Find diagonal matrix $D$ such that $A D$ is Hurwitz












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Let $A in mathbb{R}^{m times m}$. Give necessary and/or sufficient conditions for the existence of a matrix $D in mathbb{R}^{m times m}$ such that all eigenvalues of $AD$ have negative real part (i.e., $AD$ Hurwitz). Some initial thoughts ...




  1. Obviously $A, D$ must have full rank.


  2. If $-A$ is so-called $D$-stable, the result seems to follow, although this is a much stronger condition than I need as it ensures that $AD$ is Hurwitz for all positive diagonal matrices $D$ (I think theres an easy extension there to allow for $D$ to have both positive and negative diagonal elements).


  3. An equivalent formulation is that this is true if and only if there exists a $D$ and a matrix $P succ 0$ such that $P(AD)^{sf T} + (AD)P prec 0$. Letting $K = DP$ this is equivalent to the convex inequality $K^{sf T}A + AK prec 0$ in the variable $K$. If this is feasible for some $K^{star}$, you can then probably argue about the factorization $K^{star} = DP$ with $P succ 0$ and recover an appropriate choice of $D$?



Note: This matrix analysis question is motivated by a problem in systems and control theory.










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    $begingroup$


    Let $A in mathbb{R}^{m times m}$. Give necessary and/or sufficient conditions for the existence of a matrix $D in mathbb{R}^{m times m}$ such that all eigenvalues of $AD$ have negative real part (i.e., $AD$ Hurwitz). Some initial thoughts ...




    1. Obviously $A, D$ must have full rank.


    2. If $-A$ is so-called $D$-stable, the result seems to follow, although this is a much stronger condition than I need as it ensures that $AD$ is Hurwitz for all positive diagonal matrices $D$ (I think theres an easy extension there to allow for $D$ to have both positive and negative diagonal elements).


    3. An equivalent formulation is that this is true if and only if there exists a $D$ and a matrix $P succ 0$ such that $P(AD)^{sf T} + (AD)P prec 0$. Letting $K = DP$ this is equivalent to the convex inequality $K^{sf T}A + AK prec 0$ in the variable $K$. If this is feasible for some $K^{star}$, you can then probably argue about the factorization $K^{star} = DP$ with $P succ 0$ and recover an appropriate choice of $D$?



    Note: This matrix analysis question is motivated by a problem in systems and control theory.










    share|cite|improve this question











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      0












      0








      0





      $begingroup$


      Let $A in mathbb{R}^{m times m}$. Give necessary and/or sufficient conditions for the existence of a matrix $D in mathbb{R}^{m times m}$ such that all eigenvalues of $AD$ have negative real part (i.e., $AD$ Hurwitz). Some initial thoughts ...




      1. Obviously $A, D$ must have full rank.


      2. If $-A$ is so-called $D$-stable, the result seems to follow, although this is a much stronger condition than I need as it ensures that $AD$ is Hurwitz for all positive diagonal matrices $D$ (I think theres an easy extension there to allow for $D$ to have both positive and negative diagonal elements).


      3. An equivalent formulation is that this is true if and only if there exists a $D$ and a matrix $P succ 0$ such that $P(AD)^{sf T} + (AD)P prec 0$. Letting $K = DP$ this is equivalent to the convex inequality $K^{sf T}A + AK prec 0$ in the variable $K$. If this is feasible for some $K^{star}$, you can then probably argue about the factorization $K^{star} = DP$ with $P succ 0$ and recover an appropriate choice of $D$?



      Note: This matrix analysis question is motivated by a problem in systems and control theory.










      share|cite|improve this question











      $endgroup$




      Let $A in mathbb{R}^{m times m}$. Give necessary and/or sufficient conditions for the existence of a matrix $D in mathbb{R}^{m times m}$ such that all eigenvalues of $AD$ have negative real part (i.e., $AD$ Hurwitz). Some initial thoughts ...




      1. Obviously $A, D$ must have full rank.


      2. If $-A$ is so-called $D$-stable, the result seems to follow, although this is a much stronger condition than I need as it ensures that $AD$ is Hurwitz for all positive diagonal matrices $D$ (I think theres an easy extension there to allow for $D$ to have both positive and negative diagonal elements).


      3. An equivalent formulation is that this is true if and only if there exists a $D$ and a matrix $P succ 0$ such that $P(AD)^{sf T} + (AD)P prec 0$. Letting $K = DP$ this is equivalent to the convex inequality $K^{sf T}A + AK prec 0$ in the variable $K$. If this is feasible for some $K^{star}$, you can then probably argue about the factorization $K^{star} = DP$ with $P succ 0$ and recover an appropriate choice of $D$?



      Note: This matrix analysis question is motivated by a problem in systems and control theory.







      matrices matrix-equations matrix-decomposition semidefinite-programming lmis






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      edited Dec 25 '18 at 16:52









      Rodrigo de Azevedo

      13.1k41962




      13.1k41962










      asked Dec 25 '18 at 15:20









      JohnJohn

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      1238






















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          $begingroup$

          Your problem is equivalent to the state feedback problem for LTI systems, so



          $$
          dot{x} = mathcal{A},x + mathcal{B},u
          $$



          with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.






          share|cite|improve this answer









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            $begingroup$

            Your problem is equivalent to the state feedback problem for LTI systems, so



            $$
            dot{x} = mathcal{A},x + mathcal{B},u
            $$



            with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Your problem is equivalent to the state feedback problem for LTI systems, so



              $$
              dot{x} = mathcal{A},x + mathcal{B},u
              $$



              with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Your problem is equivalent to the state feedback problem for LTI systems, so



                $$
                dot{x} = mathcal{A},x + mathcal{B},u
                $$



                with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.






                share|cite|improve this answer









                $endgroup$



                Your problem is equivalent to the state feedback problem for LTI systems, so



                $$
                dot{x} = mathcal{A},x + mathcal{B},u
                $$



                with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 29 '18 at 15:15









                Kwin van der VeenKwin van der Veen

                5,6502828




                5,6502828






























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