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Matrix trace equality [closed]
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Suppose $A$ and $B$ are positive definite matrices of the same size. Prove that
beginalign
operatornameTr( A^1/2 (A^-1/2 B A^-1/2)^1/2 A^1/2) leq operatornameTr( (A^1/2 B A^1/2)^1/2).
endalign
Here the matrix square root is taken to be principle square root: in other words, the principal square root is the only positive definite root of a positive definite matrix.
linear-algebra matrices noncommutative-algebra trace
asked Jul 19 at 22:43
user401582
616
closed as off-topic by Arnaud Mortier, Isaac Browne, Piyush Divyanakar, amWhy, Trần Thúc Minh Trà Jul 20 at 11:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Isaac Browne, Piyush Divyanakar, amWhy, Trần Thúc Minh TrÃÂ
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up vote
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Suppose $A$ and $B$ are positive definite matrices of the same size. Prove that
beginalign
operatornameTr( A^1/2 (A^-1/2 B A^-1/2)^1/2 A^1/2) leq operatornameTr( (A^1/2 B A^1/2)^1/2).
endalign
Here the matrix square root is taken to be principle square root: in other words, the principal square root is the only positive definite root of a positive definite matrix.
linear-algebra matrices noncommutative-algebra trace
asked Jul 19 at 22:43
user401582
616
closed as off-topic by Arnaud Mortier, Isaac Browne, Piyush Divyanakar, amWhy, Trần Thúc Minh Trà Jul 20 at 11:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Isaac Browne, Piyush Divyanakar, amWhy, Trần Thúc Minh TrÃÂ
It should be noted that the square root of any matrix is not unique, so it is not true that $A^frac12 = A^frac12$ unless you explicitly declare it to be so.
– John Polcari
Jul 19 at 23:51
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $A$ and $B$ are positive definite matrices of the same size. Prove that
beginalign
operatornameTr( A^1/2 (A^-1/2 B A^-1/2)^1/2 A^1/2) leq operatornameTr( (A^1/2 B A^1/2)^1/2).
endalign
Here the matrix square root is taken to be principle square root: in other words, the principal square root is the only positive definite root of a positive definite matrix.
linear-algebra matrices noncommutative-algebra trace
asked Jul 19 at 22:43
user401582
616
Suppose $A$ and $B$ are positive definite matrices of the same size. Prove that
beginalign
operatornameTr( A^1/2 (A^-1/2 B A^-1/2)^1/2 A^1/2) leq operatornameTr( (A^1/2 B A^1/2)^1/2).
endalign
Here the matrix square root is taken to be principle square root: in other words, the principal square root is the only positive definite root of a positive definite matrix.
linear-algebra matrices noncommutative-algebra trace
asked Jul 19 at 22:43
user401582
616
edited Jul 19 at 23:53
asked Jul 19 at 22:43
user401582
616
asked Jul 19 at 22:43
user401582
616
asked Jul 19 at 22:43
user401582
616
616
closed as off-topic by Arnaud Mortier, Isaac Browne, Piyush Divyanakar, amWhy, Trần Thúc Minh Trà Jul 20 at 11:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Isaac Browne, Piyush Divyanakar, amWhy, Trần Thúc Minh TrÃÂ
closed as off-topic by Arnaud Mortier, Isaac Browne, Piyush Divyanakar, amWhy, Trần Thúc Minh Trà Jul 20 at 11:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Isaac Browne, Piyush Divyanakar, amWhy, Trần Thúc Minh TrÃÂ
It should be noted that the square root of any matrix is not unique, so it is not true that $A^frac12 = A^frac12$ unless you explicitly declare it to be so.
– John Polcari
Jul 19 at 23:51
add a comment |Â
It should be noted that the square root of any matrix is not unique, so it is not true that $A^frac12 = A^frac12$ unless you explicitly declare it to be so.
– John Polcari
Jul 19 at 23:51
It should be noted that the square root of any matrix is not unique, so it is not true that $A^frac12 = A^frac12$ unless you explicitly declare it to be so.
– John Polcari
Jul 19 at 23:51
It should be noted that the square root of any matrix is not unique, so it is not true that $A^frac12 = A^frac12$ unless you explicitly declare it to be so.
– John Polcari
Jul 19 at 23:51
add a comment |Â
1 Answer
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Let $X=A(A^-1/2BA^-1/2)^1/2$. Then the matrix on the LHS of the inequality is equal to $A^1/2XA^-1/2$ and the matrix on the RHS is equal to $(XX^ast)^1/2$. Hence the inequality is equivalent to $operatornametr(X)leoperatornametrleft((XX^ast)^1/2right)$, which, as explained in my answer to your previous question, can be easily proved by performing polar decomposition on $X$.
answered Jul 20 at 1:17
user1551
66.9k564122
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $X=A(A^-1/2BA^-1/2)^1/2$. Then the matrix on the LHS of the inequality is equal to $A^1/2XA^-1/2$ and the matrix on the RHS is equal to $(XX^ast)^1/2$. Hence the inequality is equivalent to $operatornametr(X)leoperatornametrleft((XX^ast)^1/2right)$, which, as explained in my answer to your previous question, can be easily proved by performing polar decomposition on $X$.
answered Jul 20 at 1:17
user1551
66.9k564122
add a comment |Â
up vote
1
down vote
accepted
Let $X=A(A^-1/2BA^-1/2)^1/2$. Then the matrix on the LHS of the inequality is equal to $A^1/2XA^-1/2$ and the matrix on the RHS is equal to $(XX^ast)^1/2$. Hence the inequality is equivalent to $operatornametr(X)leoperatornametrleft((XX^ast)^1/2right)$, which, as explained in my answer to your previous question, can be easily proved by performing polar decomposition on $X$.
answered Jul 20 at 1:17
user1551
66.9k564122
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $X=A(A^-1/2BA^-1/2)^1/2$. Then the matrix on the LHS of the inequality is equal to $A^1/2XA^-1/2$ and the matrix on the RHS is equal to $(XX^ast)^1/2$. Hence the inequality is equivalent to $operatornametr(X)leoperatornametrleft((XX^ast)^1/2right)$, which, as explained in my answer to your previous question, can be easily proved by performing polar decomposition on $X$.
answered Jul 20 at 1:17
user1551
66.9k564122
Let $X=A(A^-1/2BA^-1/2)^1/2$. Then the matrix on the LHS of the inequality is equal to $A^1/2XA^-1/2$ and the matrix on the RHS is equal to $(XX^ast)^1/2$. Hence the inequality is equivalent to $operatornametr(X)leoperatornametrleft((XX^ast)^1/2right)$, which, as explained in my answer to your previous question, can be easily proved by performing polar decomposition on $X$.
answered Jul 20 at 1:17
user1551
66.9k564122
edited Jul 20 at 2:53
answered Jul 20 at 1:17
user1551
66.9k564122
answered Jul 20 at 1:17
user1551
66.9k564122
answered Jul 20 at 1:17
user1551
66.9k564122
66.9k564122
add a comment |Â
add a comment |Â
It should be noted that the square root of any matrix is not unique, so it is not true that $A^frac12 = A^frac12$ unless you explicitly declare it to be so.
– John Polcari
Jul 19 at 23:51