the inverse of a sum of two symmetric for schur completion?
$begingroup$
I have a up-triangulate Jacobi matrix J which can be blocked like :
$J = begin{bmatrix}A & B\ 0 & Cend{bmatrix} $
both A and C are up-triangulate, we can get Hessian matrix H by:
$H = J'J =begin{bmatrix}A' & 0\B'&C'end{bmatrix} begin{bmatrix}A & B\ 0 & Cend{bmatrix} = begin{bmatrix}A'A & A'B\B'A & B'B+C'Cend{bmatrix} $
by schur completion, I want to margin $B'B+C'C$, the equation is:
$ A'A - A'B(B'B+C'C)^{-1}B'A $
I know $C^{-1}$ is easy computed because it's a up-triangulate matrix, but unfortunately the add ruin this equation.
I have found that
$ X(X^{-1}+Y^{-1})Y = (X^{-1}+Y^{-1})^{-1} $ only when both X and Y are invertible, which B'B dose not meet
so any one have some brilliant idea to compute this equation easily?
linear-algebra matrices inverse matrix-decomposition schur-complement
asked Jan 7 at 1:56
Mr.GuoMr.Guo
307
$endgroup$
|
show 6 more comments
$begingroup$
I have a up-triangulate Jacobi matrix J which can be blocked like :
$J = begin{bmatrix}A & B\ 0 & Cend{bmatrix} $
both A and C are up-triangulate, we can get Hessian matrix H by:
$H = J'J =begin{bmatrix}A' & 0\B'&C'end{bmatrix} begin{bmatrix}A & B\ 0 & Cend{bmatrix} = begin{bmatrix}A'A & A'B\B'A & B'B+C'Cend{bmatrix} $
by schur completion, I want to margin $B'B+C'C$, the equation is:
$ A'A - A'B(B'B+C'C)^{-1}B'A $
I know $C^{-1}$ is easy computed because it's a up-triangulate matrix, but unfortunately the add ruin this equation.
I have found that
$ X(X^{-1}+Y^{-1})Y = (X^{-1}+Y^{-1})^{-1} $ only when both X and Y are invertible, which B'B dose not meet
so any one have some brilliant idea to compute this equation easily?
linear-algebra matrices inverse matrix-decomposition schur-complement
asked Jan 7 at 1:56
Mr.GuoMr.Guo
307
$endgroup$
$begingroup$
I understand most of your question, but I don't understand this: "by schur completion, I want to margin $B'B+C'C$". What are you trying to say here? What do you mean by "margin"?
$endgroup$
– Omnomnomnom
Jan 7 at 2:03
$begingroup$
@Omnomnomnom thank you for your attention , basically margin means remove some target variable in optimization problem , you can refer it from g2o paper cct.lsu.edu/~kzhang/papers/g2o.pdf , equ(25)
$endgroup$
– Mr.Guo
Jan 7 at 2:10
$begingroup$
If $B$ (or $B^TB$) has low rank you can update the cholesky decomposition $C^TC$
$endgroup$
– LinAlg
Jan 7 at 2:12
$begingroup$
@LinAlg yes B'B is not full rank , could you give me some more information about "update the cholesky decomposition" ?
$endgroup$
– Mr.Guo
Jan 7 at 2:15
$begingroup$
@LinAlg thank you for your hint, I think cholesky decomposition update should be work
$endgroup$
– Mr.Guo
Jan 7 at 2:24
|
show 6 more comments
$begingroup$
I have a up-triangulate Jacobi matrix J which can be blocked like :
$J = begin{bmatrix}A & B\ 0 & Cend{bmatrix} $
both A and C are up-triangulate, we can get Hessian matrix H by:
$H = J'J =begin{bmatrix}A' & 0\B'&C'end{bmatrix} begin{bmatrix}A & B\ 0 & Cend{bmatrix} = begin{bmatrix}A'A & A'B\B'A & B'B+C'Cend{bmatrix} $
by schur completion, I want to margin $B'B+C'C$, the equation is:
$ A'A - A'B(B'B+C'C)^{-1}B'A $
I know $C^{-1}$ is easy computed because it's a up-triangulate matrix, but unfortunately the add ruin this equation.
I have found that
$ X(X^{-1}+Y^{-1})Y = (X^{-1}+Y^{-1})^{-1} $ only when both X and Y are invertible, which B'B dose not meet
so any one have some brilliant idea to compute this equation easily?
linear-algebra matrices inverse matrix-decomposition schur-complement
asked Jan 7 at 1:56
Mr.GuoMr.Guo
307
$endgroup$
I have a up-triangulate Jacobi matrix J which can be blocked like :
$J = begin{bmatrix}A & B\ 0 & Cend{bmatrix} $
both A and C are up-triangulate, we can get Hessian matrix H by:
$H = J'J =begin{bmatrix}A' & 0\B'&C'end{bmatrix} begin{bmatrix}A & B\ 0 & Cend{bmatrix} = begin{bmatrix}A'A & A'B\B'A & B'B+C'Cend{bmatrix} $
by schur completion, I want to margin $B'B+C'C$, the equation is:
$ A'A - A'B(B'B+C'C)^{-1}B'A $
I know $C^{-1}$ is easy computed because it's a up-triangulate matrix, but unfortunately the add ruin this equation.
I have found that
$ X(X^{-1}+Y^{-1})Y = (X^{-1}+Y^{-1})^{-1} $ only when both X and Y are invertible, which B'B dose not meet
so any one have some brilliant idea to compute this equation easily?
linear-algebra matrices inverse matrix-decomposition schur-complement
linear-algebra matrices inverse matrix-decomposition schur-complement
asked Jan 7 at 1:56
Mr.GuoMr.Guo
307
asked Jan 7 at 1:56
Mr.GuoMr.Guo
307
edited Jan 7 at 2:05
Mr.Guo
asked Jan 7 at 1:56
Mr.GuoMr.Guo
307
asked Jan 7 at 1:56
Mr.GuoMr.Guo
307
asked Jan 7 at 1:56
Mr.GuoMr.Guo
307
307
$begingroup$
I understand most of your question, but I don't understand this: "by schur completion, I want to margin $B'B+C'C$". What are you trying to say here? What do you mean by "margin"?
$endgroup$
– Omnomnomnom
Jan 7 at 2:03
$begingroup$
@Omnomnomnom thank you for your attention , basically margin means remove some target variable in optimization problem , you can refer it from g2o paper cct.lsu.edu/~kzhang/papers/g2o.pdf , equ(25)
$endgroup$
– Mr.Guo
Jan 7 at 2:10
$begingroup$
If $B$ (or $B^TB$) has low rank you can update the cholesky decomposition $C^TC$
$endgroup$
– LinAlg
Jan 7 at 2:12
$begingroup$
@LinAlg yes B'B is not full rank , could you give me some more information about "update the cholesky decomposition" ?
$endgroup$
– Mr.Guo
Jan 7 at 2:15
$begingroup$
@LinAlg thank you for your hint, I think cholesky decomposition update should be work
$endgroup$
– Mr.Guo
Jan 7 at 2:24
|
show 6 more comments
$begingroup$
I understand most of your question, but I don't understand this: "by schur completion, I want to margin $B'B+C'C$". What are you trying to say here? What do you mean by "margin"?
$endgroup$
– Omnomnomnom
Jan 7 at 2:03
$begingroup$
@Omnomnomnom thank you for your attention , basically margin means remove some target variable in optimization problem , you can refer it from g2o paper cct.lsu.edu/~kzhang/papers/g2o.pdf , equ(25)
$endgroup$
– Mr.Guo
Jan 7 at 2:10
$begingroup$
If $B$ (or $B^TB$) has low rank you can update the cholesky decomposition $C^TC$
$endgroup$
– LinAlg
Jan 7 at 2:12
$begingroup$
@LinAlg yes B'B is not full rank , could you give me some more information about "update the cholesky decomposition" ?
$endgroup$
– Mr.Guo
Jan 7 at 2:15
$begingroup$
@LinAlg thank you for your hint, I think cholesky decomposition update should be work
$endgroup$
– Mr.Guo
Jan 7 at 2:24
$begingroup$
I understand most of your question, but I don't understand this: "by schur completion, I want to margin $B'B+C'C$". What are you trying to say here? What do you mean by "margin"?
$endgroup$
– Omnomnomnom
Jan 7 at 2:03
$begingroup$
I understand most of your question, but I don't understand this: "by schur completion, I want to margin $B'B+C'C$". What are you trying to say here? What do you mean by "margin"?
$endgroup$
– Omnomnomnom
Jan 7 at 2:03
$begingroup$
@Omnomnomnom thank you for your attention , basically margin means remove some target variable in optimization problem , you can refer it from g2o paper cct.lsu.edu/~kzhang/papers/g2o.pdf , equ(25)
$endgroup$
– Mr.Guo
Jan 7 at 2:10
$begingroup$
@Omnomnomnom thank you for your attention , basically margin means remove some target variable in optimization problem , you can refer it from g2o paper cct.lsu.edu/~kzhang/papers/g2o.pdf , equ(25)
$endgroup$
– Mr.Guo
Jan 7 at 2:10
$begingroup$
If $B$ (or $B^TB$) has low rank you can update the cholesky decomposition $C^TC$
$endgroup$
– LinAlg
Jan 7 at 2:12
$begingroup$
If $B$ (or $B^TB$) has low rank you can update the cholesky decomposition $C^TC$
$endgroup$
– LinAlg
Jan 7 at 2:12
$begingroup$
@LinAlg yes B'B is not full rank , could you give me some more information about "update the cholesky decomposition" ?
$endgroup$
– Mr.Guo
Jan 7 at 2:15
$begingroup$
@LinAlg yes B'B is not full rank , could you give me some more information about "update the cholesky decomposition" ?
$endgroup$
– Mr.Guo
Jan 7 at 2:15
$begingroup$
@LinAlg thank you for your hint, I think cholesky decomposition update should be work
$endgroup$
– Mr.Guo
Jan 7 at 2:24
$begingroup$
@LinAlg thank you for your hint, I think cholesky decomposition update should be work
$endgroup$
– Mr.Guo
Jan 7 at 2:24
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
To compute $X:=(B^TB+C^TC)^{-1}B^TA$, you never actually compute $(B^TB+C^TC)^{-1}$, but instead you solve $(B^TB+C^TC)X=B^TA$. Since $C$ is upper triangular, you can see $C^T$ as the Cholesky factor of $C^TC$. You can update the Cholesky factor $C^T$ to account for the low rank update $B^TB$ (via repeated rank-1 updates). That will give you a new Cholesky factor $L$ such that $LL^T = B^TB+C^TC$. Then all you need to do is solve $LL^TX = B^TA$, which is relatively easy.
answered Jan 7 at 2:58
LinAlgLinAlg
9,4991521
$endgroup$
$begingroup$
cholesky factor update is a good idea, I'm using eigen , but I didn't find the rank one update for C, it only have interface for C'C which need mutiply first, so I think @Omnomnomnom 's idea is more fitting for me :) , I have listed the detail below
$endgroup$
– Mr.Guo
Jan 7 at 3:50
$begingroup$
@Mr.Guo what do you mean by "I didn't find the rank one update for C, it only have interface for C'C which need mutiply first"?
$endgroup$
– LinAlg
Jan 7 at 4:42
$begingroup$
I mean Eigen have no good interface for up-triangulate matrix rank-one update :)
$endgroup$
– Mr.Guo
Jan 7 at 9:32
$begingroup$
@Mr.Guo I think $C$ is of the LLT type with the UpLo flag set to upper, and that you can use this procedure.
$endgroup$
– LinAlg
Jan 7 at 13:11
add a comment |
$begingroup$
hi @Omnomnomnom you have given me a better hint to solve this problem , I'd like to list it here
I have checked your equation it's correct except one little mistaken, it should be:
$ (C'C+B'B)^{-1} = (C'C)^{-1} - (C'C)^{-1}B'(I+B(C'C)^{-1}B')^{-1}B(C'C)^{-1} $
I think it's a good solution because B is (small rows * big cols) matrix, we can get
$D = (C'C)^{-1}$
$E = BDB'$
$F = (I+E)^{-1}$
D,E,F are all easy computed, so the result is :
$ (C'C+B'B)^{-1} = D - DB'FBD $
and finally,
$ result = A'A-A'B(B'B+C'C)^{-1}B'A = A'(I-B(B'B+C'C)^{-1}B')A = A'(I-BDB' + BDB'FBDB')A = A'(I-E+EFE)A$
answered Jan 7 at 3:46
Mr.GuoMr.Guo
307
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To compute $X:=(B^TB+C^TC)^{-1}B^TA$, you never actually compute $(B^TB+C^TC)^{-1}$, but instead you solve $(B^TB+C^TC)X=B^TA$. Since $C$ is upper triangular, you can see $C^T$ as the Cholesky factor of $C^TC$. You can update the Cholesky factor $C^T$ to account for the low rank update $B^TB$ (via repeated rank-1 updates). That will give you a new Cholesky factor $L$ such that $LL^T = B^TB+C^TC$. Then all you need to do is solve $LL^TX = B^TA$, which is relatively easy.
answered Jan 7 at 2:58
LinAlgLinAlg
9,4991521
$endgroup$
$begingroup$
cholesky factor update is a good idea, I'm using eigen , but I didn't find the rank one update for C, it only have interface for C'C which need mutiply first, so I think @Omnomnomnom 's idea is more fitting for me :) , I have listed the detail below
$endgroup$
– Mr.Guo
Jan 7 at 3:50
$begingroup$
@Mr.Guo what do you mean by "I didn't find the rank one update for C, it only have interface for C'C which need mutiply first"?
$endgroup$
– LinAlg
Jan 7 at 4:42
$begingroup$
I mean Eigen have no good interface for up-triangulate matrix rank-one update :)
$endgroup$
– Mr.Guo
Jan 7 at 9:32
$begingroup$
@Mr.Guo I think $C$ is of the LLT type with the UpLo flag set to upper, and that you can use this procedure.
$endgroup$
– LinAlg
Jan 7 at 13:11
add a comment |
$begingroup$
To compute $X:=(B^TB+C^TC)^{-1}B^TA$, you never actually compute $(B^TB+C^TC)^{-1}$, but instead you solve $(B^TB+C^TC)X=B^TA$. Since $C$ is upper triangular, you can see $C^T$ as the Cholesky factor of $C^TC$. You can update the Cholesky factor $C^T$ to account for the low rank update $B^TB$ (via repeated rank-1 updates). That will give you a new Cholesky factor $L$ such that $LL^T = B^TB+C^TC$. Then all you need to do is solve $LL^TX = B^TA$, which is relatively easy.
answered Jan 7 at 2:58
LinAlgLinAlg
9,4991521
$endgroup$
$begingroup$
cholesky factor update is a good idea, I'm using eigen , but I didn't find the rank one update for C, it only have interface for C'C which need mutiply first, so I think @Omnomnomnom 's idea is more fitting for me :) , I have listed the detail below
$endgroup$
– Mr.Guo
Jan 7 at 3:50
$begingroup$
@Mr.Guo what do you mean by "I didn't find the rank one update for C, it only have interface for C'C which need mutiply first"?
$endgroup$
– LinAlg
Jan 7 at 4:42
$begingroup$
I mean Eigen have no good interface for up-triangulate matrix rank-one update :)
$endgroup$
– Mr.Guo
Jan 7 at 9:32
$begingroup$
@Mr.Guo I think $C$ is of the LLT type with the UpLo flag set to upper, and that you can use this procedure.
$endgroup$
– LinAlg
Jan 7 at 13:11
add a comment |
$begingroup$
To compute $X:=(B^TB+C^TC)^{-1}B^TA$, you never actually compute $(B^TB+C^TC)^{-1}$, but instead you solve $(B^TB+C^TC)X=B^TA$. Since $C$ is upper triangular, you can see $C^T$ as the Cholesky factor of $C^TC$. You can update the Cholesky factor $C^T$ to account for the low rank update $B^TB$ (via repeated rank-1 updates). That will give you a new Cholesky factor $L$ such that $LL^T = B^TB+C^TC$. Then all you need to do is solve $LL^TX = B^TA$, which is relatively easy.
answered Jan 7 at 2:58
LinAlgLinAlg
9,4991521
$endgroup$
To compute $X:=(B^TB+C^TC)^{-1}B^TA$, you never actually compute $(B^TB+C^TC)^{-1}$, but instead you solve $(B^TB+C^TC)X=B^TA$. Since $C$ is upper triangular, you can see $C^T$ as the Cholesky factor of $C^TC$. You can update the Cholesky factor $C^T$ to account for the low rank update $B^TB$ (via repeated rank-1 updates). That will give you a new Cholesky factor $L$ such that $LL^T = B^TB+C^TC$. Then all you need to do is solve $LL^TX = B^TA$, which is relatively easy.
answered Jan 7 at 2:58
LinAlgLinAlg
9,4991521
answered Jan 7 at 2:58
LinAlgLinAlg
9,4991521
answered Jan 7 at 2:58
LinAlgLinAlg
9,4991521
answered Jan 7 at 2:58
LinAlgLinAlg
9,4991521
9,4991521
$begingroup$
cholesky factor update is a good idea, I'm using eigen , but I didn't find the rank one update for C, it only have interface for C'C which need mutiply first, so I think @Omnomnomnom 's idea is more fitting for me :) , I have listed the detail below
$endgroup$
– Mr.Guo
Jan 7 at 3:50
$begingroup$
@Mr.Guo what do you mean by "I didn't find the rank one update for C, it only have interface for C'C which need mutiply first"?
$endgroup$
– LinAlg
Jan 7 at 4:42
$begingroup$
I mean Eigen have no good interface for up-triangulate matrix rank-one update :)
$endgroup$
– Mr.Guo
Jan 7 at 9:32
$begingroup$
@Mr.Guo I think $C$ is of the LLT type with the UpLo flag set to upper, and that you can use this procedure.
$endgroup$
– LinAlg
Jan 7 at 13:11
add a comment |
$begingroup$
cholesky factor update is a good idea, I'm using eigen , but I didn't find the rank one update for C, it only have interface for C'C which need mutiply first, so I think @Omnomnomnom 's idea is more fitting for me :) , I have listed the detail below
$endgroup$
– Mr.Guo
Jan 7 at 3:50
$begingroup$
@Mr.Guo what do you mean by "I didn't find the rank one update for C, it only have interface for C'C which need mutiply first"?
$endgroup$
– LinAlg
Jan 7 at 4:42
$begingroup$
I mean Eigen have no good interface for up-triangulate matrix rank-one update :)
$endgroup$
– Mr.Guo
Jan 7 at 9:32
$begingroup$
@Mr.Guo I think $C$ is of the LLT type with the UpLo flag set to upper, and that you can use this procedure.
$endgroup$
– LinAlg
Jan 7 at 13:11
$begingroup$
cholesky factor update is a good idea, I'm using eigen , but I didn't find the rank one update for C, it only have interface for C'C which need mutiply first, so I think @Omnomnomnom 's idea is more fitting for me :) , I have listed the detail below
$endgroup$
– Mr.Guo
Jan 7 at 3:50
$begingroup$
cholesky factor update is a good idea, I'm using eigen , but I didn't find the rank one update for C, it only have interface for C'C which need mutiply first, so I think @Omnomnomnom 's idea is more fitting for me :) , I have listed the detail below
$endgroup$
– Mr.Guo
Jan 7 at 3:50
$begingroup$
@Mr.Guo what do you mean by "I didn't find the rank one update for C, it only have interface for C'C which need mutiply first"?
$endgroup$
– LinAlg
Jan 7 at 4:42
$begingroup$
@Mr.Guo what do you mean by "I didn't find the rank one update for C, it only have interface for C'C which need mutiply first"?
$endgroup$
– LinAlg
Jan 7 at 4:42
$begingroup$
I mean Eigen have no good interface for up-triangulate matrix rank-one update :)
$endgroup$
– Mr.Guo
Jan 7 at 9:32
$begingroup$
I mean Eigen have no good interface for up-triangulate matrix rank-one update :)
$endgroup$
– Mr.Guo
Jan 7 at 9:32
$begingroup$
@Mr.Guo I think $C$ is of the LLT type with the UpLo flag set to upper, and that you can use this procedure.
$endgroup$
– LinAlg
Jan 7 at 13:11
$begingroup$
@Mr.Guo I think $C$ is of the LLT type with the UpLo flag set to upper, and that you can use this procedure.
$endgroup$
– LinAlg
Jan 7 at 13:11
add a comment |
$begingroup$
hi @Omnomnomnom you have given me a better hint to solve this problem , I'd like to list it here
I have checked your equation it's correct except one little mistaken, it should be:
$ (C'C+B'B)^{-1} = (C'C)^{-1} - (C'C)^{-1}B'(I+B(C'C)^{-1}B')^{-1}B(C'C)^{-1} $
I think it's a good solution because B is (small rows * big cols) matrix, we can get
$D = (C'C)^{-1}$
$E = BDB'$
$F = (I+E)^{-1}$
D,E,F are all easy computed, so the result is :
$ (C'C+B'B)^{-1} = D - DB'FBD $
and finally,
$ result = A'A-A'B(B'B+C'C)^{-1}B'A = A'(I-B(B'B+C'C)^{-1}B')A = A'(I-BDB' + BDB'FBDB')A = A'(I-E+EFE)A$
answered Jan 7 at 3:46
Mr.GuoMr.Guo
307
$endgroup$
add a comment |
$begingroup$
hi @Omnomnomnom you have given me a better hint to solve this problem , I'd like to list it here
I have checked your equation it's correct except one little mistaken, it should be:
$ (C'C+B'B)^{-1} = (C'C)^{-1} - (C'C)^{-1}B'(I+B(C'C)^{-1}B')^{-1}B(C'C)^{-1} $
I think it's a good solution because B is (small rows * big cols) matrix, we can get
$D = (C'C)^{-1}$
$E = BDB'$
$F = (I+E)^{-1}$
D,E,F are all easy computed, so the result is :
$ (C'C+B'B)^{-1} = D - DB'FBD $
and finally,
$ result = A'A-A'B(B'B+C'C)^{-1}B'A = A'(I-B(B'B+C'C)^{-1}B')A = A'(I-BDB' + BDB'FBDB')A = A'(I-E+EFE)A$
answered Jan 7 at 3:46
Mr.GuoMr.Guo
307
$endgroup$
add a comment |
$begingroup$
hi @Omnomnomnom you have given me a better hint to solve this problem , I'd like to list it here
I have checked your equation it's correct except one little mistaken, it should be:
$ (C'C+B'B)^{-1} = (C'C)^{-1} - (C'C)^{-1}B'(I+B(C'C)^{-1}B')^{-1}B(C'C)^{-1} $
I think it's a good solution because B is (small rows * big cols) matrix, we can get
$D = (C'C)^{-1}$
$E = BDB'$
$F = (I+E)^{-1}$
D,E,F are all easy computed, so the result is :
$ (C'C+B'B)^{-1} = D - DB'FBD $
and finally,
$ result = A'A-A'B(B'B+C'C)^{-1}B'A = A'(I-B(B'B+C'C)^{-1}B')A = A'(I-BDB' + BDB'FBDB')A = A'(I-E+EFE)A$
answered Jan 7 at 3:46
Mr.GuoMr.Guo
307
$endgroup$
hi @Omnomnomnom you have given me a better hint to solve this problem , I'd like to list it here
I have checked your equation it's correct except one little mistaken, it should be:
$ (C'C+B'B)^{-1} = (C'C)^{-1} - (C'C)^{-1}B'(I+B(C'C)^{-1}B')^{-1}B(C'C)^{-1} $
I think it's a good solution because B is (small rows * big cols) matrix, we can get
$D = (C'C)^{-1}$
$E = BDB'$
$F = (I+E)^{-1}$
D,E,F are all easy computed, so the result is :
$ (C'C+B'B)^{-1} = D - DB'FBD $
and finally,
$ result = A'A-A'B(B'B+C'C)^{-1}B'A = A'(I-B(B'B+C'C)^{-1}B')A = A'(I-BDB' + BDB'FBDB')A = A'(I-E+EFE)A$
answered Jan 7 at 3:46
Mr.GuoMr.Guo
307
edited Jan 7 at 4:11
answered Jan 7 at 3:46
Mr.GuoMr.Guo
307
answered Jan 7 at 3:46
Mr.GuoMr.Guo
307
answered Jan 7 at 3:46
Mr.GuoMr.Guo
307
307
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$begingroup$
I understand most of your question, but I don't understand this: "by schur completion, I want to margin $B'B+C'C$". What are you trying to say here? What do you mean by "margin"?
$endgroup$
– Omnomnomnom
Jan 7 at 2:03
$begingroup$
@Omnomnomnom thank you for your attention , basically margin means remove some target variable in optimization problem , you can refer it from g2o paper cct.lsu.edu/~kzhang/papers/g2o.pdf , equ(25)
$endgroup$
– Mr.Guo
Jan 7 at 2:10
$begingroup$
If $B$ (or $B^TB$) has low rank you can update the cholesky decomposition $C^TC$
$endgroup$
– LinAlg
Jan 7 at 2:12
$begingroup$
@LinAlg yes B'B is not full rank , could you give me some more information about "update the cholesky decomposition" ?
$endgroup$
– Mr.Guo
Jan 7 at 2:15
$begingroup$
@LinAlg thank you for your hint, I think cholesky decomposition update should be work
$endgroup$
– Mr.Guo
Jan 7 at 2:24