Is square root of a matrix necessarily Cholesky decomposition? Or other decomposition satisfies?
If $a$ is a scalar, $a = sqrt{a}sqrt{a}$, then if A is a matrix, $A=sqrt{A}sqrt{A}$, now I think the square root operator is applied to the matrix instead of a scalar.
Is square root of a matrix necessarily Cholesky decomposition? Or other decomposition satisfies, like QR decomposition? Because $A=QR$ where Q and R are two terms like $sqrt{A}sqrt{A}$. Or does the square roots have to satisfy the condition that one is the other's conjugate transpose as in Cholesky decomposition?
matrix-calculus matrix-decomposition
asked Nov 19 '18 at 1:51
drerD
1519
add a comment |
If $a$ is a scalar, $a = sqrt{a}sqrt{a}$, then if A is a matrix, $A=sqrt{A}sqrt{A}$, now I think the square root operator is applied to the matrix instead of a scalar.
Is square root of a matrix necessarily Cholesky decomposition? Or other decomposition satisfies, like QR decomposition? Because $A=QR$ where Q and R are two terms like $sqrt{A}sqrt{A}$. Or does the square roots have to satisfy the condition that one is the other's conjugate transpose as in Cholesky decomposition?
matrix-calculus matrix-decomposition
asked Nov 19 '18 at 1:51
drerD
1519
add a comment |
If $a$ is a scalar, $a = sqrt{a}sqrt{a}$, then if A is a matrix, $A=sqrt{A}sqrt{A}$, now I think the square root operator is applied to the matrix instead of a scalar.
Is square root of a matrix necessarily Cholesky decomposition? Or other decomposition satisfies, like QR decomposition? Because $A=QR$ where Q and R are two terms like $sqrt{A}sqrt{A}$. Or does the square roots have to satisfy the condition that one is the other's conjugate transpose as in Cholesky decomposition?
matrix-calculus matrix-decomposition
asked Nov 19 '18 at 1:51
drerD
1519
If $a$ is a scalar, $a = sqrt{a}sqrt{a}$, then if A is a matrix, $A=sqrt{A}sqrt{A}$, now I think the square root operator is applied to the matrix instead of a scalar.
Is square root of a matrix necessarily Cholesky decomposition? Or other decomposition satisfies, like QR decomposition? Because $A=QR$ where Q and R are two terms like $sqrt{A}sqrt{A}$. Or does the square roots have to satisfy the condition that one is the other's conjugate transpose as in Cholesky decomposition?
matrix-calculus matrix-decomposition
matrix-calculus matrix-decomposition
asked Nov 19 '18 at 1:51
drerD
1519
asked Nov 19 '18 at 1:51
drerD
1519
asked Nov 19 '18 at 1:51
drerD
1519
asked Nov 19 '18 at 1:51
drerD
1519
asked Nov 19 '18 at 1:51
drerD
1519
1519
add a comment |
add a comment |
1 Answer
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If $B$ can at all be called a square root of $A$, then $B^2=A$, whether we are taking about matrices or anything else. For matrices, there can be infinitely many distinct square roots (even for the zero matrix), but they all satisfy the same equation and no other conditions are necessary.
answered Nov 19 '18 at 1:57
Matt Samuel
37.1k63465
the $B^2=A$ definition makes sense, I got a question, do B's have to be equal? Because Cholesky decomposition $L, L*$ aren't exactly equal.
– drerD
Nov 19 '18 at 17:06
@user Yes, they have to be equal. It has nothing to do with Cholesky decomposition.
– Matt Samuel
Nov 19 '18 at 17:10
The reason I kept referring to Cholesky decomposition is that I've read about unscented Kalman filter, and it finds sigma points by finding the $sqrt{Sigma}$ and $sqrt{Sigma}$ is always said to be calculated by Cholesky decomposition. Is it because this square root is just a symbolic representation, and not exactly the square root B in $B^2 =A$?
– drerD
Nov 19 '18 at 18:50
@user Does the matrix have real entries?
– Matt Samuel
Nov 19 '18 at 18:59
@user en.wikipedia.org/wiki/Square_root_of_a_matrix Confusingly, the matrix in the Cholesky decomposition is called a square root even though it isn't.
– Matt Samuel
Nov 19 '18 at 19:05
|
show 1 more comment
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If $B$ can at all be called a square root of $A$, then $B^2=A$, whether we are taking about matrices or anything else. For matrices, there can be infinitely many distinct square roots (even for the zero matrix), but they all satisfy the same equation and no other conditions are necessary.
answered Nov 19 '18 at 1:57
Matt Samuel
37.1k63465
the $B^2=A$ definition makes sense, I got a question, do B's have to be equal? Because Cholesky decomposition $L, L*$ aren't exactly equal.
– drerD
Nov 19 '18 at 17:06
@user Yes, they have to be equal. It has nothing to do with Cholesky decomposition.
– Matt Samuel
Nov 19 '18 at 17:10
The reason I kept referring to Cholesky decomposition is that I've read about unscented Kalman filter, and it finds sigma points by finding the $sqrt{Sigma}$ and $sqrt{Sigma}$ is always said to be calculated by Cholesky decomposition. Is it because this square root is just a symbolic representation, and not exactly the square root B in $B^2 =A$?
– drerD
Nov 19 '18 at 18:50
@user Does the matrix have real entries?
– Matt Samuel
Nov 19 '18 at 18:59
@user en.wikipedia.org/wiki/Square_root_of_a_matrix Confusingly, the matrix in the Cholesky decomposition is called a square root even though it isn't.
– Matt Samuel
Nov 19 '18 at 19:05
|
show 1 more comment
If $B$ can at all be called a square root of $A$, then $B^2=A$, whether we are taking about matrices or anything else. For matrices, there can be infinitely many distinct square roots (even for the zero matrix), but they all satisfy the same equation and no other conditions are necessary.
answered Nov 19 '18 at 1:57
Matt Samuel
37.1k63465
the $B^2=A$ definition makes sense, I got a question, do B's have to be equal? Because Cholesky decomposition $L, L*$ aren't exactly equal.
– drerD
Nov 19 '18 at 17:06
@user Yes, they have to be equal. It has nothing to do with Cholesky decomposition.
– Matt Samuel
Nov 19 '18 at 17:10
The reason I kept referring to Cholesky decomposition is that I've read about unscented Kalman filter, and it finds sigma points by finding the $sqrt{Sigma}$ and $sqrt{Sigma}$ is always said to be calculated by Cholesky decomposition. Is it because this square root is just a symbolic representation, and not exactly the square root B in $B^2 =A$?
– drerD
Nov 19 '18 at 18:50
@user Does the matrix have real entries?
– Matt Samuel
Nov 19 '18 at 18:59
@user en.wikipedia.org/wiki/Square_root_of_a_matrix Confusingly, the matrix in the Cholesky decomposition is called a square root even though it isn't.
– Matt Samuel
Nov 19 '18 at 19:05
|
show 1 more comment
If $B$ can at all be called a square root of $A$, then $B^2=A$, whether we are taking about matrices or anything else. For matrices, there can be infinitely many distinct square roots (even for the zero matrix), but they all satisfy the same equation and no other conditions are necessary.
answered Nov 19 '18 at 1:57
Matt Samuel
37.1k63465
If $B$ can at all be called a square root of $A$, then $B^2=A$, whether we are taking about matrices or anything else. For matrices, there can be infinitely many distinct square roots (even for the zero matrix), but they all satisfy the same equation and no other conditions are necessary.
answered Nov 19 '18 at 1:57
Matt Samuel
37.1k63465
answered Nov 19 '18 at 1:57
Matt Samuel
37.1k63465
answered Nov 19 '18 at 1:57
Matt Samuel
37.1k63465
answered Nov 19 '18 at 1:57
Matt Samuel
37.1k63465
37.1k63465
the $B^2=A$ definition makes sense, I got a question, do B's have to be equal? Because Cholesky decomposition $L, L*$ aren't exactly equal.
– drerD
Nov 19 '18 at 17:06
@user Yes, they have to be equal. It has nothing to do with Cholesky decomposition.
– Matt Samuel
Nov 19 '18 at 17:10
The reason I kept referring to Cholesky decomposition is that I've read about unscented Kalman filter, and it finds sigma points by finding the $sqrt{Sigma}$ and $sqrt{Sigma}$ is always said to be calculated by Cholesky decomposition. Is it because this square root is just a symbolic representation, and not exactly the square root B in $B^2 =A$?
– drerD
Nov 19 '18 at 18:50
@user Does the matrix have real entries?
– Matt Samuel
Nov 19 '18 at 18:59
@user en.wikipedia.org/wiki/Square_root_of_a_matrix Confusingly, the matrix in the Cholesky decomposition is called a square root even though it isn't.
– Matt Samuel
Nov 19 '18 at 19:05
|
show 1 more comment
the $B^2=A$ definition makes sense, I got a question, do B's have to be equal? Because Cholesky decomposition $L, L*$ aren't exactly equal.
– drerD
Nov 19 '18 at 17:06
@user Yes, they have to be equal. It has nothing to do with Cholesky decomposition.
– Matt Samuel
Nov 19 '18 at 17:10
The reason I kept referring to Cholesky decomposition is that I've read about unscented Kalman filter, and it finds sigma points by finding the $sqrt{Sigma}$ and $sqrt{Sigma}$ is always said to be calculated by Cholesky decomposition. Is it because this square root is just a symbolic representation, and not exactly the square root B in $B^2 =A$?
– drerD
Nov 19 '18 at 18:50
@user Does the matrix have real entries?
– Matt Samuel
Nov 19 '18 at 18:59
@user en.wikipedia.org/wiki/Square_root_of_a_matrix Confusingly, the matrix in the Cholesky decomposition is called a square root even though it isn't.
– Matt Samuel
Nov 19 '18 at 19:05
the $B^2=A$ definition makes sense, I got a question, do B's have to be equal? Because Cholesky decomposition $L, L*$ aren't exactly equal.
– drerD
Nov 19 '18 at 17:06
the $B^2=A$ definition makes sense, I got a question, do B's have to be equal? Because Cholesky decomposition $L, L*$ aren't exactly equal.
– drerD
Nov 19 '18 at 17:06
@user Yes, they have to be equal. It has nothing to do with Cholesky decomposition.
– Matt Samuel
Nov 19 '18 at 17:10
@user Yes, they have to be equal. It has nothing to do with Cholesky decomposition.
– Matt Samuel
Nov 19 '18 at 17:10
The reason I kept referring to Cholesky decomposition is that I've read about unscented Kalman filter, and it finds sigma points by finding the $sqrt{Sigma}$ and $sqrt{Sigma}$ is always said to be calculated by Cholesky decomposition. Is it because this square root is just a symbolic representation, and not exactly the square root B in $B^2 =A$?
– drerD
Nov 19 '18 at 18:50
The reason I kept referring to Cholesky decomposition is that I've read about unscented Kalman filter, and it finds sigma points by finding the $sqrt{Sigma}$ and $sqrt{Sigma}$ is always said to be calculated by Cholesky decomposition. Is it because this square root is just a symbolic representation, and not exactly the square root B in $B^2 =A$?
– drerD
Nov 19 '18 at 18:50
@user Does the matrix have real entries?
– Matt Samuel
Nov 19 '18 at 18:59
@user Does the matrix have real entries?
– Matt Samuel
Nov 19 '18 at 18:59
@user en.wikipedia.org/wiki/Square_root_of_a_matrix Confusingly, the matrix in the Cholesky decomposition is called a square root even though it isn't.
– Matt Samuel
Nov 19 '18 at 19:05
@user en.wikipedia.org/wiki/Square_root_of_a_matrix Confusingly, the matrix in the Cholesky decomposition is called a square root even though it isn't.
– Matt Samuel
Nov 19 '18 at 19:05
|
show 1 more comment
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