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Let $P_1 = \sum_{k=1}^m A_kA_k^T$ and $P_0 = \sum_{k=1}^m c_k(t)\cdot A_kA_k^T$ with $c_k(t) \in [a,b]$ where $a,b \in (0,1)$. We will also assume that $P_1$ has full rank. Then it is obvious that $P_0,P_1$ are symmetric positive definite matrices.

Question:

I want to find $Q_1,Q_0$ such that $$ \begin{cases} Q_1 + Q_0 = P_1\\ Q_1\cdot Q_0 = P_0\end{cases}$$ Do these matrices, $Q_1,Q_0$ exist? If so, are their eigenvalues with positive real part?

My attempt:

Since $P_1,P_0$ commute we will treat them as scalars ... Hence let $ Q_1 = P_1 - Q_0$ then $$ (P_1 - Q_0)\cdot Q_0 = P_0 \iff Q_0^2 - P_1\cdot Q_0 + P_0 = 0_{n\times 1}$$ From here follows $$ Q_0 = \frac{1}{2} \cdot \left( P_1 + \left( P_1^2 - 4\cdot P_0\right)^{\frac{1}{2}}\right)$$

$$ Q_1 = \frac{1}{2} \cdot \left( P_1 - \left( P_1^2 - 4\cdot P_0\right)^{\frac{1}{2}}\right)$$ If this is true, all that is left is to decide whether the eigenvalues of $Q_1, Q_0$ have positive real part or not ... Can some one confirm this and give some ideas regarding how to proceed for prooving the eigenvalue property?

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$P_0$ and $P_1$ are diagonalizable and they commute. Therefore, they are simultaneously diagonalizable, i.e. there is an invertible matrix $S$ such that $P_0 = SD_0S^{-1}$ and $P_1 = SD_1S^{-1}$ with diagonal matrices $D_0$ and $D_1.$ Now we set $Q_0=SC_0S^{-1}$ and $Q_1=SC_1S^{-1}$ and we get \begin{eqnarray*} Q_1+Q_0 &=& SC_0S^{-1}+SC_1S^{-1} = S(C_0+C_1)S^{-1} \\ P_1 &=& \phantom{SC_0S^{-1}+SC_1S^{-1} = } SD_1S^{-1} \\ Q_1Q_0 &=& SC_0S^{-1}\cdot SC_1S^{-1} = SC_0C_1S^{-1} \\ P_0 &=& \phantom{SC_0S^{-1}\cdot SC_1S^{-1} = } SD_0S^{-1} \end{eqnarray*} So it is obviously sufficient to solve $$ C_0+C_1 = D_1 \\ C_0C_1 = D_0 $$ for $C_0$ and $C_1.$ $C_0$ and $C_1$ in turn are diagonal matrices which can be found by looking at each diagonal element of $D_0$ and $D_1$ in isolation. It can now easily be verified that $C_0$ and $C_1$ are diagonal matrices with elements that have a positive real part.

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  • $\begingroup$ Sir, thank you very much! Certainly, this is what I was hoping to be true! I did not know thou about that convenient, beautiful property of simultaneous diagonalization ... shame on me! Can you please give a reference to that property? $\endgroup$ – C Marius Mar 4 at 11:48
  • $\begingroup$ @CMarius The proof can be found here. The statement can also be found here $\endgroup$ – Reinhard Meier Mar 4 at 12:49

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