Diagonalization of stochastic matrices
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Can a stochastic matrix be written as $V^-1 D V $? V is an invertible matrix and D is diagonal. I think so but I can't think of a good proof.
Also, the left eigenvectors and right eigenvectors are not necessarily the same so that the right eigenvectors are only orthogonal with respect to a measure/matrix. That is, if $v_i$ are right eigenvectors then $v_i^dagger V^dagger V v_j=delta_ij$. Is there an easy way to find out what this matrix is?
For clarity, a stochastic matrix, $M$, is one for which $sum_i M_ij=1$ for all $j$.
linear-algebra matrices diagonalization stochastic-matrices
edited Aug 17 at 7:34
Rodrigo de Azevedo
12.6k41751
asked Oct 10 '14 at 22:33
sebastianspiegel
12614
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Can a stochastic matrix be written as $V^-1 D V $? V is an invertible matrix and D is diagonal. I think so but I can't think of a good proof.
Also, the left eigenvectors and right eigenvectors are not necessarily the same so that the right eigenvectors are only orthogonal with respect to a measure/matrix. That is, if $v_i$ are right eigenvectors then $v_i^dagger V^dagger V v_j=delta_ij$. Is there an easy way to find out what this matrix is?
For clarity, a stochastic matrix, $M$, is one for which $sum_i M_ij=1$ for all $j$.
linear-algebra matrices diagonalization stochastic-matrices
edited Aug 17 at 7:34
Rodrigo de Azevedo
12.6k41751
asked Oct 10 '14 at 22:33
sebastianspiegel
12614
Maybe this helps? mathoverflow.net/questions/51887/…
– thanasissdr
Oct 10 '14 at 22:39
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can a stochastic matrix be written as $V^-1 D V $? V is an invertible matrix and D is diagonal. I think so but I can't think of a good proof.
Also, the left eigenvectors and right eigenvectors are not necessarily the same so that the right eigenvectors are only orthogonal with respect to a measure/matrix. That is, if $v_i$ are right eigenvectors then $v_i^dagger V^dagger V v_j=delta_ij$. Is there an easy way to find out what this matrix is?
For clarity, a stochastic matrix, $M$, is one for which $sum_i M_ij=1$ for all $j$.
linear-algebra matrices diagonalization stochastic-matrices
edited Aug 17 at 7:34
Rodrigo de Azevedo
12.6k41751
asked Oct 10 '14 at 22:33
sebastianspiegel
12614
Can a stochastic matrix be written as $V^-1 D V $? V is an invertible matrix and D is diagonal. I think so but I can't think of a good proof.
Also, the left eigenvectors and right eigenvectors are not necessarily the same so that the right eigenvectors are only orthogonal with respect to a measure/matrix. That is, if $v_i$ are right eigenvectors then $v_i^dagger V^dagger V v_j=delta_ij$. Is there an easy way to find out what this matrix is?
For clarity, a stochastic matrix, $M$, is one for which $sum_i M_ij=1$ for all $j$.
linear-algebra matrices diagonalization stochastic-matrices
edited Aug 17 at 7:34
Rodrigo de Azevedo
12.6k41751
asked Oct 10 '14 at 22:33
sebastianspiegel
12614
edited Aug 17 at 7:34
Rodrigo de Azevedo
12.6k41751
edited Aug 17 at 7:34
Rodrigo de Azevedo
12.6k41751
edited Aug 17 at 7:34
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Oct 10 '14 at 22:33
sebastianspiegel
12614
asked Oct 10 '14 at 22:33
sebastianspiegel
12614
asked Oct 10 '14 at 22:33
sebastianspiegel
12614
12614
Maybe this helps? mathoverflow.net/questions/51887/…
– thanasissdr
Oct 10 '14 at 22:39
add a comment |Â
Maybe this helps? mathoverflow.net/questions/51887/…
– thanasissdr
Oct 10 '14 at 22:39
Maybe this helps? mathoverflow.net/questions/51887/…
– thanasissdr
Oct 10 '14 at 22:39
Maybe this helps? mathoverflow.net/questions/51887/…
– thanasissdr
Oct 10 '14 at 22:39
add a comment |Â
1 Answer
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I want to partially answer my own questions. If the stochastic matrix satisfies detailed balance then the matrix that we take the inner product respect to is given by $sum_ijdelta_ijfrac1pi_i$. where $pi_i$ is the equilibrium distribution. Detailed balance says that $M_ij pi_j=M_jipi_i$. This is because the matrix $sum_ijdelta_ijfrac1pi_i M$ is Hermitian and thus has real eigenvalues. Hence $v_i^dagger sum_ijdelta_ijfrac1pi_i M v_j=lambda_i v_i^dagger sum_ijdelta_ijfrac1pi_i v_j=lambda_j v_i^dagger sum_ijdelta_ijfrac1pi_i v_j$ and so if the eigenvalues differ, then the eigenvectors are orthogonal.
answered Oct 10 '14 at 22:49
sebastianspiegel
12614
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I want to partially answer my own questions. If the stochastic matrix satisfies detailed balance then the matrix that we take the inner product respect to is given by $sum_ijdelta_ijfrac1pi_i$. where $pi_i$ is the equilibrium distribution. Detailed balance says that $M_ij pi_j=M_jipi_i$. This is because the matrix $sum_ijdelta_ijfrac1pi_i M$ is Hermitian and thus has real eigenvalues. Hence $v_i^dagger sum_ijdelta_ijfrac1pi_i M v_j=lambda_i v_i^dagger sum_ijdelta_ijfrac1pi_i v_j=lambda_j v_i^dagger sum_ijdelta_ijfrac1pi_i v_j$ and so if the eigenvalues differ, then the eigenvectors are orthogonal.
answered Oct 10 '14 at 22:49
sebastianspiegel
12614
add a comment |Â
up vote
0
down vote
I want to partially answer my own questions. If the stochastic matrix satisfies detailed balance then the matrix that we take the inner product respect to is given by $sum_ijdelta_ijfrac1pi_i$. where $pi_i$ is the equilibrium distribution. Detailed balance says that $M_ij pi_j=M_jipi_i$. This is because the matrix $sum_ijdelta_ijfrac1pi_i M$ is Hermitian and thus has real eigenvalues. Hence $v_i^dagger sum_ijdelta_ijfrac1pi_i M v_j=lambda_i v_i^dagger sum_ijdelta_ijfrac1pi_i v_j=lambda_j v_i^dagger sum_ijdelta_ijfrac1pi_i v_j$ and so if the eigenvalues differ, then the eigenvectors are orthogonal.
answered Oct 10 '14 at 22:49
sebastianspiegel
12614
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I want to partially answer my own questions. If the stochastic matrix satisfies detailed balance then the matrix that we take the inner product respect to is given by $sum_ijdelta_ijfrac1pi_i$. where $pi_i$ is the equilibrium distribution. Detailed balance says that $M_ij pi_j=M_jipi_i$. This is because the matrix $sum_ijdelta_ijfrac1pi_i M$ is Hermitian and thus has real eigenvalues. Hence $v_i^dagger sum_ijdelta_ijfrac1pi_i M v_j=lambda_i v_i^dagger sum_ijdelta_ijfrac1pi_i v_j=lambda_j v_i^dagger sum_ijdelta_ijfrac1pi_i v_j$ and so if the eigenvalues differ, then the eigenvectors are orthogonal.
answered Oct 10 '14 at 22:49
sebastianspiegel
12614
I want to partially answer my own questions. If the stochastic matrix satisfies detailed balance then the matrix that we take the inner product respect to is given by $sum_ijdelta_ijfrac1pi_i$. where $pi_i$ is the equilibrium distribution. Detailed balance says that $M_ij pi_j=M_jipi_i$. This is because the matrix $sum_ijdelta_ijfrac1pi_i M$ is Hermitian and thus has real eigenvalues. Hence $v_i^dagger sum_ijdelta_ijfrac1pi_i M v_j=lambda_i v_i^dagger sum_ijdelta_ijfrac1pi_i v_j=lambda_j v_i^dagger sum_ijdelta_ijfrac1pi_i v_j$ and so if the eigenvalues differ, then the eigenvectors are orthogonal.
answered Oct 10 '14 at 22:49
sebastianspiegel
12614
answered Oct 10 '14 at 22:49
sebastianspiegel
12614
answered Oct 10 '14 at 22:49
sebastianspiegel
12614
answered Oct 10 '14 at 22:49
sebastianspiegel
12614
12614
add a comment |Â
add a comment |Â
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Maybe this helps? mathoverflow.net/questions/51887/…
– thanasissdr
Oct 10 '14 at 22:39