Positive definiteness of difference of inverse matrices
Let $A$ and $B$ be two $n times n$ symmetric and positive definite matrices.
If $A prec B$, then is it true that $B^{-1} prec A^{-1}$?
Here, $A prec B$ means that $B-A$ is positive definite.
linear-algebra matrices inverse positive-definite
edited Aug 22 '17 at 17:47
Rodrigo de Azevedo
12.8k41855
asked Aug 22 '17 at 16:36
Sudipta RoySudipta Roy
29518
add a comment |
Let $A$ and $B$ be two $n times n$ symmetric and positive definite matrices.
If $A prec B$, then is it true that $B^{-1} prec A^{-1}$?
Here, $A prec B$ means that $B-A$ is positive definite.
linear-algebra matrices inverse positive-definite
edited Aug 22 '17 at 17:47
Rodrigo de Azevedo
12.8k41855
asked Aug 22 '17 at 16:36
Sudipta RoySudipta Roy
29518
add a comment |
Let $A$ and $B$ be two $n times n$ symmetric and positive definite matrices.
If $A prec B$, then is it true that $B^{-1} prec A^{-1}$?
Here, $A prec B$ means that $B-A$ is positive definite.
linear-algebra matrices inverse positive-definite
edited Aug 22 '17 at 17:47
Rodrigo de Azevedo
12.8k41855
asked Aug 22 '17 at 16:36
Sudipta RoySudipta Roy
29518
Let $A$ and $B$ be two $n times n$ symmetric and positive definite matrices.
If $A prec B$, then is it true that $B^{-1} prec A^{-1}$?
Here, $A prec B$ means that $B-A$ is positive definite.
linear-algebra matrices inverse positive-definite
linear-algebra matrices inverse positive-definite
edited Aug 22 '17 at 17:47
Rodrigo de Azevedo
12.8k41855
asked Aug 22 '17 at 16:36
Sudipta RoySudipta Roy
29518
edited Aug 22 '17 at 17:47
Rodrigo de Azevedo
12.8k41855
asked Aug 22 '17 at 16:36
Sudipta RoySudipta Roy
29518
edited Aug 22 '17 at 17:47
Rodrigo de Azevedo
12.8k41855
edited Aug 22 '17 at 17:47
Rodrigo de Azevedo
12.8k41855
edited Aug 22 '17 at 17:47
Rodrigo de Azevedo
12.8k41855
12.8k41855
asked Aug 22 '17 at 16:36
Sudipta RoySudipta Roy
29518
asked Aug 22 '17 at 16:36
Sudipta RoySudipta Roy
29518
asked Aug 22 '17 at 16:36
Sudipta RoySudipta Roy
29518
29518
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
HINT:
Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)
Then, $det(A^{-1}-mu B^{-1})=0$
$Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$
$Rightarrow det(B-mu A)=0$
So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.
As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)
answered Aug 23 '17 at 15:43
StubbornAtomStubbornAtom
5,39411138
exercise from what book?
– becko
Jun 4 '18 at 15:56
1
@becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
– StubbornAtom
Jun 4 '18 at 16:03
I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
– StubbornAtom
Sep 15 '18 at 7:01
add a comment |
For ease of notation I'll use $ge$ instead of $succeq$ and so on.
Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$
Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$
Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$
Proof.
- Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.
- Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$
Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$
Proof.
begin{align*}
Ale B &Rightarrow B-A ge 0 \
&Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
&Rightarrow A^{-1/2}BA^{-1/2} ge I\
&Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
end{align*}
hence
begin{align*}
B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
&= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
&le A^{-1/2}IA^{-1/2}\
&= A^{-1}.
end{align*}
answered Nov 29 '18 at 14:09
EpiousiosEpiousios
1,545622
add a comment |
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2 Answers
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2 Answers
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HINT:
Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)
Then, $det(A^{-1}-mu B^{-1})=0$
$Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$
$Rightarrow det(B-mu A)=0$
So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.
As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)
answered Aug 23 '17 at 15:43
StubbornAtomStubbornAtom
5,39411138
exercise from what book?
– becko
Jun 4 '18 at 15:56
1
@becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
– StubbornAtom
Jun 4 '18 at 16:03
I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
– StubbornAtom
Sep 15 '18 at 7:01
add a comment |
HINT:
Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)
Then, $det(A^{-1}-mu B^{-1})=0$
$Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$
$Rightarrow det(B-mu A)=0$
So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.
As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)
answered Aug 23 '17 at 15:43
StubbornAtomStubbornAtom
5,39411138
exercise from what book?
– becko
Jun 4 '18 at 15:56
1
@becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
– StubbornAtom
Jun 4 '18 at 16:03
I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
– StubbornAtom
Sep 15 '18 at 7:01
add a comment |
HINT:
Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)
Then, $det(A^{-1}-mu B^{-1})=0$
$Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$
$Rightarrow det(B-mu A)=0$
So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.
As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)
answered Aug 23 '17 at 15:43
StubbornAtomStubbornAtom
5,39411138
HINT:
Try proving that if $A$ is positive definite and $B-A$ is non-negative definite, then $det(B-lambda A)=0$ has all its roots $lambdage 1$ and conversely if $lambdage1$, then $B-A$ is non-negative definite. (it's an exercise from a book by the way)
Then, $det(A^{-1}-mu B^{-1})=0$
$Rightarrow det Bdet(A^{-1}-mu B^{-1})det Age0$ $quad($ as $A$ and $B$ are both p.d., $det A,det B>0)$
$Rightarrow det(B-mu A)=0$
So the roots of $det(A^{-1}-mu B^{-1})=0$ are the same as the roots of $det(B-mu A)=0$.
As $B-A$ is n.n.d. we have $muge1$, which in turn makes $A^{-1}-B^{-1}$ n.n.d. (by the claim above)
answered Aug 23 '17 at 15:43
StubbornAtomStubbornAtom
5,39411138
answered Aug 23 '17 at 15:43
StubbornAtomStubbornAtom
5,39411138
answered Aug 23 '17 at 15:43
StubbornAtomStubbornAtom
5,39411138
answered Aug 23 '17 at 15:43
StubbornAtomStubbornAtom
5,39411138
5,39411138
exercise from what book?
– becko
Jun 4 '18 at 15:56
1
@becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
– StubbornAtom
Jun 4 '18 at 16:03
I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
– StubbornAtom
Sep 15 '18 at 7:01
add a comment |
exercise from what book?
– becko
Jun 4 '18 at 15:56
1
@becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
– StubbornAtom
Jun 4 '18 at 16:03
I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
– StubbornAtom
Sep 15 '18 at 7:01
exercise from what book?
– becko
Jun 4 '18 at 15:56
exercise from what book?
– becko
Jun 4 '18 at 15:56
1
1
@becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
– StubbornAtom
Jun 4 '18 at 16:03
@becko Linear Statistical Inference and it's Applications by C.R. Rao (not a linear algebra book).
– StubbornAtom
Jun 4 '18 at 16:03
I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
– StubbornAtom
Sep 15 '18 at 7:01
I had asked a question on the proof of the claim made here: math.stackexchange.com/q/2671111/321264
– StubbornAtom
Sep 15 '18 at 7:01
add a comment |
For ease of notation I'll use $ge$ instead of $succeq$ and so on.
Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$
Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$
Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$
Proof.
- Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.
- Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$
Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$
Proof.
begin{align*}
Ale B &Rightarrow B-A ge 0 \
&Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
&Rightarrow A^{-1/2}BA^{-1/2} ge I\
&Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
end{align*}
hence
begin{align*}
B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
&= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
&le A^{-1/2}IA^{-1/2}\
&= A^{-1}.
end{align*}
answered Nov 29 '18 at 14:09
EpiousiosEpiousios
1,545622
add a comment |
For ease of notation I'll use $ge$ instead of $succeq$ and so on.
Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$
Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$
Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$
Proof.
- Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.
- Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$
Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$
Proof.
begin{align*}
Ale B &Rightarrow B-A ge 0 \
&Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
&Rightarrow A^{-1/2}BA^{-1/2} ge I\
&Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
end{align*}
hence
begin{align*}
B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
&= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
&le A^{-1/2}IA^{-1/2}\
&= A^{-1}.
end{align*}
answered Nov 29 '18 at 14:09
EpiousiosEpiousios
1,545622
add a comment |
For ease of notation I'll use $ge$ instead of $succeq$ and so on.
Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$
Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$
Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$
Proof.
- Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.
- Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$
Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$
Proof.
begin{align*}
Ale B &Rightarrow B-A ge 0 \
&Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
&Rightarrow A^{-1/2}BA^{-1/2} ge I\
&Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
end{align*}
hence
begin{align*}
B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
&= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
&le A^{-1/2}IA^{-1/2}\
&= A^{-1}.
end{align*}
answered Nov 29 '18 at 14:09
EpiousiosEpiousios
1,545622
For ease of notation I'll use $ge$ instead of $succeq$ and so on.
Lemma 1: $A le B Rightarrow C^T A C le C^T B C$ for any conformable matrix $C.$
Proof. We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x ge 0$ for any conformable vector $x$ so that $C^T(B-A)C ge 0.$
Lemma 2: $I preceq B Rightarrow$ $B$ is invertible and $B^{-1} preceq I.$
Proof.
- Since $B$ is symmetric we can find an orthogonal matrix $G$ and a diagonal matrix $D$ such that $B=GDG^T.$ Hence $I = G^T I G le G^T B G = D.$ Thus all eigenvalues of $B$ are $ge 1 >0,$ hence $B$ is invertible.
- Now write $B^{-1} = B^{-1/2}IB^{-1/2} le B^{-1/2}BB^{-1/2} = I,$ where $B^alpha := GD^{alpha}G^T.$
Proposition: $0 < A le B Rightarrow$ $B$ invertible and $B^{-1} le A^{-1}.$
Proof.
begin{align*}
Ale B &Rightarrow B-A ge 0 \
&Rightarrow A^{-1/2}(B-A)A^{-1/2} ge 0\
&Rightarrow A^{-1/2}BA^{-1/2} ge I\
&Rightarrow A^{1/2}B^{-1}A^{1/2} le I,\
end{align*}
hence
begin{align*}
B^{-1} &= A^{-1/2}A^{1/2}BA^{1/2}A^{-1/2}\
&= A^{-1/2}(A^{-1/2}B^{-1}A^{-1/2})^{-1}A^{-1/2}\
&le A^{-1/2}IA^{-1/2}\
&= A^{-1}.
end{align*}
answered Nov 29 '18 at 14:09
EpiousiosEpiousios
1,545622
answered Nov 29 '18 at 14:09
EpiousiosEpiousios
1,545622
answered Nov 29 '18 at 14:09
EpiousiosEpiousios
1,545622
answered Nov 29 '18 at 14:09
EpiousiosEpiousios
1,545622
1,545622
add a comment |
add a comment |
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