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For a real square matrix $S$ with all eigenvalues on the imaginary axis, does the following hold?
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Given a real square matrix $S$ which has all the eigenvalues on the imaginary axis, is it true that there exists a positive definite matrix $P=P^T>0$ such that
beginequation
S^TP+PS=0?
endequation
I found that claim in a research paper (which is not yet available online!) but I do not find a way to prove this.
While trying to do an example problem, I noticed that for a matrix
beginequation
S=beginbmatrixomega & sigma\-sigma & omegaendbmatrix,
endequation
the claim holds if $P$ is an identity matrix. Additionally, for any block-diagonal matrix $S_1$ with the diagonal matrix being in the same form as $S$, there exists a corresponding unique $P$ which satisfies the claim.
But in general, the matrix which has all eigenvalues on the imaginary axis, needs not have the same form as $S_1$, and is there a way to show that there exists a unique $P$ for which the claim holds?
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition positive-definite
asked Jul 16 at 19:04
jbgujgu
1139
add a comment |Â
up vote
2
down vote
favorite
Given a real square matrix $S$ which has all the eigenvalues on the imaginary axis, is it true that there exists a positive definite matrix $P=P^T>0$ such that
beginequation
S^TP+PS=0?
endequation
I found that claim in a research paper (which is not yet available online!) but I do not find a way to prove this.
While trying to do an example problem, I noticed that for a matrix
beginequation
S=beginbmatrixomega & sigma\-sigma & omegaendbmatrix,
endequation
the claim holds if $P$ is an identity matrix. Additionally, for any block-diagonal matrix $S_1$ with the diagonal matrix being in the same form as $S$, there exists a corresponding unique $P$ which satisfies the claim.
But in general, the matrix which has all eigenvalues on the imaginary axis, needs not have the same form as $S_1$, and is there a way to show that there exists a unique $P$ for which the claim holds?
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition positive-definite
asked Jul 16 at 19:04
jbgujgu
1139
Considering real matrices only.
– jbgujgu
Jul 16 at 19:23
The result is clearly true for skew-symmetric matrices with $P = I$.
– mechanodroid
Jul 16 at 19:31
Yeah that's obvious. But I only know that $S$ has all eigenvalues on the imaginary axis and nothing on its symmetricity.
– jbgujgu
Jul 16 at 19:34
Is $P$ also real?
– Mostafa Ayaz
Jul 16 at 19:47
Yes, the matrix $P$ is also real.
– jbgujgu
Jul 16 at 19:52
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given a real square matrix $S$ which has all the eigenvalues on the imaginary axis, is it true that there exists a positive definite matrix $P=P^T>0$ such that
beginequation
S^TP+PS=0?
endequation
I found that claim in a research paper (which is not yet available online!) but I do not find a way to prove this.
While trying to do an example problem, I noticed that for a matrix
beginequation
S=beginbmatrixomega & sigma\-sigma & omegaendbmatrix,
endequation
the claim holds if $P$ is an identity matrix. Additionally, for any block-diagonal matrix $S_1$ with the diagonal matrix being in the same form as $S$, there exists a corresponding unique $P$ which satisfies the claim.
But in general, the matrix which has all eigenvalues on the imaginary axis, needs not have the same form as $S_1$, and is there a way to show that there exists a unique $P$ for which the claim holds?
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition positive-definite
asked Jul 16 at 19:04
jbgujgu
1139
Given a real square matrix $S$ which has all the eigenvalues on the imaginary axis, is it true that there exists a positive definite matrix $P=P^T>0$ such that
beginequation
S^TP+PS=0?
endequation
I found that claim in a research paper (which is not yet available online!) but I do not find a way to prove this.
While trying to do an example problem, I noticed that for a matrix
beginequation
S=beginbmatrixomega & sigma\-sigma & omegaendbmatrix,
endequation
the claim holds if $P$ is an identity matrix. Additionally, for any block-diagonal matrix $S_1$ with the diagonal matrix being in the same form as $S$, there exists a corresponding unique $P$ which satisfies the claim.
But in general, the matrix which has all eigenvalues on the imaginary axis, needs not have the same form as $S_1$, and is there a way to show that there exists a unique $P$ for which the claim holds?
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition positive-definite
asked Jul 16 at 19:04
jbgujgu
1139
edited Jul 16 at 19:22
asked Jul 16 at 19:04
jbgujgu
1139
asked Jul 16 at 19:04
jbgujgu
1139
asked Jul 16 at 19:04
jbgujgu
1139
1139
Considering real matrices only.
– jbgujgu
Jul 16 at 19:23
The result is clearly true for skew-symmetric matrices with $P = I$.
– mechanodroid
Jul 16 at 19:31
Yeah that's obvious. But I only know that $S$ has all eigenvalues on the imaginary axis and nothing on its symmetricity.
– jbgujgu
Jul 16 at 19:34
Is $P$ also real?
– Mostafa Ayaz
Jul 16 at 19:47
Yes, the matrix $P$ is also real.
– jbgujgu
Jul 16 at 19:52
add a comment |Â
Considering real matrices only.
– jbgujgu
Jul 16 at 19:23
The result is clearly true for skew-symmetric matrices with $P = I$.
– mechanodroid
Jul 16 at 19:31
Yeah that's obvious. But I only know that $S$ has all eigenvalues on the imaginary axis and nothing on its symmetricity.
– jbgujgu
Jul 16 at 19:34
Is $P$ also real?
– Mostafa Ayaz
Jul 16 at 19:47
Yes, the matrix $P$ is also real.
– jbgujgu
Jul 16 at 19:52
Considering real matrices only.
– jbgujgu
Jul 16 at 19:23
Considering real matrices only.
– jbgujgu
Jul 16 at 19:23
The result is clearly true for skew-symmetric matrices with $P = I$.
– mechanodroid
Jul 16 at 19:31
The result is clearly true for skew-symmetric matrices with $P = I$.
– mechanodroid
Jul 16 at 19:31
Yeah that's obvious. But I only know that $S$ has all eigenvalues on the imaginary axis and nothing on its symmetricity.
– jbgujgu
Jul 16 at 19:34
Yeah that's obvious. But I only know that $S$ has all eigenvalues on the imaginary axis and nothing on its symmetricity.
– jbgujgu
Jul 16 at 19:34
Is $P$ also real?
– Mostafa Ayaz
Jul 16 at 19:47
Is $P$ also real?
– Mostafa Ayaz
Jul 16 at 19:47
Yes, the matrix $P$ is also real.
– jbgujgu
Jul 16 at 19:52
Yes, the matrix $P$ is also real.
– jbgujgu
Jul 16 at 19:52
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
The answer is no. Here $Sin M_n(mathbbR)$ where $n$ is even.
Choose $n=4,S=beginpmatrixU&I_2\0&Uendpmatrix$ where $U=beginpmatrix0&1\-1&0endpmatrix$. Let $P=[p_i,j]in M_4(mathbbR)$ s.t. $S^TP+PS=0$.
With an easy calculation, we find that $P$ is in the form $beginpmatrix0&0&0&-a\0&0&a&0\0&a&b&0\-a&0&0&bendpmatrix$; thus $P$ cannot be symmetric $>0$.
answered Jul 16 at 22:10
loup blanc
20.4k21549
add a comment |Â
up vote
3
down vote
That $S^TP+PS=0$ means $PS$ is skew-symmetric. In turn, $P^-1/2(PS)P^-1/2=P^1/2SP^-1/2$ is skew-symmetric too. Thus a necessary condition for $P$ to exist is that $S$ is diagonalisable over $mathbb C$.
This condition is also sufficient: if $S$ is diagonalisable over $mathbb C$, it admits a real Jordan form $S=M^-1KM$. As $S$ has a purely imaginary spectrum and non-real eigenvalues of a real matrix must occur in conjugate pairs, $K$ is skew-symmetric. Hence $PS=M^TKM$ is skew-symmetric for $P=M^TM>0$.
answered Jul 16 at 22:56
user1551
66.9k564122
Since $P^1/2SP^-1/2$ is real skew-symmetric matrix and any real skew-symmetric matrix is also a normal matrix and thus unitarily diagonalisable, $P^1/2SP^-1/2$ is diagonalisable. But does it imply diagonalisability of $S$?
– jbgujgu
Jul 17 at 12:50
@jbgujgu They are similar matrices. If one of them is diagonalisable, so is the other.
– user1551
Jul 17 at 21:08
add a comment |Â
up vote
0
down vote
Since $P$ is also real we have $P=P^*$ therefore $$P=P^H$$using Schur decomposition for $P$ we have$$P=UDU^H$$ where $D$ is diagonal with all the entries being real. For satisfying $S^TP+PS=0$ (or equivalently $S^HP+PS=0$) we must have:$$P^-1S^HP+S=0\UD^-1U^HS^HUDU^H+S=0\D^-1U^HS^HUDU+U^HSU=0$$defining $T=U^HSU$ we have$$D^-1TD+T=0$$or $$TD+DT=0$$ also by daggering it $$T^HD+DT^H=0$$which yields to $$D(T+T^H)+(T+T^H)D=0$$defining $V=T+T^H$ where $V$ is also hermition we obtain$$DV+VD=0$$The (ith,jth) entry of $DV$ is $d_iiv_ij$ and the (ith,jth) entry of $VD$ is $d_jjv_ij$ where$$d_iiv_ij+d_jjv_ij=0$$ which means that $v_ij=0$ for all $i,j$ since $D$ is positive definite and $d_ii>0quad,quadforall i$. This means that $V=0$ or $T^H=-T$ or equivalently $S^H=-S$ which is the only solution with $P=I$.
answered Jul 16 at 20:18
Mostafa Ayaz
8,6023630
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The answer is no. Here $Sin M_n(mathbbR)$ where $n$ is even.
Choose $n=4,S=beginpmatrixU&I_2\0&Uendpmatrix$ where $U=beginpmatrix0&1\-1&0endpmatrix$. Let $P=[p_i,j]in M_4(mathbbR)$ s.t. $S^TP+PS=0$.
With an easy calculation, we find that $P$ is in the form $beginpmatrix0&0&0&-a\0&0&a&0\0&a&b&0\-a&0&0&bendpmatrix$; thus $P$ cannot be symmetric $>0$.
answered Jul 16 at 22:10
loup blanc
20.4k21549
add a comment |Â
up vote
3
down vote
The answer is no. Here $Sin M_n(mathbbR)$ where $n$ is even.
Choose $n=4,S=beginpmatrixU&I_2\0&Uendpmatrix$ where $U=beginpmatrix0&1\-1&0endpmatrix$. Let $P=[p_i,j]in M_4(mathbbR)$ s.t. $S^TP+PS=0$.
With an easy calculation, we find that $P$ is in the form $beginpmatrix0&0&0&-a\0&0&a&0\0&a&b&0\-a&0&0&bendpmatrix$; thus $P$ cannot be symmetric $>0$.
answered Jul 16 at 22:10
loup blanc
20.4k21549
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The answer is no. Here $Sin M_n(mathbbR)$ where $n$ is even.
Choose $n=4,S=beginpmatrixU&I_2\0&Uendpmatrix$ where $U=beginpmatrix0&1\-1&0endpmatrix$. Let $P=[p_i,j]in M_4(mathbbR)$ s.t. $S^TP+PS=0$.
With an easy calculation, we find that $P$ is in the form $beginpmatrix0&0&0&-a\0&0&a&0\0&a&b&0\-a&0&0&bendpmatrix$; thus $P$ cannot be symmetric $>0$.
answered Jul 16 at 22:10
loup blanc
20.4k21549
The answer is no. Here $Sin M_n(mathbbR)$ where $n$ is even.
Choose $n=4,S=beginpmatrixU&I_2\0&Uendpmatrix$ where $U=beginpmatrix0&1\-1&0endpmatrix$. Let $P=[p_i,j]in M_4(mathbbR)$ s.t. $S^TP+PS=0$.
With an easy calculation, we find that $P$ is in the form $beginpmatrix0&0&0&-a\0&0&a&0\0&a&b&0\-a&0&0&bendpmatrix$; thus $P$ cannot be symmetric $>0$.
answered Jul 16 at 22:10
loup blanc
20.4k21549
answered Jul 16 at 22:10
loup blanc
20.4k21549
answered Jul 16 at 22:10
loup blanc
20.4k21549
answered Jul 16 at 22:10
loup blanc
20.4k21549
20.4k21549
add a comment |Â
add a comment |Â
up vote
3
down vote
That $S^TP+PS=0$ means $PS$ is skew-symmetric. In turn, $P^-1/2(PS)P^-1/2=P^1/2SP^-1/2$ is skew-symmetric too. Thus a necessary condition for $P$ to exist is that $S$ is diagonalisable over $mathbb C$.
This condition is also sufficient: if $S$ is diagonalisable over $mathbb C$, it admits a real Jordan form $S=M^-1KM$. As $S$ has a purely imaginary spectrum and non-real eigenvalues of a real matrix must occur in conjugate pairs, $K$ is skew-symmetric. Hence $PS=M^TKM$ is skew-symmetric for $P=M^TM>0$.
answered Jul 16 at 22:56
user1551
66.9k564122
Since $P^1/2SP^-1/2$ is real skew-symmetric matrix and any real skew-symmetric matrix is also a normal matrix and thus unitarily diagonalisable, $P^1/2SP^-1/2$ is diagonalisable. But does it imply diagonalisability of $S$?
– jbgujgu
Jul 17 at 12:50
@jbgujgu They are similar matrices. If one of them is diagonalisable, so is the other.
– user1551
Jul 17 at 21:08
add a comment |Â
up vote
3
down vote
That $S^TP+PS=0$ means $PS$ is skew-symmetric. In turn, $P^-1/2(PS)P^-1/2=P^1/2SP^-1/2$ is skew-symmetric too. Thus a necessary condition for $P$ to exist is that $S$ is diagonalisable over $mathbb C$.
This condition is also sufficient: if $S$ is diagonalisable over $mathbb C$, it admits a real Jordan form $S=M^-1KM$. As $S$ has a purely imaginary spectrum and non-real eigenvalues of a real matrix must occur in conjugate pairs, $K$ is skew-symmetric. Hence $PS=M^TKM$ is skew-symmetric for $P=M^TM>0$.
answered Jul 16 at 22:56
user1551
66.9k564122
Since $P^1/2SP^-1/2$ is real skew-symmetric matrix and any real skew-symmetric matrix is also a normal matrix and thus unitarily diagonalisable, $P^1/2SP^-1/2$ is diagonalisable. But does it imply diagonalisability of $S$?
– jbgujgu
Jul 17 at 12:50
@jbgujgu They are similar matrices. If one of them is diagonalisable, so is the other.
– user1551
Jul 17 at 21:08
add a comment |Â
up vote
3
down vote
up vote
3
down vote
That $S^TP+PS=0$ means $PS$ is skew-symmetric. In turn, $P^-1/2(PS)P^-1/2=P^1/2SP^-1/2$ is skew-symmetric too. Thus a necessary condition for $P$ to exist is that $S$ is diagonalisable over $mathbb C$.
This condition is also sufficient: if $S$ is diagonalisable over $mathbb C$, it admits a real Jordan form $S=M^-1KM$. As $S$ has a purely imaginary spectrum and non-real eigenvalues of a real matrix must occur in conjugate pairs, $K$ is skew-symmetric. Hence $PS=M^TKM$ is skew-symmetric for $P=M^TM>0$.
answered Jul 16 at 22:56
user1551
66.9k564122
That $S^TP+PS=0$ means $PS$ is skew-symmetric. In turn, $P^-1/2(PS)P^-1/2=P^1/2SP^-1/2$ is skew-symmetric too. Thus a necessary condition for $P$ to exist is that $S$ is diagonalisable over $mathbb C$.
This condition is also sufficient: if $S$ is diagonalisable over $mathbb C$, it admits a real Jordan form $S=M^-1KM$. As $S$ has a purely imaginary spectrum and non-real eigenvalues of a real matrix must occur in conjugate pairs, $K$ is skew-symmetric. Hence $PS=M^TKM$ is skew-symmetric for $P=M^TM>0$.
answered Jul 16 at 22:56
user1551
66.9k564122
answered Jul 16 at 22:56
user1551
66.9k564122
answered Jul 16 at 22:56
user1551
66.9k564122
answered Jul 16 at 22:56
user1551
66.9k564122
66.9k564122
Since $P^1/2SP^-1/2$ is real skew-symmetric matrix and any real skew-symmetric matrix is also a normal matrix and thus unitarily diagonalisable, $P^1/2SP^-1/2$ is diagonalisable. But does it imply diagonalisability of $S$?
– jbgujgu
Jul 17 at 12:50
@jbgujgu They are similar matrices. If one of them is diagonalisable, so is the other.
– user1551
Jul 17 at 21:08
add a comment |Â
Since $P^1/2SP^-1/2$ is real skew-symmetric matrix and any real skew-symmetric matrix is also a normal matrix and thus unitarily diagonalisable, $P^1/2SP^-1/2$ is diagonalisable. But does it imply diagonalisability of $S$?
– jbgujgu
Jul 17 at 12:50
@jbgujgu They are similar matrices. If one of them is diagonalisable, so is the other.
– user1551
Jul 17 at 21:08
Since $P^1/2SP^-1/2$ is real skew-symmetric matrix and any real skew-symmetric matrix is also a normal matrix and thus unitarily diagonalisable, $P^1/2SP^-1/2$ is diagonalisable. But does it imply diagonalisability of $S$?
– jbgujgu
Jul 17 at 12:50
Since $P^1/2SP^-1/2$ is real skew-symmetric matrix and any real skew-symmetric matrix is also a normal matrix and thus unitarily diagonalisable, $P^1/2SP^-1/2$ is diagonalisable. But does it imply diagonalisability of $S$?
– jbgujgu
Jul 17 at 12:50
@jbgujgu They are similar matrices. If one of them is diagonalisable, so is the other.
– user1551
Jul 17 at 21:08
@jbgujgu They are similar matrices. If one of them is diagonalisable, so is the other.
– user1551
Jul 17 at 21:08
add a comment |Â
up vote
0
down vote
Since $P$ is also real we have $P=P^*$ therefore $$P=P^H$$using Schur decomposition for $P$ we have$$P=UDU^H$$ where $D$ is diagonal with all the entries being real. For satisfying $S^TP+PS=0$ (or equivalently $S^HP+PS=0$) we must have:$$P^-1S^HP+S=0\UD^-1U^HS^HUDU^H+S=0\D^-1U^HS^HUDU+U^HSU=0$$defining $T=U^HSU$ we have$$D^-1TD+T=0$$or $$TD+DT=0$$ also by daggering it $$T^HD+DT^H=0$$which yields to $$D(T+T^H)+(T+T^H)D=0$$defining $V=T+T^H$ where $V$ is also hermition we obtain$$DV+VD=0$$The (ith,jth) entry of $DV$ is $d_iiv_ij$ and the (ith,jth) entry of $VD$ is $d_jjv_ij$ where$$d_iiv_ij+d_jjv_ij=0$$ which means that $v_ij=0$ for all $i,j$ since $D$ is positive definite and $d_ii>0quad,quadforall i$. This means that $V=0$ or $T^H=-T$ or equivalently $S^H=-S$ which is the only solution with $P=I$.
answered Jul 16 at 20:18
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
0
down vote
Since $P$ is also real we have $P=P^*$ therefore $$P=P^H$$using Schur decomposition for $P$ we have$$P=UDU^H$$ where $D$ is diagonal with all the entries being real. For satisfying $S^TP+PS=0$ (or equivalently $S^HP+PS=0$) we must have:$$P^-1S^HP+S=0\UD^-1U^HS^HUDU^H+S=0\D^-1U^HS^HUDU+U^HSU=0$$defining $T=U^HSU$ we have$$D^-1TD+T=0$$or $$TD+DT=0$$ also by daggering it $$T^HD+DT^H=0$$which yields to $$D(T+T^H)+(T+T^H)D=0$$defining $V=T+T^H$ where $V$ is also hermition we obtain$$DV+VD=0$$The (ith,jth) entry of $DV$ is $d_iiv_ij$ and the (ith,jth) entry of $VD$ is $d_jjv_ij$ where$$d_iiv_ij+d_jjv_ij=0$$ which means that $v_ij=0$ for all $i,j$ since $D$ is positive definite and $d_ii>0quad,quadforall i$. This means that $V=0$ or $T^H=-T$ or equivalently $S^H=-S$ which is the only solution with $P=I$.
answered Jul 16 at 20:18
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $P$ is also real we have $P=P^*$ therefore $$P=P^H$$using Schur decomposition for $P$ we have$$P=UDU^H$$ where $D$ is diagonal with all the entries being real. For satisfying $S^TP+PS=0$ (or equivalently $S^HP+PS=0$) we must have:$$P^-1S^HP+S=0\UD^-1U^HS^HUDU^H+S=0\D^-1U^HS^HUDU+U^HSU=0$$defining $T=U^HSU$ we have$$D^-1TD+T=0$$or $$TD+DT=0$$ also by daggering it $$T^HD+DT^H=0$$which yields to $$D(T+T^H)+(T+T^H)D=0$$defining $V=T+T^H$ where $V$ is also hermition we obtain$$DV+VD=0$$The (ith,jth) entry of $DV$ is $d_iiv_ij$ and the (ith,jth) entry of $VD$ is $d_jjv_ij$ where$$d_iiv_ij+d_jjv_ij=0$$ which means that $v_ij=0$ for all $i,j$ since $D$ is positive definite and $d_ii>0quad,quadforall i$. This means that $V=0$ or $T^H=-T$ or equivalently $S^H=-S$ which is the only solution with $P=I$.
answered Jul 16 at 20:18
Mostafa Ayaz
8,6023630
Since $P$ is also real we have $P=P^*$ therefore $$P=P^H$$using Schur decomposition for $P$ we have$$P=UDU^H$$ where $D$ is diagonal with all the entries being real. For satisfying $S^TP+PS=0$ (or equivalently $S^HP+PS=0$) we must have:$$P^-1S^HP+S=0\UD^-1U^HS^HUDU^H+S=0\D^-1U^HS^HUDU+U^HSU=0$$defining $T=U^HSU$ we have$$D^-1TD+T=0$$or $$TD+DT=0$$ also by daggering it $$T^HD+DT^H=0$$which yields to $$D(T+T^H)+(T+T^H)D=0$$defining $V=T+T^H$ where $V$ is also hermition we obtain$$DV+VD=0$$The (ith,jth) entry of $DV$ is $d_iiv_ij$ and the (ith,jth) entry of $VD$ is $d_jjv_ij$ where$$d_iiv_ij+d_jjv_ij=0$$ which means that $v_ij=0$ for all $i,j$ since $D$ is positive definite and $d_ii>0quad,quadforall i$. This means that $V=0$ or $T^H=-T$ or equivalently $S^H=-S$ which is the only solution with $P=I$.
answered Jul 16 at 20:18
Mostafa Ayaz
8,6023630
edited Jul 23 at 8:45
answered Jul 16 at 20:18
Mostafa Ayaz
8,6023630
answered Jul 16 at 20:18
Mostafa Ayaz
8,6023630
answered Jul 16 at 20:18
Mostafa Ayaz
8,6023630
8,6023630
add a comment |Â
add a comment |Â
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Considering real matrices only.
– jbgujgu
Jul 16 at 19:23
The result is clearly true for skew-symmetric matrices with $P = I$.
– mechanodroid
Jul 16 at 19:31
Yeah that's obvious. But I only know that $S$ has all eigenvalues on the imaginary axis and nothing on its symmetricity.
– jbgujgu
Jul 16 at 19:34
Is $P$ also real?
– Mostafa Ayaz
Jul 16 at 19:47
Yes, the matrix $P$ is also real.
– jbgujgu
Jul 16 at 19:52