the spectra of $AB$ and $BA$ are identical. Therefore the spectrum of $Z$ is identical to that of the real symmetric matrix \begin{align*} \widetilde{Z} :=&\Phi^{1/2}\left[Q^{-1}-Q^{-1}A^T\left(AQ^{-1}A^T\right)^{-1}AQ^{-1}\right]\Phi^{1/2}\\ =&\Phi^{1/2}Q^{-1/2}\left[I - Q^{-1/2}A^T(AQ^{-1}A^T)^{-1}AQ^{-1/2} \right]Q^{-1/2}\Phi^{1/2}\\ =&\Phi^{1/2}Q^{-1/2}\left(I-P\right)Q^{-1/2}\Phi^{1/2}, \end{align*} where $P := Q^{-1/2}A^T(AQ^{-1}A^T)^{-1}AQ^{-1/2}$ is an orthogonal projection (because it is real symmetric and $P^2=P$). Therefore, $I-P$ is also an orthogonal projection. As $A$ has full rank, it follows that the eigenvalues of $I-P$ are $0$ (of multiplicity $m$) and $1$ (of multiplicity $n$). Since all diagonal entries of $\Phi^{1/2}Q^{-1/2}$ do not exceed $1$, the spectral norm and in turn the spectral radius of $\widetilde{Z}$ are also $\le1$.

Now, $1$ is a repeated eigenvalue of $\widetilde{Z}$ (or $Z$) if and only if the eigenspaces of $\Phi^{1/2}Q^{-1/2}$ and $I-P$ corresponding to the eigenvalue $1$ have a two- or higher dimensional intersection. Such undesired cases can happen, for instances, when (a) $\Phi=Q$ or (b) when $J=0,\,n\ge2$ and some two among the first $n$ entries of $\Phi^{1/2}Q^{-1/2}$ are equal to $1$. However, if $n=1$ or $\Phi^{1/2}Q^{-1/2}$ has at most one diagonal entry equal to $1$, then it is guaranteed that $1$ is not a repeated eigenvalue of $Z$.

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Wow that is amazing. I didn't realize that the second term in the brackets was the form of a projection. Thank you! –  user91124 Aug 21 '13 at 18:00
    
@user91124 The second term in your square bracket is not a projection. It only becomes one after you pull out $Q^{-1/2}$ from both left and right. –  user1551 Aug 21 '13 at 18:18
    
Thanks again, I see that now. Do you have a source or citation for your proof? or did you come up with it? –  user91124 Aug 21 '13 at 20:10
    
I noticed you edited the proof. I am wondering where the $\Phi$ outside the bracket went in the first equation? –  user91124 Aug 21 '13 at 20:46
    
@user91124 The $\Phi$ does not go away. It becomes $\Phi^{1/2}\Phi^{1/2}$ and I move the second $\Phi^{1/2}$ in front of the square bracket term. Now $Z=[\cdots]\Phi^{1/2}\Phi^{1/2}$ and $\widetilde{Z}=\Phi^{1/2}[\cdots]\Phi^{1/2}$ have identical spectra because this is true for $AB$ and $BA$ for general complex square matrices $A$ and $B$. –  user1551 Aug 22 '13 at 14:46

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