Mixing
How to prove that a banded matrix is irreducible?
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I want to proof that for a certain $ninmathbb{N}$ a banded matrix, $Ainmathbb{R^{ntimes n}}$, with elements $frac{1}{h^{2}}$ on the $+1$ and $-1$ diagonals and $1+2/h^{2}$ on the main diagonal is irreducibe. Therefore i have to proof that there does not exist a permutation matrix $P$, such that $PAP^{T}$ is block upper triangular. I do not know how to begin this proof and cannot find any clear conditions for banded matrices, which imply irreducibility.
linear-algebra matrices
asked Jan 25 at 22:37
rs4rs35rs4rs35
447
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add a comment |
$begingroup$
I want to proof that for a certain $ninmathbb{N}$ a banded matrix, $Ainmathbb{R^{ntimes n}}$, with elements $frac{1}{h^{2}}$ on the $+1$ and $-1$ diagonals and $1+2/h^{2}$ on the main diagonal is irreducibe. Therefore i have to proof that there does not exist a permutation matrix $P$, such that $PAP^{T}$ is block upper triangular. I do not know how to begin this proof and cannot find any clear conditions for banded matrices, which imply irreducibility.
linear-algebra matrices
asked Jan 25 at 22:37
rs4rs35rs4rs35
447
$endgroup$
add a comment |
$begingroup$
I want to proof that for a certain $ninmathbb{N}$ a banded matrix, $Ainmathbb{R^{ntimes n}}$, with elements $frac{1}{h^{2}}$ on the $+1$ and $-1$ diagonals and $1+2/h^{2}$ on the main diagonal is irreducibe. Therefore i have to proof that there does not exist a permutation matrix $P$, such that $PAP^{T}$ is block upper triangular. I do not know how to begin this proof and cannot find any clear conditions for banded matrices, which imply irreducibility.
linear-algebra matrices
asked Jan 25 at 22:37
rs4rs35rs4rs35
447
$endgroup$
I want to proof that for a certain $ninmathbb{N}$ a banded matrix, $Ainmathbb{R^{ntimes n}}$, with elements $frac{1}{h^{2}}$ on the $+1$ and $-1$ diagonals and $1+2/h^{2}$ on the main diagonal is irreducibe. Therefore i have to proof that there does not exist a permutation matrix $P$, such that $PAP^{T}$ is block upper triangular. I do not know how to begin this proof and cannot find any clear conditions for banded matrices, which imply irreducibility.
linear-algebra matrices
linear-algebra matrices
asked Jan 25 at 22:37
rs4rs35rs4rs35
447
asked Jan 25 at 22:37
rs4rs35rs4rs35
447
asked Jan 25 at 22:37
rs4rs35rs4rs35
447
asked Jan 25 at 22:37
rs4rs35rs4rs35
447
asked Jan 25 at 22:37
rs4rs35rs4rs35
447
447
add a comment |
add a comment |
1 Answer
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Hint: Applying the permutation similarity $A mapsto PAP^T$ amounts to moving the $i,j$ entry of $A$ to the $pi(i),pi(j)$ position for some permutation $pi:{1,dots,n} to {1,dots,n}$. Note that the $(i,j)$ entry is above the diagonal if and only if $jgeq i$.
Suppose that $pi$ is a permutation such that the corresponding $PAP^T$ is upper triangular. Since $a_{i,i+1}$ is non-zero for $i=1,dots,n-1$, we must have $pi(i+1) geq pi(i)$ for $i = 1,dots,n$. On the other hand, since $a_{i-1,i}$ is non-zero for $i = 2,dots,n$, we must have $pi(i-1) leq pi(i)$ for $i = 2,dots,n$.
Argue that these conditions cannot hold simultaneously.
answered Jan 25 at 22:51
OmnomnomnomOmnomnomnom
129k792185
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1 Answer
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1 Answer
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active
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$begingroup$
Hint: Applying the permutation similarity $A mapsto PAP^T$ amounts to moving the $i,j$ entry of $A$ to the $pi(i),pi(j)$ position for some permutation $pi:{1,dots,n} to {1,dots,n}$. Note that the $(i,j)$ entry is above the diagonal if and only if $jgeq i$.
Suppose that $pi$ is a permutation such that the corresponding $PAP^T$ is upper triangular. Since $a_{i,i+1}$ is non-zero for $i=1,dots,n-1$, we must have $pi(i+1) geq pi(i)$ for $i = 1,dots,n$. On the other hand, since $a_{i-1,i}$ is non-zero for $i = 2,dots,n$, we must have $pi(i-1) leq pi(i)$ for $i = 2,dots,n$.
Argue that these conditions cannot hold simultaneously.
answered Jan 25 at 22:51
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
add a comment |
$begingroup$
Hint: Applying the permutation similarity $A mapsto PAP^T$ amounts to moving the $i,j$ entry of $A$ to the $pi(i),pi(j)$ position for some permutation $pi:{1,dots,n} to {1,dots,n}$. Note that the $(i,j)$ entry is above the diagonal if and only if $jgeq i$.
Suppose that $pi$ is a permutation such that the corresponding $PAP^T$ is upper triangular. Since $a_{i,i+1}$ is non-zero for $i=1,dots,n-1$, we must have $pi(i+1) geq pi(i)$ for $i = 1,dots,n$. On the other hand, since $a_{i-1,i}$ is non-zero for $i = 2,dots,n$, we must have $pi(i-1) leq pi(i)$ for $i = 2,dots,n$.
Argue that these conditions cannot hold simultaneously.
answered Jan 25 at 22:51
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
add a comment |
$begingroup$
Hint: Applying the permutation similarity $A mapsto PAP^T$ amounts to moving the $i,j$ entry of $A$ to the $pi(i),pi(j)$ position for some permutation $pi:{1,dots,n} to {1,dots,n}$. Note that the $(i,j)$ entry is above the diagonal if and only if $jgeq i$.
Suppose that $pi$ is a permutation such that the corresponding $PAP^T$ is upper triangular. Since $a_{i,i+1}$ is non-zero for $i=1,dots,n-1$, we must have $pi(i+1) geq pi(i)$ for $i = 1,dots,n$. On the other hand, since $a_{i-1,i}$ is non-zero for $i = 2,dots,n$, we must have $pi(i-1) leq pi(i)$ for $i = 2,dots,n$.
Argue that these conditions cannot hold simultaneously.
answered Jan 25 at 22:51
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
Hint: Applying the permutation similarity $A mapsto PAP^T$ amounts to moving the $i,j$ entry of $A$ to the $pi(i),pi(j)$ position for some permutation $pi:{1,dots,n} to {1,dots,n}$. Note that the $(i,j)$ entry is above the diagonal if and only if $jgeq i$.
Suppose that $pi$ is a permutation such that the corresponding $PAP^T$ is upper triangular. Since $a_{i,i+1}$ is non-zero for $i=1,dots,n-1$, we must have $pi(i+1) geq pi(i)$ for $i = 1,dots,n$. On the other hand, since $a_{i-1,i}$ is non-zero for $i = 2,dots,n$, we must have $pi(i-1) leq pi(i)$ for $i = 2,dots,n$.
Argue that these conditions cannot hold simultaneously.
answered Jan 25 at 22:51
OmnomnomnomOmnomnomnom
129k792185
answered Jan 25 at 22:51
OmnomnomnomOmnomnomnom
129k792185
answered Jan 25 at 22:51
OmnomnomnomOmnomnomnom
129k792185
answered Jan 25 at 22:51
OmnomnomnomOmnomnomnom
129k792185
129k792185
add a comment |
add a comment |
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