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For what kind of matrix $A$, there is a (symmetric) positive definite matrix $B$ such that $BA$ is symmetric
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
asked Nov 21 '18 at 15:25
Tony B
776418
add a comment |
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
asked Nov 21 '18 at 15:25
Tony B
776418
add a comment |
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
asked Nov 21 '18 at 15:25
Tony B
776418
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
linear-algebra matrices matrix-equations positive-definite symmetric-matrices
asked Nov 21 '18 at 15:25
Tony B
776418
asked Nov 21 '18 at 15:25
Tony B
776418
asked Nov 21 '18 at 15:25
Tony B
776418
asked Nov 21 '18 at 15:25
Tony B
776418
asked Nov 21 '18 at 15:25
Tony B
776418
776418
add a comment |
add a comment |
1 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.
answered Nov 21 '18 at 15:56
user1551
71.7k566125
add a comment |
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1 Answer
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1 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.
answered Nov 21 '18 at 15:56
user1551
71.7k566125
add a comment |
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.
answered Nov 21 '18 at 15:56
user1551
71.7k566125
add a comment |
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.
answered Nov 21 '18 at 15:56
user1551
71.7k566125
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.
answered Nov 21 '18 at 15:56
user1551
71.7k566125
answered Nov 21 '18 at 15:56
user1551
71.7k566125
answered Nov 21 '18 at 15:56
user1551
71.7k566125
answered Nov 21 '18 at 15:56
user1551
71.7k566125
71.7k566125
add a comment |
add a comment |
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