For what kind of matrix $A$, there is a (symmetric) positive definite matrix $B$ such that $BA$ is symmetric












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Let $A$ be a $ntimes n$ real matrix. My ultimate goal is to find a sufficient condition on $A$ such that all the eigenvalues of $A$ are real.



Therefore, I want $A$ to be self-adjoint with respect to some inner-product. A general inner product associated with a positive definite matrix $B$ is given by $<x,y>=x^tBy$. Then $A$ is self-adjoint if $A^tB=BA$, i.e. $BA$ is symmetric.



So my problem reduces to "for what kind of matrix $A$, there is a (symmetric) positive definite matrix $B$ such that $BA$ is symmetric".



My idea: Given $A$, we want to find a positive definite $ntimes n$ matrix $B$ by solving the equation $A^tB=BA$. There are $frac{n(n+1)}{2}$ variables (not $n^2$) as $B$ is symmetric. There are $n^2$ equations but it seems that only $frac{n(n-1)}{2}$ of them are independent. Therefore, the solutions should be abundant as there are more variables than equations. However, I don't know how to make sure the solution is positive definite. It is possible that the solution space does not intersect with the positive cone of positive definite matrices, except at $0$.










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    Let $A$ be a $ntimes n$ real matrix. My ultimate goal is to find a sufficient condition on $A$ such that all the eigenvalues of $A$ are real.



    Therefore, I want $A$ to be self-adjoint with respect to some inner-product. A general inner product associated with a positive definite matrix $B$ is given by $<x,y>=x^tBy$. Then $A$ is self-adjoint if $A^tB=BA$, i.e. $BA$ is symmetric.



    So my problem reduces to "for what kind of matrix $A$, there is a (symmetric) positive definite matrix $B$ such that $BA$ is symmetric".



    My idea: Given $A$, we want to find a positive definite $ntimes n$ matrix $B$ by solving the equation $A^tB=BA$. There are $frac{n(n+1)}{2}$ variables (not $n^2$) as $B$ is symmetric. There are $n^2$ equations but it seems that only $frac{n(n-1)}{2}$ of them are independent. Therefore, the solutions should be abundant as there are more variables than equations. However, I don't know how to make sure the solution is positive definite. It is possible that the solution space does not intersect with the positive cone of positive definite matrices, except at $0$.










    share|cite|improve this question

























      0












      0








      0







      Let $A$ be a $ntimes n$ real matrix. My ultimate goal is to find a sufficient condition on $A$ such that all the eigenvalues of $A$ are real.



      Therefore, I want $A$ to be self-adjoint with respect to some inner-product. A general inner product associated with a positive definite matrix $B$ is given by $<x,y>=x^tBy$. Then $A$ is self-adjoint if $A^tB=BA$, i.e. $BA$ is symmetric.



      So my problem reduces to "for what kind of matrix $A$, there is a (symmetric) positive definite matrix $B$ such that $BA$ is symmetric".



      My idea: Given $A$, we want to find a positive definite $ntimes n$ matrix $B$ by solving the equation $A^tB=BA$. There are $frac{n(n+1)}{2}$ variables (not $n^2$) as $B$ is symmetric. There are $n^2$ equations but it seems that only $frac{n(n-1)}{2}$ of them are independent. Therefore, the solutions should be abundant as there are more variables than equations. However, I don't know how to make sure the solution is positive definite. It is possible that the solution space does not intersect with the positive cone of positive definite matrices, except at $0$.










      share|cite|improve this question













      Let $A$ be a $ntimes n$ real matrix. My ultimate goal is to find a sufficient condition on $A$ such that all the eigenvalues of $A$ are real.



      Therefore, I want $A$ to be self-adjoint with respect to some inner-product. A general inner product associated with a positive definite matrix $B$ is given by $<x,y>=x^tBy$. Then $A$ is self-adjoint if $A^tB=BA$, i.e. $BA$ is symmetric.



      So my problem reduces to "for what kind of matrix $A$, there is a (symmetric) positive definite matrix $B$ such that $BA$ is symmetric".



      My idea: Given $A$, we want to find a positive definite $ntimes n$ matrix $B$ by solving the equation $A^tB=BA$. There are $frac{n(n+1)}{2}$ variables (not $n^2$) as $B$ is symmetric. There are $n^2$ equations but it seems that only $frac{n(n-1)}{2}$ of them are independent. Therefore, the solutions should be abundant as there are more variables than equations. However, I don't know how to make sure the solution is positive definite. It is possible that the solution space does not intersect with the positive cone of positive definite matrices, except at $0$.







      linear-algebra matrices matrix-equations positive-definite symmetric-matrices






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      asked Nov 21 '18 at 15:25









      Tony B

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          Indeed, such a matrix $B$ exists if and only if $A$ has a real spectrum.



          If $A$ has a real spectrum, then $A=P^{-1}DP$ for some real matrix $P$ and diagonal matrix $D$. Therefore, $B:=P^TP$ is positive definite and $BA=P^TDP$ is symmetric.



          Conversely, suppose $S:=BA$ is symmetric for some positive definite $B$. Then $A=B^{-1}S$ is similar to $B^{-1/2}SB^{-1/2}$ and hence it has a real spectrum.






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            Indeed, such a matrix $B$ exists if and only if $A$ has a real spectrum.



            If $A$ has a real spectrum, then $A=P^{-1}DP$ for some real matrix $P$ and diagonal matrix $D$. Therefore, $B:=P^TP$ is positive definite and $BA=P^TDP$ is symmetric.



            Conversely, suppose $S:=BA$ is symmetric for some positive definite $B$. Then $A=B^{-1}S$ is similar to $B^{-1/2}SB^{-1/2}$ and hence it has a real spectrum.






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              0














              Indeed, such a matrix $B$ exists if and only if $A$ has a real spectrum.



              If $A$ has a real spectrum, then $A=P^{-1}DP$ for some real matrix $P$ and diagonal matrix $D$. Therefore, $B:=P^TP$ is positive definite and $BA=P^TDP$ is symmetric.



              Conversely, suppose $S:=BA$ is symmetric for some positive definite $B$. Then $A=B^{-1}S$ is similar to $B^{-1/2}SB^{-1/2}$ and hence it has a real spectrum.






              share|cite|improve this answer
























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                Indeed, such a matrix $B$ exists if and only if $A$ has a real spectrum.



                If $A$ has a real spectrum, then $A=P^{-1}DP$ for some real matrix $P$ and diagonal matrix $D$. Therefore, $B:=P^TP$ is positive definite and $BA=P^TDP$ is symmetric.



                Conversely, suppose $S:=BA$ is symmetric for some positive definite $B$. Then $A=B^{-1}S$ is similar to $B^{-1/2}SB^{-1/2}$ and hence it has a real spectrum.






                share|cite|improve this answer












                Indeed, such a matrix $B$ exists if and only if $A$ has a real spectrum.



                If $A$ has a real spectrum, then $A=P^{-1}DP$ for some real matrix $P$ and diagonal matrix $D$. Therefore, $B:=P^TP$ is positive definite and $BA=P^TDP$ is symmetric.



                Conversely, suppose $S:=BA$ is symmetric for some positive definite $B$. Then $A=B^{-1}S$ is similar to $B^{-1/2}SB^{-1/2}$ and hence it has a real spectrum.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 '18 at 15:56









                user1551

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                71.7k566125






























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