Constraints to make a matrix triangular [closed]
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I have a rank one matrix given below
$$A = bc^T$$
where $A in R^{n times n}$, $b in R^n$ and $c in R^n$. I have where the matrix is made of rank components where each component should be a lower triangular matrix. How should I put constraints on $b$ and $c$ such that $A$ is a lower triangular matrix? and how do I solve them?
linear-algebra matrices optimization
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closed as off-topic by Saad, KReiser, user10354138, BigbearZzz, Paul Frost Dec 16 '18 at 13:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$begingroup$
I have a rank one matrix given below
$$A = bc^T$$
where $A in R^{n times n}$, $b in R^n$ and $c in R^n$. I have where the matrix is made of rank components where each component should be a lower triangular matrix. How should I put constraints on $b$ and $c$ such that $A$ is a lower triangular matrix? and how do I solve them?
linear-algebra matrices optimization
$endgroup$
closed as off-topic by Saad, KReiser, user10354138, BigbearZzz, Paul Frost Dec 16 '18 at 13:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, user10354138, BigbearZzz, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have a rank one matrix given below
$$A = bc^T$$
where $A in R^{n times n}$, $b in R^n$ and $c in R^n$. I have where the matrix is made of rank components where each component should be a lower triangular matrix. How should I put constraints on $b$ and $c$ such that $A$ is a lower triangular matrix? and how do I solve them?
linear-algebra matrices optimization
$endgroup$
I have a rank one matrix given below
$$A = bc^T$$
where $A in R^{n times n}$, $b in R^n$ and $c in R^n$. I have where the matrix is made of rank components where each component should be a lower triangular matrix. How should I put constraints on $b$ and $c$ such that $A$ is a lower triangular matrix? and how do I solve them?
linear-algebra matrices optimization
linear-algebra matrices optimization
asked Nov 27 '18 at 4:46
Dushyant SahooDushyant Sahoo
528
528
closed as off-topic by Saad, KReiser, user10354138, BigbearZzz, Paul Frost Dec 16 '18 at 13:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, user10354138, BigbearZzz, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, KReiser, user10354138, BigbearZzz, Paul Frost Dec 16 '18 at 13:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, KReiser, user10354138, BigbearZzz, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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3 Answers
3
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oldest
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Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.
Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?
I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.
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Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.
If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.
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add a comment |
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We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.
Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?
I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.
$endgroup$
add a comment |
$begingroup$
Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.
Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?
I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.
$endgroup$
add a comment |
$begingroup$
Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.
Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?
I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.
$endgroup$
Hint: $A_{ij} = b_i c_j$ and you want $A_{ij} = 0$ whenever $j > i$.
Hint: alternatively, note that the columns of $A$ are multiples of $b$. How can this be lower triangular?
I think you need either $b_1 = cdots = b_{n-1} = 0$ or $c_2 = cdots = c_n = 0$.
answered Nov 27 '18 at 4:53
angryavianangryavian
40.5k23280
40.5k23280
add a comment |
add a comment |
$begingroup$
Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.
If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.
$endgroup$
add a comment |
$begingroup$
Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.
If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.
$endgroup$
add a comment |
$begingroup$
Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.
If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.
$endgroup$
Let $j$ be the index (if exists) such that $c_jne0=c_{j+1}=cdots=c_n$. That is, let $c_j$ be the last nonzero entry of $c$.
If $A=bc^T$ is lower triangular, we must have $a_{1j}=a_{2j}=cdots=a_{j-1,j}=0$. This imposes some necessary conditions on the entries of $b$. One may verify that these conditions are also sufficient for $bc^T$ to be lower triangular.
answered Nov 27 '18 at 5:10
user1551user1551
72.4k566127
72.4k566127
add a comment |
add a comment |
$begingroup$
We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.
$endgroup$
add a comment |
$begingroup$
We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.
$endgroup$
add a comment |
$begingroup$
We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.
$endgroup$
We have $$a_{ij}=b_ic_j$$therefore $$text{if }b_ine 0 text{ then }c_j=0,j>i$$the algorithm then goes like this: for any vector $b$, find $i$ as the maximum index for which $b_i=0$. Then any vector $c$ for which $c_j=0$ for all $j>i$ is qualified to make an upper-triangular matrix in companion to $b$ in form of $bc^T$.
answered Nov 29 '18 at 10:47
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
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