Recurrence Relation with multiple variables












2












$begingroup$


Consider the following recurrence relation with two variables and similar bounding conditions:



begin{align*}
x_{1,t} &= frac{1}{2-frac{2}{n}} x_{1,t-1} + frac{1}{2} x_{2,t-1} \
x_{i,t} &= frac{1}{2} x_{i-1,t-1} + frac{1}{2} x_{i+1,t-1} & 1 < i < n \
x_{n,t} &= frac{1}{2} x_{n-1,t-1} + frac{1}{2-frac{2}{n}} x_{n,t-1}
end{align*}



$x_{i,0} = 1 $ for all $1 leq i leq n$.
Is there any standard method to solve these?






analysis recurrence-relations recursion







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you know Markov Chains ?
    $endgroup$
    – G Cab
    Dec 5 '18 at 15:46










  • $begingroup$
    Yes, but $frac{1}{2-frac{2}{n}}$ is larger than $frac{1}{2}$ and hence it's not stochastic anymore.
    $endgroup$
    – Jannik
    Dec 5 '18 at 15:48










  • $begingroup$
    well, then you are going to have non-stochastic vectors $x$. The principle of having a vectorial recurrence $bf{ x}(t)= bf M bf {x}(t-1)$ remains.
    $endgroup$
    – G Cab
    Dec 5 '18 at 16:02












  • $begingroup$
    That's true but this matrix has a strange behavior I cannot predict (it's pretty hard to calculate). That's why I'm asking for a different solution :-)
    $endgroup$
    – Jannik
    Dec 5 '18 at 16:03










  • $begingroup$
    That's why it is recommended that OP indicates what he has done and what is the obstacle. In fact, even for small $n$ the matrix diagonalize "strangely". So you have better to reformulate your post asking on how to diagonalize that matrix.
    $endgroup$
    – G Cab
    Dec 5 '18 at 16:28
















2












$begingroup$


Consider the following recurrence relation with two variables and similar bounding conditions:



begin{align*}
x_{1,t} &= frac{1}{2-frac{2}{n}} x_{1,t-1} + frac{1}{2} x_{2,t-1} \
x_{i,t} &= frac{1}{2} x_{i-1,t-1} + frac{1}{2} x_{i+1,t-1} & 1 < i < n \
x_{n,t} &= frac{1}{2} x_{n-1,t-1} + frac{1}{2-frac{2}{n}} x_{n,t-1}
end{align*}



$x_{i,0} = 1 $ for all $1 leq i leq n$.
Is there any standard method to solve these?






analysis recurrence-relations recursion







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you know Markov Chains ?
    $endgroup$
    – G Cab
    Dec 5 '18 at 15:46










  • $begingroup$
    Yes, but $frac{1}{2-frac{2}{n}}$ is larger than $frac{1}{2}$ and hence it's not stochastic anymore.
    $endgroup$
    – Jannik
    Dec 5 '18 at 15:48










  • $begingroup$
    well, then you are going to have non-stochastic vectors $x$. The principle of having a vectorial recurrence $bf{ x}(t)= bf M bf {x}(t-1)$ remains.
    $endgroup$
    – G Cab
    Dec 5 '18 at 16:02












  • $begingroup$
    That's true but this matrix has a strange behavior I cannot predict (it's pretty hard to calculate). That's why I'm asking for a different solution :-)
    $endgroup$
    – Jannik
    Dec 5 '18 at 16:03










  • $begingroup$
    That's why it is recommended that OP indicates what he has done and what is the obstacle. In fact, even for small $n$ the matrix diagonalize "strangely". So you have better to reformulate your post asking on how to diagonalize that matrix.
    $endgroup$
    – G Cab
    Dec 5 '18 at 16:28














2












2








2





$begingroup$


Consider the following recurrence relation with two variables and similar bounding conditions:



begin{align*}
x_{1,t} &= frac{1}{2-frac{2}{n}} x_{1,t-1} + frac{1}{2} x_{2,t-1} \
x_{i,t} &= frac{1}{2} x_{i-1,t-1} + frac{1}{2} x_{i+1,t-1} & 1 < i < n \
x_{n,t} &= frac{1}{2} x_{n-1,t-1} + frac{1}{2-frac{2}{n}} x_{n,t-1}
end{align*}



$x_{i,0} = 1 $ for all $1 leq i leq n$.
Is there any standard method to solve these?






analysis recurrence-relations recursion







share|cite|improve this question











$endgroup$




Consider the following recurrence relation with two variables and similar bounding conditions:



begin{align*}
x_{1,t} &= frac{1}{2-frac{2}{n}} x_{1,t-1} + frac{1}{2} x_{2,t-1} \
x_{i,t} &= frac{1}{2} x_{i-1,t-1} + frac{1}{2} x_{i+1,t-1} & 1 < i < n \
x_{n,t} &= frac{1}{2} x_{n-1,t-1} + frac{1}{2-frac{2}{n}} x_{n,t-1}
end{align*}



$x_{i,0} = 1 $ for all $1 leq i leq n$.
Is there any standard method to solve these?






analysis recurrence-relations recursion




analysis recurrence-relations recursion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 14:53







Jannik

















asked Dec 5 '18 at 14:44









JannikJannik

587




587












  • $begingroup$
    do you know Markov Chains ?
    $endgroup$
    – G Cab
    Dec 5 '18 at 15:46










  • $begingroup$
    Yes, but $frac{1}{2-frac{2}{n}}$ is larger than $frac{1}{2}$ and hence it's not stochastic anymore.
    $endgroup$
    – Jannik
    Dec 5 '18 at 15:48










  • $begingroup$
    well, then you are going to have non-stochastic vectors $x$. The principle of having a vectorial recurrence $bf{ x}(t)= bf M bf {x}(t-1)$ remains.
    $endgroup$
    – G Cab
    Dec 5 '18 at 16:02












  • $begingroup$
    That's true but this matrix has a strange behavior I cannot predict (it's pretty hard to calculate). That's why I'm asking for a different solution :-)
    $endgroup$
    – Jannik
    Dec 5 '18 at 16:03










  • $begingroup$
    That's why it is recommended that OP indicates what he has done and what is the obstacle. In fact, even for small $n$ the matrix diagonalize "strangely". So you have better to reformulate your post asking on how to diagonalize that matrix.
    $endgroup$
    – G Cab
    Dec 5 '18 at 16:28


















  • $begingroup$
    do you know Markov Chains ?
    $endgroup$
    – G Cab
    Dec 5 '18 at 15:46










  • $begingroup$
    Yes, but $frac{1}{2-frac{2}{n}}$ is larger than $frac{1}{2}$ and hence it's not stochastic anymore.
    $endgroup$
    – Jannik
    Dec 5 '18 at 15:48










  • $begingroup$
    well, then you are going to have non-stochastic vectors $x$. The principle of having a vectorial recurrence $bf{ x}(t)= bf M bf {x}(t-1)$ remains.
    $endgroup$
    – G Cab
    Dec 5 '18 at 16:02












  • $begingroup$
    That's true but this matrix has a strange behavior I cannot predict (it's pretty hard to calculate). That's why I'm asking for a different solution :-)
    $endgroup$
    – Jannik
    Dec 5 '18 at 16:03










  • $begingroup$
    That's why it is recommended that OP indicates what he has done and what is the obstacle. In fact, even for small $n$ the matrix diagonalize "strangely". So you have better to reformulate your post asking on how to diagonalize that matrix.
    $endgroup$
    – G Cab
    Dec 5 '18 at 16:28
















$begingroup$
do you know Markov Chains ?
$endgroup$
– G Cab
Dec 5 '18 at 15:46




$begingroup$
do you know Markov Chains ?
$endgroup$
– G Cab
Dec 5 '18 at 15:46












$begingroup$
Yes, but $frac{1}{2-frac{2}{n}}$ is larger than $frac{1}{2}$ and hence it's not stochastic anymore.
$endgroup$
– Jannik
Dec 5 '18 at 15:48




$begingroup$
Yes, but $frac{1}{2-frac{2}{n}}$ is larger than $frac{1}{2}$ and hence it's not stochastic anymore.
$endgroup$
– Jannik
Dec 5 '18 at 15:48












$begingroup$
well, then you are going to have non-stochastic vectors $x$. The principle of having a vectorial recurrence $bf{ x}(t)= bf M bf {x}(t-1)$ remains.
$endgroup$
– G Cab
Dec 5 '18 at 16:02






$begingroup$
well, then you are going to have non-stochastic vectors $x$. The principle of having a vectorial recurrence $bf{ x}(t)= bf M bf {x}(t-1)$ remains.
$endgroup$
– G Cab
Dec 5 '18 at 16:02














$begingroup$
That's true but this matrix has a strange behavior I cannot predict (it's pretty hard to calculate). That's why I'm asking for a different solution :-)
$endgroup$
– Jannik
Dec 5 '18 at 16:03




$begingroup$
That's true but this matrix has a strange behavior I cannot predict (it's pretty hard to calculate). That's why I'm asking for a different solution :-)
$endgroup$
– Jannik
Dec 5 '18 at 16:03












$begingroup$
That's why it is recommended that OP indicates what he has done and what is the obstacle. In fact, even for small $n$ the matrix diagonalize "strangely". So you have better to reformulate your post asking on how to diagonalize that matrix.
$endgroup$
– G Cab
Dec 5 '18 at 16:28




$begingroup$
That's why it is recommended that OP indicates what he has done and what is the obstacle. In fact, even for small $n$ the matrix diagonalize "strangely". So you have better to reformulate your post asking on how to diagonalize that matrix.
$endgroup$
– G Cab
Dec 5 '18 at 16:28










1 Answer
1






active

oldest

votes


















1












$begingroup$

Your system reads as
$$
{bf x}(t) = {1 over 2};left( {matrix{
{1 + {1 over {n - 1}}} & 1 & 0 & cdots & 0 cr
1 & 0 & 1 & cdots & 0 cr
vdots & ddots & ddots & ddots & vdots cr
0 & cdots & 1 & 0 & 1 cr
0 & cdots & 0 & 1 & {1 + {1 over {n - 1}}} cr
} } right);{bf x}(t - 1)
$$



So it is a random walk, with a stop at the ends, and with the
immission (always at the ends) of a influx
given as constant percentage increase of the population therein.

And the increase percentage lowers at growing of the vector dimension.



Since the matrix diagonalization is not neat, you are probably looking for
an approximation. If so with respect to what ?



-- reply to your comment --



The matrix above (call it $bf M$) is symmetric, either wrt the diagonal and wrt the antidiagonal , i.e.
$$
{bf M} = {bf M}^{,T} = {bf J};{bf M};{bf J}
$$
where ${bf J}$ is the exchange matrix.

Thus the matrix is centrosymmetric and in particular symmetric tridiagonal,
and only almost-Toeplitz symmetric.



As such:

- $bf M$ can be diagonalized by an orthogonal matrix $bf Q$;

- all its eigenvalues are real and distinct;

- if $bf v$ is an eigenvector so is its symmetric ${bf J}{bf v}$;

- since $Q$ is orthogonal, it will contain the eigenvectors.



For an eigenvector we surely have that the ratio of the components is constant, and v.v. a vector for which the ratio remains constant must be an eigenvector.

A calculation for small $n$ reveals that the e.v. associated to the higher eigenvalue
has all the components positive and is in fact symmetric.

But I couldn't yet reach to a demonstration of that, nor of a bound in such a ratio.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Good lower and upper bounds for the values of $x_{i,t}$ would be nice. Perfect would be to prove that the ratio$(x_{1,t}/x_{i,t})$ is bounded by a constant for every $i$
    $endgroup$
    – Jannik
    Dec 5 '18 at 17:30












  • $begingroup$
    I thank you anyway for your time and hints. Actually you could help me solving a different problem because I weren't familiar with the theory of tridiagonal matrices. Maybe I can still find something that helps me out here. Thanks a lot :-)
    $endgroup$
    – Jannik
    Dec 6 '18 at 10:40








  • 1




    $begingroup$
    @Jannik: glad to have offered a minimal help. I could find on the net some papers dealing with the diagonalization of a symmetric tridiagonal matrix, but mainly relevant to the algorithms for that. But could not find yet a good analytical hint about the values of the eigenvectors
    $endgroup$
    – G Cab
    Dec 6 '18 at 18:44











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Your system reads as
$$
{bf x}(t) = {1 over 2};left( {matrix{
{1 + {1 over {n - 1}}} & 1 & 0 & cdots & 0 cr
1 & 0 & 1 & cdots & 0 cr
vdots & ddots & ddots & ddots & vdots cr
0 & cdots & 1 & 0 & 1 cr
0 & cdots & 0 & 1 & {1 + {1 over {n - 1}}} cr
} } right);{bf x}(t - 1)
$$



So it is a random walk, with a stop at the ends, and with the
immission (always at the ends) of a influx
given as constant percentage increase of the population therein.

And the increase percentage lowers at growing of the vector dimension.



Since the matrix diagonalization is not neat, you are probably looking for
an approximation. If so with respect to what ?



-- reply to your comment --



The matrix above (call it $bf M$) is symmetric, either wrt the diagonal and wrt the antidiagonal , i.e.
$$
{bf M} = {bf M}^{,T} = {bf J};{bf M};{bf J}
$$
where ${bf J}$ is the exchange matrix.

Thus the matrix is centrosymmetric and in particular symmetric tridiagonal,
and only almost-Toeplitz symmetric.



As such:

- $bf M$ can be diagonalized by an orthogonal matrix $bf Q$;

- all its eigenvalues are real and distinct;

- if $bf v$ is an eigenvector so is its symmetric ${bf J}{bf v}$;

- since $Q$ is orthogonal, it will contain the eigenvectors.



For an eigenvector we surely have that the ratio of the components is constant, and v.v. a vector for which the ratio remains constant must be an eigenvector.

A calculation for small $n$ reveals that the e.v. associated to the higher eigenvalue
has all the components positive and is in fact symmetric.

But I couldn't yet reach to a demonstration of that, nor of a bound in such a ratio.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Good lower and upper bounds for the values of $x_{i,t}$ would be nice. Perfect would be to prove that the ratio$(x_{1,t}/x_{i,t})$ is bounded by a constant for every $i$
    $endgroup$
    – Jannik
    Dec 5 '18 at 17:30












  • $begingroup$
    I thank you anyway for your time and hints. Actually you could help me solving a different problem because I weren't familiar with the theory of tridiagonal matrices. Maybe I can still find something that helps me out here. Thanks a lot :-)
    $endgroup$
    – Jannik
    Dec 6 '18 at 10:40








  • 1




    $begingroup$
    @Jannik: glad to have offered a minimal help. I could find on the net some papers dealing with the diagonalization of a symmetric tridiagonal matrix, but mainly relevant to the algorithms for that. But could not find yet a good analytical hint about the values of the eigenvectors
    $endgroup$
    – G Cab
    Dec 6 '18 at 18:44
















1












$begingroup$

Your system reads as
$$
{bf x}(t) = {1 over 2};left( {matrix{
{1 + {1 over {n - 1}}} & 1 & 0 & cdots & 0 cr
1 & 0 & 1 & cdots & 0 cr
vdots & ddots & ddots & ddots & vdots cr
0 & cdots & 1 & 0 & 1 cr
0 & cdots & 0 & 1 & {1 + {1 over {n - 1}}} cr
} } right);{bf x}(t - 1)
$$



So it is a random walk, with a stop at the ends, and with the
immission (always at the ends) of a influx
given as constant percentage increase of the population therein.

And the increase percentage lowers at growing of the vector dimension.



Since the matrix diagonalization is not neat, you are probably looking for
an approximation. If so with respect to what ?



-- reply to your comment --



The matrix above (call it $bf M$) is symmetric, either wrt the diagonal and wrt the antidiagonal , i.e.
$$
{bf M} = {bf M}^{,T} = {bf J};{bf M};{bf J}
$$
where ${bf J}$ is the exchange matrix.

Thus the matrix is centrosymmetric and in particular symmetric tridiagonal,
and only almost-Toeplitz symmetric.



As such:

- $bf M$ can be diagonalized by an orthogonal matrix $bf Q$;

- all its eigenvalues are real and distinct;

- if $bf v$ is an eigenvector so is its symmetric ${bf J}{bf v}$;

- since $Q$ is orthogonal, it will contain the eigenvectors.



For an eigenvector we surely have that the ratio of the components is constant, and v.v. a vector for which the ratio remains constant must be an eigenvector.

A calculation for small $n$ reveals that the e.v. associated to the higher eigenvalue
has all the components positive and is in fact symmetric.

But I couldn't yet reach to a demonstration of that, nor of a bound in such a ratio.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Good lower and upper bounds for the values of $x_{i,t}$ would be nice. Perfect would be to prove that the ratio$(x_{1,t}/x_{i,t})$ is bounded by a constant for every $i$
    $endgroup$
    – Jannik
    Dec 5 '18 at 17:30












  • $begingroup$
    I thank you anyway for your time and hints. Actually you could help me solving a different problem because I weren't familiar with the theory of tridiagonal matrices. Maybe I can still find something that helps me out here. Thanks a lot :-)
    $endgroup$
    – Jannik
    Dec 6 '18 at 10:40








  • 1




    $begingroup$
    @Jannik: glad to have offered a minimal help. I could find on the net some papers dealing with the diagonalization of a symmetric tridiagonal matrix, but mainly relevant to the algorithms for that. But could not find yet a good analytical hint about the values of the eigenvectors
    $endgroup$
    – G Cab
    Dec 6 '18 at 18:44














1












1








1





$begingroup$

Your system reads as
$$
{bf x}(t) = {1 over 2};left( {matrix{
{1 + {1 over {n - 1}}} & 1 & 0 & cdots & 0 cr
1 & 0 & 1 & cdots & 0 cr
vdots & ddots & ddots & ddots & vdots cr
0 & cdots & 1 & 0 & 1 cr
0 & cdots & 0 & 1 & {1 + {1 over {n - 1}}} cr
} } right);{bf x}(t - 1)
$$



So it is a random walk, with a stop at the ends, and with the
immission (always at the ends) of a influx
given as constant percentage increase of the population therein.

And the increase percentage lowers at growing of the vector dimension.



Since the matrix diagonalization is not neat, you are probably looking for
an approximation. If so with respect to what ?



-- reply to your comment --



The matrix above (call it $bf M$) is symmetric, either wrt the diagonal and wrt the antidiagonal , i.e.
$$
{bf M} = {bf M}^{,T} = {bf J};{bf M};{bf J}
$$
where ${bf J}$ is the exchange matrix.

Thus the matrix is centrosymmetric and in particular symmetric tridiagonal,
and only almost-Toeplitz symmetric.



As such:

- $bf M$ can be diagonalized by an orthogonal matrix $bf Q$;

- all its eigenvalues are real and distinct;

- if $bf v$ is an eigenvector so is its symmetric ${bf J}{bf v}$;

- since $Q$ is orthogonal, it will contain the eigenvectors.



For an eigenvector we surely have that the ratio of the components is constant, and v.v. a vector for which the ratio remains constant must be an eigenvector.

A calculation for small $n$ reveals that the e.v. associated to the higher eigenvalue
has all the components positive and is in fact symmetric.

But I couldn't yet reach to a demonstration of that, nor of a bound in such a ratio.






share|cite|improve this answer











$endgroup$



Your system reads as
$$
{bf x}(t) = {1 over 2};left( {matrix{
{1 + {1 over {n - 1}}} & 1 & 0 & cdots & 0 cr
1 & 0 & 1 & cdots & 0 cr
vdots & ddots & ddots & ddots & vdots cr
0 & cdots & 1 & 0 & 1 cr
0 & cdots & 0 & 1 & {1 + {1 over {n - 1}}} cr
} } right);{bf x}(t - 1)
$$



So it is a random walk, with a stop at the ends, and with the
immission (always at the ends) of a influx
given as constant percentage increase of the population therein.

And the increase percentage lowers at growing of the vector dimension.



Since the matrix diagonalization is not neat, you are probably looking for
an approximation. If so with respect to what ?



-- reply to your comment --



The matrix above (call it $bf M$) is symmetric, either wrt the diagonal and wrt the antidiagonal , i.e.
$$
{bf M} = {bf M}^{,T} = {bf J};{bf M};{bf J}
$$
where ${bf J}$ is the exchange matrix.

Thus the matrix is centrosymmetric and in particular symmetric tridiagonal,
and only almost-Toeplitz symmetric.



As such:

- $bf M$ can be diagonalized by an orthogonal matrix $bf Q$;

- all its eigenvalues are real and distinct;

- if $bf v$ is an eigenvector so is its symmetric ${bf J}{bf v}$;

- since $Q$ is orthogonal, it will contain the eigenvectors.



For an eigenvector we surely have that the ratio of the components is constant, and v.v. a vector for which the ratio remains constant must be an eigenvector.

A calculation for small $n$ reveals that the e.v. associated to the higher eigenvalue
has all the components positive and is in fact symmetric.

But I couldn't yet reach to a demonstration of that, nor of a bound in such a ratio.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 18:55

























answered Dec 5 '18 at 17:20









G CabG Cab

18.5k31237




18.5k31237












  • $begingroup$
    Good lower and upper bounds for the values of $x_{i,t}$ would be nice. Perfect would be to prove that the ratio$(x_{1,t}/x_{i,t})$ is bounded by a constant for every $i$
    $endgroup$
    – Jannik
    Dec 5 '18 at 17:30












  • $begingroup$
    I thank you anyway for your time and hints. Actually you could help me solving a different problem because I weren't familiar with the theory of tridiagonal matrices. Maybe I can still find something that helps me out here. Thanks a lot :-)
    $endgroup$
    – Jannik
    Dec 6 '18 at 10:40








  • 1




    $begingroup$
    @Jannik: glad to have offered a minimal help. I could find on the net some papers dealing with the diagonalization of a symmetric tridiagonal matrix, but mainly relevant to the algorithms for that. But could not find yet a good analytical hint about the values of the eigenvectors
    $endgroup$
    – G Cab
    Dec 6 '18 at 18:44


















  • $begingroup$
    Good lower and upper bounds for the values of $x_{i,t}$ would be nice. Perfect would be to prove that the ratio$(x_{1,t}/x_{i,t})$ is bounded by a constant for every $i$
    $endgroup$
    – Jannik
    Dec 5 '18 at 17:30












  • $begingroup$
    I thank you anyway for your time and hints. Actually you could help me solving a different problem because I weren't familiar with the theory of tridiagonal matrices. Maybe I can still find something that helps me out here. Thanks a lot :-)
    $endgroup$
    – Jannik
    Dec 6 '18 at 10:40








  • 1




    $begingroup$
    @Jannik: glad to have offered a minimal help. I could find on the net some papers dealing with the diagonalization of a symmetric tridiagonal matrix, but mainly relevant to the algorithms for that. But could not find yet a good analytical hint about the values of the eigenvectors
    $endgroup$
    – G Cab
    Dec 6 '18 at 18:44
















$begingroup$
Good lower and upper bounds for the values of $x_{i,t}$ would be nice. Perfect would be to prove that the ratio$(x_{1,t}/x_{i,t})$ is bounded by a constant for every $i$
$endgroup$
– Jannik
Dec 5 '18 at 17:30






$begingroup$
Good lower and upper bounds for the values of $x_{i,t}$ would be nice. Perfect would be to prove that the ratio$(x_{1,t}/x_{i,t})$ is bounded by a constant for every $i$
$endgroup$
– Jannik
Dec 5 '18 at 17:30














$begingroup$
I thank you anyway for your time and hints. Actually you could help me solving a different problem because I weren't familiar with the theory of tridiagonal matrices. Maybe I can still find something that helps me out here. Thanks a lot :-)
$endgroup$
– Jannik
Dec 6 '18 at 10:40






$begingroup$
I thank you anyway for your time and hints. Actually you could help me solving a different problem because I weren't familiar with the theory of tridiagonal matrices. Maybe I can still find something that helps me out here. Thanks a lot :-)
$endgroup$
– Jannik
Dec 6 '18 at 10:40






1




1




$begingroup$
@Jannik: glad to have offered a minimal help. I could find on the net some papers dealing with the diagonalization of a symmetric tridiagonal matrix, but mainly relevant to the algorithms for that. But could not find yet a good analytical hint about the values of the eigenvectors
$endgroup$
– G Cab
Dec 6 '18 at 18:44




$begingroup$
@Jannik: glad to have offered a minimal help. I could find on the net some papers dealing with the diagonalization of a symmetric tridiagonal matrix, but mainly relevant to the algorithms for that. But could not find yet a good analytical hint about the values of the eigenvectors
$endgroup$
– G Cab
Dec 6 '18 at 18:44


















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