Mixing
How is A= UL different from A = LU
$begingroup$
Using Guassian elimination introducing zeros into the columns of $A$ in the
order, $n: - 1:2$ and producing the factorization $A = UL$ where $U$ is unit upper triangular and $L$ is lower triangular. How is this different from $A = LU.$ Can I relate the two procedures?
linear-algebra
edited Jan 30 at 14:03
Vinyl_cape_jawa
3,33011433
asked Jan 30 at 13:34
Usman AshrafUsman Ashraf
13
$endgroup$
add a comment |
$begingroup$
Using Guassian elimination introducing zeros into the columns of $A$ in the
order, $n: - 1:2$ and producing the factorization $A = UL$ where $U$ is unit upper triangular and $L$ is lower triangular. How is this different from $A = LU.$ Can I relate the two procedures?
linear-algebra
edited Jan 30 at 14:03
Vinyl_cape_jawa
3,33011433
asked Jan 30 at 13:34
Usman AshrafUsman Ashraf
13
$endgroup$
$begingroup$
Does "unit upper triangular" mean "ones on the diagonal"?
$endgroup$
– John Hughes
Jan 30 at 14:11
add a comment |
$begingroup$
Using Guassian elimination introducing zeros into the columns of $A$ in the
order, $n: - 1:2$ and producing the factorization $A = UL$ where $U$ is unit upper triangular and $L$ is lower triangular. How is this different from $A = LU.$ Can I relate the two procedures?
linear-algebra
edited Jan 30 at 14:03
Vinyl_cape_jawa
3,33011433
asked Jan 30 at 13:34
Usman AshrafUsman Ashraf
13
$endgroup$
Using Guassian elimination introducing zeros into the columns of $A$ in the
order, $n: - 1:2$ and producing the factorization $A = UL$ where $U$ is unit upper triangular and $L$ is lower triangular. How is this different from $A = LU.$ Can I relate the two procedures?
linear-algebra
linear-algebra
edited Jan 30 at 14:03
Vinyl_cape_jawa
3,33011433
asked Jan 30 at 13:34
Usman AshrafUsman Ashraf
13
edited Jan 30 at 14:03
Vinyl_cape_jawa
3,33011433
asked Jan 30 at 13:34
Usman AshrafUsman Ashraf
13
edited Jan 30 at 14:03
Vinyl_cape_jawa
3,33011433
edited Jan 30 at 14:03
Vinyl_cape_jawa
3,33011433
edited Jan 30 at 14:03
Vinyl_cape_jawa
3,33011433
3,33011433
asked Jan 30 at 13:34
Usman AshrafUsman Ashraf
13
asked Jan 30 at 13:34
Usman AshrafUsman Ashraf
13
asked Jan 30 at 13:34
Usman AshrafUsman Ashraf
13
13
$begingroup$
Does "unit upper triangular" mean "ones on the diagonal"?
$endgroup$
– John Hughes
Jan 30 at 14:11
add a comment |
$begingroup$
Does "unit upper triangular" mean "ones on the diagonal"?
$endgroup$
– John Hughes
Jan 30 at 14:11
$begingroup$
Does "unit upper triangular" mean "ones on the diagonal"?
$endgroup$
– John Hughes
Jan 30 at 14:11
$begingroup$
Does "unit upper triangular" mean "ones on the diagonal"?
$endgroup$
– John Hughes
Jan 30 at 14:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 2 \
0 & 1}
pmatrix{-1 & 0 \
1 & 1} \
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 0 \
1 & 1}
pmatrix{1 & 2 \
0 & -1}
$$
So evidently the two approaches produce different results.
One (weak) relation is that if you start with
$$
A = L U
$$
and transpose everything, you get
$$
A^t = U^t L^t
$$
and because $U^t$ is lower-triangular, and $L^t$ is upper triangular, you get
$$
A^t = L' U',
$$
but this time $L'$ has the ones on the diagonal rather than $U$ having ones on the diagonal. I guess that's really a relationship between two different versions of LU factorization, rather than one being UL factorization.
answered Jan 30 at 14:18
John HughesJohn Hughes
65.2k24293
$endgroup$
$begingroup$
Transpose shuffles upper/lowerness, but inverse preserves it.
$endgroup$
– mathreadler
Jan 30 at 14:27
add a comment |
$begingroup$
Observe that if $$A = LU$$
then (assuming $A^{-1}$ exists.) :
$$A^{-1} = U^{-1}L^{-1}$$
We can verify this by hand if you don't believe it.
What remains is to show that $U^{-1}$ is also upper triangular and $L^{-1}$ is also lower triangular if $U,L$ are, respectively.
answered Jan 30 at 14:26
mathreadlermathreadler
15.4k72263
$endgroup$
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
$begingroup$
$$
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 2 \
0 & 1}
pmatrix{-1 & 0 \
1 & 1} \
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 0 \
1 & 1}
pmatrix{1 & 2 \
0 & -1}
$$
So evidently the two approaches produce different results.
One (weak) relation is that if you start with
$$
A = L U
$$
and transpose everything, you get
$$
A^t = U^t L^t
$$
and because $U^t$ is lower-triangular, and $L^t$ is upper triangular, you get
$$
A^t = L' U',
$$
but this time $L'$ has the ones on the diagonal rather than $U$ having ones on the diagonal. I guess that's really a relationship between two different versions of LU factorization, rather than one being UL factorization.
answered Jan 30 at 14:18
John HughesJohn Hughes
65.2k24293
$endgroup$
$begingroup$
Transpose shuffles upper/lowerness, but inverse preserves it.
$endgroup$
– mathreadler
Jan 30 at 14:27
add a comment |
$begingroup$
$$
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 2 \
0 & 1}
pmatrix{-1 & 0 \
1 & 1} \
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 0 \
1 & 1}
pmatrix{1 & 2 \
0 & -1}
$$
So evidently the two approaches produce different results.
One (weak) relation is that if you start with
$$
A = L U
$$
and transpose everything, you get
$$
A^t = U^t L^t
$$
and because $U^t$ is lower-triangular, and $L^t$ is upper triangular, you get
$$
A^t = L' U',
$$
but this time $L'$ has the ones on the diagonal rather than $U$ having ones on the diagonal. I guess that's really a relationship between two different versions of LU factorization, rather than one being UL factorization.
answered Jan 30 at 14:18
John HughesJohn Hughes
65.2k24293
$endgroup$
$begingroup$
Transpose shuffles upper/lowerness, but inverse preserves it.
$endgroup$
– mathreadler
Jan 30 at 14:27
add a comment |
$begingroup$
$$
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 2 \
0 & 1}
pmatrix{-1 & 0 \
1 & 1} \
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 0 \
1 & 1}
pmatrix{1 & 2 \
0 & -1}
$$
So evidently the two approaches produce different results.
One (weak) relation is that if you start with
$$
A = L U
$$
and transpose everything, you get
$$
A^t = U^t L^t
$$
and because $U^t$ is lower-triangular, and $L^t$ is upper triangular, you get
$$
A^t = L' U',
$$
but this time $L'$ has the ones on the diagonal rather than $U$ having ones on the diagonal. I guess that's really a relationship between two different versions of LU factorization, rather than one being UL factorization.
answered Jan 30 at 14:18
John HughesJohn Hughes
65.2k24293
$endgroup$
$$
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 2 \
0 & 1}
pmatrix{-1 & 0 \
1 & 1} \
pmatrix{
1 & 2 \
1 & 1 } =
pmatrix{1 & 0 \
1 & 1}
pmatrix{1 & 2 \
0 & -1}
$$
So evidently the two approaches produce different results.
One (weak) relation is that if you start with
$$
A = L U
$$
and transpose everything, you get
$$
A^t = U^t L^t
$$
and because $U^t$ is lower-triangular, and $L^t$ is upper triangular, you get
$$
A^t = L' U',
$$
but this time $L'$ has the ones on the diagonal rather than $U$ having ones on the diagonal. I guess that's really a relationship between two different versions of LU factorization, rather than one being UL factorization.
answered Jan 30 at 14:18
John HughesJohn Hughes
65.2k24293
answered Jan 30 at 14:18
John HughesJohn Hughes
65.2k24293
answered Jan 30 at 14:18
John HughesJohn Hughes
65.2k24293
answered Jan 30 at 14:18
John HughesJohn Hughes
65.2k24293
65.2k24293
$begingroup$
Transpose shuffles upper/lowerness, but inverse preserves it.
$endgroup$
– mathreadler
Jan 30 at 14:27
add a comment |
$begingroup$
Transpose shuffles upper/lowerness, but inverse preserves it.
$endgroup$
– mathreadler
Jan 30 at 14:27
$begingroup$
Transpose shuffles upper/lowerness, but inverse preserves it.
$endgroup$
– mathreadler
Jan 30 at 14:27
$begingroup$
Transpose shuffles upper/lowerness, but inverse preserves it.
$endgroup$
– mathreadler
Jan 30 at 14:27
add a comment |
$begingroup$
Observe that if $$A = LU$$
then (assuming $A^{-1}$ exists.) :
$$A^{-1} = U^{-1}L^{-1}$$
We can verify this by hand if you don't believe it.
What remains is to show that $U^{-1}$ is also upper triangular and $L^{-1}$ is also lower triangular if $U,L$ are, respectively.
answered Jan 30 at 14:26
mathreadlermathreadler
15.4k72263
$endgroup$
add a comment |
$begingroup$
Observe that if $$A = LU$$
then (assuming $A^{-1}$ exists.) :
$$A^{-1} = U^{-1}L^{-1}$$
We can verify this by hand if you don't believe it.
What remains is to show that $U^{-1}$ is also upper triangular and $L^{-1}$ is also lower triangular if $U,L$ are, respectively.
answered Jan 30 at 14:26
mathreadlermathreadler
15.4k72263
$endgroup$
add a comment |
$begingroup$
Observe that if $$A = LU$$
then (assuming $A^{-1}$ exists.) :
$$A^{-1} = U^{-1}L^{-1}$$
We can verify this by hand if you don't believe it.
What remains is to show that $U^{-1}$ is also upper triangular and $L^{-1}$ is also lower triangular if $U,L$ are, respectively.
answered Jan 30 at 14:26
mathreadlermathreadler
15.4k72263
$endgroup$
Observe that if $$A = LU$$
then (assuming $A^{-1}$ exists.) :
$$A^{-1} = U^{-1}L^{-1}$$
We can verify this by hand if you don't believe it.
What remains is to show that $U^{-1}$ is also upper triangular and $L^{-1}$ is also lower triangular if $U,L$ are, respectively.
answered Jan 30 at 14:26
mathreadlermathreadler
15.4k72263
answered Jan 30 at 14:26
mathreadlermathreadler
15.4k72263
answered Jan 30 at 14:26
mathreadlermathreadler
15.4k72263
answered Jan 30 at 14:26
mathreadlermathreadler
15.4k72263
15.4k72263
add a comment |
add a comment |
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$begingroup$
Does "unit upper triangular" mean "ones on the diagonal"?
$endgroup$
– John Hughes
Jan 30 at 14:11