LU decomposition of a matrix given LU decomposition of its blocks.
Suppose $A, B, C$ are $ntimes n$ matrices. Let $A = L_1U_1$ and $D = L_2U_2$. Then what is the LU decomposition of $$begin{bmatrix} A&B\ 0&Dend{bmatrix}$$
How to find this?
I am able to find $$begin{bmatrix} A&B\ 0&Dend{bmatrix} = begin{bmatrix} L_1&0\ 0&L_2end{bmatrix}times begin{bmatrix} U_1&X\ 0&U_2end{bmatrix}$$but this means $L_1X = B$ which might not be the case. How to look for this?
linear-algebra matrix-calculus lu-decomposition
asked Nov 24 at 11:36
Mittal G
1,188515
add a comment |
Suppose $A, B, C$ are $ntimes n$ matrices. Let $A = L_1U_1$ and $D = L_2U_2$. Then what is the LU decomposition of $$begin{bmatrix} A&B\ 0&Dend{bmatrix}$$
How to find this?
I am able to find $$begin{bmatrix} A&B\ 0&Dend{bmatrix} = begin{bmatrix} L_1&0\ 0&L_2end{bmatrix}times begin{bmatrix} U_1&X\ 0&U_2end{bmatrix}$$but this means $L_1X = B$ which might not be the case. How to look for this?
linear-algebra matrix-calculus lu-decomposition
asked Nov 24 at 11:36
Mittal G
1,188515
Why might that not be the case? In the first sentence, do you mean $A,B,D$ are $n times n$ matrices?
– Viktor Glombik
Nov 24 at 11:39
Because B can have a different LU decomposition.
– Mittal G
Nov 24 at 11:41
If $L_1$ is invertible then such a matrix $X$ exists.
– Mittal G
Nov 24 at 11:42
@Viktor Glombik Yes all are $ntimes n$ matrices.
– Mittal G
Nov 24 at 11:51
add a comment |
Suppose $A, B, C$ are $ntimes n$ matrices. Let $A = L_1U_1$ and $D = L_2U_2$. Then what is the LU decomposition of $$begin{bmatrix} A&B\ 0&Dend{bmatrix}$$
How to find this?
I am able to find $$begin{bmatrix} A&B\ 0&Dend{bmatrix} = begin{bmatrix} L_1&0\ 0&L_2end{bmatrix}times begin{bmatrix} U_1&X\ 0&U_2end{bmatrix}$$but this means $L_1X = B$ which might not be the case. How to look for this?
linear-algebra matrix-calculus lu-decomposition
asked Nov 24 at 11:36
Mittal G
1,188515
Suppose $A, B, C$ are $ntimes n$ matrices. Let $A = L_1U_1$ and $D = L_2U_2$. Then what is the LU decomposition of $$begin{bmatrix} A&B\ 0&Dend{bmatrix}$$
How to find this?
I am able to find $$begin{bmatrix} A&B\ 0&Dend{bmatrix} = begin{bmatrix} L_1&0\ 0&L_2end{bmatrix}times begin{bmatrix} U_1&X\ 0&U_2end{bmatrix}$$but this means $L_1X = B$ which might not be the case. How to look for this?
linear-algebra matrix-calculus lu-decomposition
linear-algebra matrix-calculus lu-decomposition
asked Nov 24 at 11:36
Mittal G
1,188515
asked Nov 24 at 11:36
Mittal G
1,188515
asked Nov 24 at 11:36
Mittal G
1,188515
asked Nov 24 at 11:36
Mittal G
1,188515
asked Nov 24 at 11:36
Mittal G
1,188515
1,188515
Why might that not be the case? In the first sentence, do you mean $A,B,D$ are $n times n$ matrices?
– Viktor Glombik
Nov 24 at 11:39
Because B can have a different LU decomposition.
– Mittal G
Nov 24 at 11:41
If $L_1$ is invertible then such a matrix $X$ exists.
– Mittal G
Nov 24 at 11:42
@Viktor Glombik Yes all are $ntimes n$ matrices.
– Mittal G
Nov 24 at 11:51
add a comment |
Why might that not be the case? In the first sentence, do you mean $A,B,D$ are $n times n$ matrices?
– Viktor Glombik
Nov 24 at 11:39
Because B can have a different LU decomposition.
– Mittal G
Nov 24 at 11:41
If $L_1$ is invertible then such a matrix $X$ exists.
– Mittal G
Nov 24 at 11:42
@Viktor Glombik Yes all are $ntimes n$ matrices.
– Mittal G
Nov 24 at 11:51
Why might that not be the case? In the first sentence, do you mean $A,B,D$ are $n times n$ matrices?
– Viktor Glombik
Nov 24 at 11:39
Why might that not be the case? In the first sentence, do you mean $A,B,D$ are $n times n$ matrices?
– Viktor Glombik
Nov 24 at 11:39
Because B can have a different LU decomposition.
– Mittal G
Nov 24 at 11:41
Because B can have a different LU decomposition.
– Mittal G
Nov 24 at 11:41
If $L_1$ is invertible then such a matrix $X$ exists.
– Mittal G
Nov 24 at 11:42
If $L_1$ is invertible then such a matrix $X$ exists.
– Mittal G
Nov 24 at 11:42
@Viktor Glombik Yes all are $ntimes n$ matrices.
– Mittal G
Nov 24 at 11:51
@Viktor Glombik Yes all are $ntimes n$ matrices.
– Mittal G
Nov 24 at 11:51
add a comment |
1 Answer
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If $rank(L_1)=rank([L_1,B])$, then $L_1L_1^+B=B$ and one has the solution
$diag(L_1,L_2)begin{pmatrix}U_1&L_1^+B\0&U_2end{pmatrix}$.
Otherwise, I'm not sure there is a solution in closed form.
answered Nov 25 at 12:06
loup blanc
22.5k21750
add a comment |
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1 Answer
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1 Answer
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oldest
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If $rank(L_1)=rank([L_1,B])$, then $L_1L_1^+B=B$ and one has the solution
$diag(L_1,L_2)begin{pmatrix}U_1&L_1^+B\0&U_2end{pmatrix}$.
Otherwise, I'm not sure there is a solution in closed form.
answered Nov 25 at 12:06
loup blanc
22.5k21750
add a comment |
If $rank(L_1)=rank([L_1,B])$, then $L_1L_1^+B=B$ and one has the solution
$diag(L_1,L_2)begin{pmatrix}U_1&L_1^+B\0&U_2end{pmatrix}$.
Otherwise, I'm not sure there is a solution in closed form.
answered Nov 25 at 12:06
loup blanc
22.5k21750
add a comment |
If $rank(L_1)=rank([L_1,B])$, then $L_1L_1^+B=B$ and one has the solution
$diag(L_1,L_2)begin{pmatrix}U_1&L_1^+B\0&U_2end{pmatrix}$.
Otherwise, I'm not sure there is a solution in closed form.
answered Nov 25 at 12:06
loup blanc
22.5k21750
If $rank(L_1)=rank([L_1,B])$, then $L_1L_1^+B=B$ and one has the solution
$diag(L_1,L_2)begin{pmatrix}U_1&L_1^+B\0&U_2end{pmatrix}$.
Otherwise, I'm not sure there is a solution in closed form.
answered Nov 25 at 12:06
loup blanc
22.5k21750
answered Nov 25 at 12:06
loup blanc
22.5k21750
answered Nov 25 at 12:06
loup blanc
22.5k21750
answered Nov 25 at 12:06
loup blanc
22.5k21750
22.5k21750
add a comment |
add a comment |
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Why might that not be the case? In the first sentence, do you mean $A,B,D$ are $n times n$ matrices?
– Viktor Glombik
Nov 24 at 11:39
Because B can have a different LU decomposition.
– Mittal G
Nov 24 at 11:41
If $L_1$ is invertible then such a matrix $X$ exists.
– Mittal G
Nov 24 at 11:42
@Viktor Glombik Yes all are $ntimes n$ matrices.
– Mittal G
Nov 24 at 11:51