Exisitence of the Solution to the Linear Matrix Inequality

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Suppose $A$ is an arbitrary invertible matrix. Does there always exist a square matrix $H$ (does not have to be symmetric), such that $H^TA+A^TH$ is strictly positive definite?



I know as long as $A$ and $-A^T$ do not have common eigenvalue, by the existence of the solution to Sylvester equation, it can be concluded that we can always find such $H$. But does this hold for the general case?




linear-algebra matrix-equations sylvester-equation




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    Suppose $A$ is an arbitrary invertible matrix. Does there always exist a square matrix $H$ (does not have to be symmetric), such that $H^TA+A^TH$ is strictly positive definite?



    I know as long as $A$ and $-A^T$ do not have common eigenvalue, by the existence of the solution to Sylvester equation, it can be concluded that we can always find such $H$. But does this hold for the general case?




    linear-algebra matrix-equations sylvester-equation




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      up vote
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      Suppose $A$ is an arbitrary invertible matrix. Does there always exist a square matrix $H$ (does not have to be symmetric), such that $H^TA+A^TH$ is strictly positive definite?



      I know as long as $A$ and $-A^T$ do not have common eigenvalue, by the existence of the solution to Sylvester equation, it can be concluded that we can always find such $H$. But does this hold for the general case?




      linear-algebra matrix-equations sylvester-equation




      share|cite|improve this question











      Suppose $A$ is an arbitrary invertible matrix. Does there always exist a square matrix $H$ (does not have to be symmetric), such that $H^TA+A^TH$ is strictly positive definite?



      I know as long as $A$ and $-A^T$ do not have common eigenvalue, by the existence of the solution to Sylvester equation, it can be concluded that we can always find such $H$. But does this hold for the general case?





      linear-algebra matrix-equations sylvester-equation





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          One can take $H=A$. In this case $H^TA+A^TH=A^TA+A^TA=2A^TA$.



          Since $A$ is invertible, $Ax=0$ iff $x=0$, thus, for any $xne 0$
          $$
          x^T (2A^TA) x= 2(Ax)^T (Ax)= 2|Ax|_2^2>0.
          $$
          It means that $H^TA+A^TH=2A^TA$ is strictly positive definite.






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            up vote
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            down vote



            accepted










            One can take $H=A$. In this case $H^TA+A^TH=A^TA+A^TA=2A^TA$.



            Since $A$ is invertible, $Ax=0$ iff $x=0$, thus, for any $xne 0$
            $$
            x^T (2A^TA) x= 2(Ax)^T (Ax)= 2|Ax|_2^2>0.
            $$
            It means that $H^TA+A^TH=2A^TA$ is strictly positive definite.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              One can take $H=A$. In this case $H^TA+A^TH=A^TA+A^TA=2A^TA$.



              Since $A$ is invertible, $Ax=0$ iff $x=0$, thus, for any $xne 0$
              $$
              x^T (2A^TA) x= 2(Ax)^T (Ax)= 2|Ax|_2^2>0.
              $$
              It means that $H^TA+A^TH=2A^TA$ is strictly positive definite.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                One can take $H=A$. In this case $H^TA+A^TH=A^TA+A^TA=2A^TA$.



                Since $A$ is invertible, $Ax=0$ iff $x=0$, thus, for any $xne 0$
                $$
                x^T (2A^TA) x= 2(Ax)^T (Ax)= 2|Ax|_2^2>0.
                $$
                It means that $H^TA+A^TH=2A^TA$ is strictly positive definite.






                share|cite|improve this answer













                One can take $H=A$. In this case $H^TA+A^TH=A^TA+A^TA=2A^TA$.



                Since $A$ is invertible, $Ax=0$ iff $x=0$, thus, for any $xne 0$
                $$
                x^T (2A^TA) x= 2(Ax)^T (Ax)= 2|Ax|_2^2>0.
                $$
                It means that $H^TA+A^TH=2A^TA$ is strictly positive definite.







                share|cite|improve this answer













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                share|cite|improve this answer











                answered 2 days ago









                AVK

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