Lower and Upper Triangular Matrices
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$A$ is an $ntimes n$ matrix and $L$ is an $n times n$ nonsingular lower triangular matrix. How can I prove that if $LA$ is lower triangular, then $A$ is lower triangular?
How can I do the same for upper triangular matrix, $B$ is $ntimes n$ and $Z$ is $ntimes n$ nonsingular upper triangular matrix. If $ZB$ is upper triangular, then $B$ is upper triangular?
linear-algebra matrices
edited Jan 25 '15 at 19:33
apnorton
15.2k33797
asked Jan 25 '15 at 19:31
liujmliujm
559
$endgroup$
add a comment |
$begingroup$
$A$ is an $ntimes n$ matrix and $L$ is an $n times n$ nonsingular lower triangular matrix. How can I prove that if $LA$ is lower triangular, then $A$ is lower triangular?
How can I do the same for upper triangular matrix, $B$ is $ntimes n$ and $Z$ is $ntimes n$ nonsingular upper triangular matrix. If $ZB$ is upper triangular, then $B$ is upper triangular?
linear-algebra matrices
edited Jan 25 '15 at 19:33
apnorton
15.2k33797
asked Jan 25 '15 at 19:31
liujmliujm
559
$endgroup$
1
$begingroup$
Welcome to Math.SE! To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
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– apnorton
Jan 25 '15 at 19:34
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Try using the definition of matrix multiplication for the elements above the diagonal and below. You will easily see which become zero.
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– Michael Angelo
Jan 25 '15 at 19:41
2
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Do you know already that the inverse of a nonsingular lower triangular matrix is also lower triangular? If so, you can simply write $A=L^{-1}cdot LA$.
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– Greg Martin
Jan 25 '15 at 19:56
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Is this question not answered? Maybe @GregMartin show type up his comment in an answer so it can be accepted.
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– Viktor Glombik
Dec 23 '18 at 11:27
add a comment |
$begingroup$
$A$ is an $ntimes n$ matrix and $L$ is an $n times n$ nonsingular lower triangular matrix. How can I prove that if $LA$ is lower triangular, then $A$ is lower triangular?
How can I do the same for upper triangular matrix, $B$ is $ntimes n$ and $Z$ is $ntimes n$ nonsingular upper triangular matrix. If $ZB$ is upper triangular, then $B$ is upper triangular?
linear-algebra matrices
edited Jan 25 '15 at 19:33
apnorton
15.2k33797
asked Jan 25 '15 at 19:31
liujmliujm
559
$endgroup$
$A$ is an $ntimes n$ matrix and $L$ is an $n times n$ nonsingular lower triangular matrix. How can I prove that if $LA$ is lower triangular, then $A$ is lower triangular?
How can I do the same for upper triangular matrix, $B$ is $ntimes n$ and $Z$ is $ntimes n$ nonsingular upper triangular matrix. If $ZB$ is upper triangular, then $B$ is upper triangular?
linear-algebra matrices
linear-algebra matrices
edited Jan 25 '15 at 19:33
apnorton
15.2k33797
asked Jan 25 '15 at 19:31
liujmliujm
559
edited Jan 25 '15 at 19:33
apnorton
15.2k33797
asked Jan 25 '15 at 19:31
liujmliujm
559
edited Jan 25 '15 at 19:33
apnorton
15.2k33797
edited Jan 25 '15 at 19:33
apnorton
15.2k33797
edited Jan 25 '15 at 19:33
apnorton
15.2k33797
15.2k33797
asked Jan 25 '15 at 19:31
liujmliujm
559
asked Jan 25 '15 at 19:31
liujmliujm
559
asked Jan 25 '15 at 19:31
liujmliujm
559
559
1
$begingroup$
Welcome to Math.SE! To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
$endgroup$
– apnorton
Jan 25 '15 at 19:34
$begingroup$
Try using the definition of matrix multiplication for the elements above the diagonal and below. You will easily see which become zero.
$endgroup$
– Michael Angelo
Jan 25 '15 at 19:41
2
$begingroup$
Do you know already that the inverse of a nonsingular lower triangular matrix is also lower triangular? If so, you can simply write $A=L^{-1}cdot LA$.
$endgroup$
– Greg Martin
Jan 25 '15 at 19:56
$begingroup$
Is this question not answered? Maybe @GregMartin show type up his comment in an answer so it can be accepted.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 11:27
add a comment |
1
$begingroup$
Welcome to Math.SE! To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
$endgroup$
– apnorton
Jan 25 '15 at 19:34
$begingroup$
Try using the definition of matrix multiplication for the elements above the diagonal and below. You will easily see which become zero.
$endgroup$
– Michael Angelo
Jan 25 '15 at 19:41
2
$begingroup$
Do you know already that the inverse of a nonsingular lower triangular matrix is also lower triangular? If so, you can simply write $A=L^{-1}cdot LA$.
$endgroup$
– Greg Martin
Jan 25 '15 at 19:56
$begingroup$
Is this question not answered? Maybe @GregMartin show type up his comment in an answer so it can be accepted.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 11:27
1
1
$begingroup$
Welcome to Math.SE! To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
$endgroup$
– apnorton
Jan 25 '15 at 19:34
$begingroup$
Welcome to Math.SE! To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
$endgroup$
– apnorton
Jan 25 '15 at 19:34
$begingroup$
Try using the definition of matrix multiplication for the elements above the diagonal and below. You will easily see which become zero.
$endgroup$
– Michael Angelo
Jan 25 '15 at 19:41
$begingroup$
Try using the definition of matrix multiplication for the elements above the diagonal and below. You will easily see which become zero.
$endgroup$
– Michael Angelo
Jan 25 '15 at 19:41
2
2
$begingroup$
Do you know already that the inverse of a nonsingular lower triangular matrix is also lower triangular? If so, you can simply write $A=L^{-1}cdot LA$.
$endgroup$
– Greg Martin
Jan 25 '15 at 19:56
$begingroup$
Do you know already that the inverse of a nonsingular lower triangular matrix is also lower triangular? If so, you can simply write $A=L^{-1}cdot LA$.
$endgroup$
– Greg Martin
Jan 25 '15 at 19:56
$begingroup$
Is this question not answered? Maybe @GregMartin show type up his comment in an answer so it can be accepted.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 11:27
$begingroup$
Is this question not answered? Maybe @GregMartin show type up his comment in an answer so it can be accepted.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 11:27
add a comment |
1 Answer
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votes
$begingroup$
It is easy to show that the product of two lower triangular matrices is again lower triangular. It's also a standard lemma that the inverse of a lower triangular matrix is again lower triangular (you can prove this by examining the usual method of finding a matrix inverse using an augmented matrix).
Given these facts, we can quickly deduce that if both $L$ and $LA$ are lower triangular, with $L$ nonsingular, then
$$
A = L^{-1}(LA)
$$
is a product of two lower triangular matrices and thus is itself lower triangular.
The same exact proof, with "lower" replaced everywhere by "upper", works for the other desired case.
answered Dec 23 '18 at 17:08
Greg MartinGreg Martin
36k23565
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add a comment |
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$begingroup$
It is easy to show that the product of two lower triangular matrices is again lower triangular. It's also a standard lemma that the inverse of a lower triangular matrix is again lower triangular (you can prove this by examining the usual method of finding a matrix inverse using an augmented matrix).
Given these facts, we can quickly deduce that if both $L$ and $LA$ are lower triangular, with $L$ nonsingular, then
$$
A = L^{-1}(LA)
$$
is a product of two lower triangular matrices and thus is itself lower triangular.
The same exact proof, with "lower" replaced everywhere by "upper", works for the other desired case.
answered Dec 23 '18 at 17:08
Greg MartinGreg Martin
36k23565
$endgroup$
add a comment |
$begingroup$
It is easy to show that the product of two lower triangular matrices is again lower triangular. It's also a standard lemma that the inverse of a lower triangular matrix is again lower triangular (you can prove this by examining the usual method of finding a matrix inverse using an augmented matrix).
Given these facts, we can quickly deduce that if both $L$ and $LA$ are lower triangular, with $L$ nonsingular, then
$$
A = L^{-1}(LA)
$$
is a product of two lower triangular matrices and thus is itself lower triangular.
The same exact proof, with "lower" replaced everywhere by "upper", works for the other desired case.
answered Dec 23 '18 at 17:08
Greg MartinGreg Martin
36k23565
$endgroup$
add a comment |
$begingroup$
It is easy to show that the product of two lower triangular matrices is again lower triangular. It's also a standard lemma that the inverse of a lower triangular matrix is again lower triangular (you can prove this by examining the usual method of finding a matrix inverse using an augmented matrix).
Given these facts, we can quickly deduce that if both $L$ and $LA$ are lower triangular, with $L$ nonsingular, then
$$
A = L^{-1}(LA)
$$
is a product of two lower triangular matrices and thus is itself lower triangular.
The same exact proof, with "lower" replaced everywhere by "upper", works for the other desired case.
answered Dec 23 '18 at 17:08
Greg MartinGreg Martin
36k23565
$endgroup$
It is easy to show that the product of two lower triangular matrices is again lower triangular. It's also a standard lemma that the inverse of a lower triangular matrix is again lower triangular (you can prove this by examining the usual method of finding a matrix inverse using an augmented matrix).
Given these facts, we can quickly deduce that if both $L$ and $LA$ are lower triangular, with $L$ nonsingular, then
$$
A = L^{-1}(LA)
$$
is a product of two lower triangular matrices and thus is itself lower triangular.
The same exact proof, with "lower" replaced everywhere by "upper", works for the other desired case.
answered Dec 23 '18 at 17:08
Greg MartinGreg Martin
36k23565
answered Dec 23 '18 at 17:08
Greg MartinGreg Martin
36k23565
answered Dec 23 '18 at 17:08
Greg MartinGreg Martin
36k23565
answered Dec 23 '18 at 17:08
Greg MartinGreg Martin
36k23565
36k23565
add a comment |
add a comment |
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$begingroup$
Welcome to Math.SE! To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
$endgroup$
– apnorton
Jan 25 '15 at 19:34
$begingroup$
Try using the definition of matrix multiplication for the elements above the diagonal and below. You will easily see which become zero.
$endgroup$
– Michael Angelo
Jan 25 '15 at 19:41
2
$begingroup$
Do you know already that the inverse of a nonsingular lower triangular matrix is also lower triangular? If so, you can simply write $A=L^{-1}cdot LA$.
$endgroup$
– Greg Martin
Jan 25 '15 at 19:56
$begingroup$
Is this question not answered? Maybe @GregMartin show type up his comment in an answer so it can be accepted.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 11:27