Cholesky and $P^T LDL^T P$ decompositions not corresponding
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I am using the Eigen library to factorize a matrix $A$ both using the Cholesky and the $LDL^T$ factorization.
Basically, I correctly compute a matrix $C$ and check that $$A=CC^T,$$ then I correctly compute $P$, $L$, and $D$, and check that $$A=P^TLDL^TP,$$ where $P$ is a permutation matrix generated by Eigen. However if $D=sqrt D(sqrt D)^T$, $$Cneq P^TLsqrt D.$$
This is really counterintuitive for me. Is there any reason about this?
Thanks a lot.
linear-algebra matrices matrix-decomposition
asked Nov 13 '18 at 14:50
marcomarco
112
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add a comment |
$begingroup$
I am using the Eigen library to factorize a matrix $A$ both using the Cholesky and the $LDL^T$ factorization.
Basically, I correctly compute a matrix $C$ and check that $$A=CC^T,$$ then I correctly compute $P$, $L$, and $D$, and check that $$A=P^TLDL^TP,$$ where $P$ is a permutation matrix generated by Eigen. However if $D=sqrt D(sqrt D)^T$, $$Cneq P^TLsqrt D.$$
This is really counterintuitive for me. Is there any reason about this?
Thanks a lot.
linear-algebra matrices matrix-decomposition
asked Nov 13 '18 at 14:50
marcomarco
112
$endgroup$
$begingroup$
In general, $C C^T = B B^T$ does not imply that $C=B$. For example, if $Q$ is a rotation, $C C^T = C Q Q^T C^T = (CQ) (CQ)^T$, and, in general, we do not have $C = CQ$.
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– copper.hat
Nov 13 '18 at 14:54
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@copper.hat, thanks a lot for your answer, I did not think to see this problem in this way. Then, I ask myself what is the meaning of the uniqueness of the Cholesky decomposition for positive definite matrices. Is it unique apart of a rotation of $L$?
$endgroup$
– marco
Nov 13 '18 at 15:56
add a comment |
$begingroup$
I am using the Eigen library to factorize a matrix $A$ both using the Cholesky and the $LDL^T$ factorization.
Basically, I correctly compute a matrix $C$ and check that $$A=CC^T,$$ then I correctly compute $P$, $L$, and $D$, and check that $$A=P^TLDL^TP,$$ where $P$ is a permutation matrix generated by Eigen. However if $D=sqrt D(sqrt D)^T$, $$Cneq P^TLsqrt D.$$
This is really counterintuitive for me. Is there any reason about this?
Thanks a lot.
linear-algebra matrices matrix-decomposition
asked Nov 13 '18 at 14:50
marcomarco
112
$endgroup$
I am using the Eigen library to factorize a matrix $A$ both using the Cholesky and the $LDL^T$ factorization.
Basically, I correctly compute a matrix $C$ and check that $$A=CC^T,$$ then I correctly compute $P$, $L$, and $D$, and check that $$A=P^TLDL^TP,$$ where $P$ is a permutation matrix generated by Eigen. However if $D=sqrt D(sqrt D)^T$, $$Cneq P^TLsqrt D.$$
This is really counterintuitive for me. Is there any reason about this?
Thanks a lot.
linear-algebra matrices matrix-decomposition
linear-algebra matrices matrix-decomposition
asked Nov 13 '18 at 14:50
marcomarco
112
asked Nov 13 '18 at 14:50
marcomarco
112
asked Nov 13 '18 at 14:50
marcomarco
112
asked Nov 13 '18 at 14:50
marcomarco
112
asked Nov 13 '18 at 14:50
marcomarco
112
112
$begingroup$
In general, $C C^T = B B^T$ does not imply that $C=B$. For example, if $Q$ is a rotation, $C C^T = C Q Q^T C^T = (CQ) (CQ)^T$, and, in general, we do not have $C = CQ$.
$endgroup$
– copper.hat
Nov 13 '18 at 14:54
$begingroup$
@copper.hat, thanks a lot for your answer, I did not think to see this problem in this way. Then, I ask myself what is the meaning of the uniqueness of the Cholesky decomposition for positive definite matrices. Is it unique apart of a rotation of $L$?
$endgroup$
– marco
Nov 13 '18 at 15:56
add a comment |
$begingroup$
In general, $C C^T = B B^T$ does not imply that $C=B$. For example, if $Q$ is a rotation, $C C^T = C Q Q^T C^T = (CQ) (CQ)^T$, and, in general, we do not have $C = CQ$.
$endgroup$
– copper.hat
Nov 13 '18 at 14:54
$begingroup$
@copper.hat, thanks a lot for your answer, I did not think to see this problem in this way. Then, I ask myself what is the meaning of the uniqueness of the Cholesky decomposition for positive definite matrices. Is it unique apart of a rotation of $L$?
$endgroup$
– marco
Nov 13 '18 at 15:56
$begingroup$
In general, $C C^T = B B^T$ does not imply that $C=B$. For example, if $Q$ is a rotation, $C C^T = C Q Q^T C^T = (CQ) (CQ)^T$, and, in general, we do not have $C = CQ$.
$endgroup$
– copper.hat
Nov 13 '18 at 14:54
$begingroup$
In general, $C C^T = B B^T$ does not imply that $C=B$. For example, if $Q$ is a rotation, $C C^T = C Q Q^T C^T = (CQ) (CQ)^T$, and, in general, we do not have $C = CQ$.
$endgroup$
– copper.hat
Nov 13 '18 at 14:54
$begingroup$
@copper.hat, thanks a lot for your answer, I did not think to see this problem in this way. Then, I ask myself what is the meaning of the uniqueness of the Cholesky decomposition for positive definite matrices. Is it unique apart of a rotation of $L$?
$endgroup$
– marco
Nov 13 '18 at 15:56
$begingroup$
@copper.hat, thanks a lot for your answer, I did not think to see this problem in this way. Then, I ask myself what is the meaning of the uniqueness of the Cholesky decomposition for positive definite matrices. Is it unique apart of a rotation of $L$?
$endgroup$
– marco
Nov 13 '18 at 15:56
add a comment |
1 Answer
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$begingroup$
As copper.hat said in his comment, we generally cannot infer that $C=B$ if we only know that $CC^T=BB^T$.
However, to answer the question in OP's comment: this fact does not violate the uniqueness of the Cholesky factorization because to call $A=CC^T$ a "Cholesky factorization", we need that $C$ is lower-triangular; presumably the $P^TLsqrt{D}$ that arose from the Eigen calculation was not lower-triangular.
(In particular, the uniqueness of Cholesky decomposition is a consequence of the uniqueness of $LU$ factorization.)
answered Dec 19 '18 at 5:16
aleph_twoaleph_two
24912
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add a comment |
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$begingroup$
As copper.hat said in his comment, we generally cannot infer that $C=B$ if we only know that $CC^T=BB^T$.
However, to answer the question in OP's comment: this fact does not violate the uniqueness of the Cholesky factorization because to call $A=CC^T$ a "Cholesky factorization", we need that $C$ is lower-triangular; presumably the $P^TLsqrt{D}$ that arose from the Eigen calculation was not lower-triangular.
(In particular, the uniqueness of Cholesky decomposition is a consequence of the uniqueness of $LU$ factorization.)
answered Dec 19 '18 at 5:16
aleph_twoaleph_two
24912
$endgroup$
add a comment |
$begingroup$
As copper.hat said in his comment, we generally cannot infer that $C=B$ if we only know that $CC^T=BB^T$.
However, to answer the question in OP's comment: this fact does not violate the uniqueness of the Cholesky factorization because to call $A=CC^T$ a "Cholesky factorization", we need that $C$ is lower-triangular; presumably the $P^TLsqrt{D}$ that arose from the Eigen calculation was not lower-triangular.
(In particular, the uniqueness of Cholesky decomposition is a consequence of the uniqueness of $LU$ factorization.)
answered Dec 19 '18 at 5:16
aleph_twoaleph_two
24912
$endgroup$
add a comment |
$begingroup$
As copper.hat said in his comment, we generally cannot infer that $C=B$ if we only know that $CC^T=BB^T$.
However, to answer the question in OP's comment: this fact does not violate the uniqueness of the Cholesky factorization because to call $A=CC^T$ a "Cholesky factorization", we need that $C$ is lower-triangular; presumably the $P^TLsqrt{D}$ that arose from the Eigen calculation was not lower-triangular.
(In particular, the uniqueness of Cholesky decomposition is a consequence of the uniqueness of $LU$ factorization.)
answered Dec 19 '18 at 5:16
aleph_twoaleph_two
24912
$endgroup$
As copper.hat said in his comment, we generally cannot infer that $C=B$ if we only know that $CC^T=BB^T$.
However, to answer the question in OP's comment: this fact does not violate the uniqueness of the Cholesky factorization because to call $A=CC^T$ a "Cholesky factorization", we need that $C$ is lower-triangular; presumably the $P^TLsqrt{D}$ that arose from the Eigen calculation was not lower-triangular.
(In particular, the uniqueness of Cholesky decomposition is a consequence of the uniqueness of $LU$ factorization.)
answered Dec 19 '18 at 5:16
aleph_twoaleph_two
24912
answered Dec 19 '18 at 5:16
aleph_twoaleph_two
24912
answered Dec 19 '18 at 5:16
aleph_twoaleph_two
24912
answered Dec 19 '18 at 5:16
aleph_twoaleph_two
24912
24912
add a comment |
add a comment |
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$begingroup$
In general, $C C^T = B B^T$ does not imply that $C=B$. For example, if $Q$ is a rotation, $C C^T = C Q Q^T C^T = (CQ) (CQ)^T$, and, in general, we do not have $C = CQ$.
$endgroup$
– copper.hat
Nov 13 '18 at 14:54
$begingroup$
@copper.hat, thanks a lot for your answer, I did not think to see this problem in this way. Then, I ask myself what is the meaning of the uniqueness of the Cholesky decomposition for positive definite matrices. Is it unique apart of a rotation of $L$?
$endgroup$
– marco
Nov 13 '18 at 15:56