Matrix equation with symmetric and positive definite matrices

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Let matrices $P_0$ and $P_1$ be symmetric and positive definite.

Is it possible to find a symmetric matrix $S$ such that $SP_0S=P_1$?

If $P_0=I$, this is always possible: $S=sqrtP_1$.

matrices matrix-equations positive-definite symmetric-matrices

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edited Jul 26 at 16:12

Rodrigo de Azevedo

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asked Jul 25 at 20:33

Pete

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up vote
4
down vote

favorite

2

Let matrices $P_0$ and $P_1$ be symmetric and positive definite.

Is it possible to find a symmetric matrix $S$ such that $SP_0S=P_1$?

If $P_0=I$, this is always possible: $S=sqrtP_1$.

matrices matrix-equations positive-definite symmetric-matrices

share|cite|improve this question

edited Jul 26 at 16:12

Rodrigo de Azevedo

12.6k41751

asked Jul 25 at 20:33

Pete

1306

add a comment | 

up vote
4
down vote

favorite

2

up vote
4
down vote

favorite

2

2

Let matrices $P_0$ and $P_1$ be symmetric and positive definite.

Is it possible to find a symmetric matrix $S$ such that $SP_0S=P_1$?

If $P_0=I$, this is always possible: $S=sqrtP_1$.

matrices matrix-equations positive-definite symmetric-matrices

share|cite|improve this question

edited Jul 26 at 16:12

Rodrigo de Azevedo

12.6k41751

asked Jul 25 at 20:33

Pete

1306

Let matrices $P_0$ and $P_1$ be symmetric and positive definite.

Is it possible to find a symmetric matrix $S$ such that $SP_0S=P_1$?

If $P_0=I$, this is always possible: $S=sqrtP_1$.

matrices matrix-equations positive-definite symmetric-matrices

share|cite|improve this question

edited Jul 26 at 16:12

Rodrigo de Azevedo

12.6k41751

asked Jul 25 at 20:33

Pete

1306

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 16:12

Rodrigo de Azevedo

12.6k41751

edited Jul 26 at 16:12

Rodrigo de Azevedo

12.6k41751

edited Jul 26 at 16:12

Rodrigo de Azevedo

12.6k41751

12.6k41751

asked Jul 25 at 20:33

Pete

1306

asked Jul 25 at 20:33

Pete

1306

asked Jul 25 at 20:33

Pete

1306

1306

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

This is quite surprising!
In general, I would suggest you to refrain from making conjectures based on the fact that the statement holds for the identity matrix.
You might use some of the very strong hidden properties of the identity matrix (e.g., it commutes with everything) when you check the result on it.

However, in this particular case, you are spot on!

Solution: $S:= P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2$.

First of all, products of the form $ABA$ with $A,B$ symmetric, are also symmetric. So $S$ is symmetric.

Moreover,
$SP_0S =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 P_0 P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2P_0^1/2P_1P_0^1/2P_0^-1/2 =
P_1$

share|cite|improve this answer

edited Jul 26 at 16:38

answered Jul 26 at 14:34

A. Pongrácz

1,802115

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

This is quite surprising!
In general, I would suggest you to refrain from making conjectures based on the fact that the statement holds for the identity matrix.
You might use some of the very strong hidden properties of the identity matrix (e.g., it commutes with everything) when you check the result on it.

However, in this particular case, you are spot on!

Solution: $S:= P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2$.

First of all, products of the form $ABA$ with $A,B$ symmetric, are also symmetric. So $S$ is symmetric.

Moreover,
$SP_0S =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 P_0 P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2P_0^1/2P_1P_0^1/2P_0^-1/2 =
P_1$

share|cite|improve this answer

edited Jul 26 at 16:38

answered Jul 26 at 14:34

A. Pongrácz

1,802115

add a comment | 

up vote
2
down vote



accepted

This is quite surprising!
In general, I would suggest you to refrain from making conjectures based on the fact that the statement holds for the identity matrix.
You might use some of the very strong hidden properties of the identity matrix (e.g., it commutes with everything) when you check the result on it.

However, in this particular case, you are spot on!

Solution: $S:= P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2$.

First of all, products of the form $ABA$ with $A,B$ symmetric, are also symmetric. So $S$ is symmetric.

Moreover,
$SP_0S =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 P_0 P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2P_0^1/2P_1P_0^1/2P_0^-1/2 =
P_1$

share|cite|improve this answer

edited Jul 26 at 16:38

answered Jul 26 at 14:34

A. Pongrácz

1,802115

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

This is quite surprising!
In general, I would suggest you to refrain from making conjectures based on the fact that the statement holds for the identity matrix.
You might use some of the very strong hidden properties of the identity matrix (e.g., it commutes with everything) when you check the result on it.

However, in this particular case, you are spot on!

Solution: $S:= P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2$.

First of all, products of the form $ABA$ with $A,B$ symmetric, are also symmetric. So $S$ is symmetric.

Moreover,
$SP_0S =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 P_0 P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2P_0^1/2P_1P_0^1/2P_0^-1/2 =
P_1$

share|cite|improve this answer

edited Jul 26 at 16:38

answered Jul 26 at 14:34

A. Pongrácz

1,802115

This is quite surprising!
In general, I would suggest you to refrain from making conjectures based on the fact that the statement holds for the identity matrix.
You might use some of the very strong hidden properties of the identity matrix (e.g., it commutes with everything) when you check the result on it.

However, in this particular case, you are spot on!

Solution: $S:= P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2$.

First of all, products of the form $ABA$ with $A,B$ symmetric, are also symmetric. So $S$ is symmetric.

Moreover,
$SP_0S =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 P_0 P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2(P_0^1/2P_1P_0^1/2)^1/2(P_0^1/2P_1P_0^1/2)^1/2P_0^-1/2 =
P_0^-1/2P_0^1/2P_1P_0^1/2P_0^-1/2 =
P_1$

share|cite|improve this answer

edited Jul 26 at 16:38

answered Jul 26 at 14:34

A. Pongrácz

1,802115

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 16:38

edited Jul 26 at 16:38

edited Jul 26 at 16:38

answered Jul 26 at 14:34

A. Pongrácz

1,802115

answered Jul 26 at 14:34

A. Pongrácz

1,802115

answered Jul 26 at 14:34

A. Pongrácz

1,802115

1,802115

add a comment | 

add a comment | 

 

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