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Eigenvalues of product of positive semidefinite matrices are greater or equal to zero
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Show that if $A,B in mathcal M_n(mathbbR)$ are positive semidefinite and $lambda$ is an eigenvalue of $AB$ then $lambda geq 0$.
I don't really know what to do here. If $AB$ was semidefinite positive we would be done but the product of positive semidefinite matrices doesn't have to be positive semidefinite itself.
matrices eigenvalues-eigenvectors
asked Sep 8 at 23:17
Yagger
6841315
add a comment |Â
up vote
2
down vote
favorite
Show that if $A,B in mathcal M_n(mathbbR)$ are positive semidefinite and $lambda$ is an eigenvalue of $AB$ then $lambda geq 0$.
I don't really know what to do here. If $AB$ was semidefinite positive we would be done but the product of positive semidefinite matrices doesn't have to be positive semidefinite itself.
matrices eigenvalues-eigenvectors
asked Sep 8 at 23:17
Yagger
6841315
are $A,B$ symmetric ?
– Ahmad Bazzi
Sep 8 at 23:22
@AhmadBazzi no they are not.
– Yagger
Sep 8 at 23:24
@Yagger Usually, "positive semidefinite" implies symmetric.
– amsmath
Sep 8 at 23:27
@AhmadBazzi they are symetric
– Yagger
Sep 8 at 23:37
I suspected @Yagger .. I have posted an answer when they are symmetric using square roots of matrices
– Ahmad Bazzi
Sep 8 at 23:39
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Show that if $A,B in mathcal M_n(mathbbR)$ are positive semidefinite and $lambda$ is an eigenvalue of $AB$ then $lambda geq 0$.
I don't really know what to do here. If $AB$ was semidefinite positive we would be done but the product of positive semidefinite matrices doesn't have to be positive semidefinite itself.
matrices eigenvalues-eigenvectors
asked Sep 8 at 23:17
Yagger
6841315
Show that if $A,B in mathcal M_n(mathbbR)$ are positive semidefinite and $lambda$ is an eigenvalue of $AB$ then $lambda geq 0$.
I don't really know what to do here. If $AB$ was semidefinite positive we would be done but the product of positive semidefinite matrices doesn't have to be positive semidefinite itself.
matrices eigenvalues-eigenvectors
matrices eigenvalues-eigenvectors
asked Sep 8 at 23:17
Yagger
6841315
asked Sep 8 at 23:17
Yagger
6841315
asked Sep 8 at 23:17
Yagger
6841315
asked Sep 8 at 23:17
Yagger
6841315
asked Sep 8 at 23:17
Yagger
6841315
6841315
are $A,B$ symmetric ?
– Ahmad Bazzi
Sep 8 at 23:22
@AhmadBazzi no they are not.
– Yagger
Sep 8 at 23:24
@Yagger Usually, "positive semidefinite" implies symmetric.
– amsmath
Sep 8 at 23:27
@AhmadBazzi they are symetric
– Yagger
Sep 8 at 23:37
I suspected @Yagger .. I have posted an answer when they are symmetric using square roots of matrices
– Ahmad Bazzi
Sep 8 at 23:39
add a comment |Â
are $A,B$ symmetric ?
– Ahmad Bazzi
Sep 8 at 23:22
@AhmadBazzi no they are not.
– Yagger
Sep 8 at 23:24
@Yagger Usually, "positive semidefinite" implies symmetric.
– amsmath
Sep 8 at 23:27
@AhmadBazzi they are symetric
– Yagger
Sep 8 at 23:37
I suspected @Yagger .. I have posted an answer when they are symmetric using square roots of matrices
– Ahmad Bazzi
Sep 8 at 23:39
are $A,B$ symmetric ?
– Ahmad Bazzi
Sep 8 at 23:22
are $A,B$ symmetric ?
– Ahmad Bazzi
Sep 8 at 23:22
@AhmadBazzi no they are not.
– Yagger
Sep 8 at 23:24
@AhmadBazzi no they are not.
– Yagger
Sep 8 at 23:24
@Yagger Usually, "positive semidefinite" implies symmetric.
– amsmath
Sep 8 at 23:27
@Yagger Usually, "positive semidefinite" implies symmetric.
– amsmath
Sep 8 at 23:27
@AhmadBazzi they are symetric
– Yagger
Sep 8 at 23:37
@AhmadBazzi they are symetric
– Yagger
Sep 8 at 23:37
I suspected @Yagger .. I have posted an answer when they are symmetric using square roots of matrices
– Ahmad Bazzi
Sep 8 at 23:39
I suspected @Yagger .. I have posted an answer when they are symmetric using square roots of matrices
– Ahmad Bazzi
Sep 8 at 23:39
add a comment |Â
2 Answers
2
active
oldest
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up vote
3
down vote
accepted
If $ABx = lambda x$, $xneq 0$, then $langle ABx,Bxrangle = lambda langle x,Bxrangle$. Now, both $langle ABx,Bxrangle$ and $langle x,Bxrangle$ are non-negative. Hence, if $langle x,Bxrangle > 0$, it follows that $lambdage 0$. If $langle x,Bxrangle = 0$, then $Bx = 0$ and thus $lambda = 0$.
answered Sep 8 at 23:27
amsmath
2,406114
1
For OP's sake, this also holds if you expand the definition of positive-semidefinite to non-symmetric matrices, as $x^TBxgeq 0Leftrightarrow x^TB^Txgeq 0$.
– Alex R.
Sep 8 at 23:30
I agree with @AlexR. (+1 Alex) Btw, you got me cracking "For OP's sake.."
– Ahmad Bazzi
Sep 8 at 23:32
@Alex: But what if $langle ABx,Bxrangle = langle Bx,xrangle = 0$? Then $lambda$ can be anything.
– amsmath
Sep 8 at 23:33
I misread the problem, the are symmetric. Sorry for the error.
– Yagger
Sep 8 at 23:36
add a comment |Â
up vote
2
down vote
Let $A$ and $B$ be positive semidefinite real symmetric matrices.
Denote $sqrtA$ the square root of $A$, i.e.
beginequation
A = sqrtAsqrtA
endequation
where $sqrtA$ is also PSD.
Notice that $$sqrtA B sqrtA$$ is symmetric and positive semidefinite. Also, $$AB = sqrtA(sqrtA B)$$ and $$(sqrtAB)sqrtA$$ have the same nonzero eigenvalues.
answered Sep 8 at 23:39
Ahmad Bazzi
4,7401623
That's an original approach! +1
– Yagger
Sep 8 at 23:49
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If $ABx = lambda x$, $xneq 0$, then $langle ABx,Bxrangle = lambda langle x,Bxrangle$. Now, both $langle ABx,Bxrangle$ and $langle x,Bxrangle$ are non-negative. Hence, if $langle x,Bxrangle > 0$, it follows that $lambdage 0$. If $langle x,Bxrangle = 0$, then $Bx = 0$ and thus $lambda = 0$.
answered Sep 8 at 23:27
amsmath
2,406114
1
For OP's sake, this also holds if you expand the definition of positive-semidefinite to non-symmetric matrices, as $x^TBxgeq 0Leftrightarrow x^TB^Txgeq 0$.
– Alex R.
Sep 8 at 23:30
I agree with @AlexR. (+1 Alex) Btw, you got me cracking "For OP's sake.."
– Ahmad Bazzi
Sep 8 at 23:32
@Alex: But what if $langle ABx,Bxrangle = langle Bx,xrangle = 0$? Then $lambda$ can be anything.
– amsmath
Sep 8 at 23:33
I misread the problem, the are symmetric. Sorry for the error.
– Yagger
Sep 8 at 23:36
add a comment |Â
up vote
3
down vote
accepted
If $ABx = lambda x$, $xneq 0$, then $langle ABx,Bxrangle = lambda langle x,Bxrangle$. Now, both $langle ABx,Bxrangle$ and $langle x,Bxrangle$ are non-negative. Hence, if $langle x,Bxrangle > 0$, it follows that $lambdage 0$. If $langle x,Bxrangle = 0$, then $Bx = 0$ and thus $lambda = 0$.
answered Sep 8 at 23:27
amsmath
2,406114
1
For OP's sake, this also holds if you expand the definition of positive-semidefinite to non-symmetric matrices, as $x^TBxgeq 0Leftrightarrow x^TB^Txgeq 0$.
– Alex R.
Sep 8 at 23:30
I agree with @AlexR. (+1 Alex) Btw, you got me cracking "For OP's sake.."
– Ahmad Bazzi
Sep 8 at 23:32
@Alex: But what if $langle ABx,Bxrangle = langle Bx,xrangle = 0$? Then $lambda$ can be anything.
– amsmath
Sep 8 at 23:33
I misread the problem, the are symmetric. Sorry for the error.
– Yagger
Sep 8 at 23:36
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If $ABx = lambda x$, $xneq 0$, then $langle ABx,Bxrangle = lambda langle x,Bxrangle$. Now, both $langle ABx,Bxrangle$ and $langle x,Bxrangle$ are non-negative. Hence, if $langle x,Bxrangle > 0$, it follows that $lambdage 0$. If $langle x,Bxrangle = 0$, then $Bx = 0$ and thus $lambda = 0$.
answered Sep 8 at 23:27
amsmath
2,406114
If $ABx = lambda x$, $xneq 0$, then $langle ABx,Bxrangle = lambda langle x,Bxrangle$. Now, both $langle ABx,Bxrangle$ and $langle x,Bxrangle$ are non-negative. Hence, if $langle x,Bxrangle > 0$, it follows that $lambdage 0$. If $langle x,Bxrangle = 0$, then $Bx = 0$ and thus $lambda = 0$.
answered Sep 8 at 23:27
amsmath
2,406114
edited Sep 8 at 23:32
answered Sep 8 at 23:27
amsmath
2,406114
answered Sep 8 at 23:27
amsmath
2,406114
answered Sep 8 at 23:27
amsmath
2,406114
2,406114
1
For OP's sake, this also holds if you expand the definition of positive-semidefinite to non-symmetric matrices, as $x^TBxgeq 0Leftrightarrow x^TB^Txgeq 0$.
– Alex R.
Sep 8 at 23:30
I agree with @AlexR. (+1 Alex) Btw, you got me cracking "For OP's sake.."
– Ahmad Bazzi
Sep 8 at 23:32
@Alex: But what if $langle ABx,Bxrangle = langle Bx,xrangle = 0$? Then $lambda$ can be anything.
– amsmath
Sep 8 at 23:33
I misread the problem, the are symmetric. Sorry for the error.
– Yagger
Sep 8 at 23:36
add a comment |Â
1
For OP's sake, this also holds if you expand the definition of positive-semidefinite to non-symmetric matrices, as $x^TBxgeq 0Leftrightarrow x^TB^Txgeq 0$.
– Alex R.
Sep 8 at 23:30
I agree with @AlexR. (+1 Alex) Btw, you got me cracking "For OP's sake.."
– Ahmad Bazzi
Sep 8 at 23:32
@Alex: But what if $langle ABx,Bxrangle = langle Bx,xrangle = 0$? Then $lambda$ can be anything.
– amsmath
Sep 8 at 23:33
I misread the problem, the are symmetric. Sorry for the error.
– Yagger
Sep 8 at 23:36
1
1
For OP's sake, this also holds if you expand the definition of positive-semidefinite to non-symmetric matrices, as $x^TBxgeq 0Leftrightarrow x^TB^Txgeq 0$.
– Alex R.
Sep 8 at 23:30
For OP's sake, this also holds if you expand the definition of positive-semidefinite to non-symmetric matrices, as $x^TBxgeq 0Leftrightarrow x^TB^Txgeq 0$.
– Alex R.
Sep 8 at 23:30
I agree with @AlexR. (+1 Alex) Btw, you got me cracking "For OP's sake.."
– Ahmad Bazzi
Sep 8 at 23:32
I agree with @AlexR. (+1 Alex) Btw, you got me cracking "For OP's sake.."
– Ahmad Bazzi
Sep 8 at 23:32
@Alex: But what if $langle ABx,Bxrangle = langle Bx,xrangle = 0$? Then $lambda$ can be anything.
– amsmath
Sep 8 at 23:33
@Alex: But what if $langle ABx,Bxrangle = langle Bx,xrangle = 0$? Then $lambda$ can be anything.
– amsmath
Sep 8 at 23:33
I misread the problem, the are symmetric. Sorry for the error.
– Yagger
Sep 8 at 23:36
I misread the problem, the are symmetric. Sorry for the error.
– Yagger
Sep 8 at 23:36
add a comment |Â
up vote
2
down vote
Let $A$ and $B$ be positive semidefinite real symmetric matrices.
Denote $sqrtA$ the square root of $A$, i.e.
beginequation
A = sqrtAsqrtA
endequation
where $sqrtA$ is also PSD.
Notice that $$sqrtA B sqrtA$$ is symmetric and positive semidefinite. Also, $$AB = sqrtA(sqrtA B)$$ and $$(sqrtAB)sqrtA$$ have the same nonzero eigenvalues.
answered Sep 8 at 23:39
Ahmad Bazzi
4,7401623
That's an original approach! +1
– Yagger
Sep 8 at 23:49
add a comment |Â
up vote
2
down vote
Let $A$ and $B$ be positive semidefinite real symmetric matrices.
Denote $sqrtA$ the square root of $A$, i.e.
beginequation
A = sqrtAsqrtA
endequation
where $sqrtA$ is also PSD.
Notice that $$sqrtA B sqrtA$$ is symmetric and positive semidefinite. Also, $$AB = sqrtA(sqrtA B)$$ and $$(sqrtAB)sqrtA$$ have the same nonzero eigenvalues.
answered Sep 8 at 23:39
Ahmad Bazzi
4,7401623
That's an original approach! +1
– Yagger
Sep 8 at 23:49
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $A$ and $B$ be positive semidefinite real symmetric matrices.
Denote $sqrtA$ the square root of $A$, i.e.
beginequation
A = sqrtAsqrtA
endequation
where $sqrtA$ is also PSD.
Notice that $$sqrtA B sqrtA$$ is symmetric and positive semidefinite. Also, $$AB = sqrtA(sqrtA B)$$ and $$(sqrtAB)sqrtA$$ have the same nonzero eigenvalues.
answered Sep 8 at 23:39
Ahmad Bazzi
4,7401623
Let $A$ and $B$ be positive semidefinite real symmetric matrices.
Denote $sqrtA$ the square root of $A$, i.e.
beginequation
A = sqrtAsqrtA
endequation
where $sqrtA$ is also PSD.
Notice that $$sqrtA B sqrtA$$ is symmetric and positive semidefinite. Also, $$AB = sqrtA(sqrtA B)$$ and $$(sqrtAB)sqrtA$$ have the same nonzero eigenvalues.
answered Sep 8 at 23:39
Ahmad Bazzi
4,7401623
answered Sep 8 at 23:39
Ahmad Bazzi
4,7401623
answered Sep 8 at 23:39
Ahmad Bazzi
4,7401623
answered Sep 8 at 23:39
Ahmad Bazzi
4,7401623
4,7401623
That's an original approach! +1
– Yagger
Sep 8 at 23:49
add a comment |Â
That's an original approach! +1
– Yagger
Sep 8 at 23:49
That's an original approach! +1
– Yagger
Sep 8 at 23:49
That's an original approach! +1
– Yagger
Sep 8 at 23:49
add a comment |Â
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are $A,B$ symmetric ?
– Ahmad Bazzi
Sep 8 at 23:22
@AhmadBazzi no they are not.
– Yagger
Sep 8 at 23:24
@Yagger Usually, "positive semidefinite" implies symmetric.
– amsmath
Sep 8 at 23:27
@AhmadBazzi they are symetric
– Yagger
Sep 8 at 23:37
I suspected @Yagger .. I have posted an answer when they are symmetric using square roots of matrices
– Ahmad Bazzi
Sep 8 at 23:39