Invariant factors and Jordan reduction : how to find the adapted basis?
$begingroup$
I have a question about the Jordan reduction using the module theory and especially the invariant factors. If we have a vector space $E$ of dimension $n$ over a field $k$, and $f in End(E)$ then it's also a $k[X]$-module for the operation : $P.v = P(f)(v)$. And of course it's a finitely generated module, and then :
$E cong frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$
as a $k[X]$-module, and $f$ is the multiplication by $X$ for $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$. Then, $E=E_1 bigoplus ... bigoplus E_n$ and on a good basis of E, the matrix of $f$ is the compagnon matrix of $P_i$ on each subspace $E_i$ which is stable by $f$.
($P_i$ can be find as the invariant factors of $det(Mat(f) - XId)$)
And then, if $P_i$ have the form $(X-a)^r$, then we can find a good basis of E when the matrix of $f$ has the Jordan form.
But, this is my problem : how to find the good basis ? Cause on $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$, we know the basis on which the matrix of the multiplication by $X$ have the Jordan form, but actually, to determine it on $E$, we should know, in the explicit way, the isomorphism between $E$ and $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$, and it's seems very complicated, in a practical way...
When we want to apply this principle in a concrete way (for example for the Linear difference equation), we make a change of basis to transform the matrix in order to obtain a matrix which have the jordan form, but we should also know the change of basis if we want to solve the problem, finally...
So, how to do ? And, are we able, from this application of invariant factors, to find the basis adapted to the jordan form ?
linear-algebra abstract-algebra jordan-normal-form
asked Jan 3 at 12:40
ChocoSavourChocoSavour
33118
$endgroup$
add a comment |
$begingroup$
I have a question about the Jordan reduction using the module theory and especially the invariant factors. If we have a vector space $E$ of dimension $n$ over a field $k$, and $f in End(E)$ then it's also a $k[X]$-module for the operation : $P.v = P(f)(v)$. And of course it's a finitely generated module, and then :
$E cong frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$
as a $k[X]$-module, and $f$ is the multiplication by $X$ for $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$. Then, $E=E_1 bigoplus ... bigoplus E_n$ and on a good basis of E, the matrix of $f$ is the compagnon matrix of $P_i$ on each subspace $E_i$ which is stable by $f$.
($P_i$ can be find as the invariant factors of $det(Mat(f) - XId)$)
And then, if $P_i$ have the form $(X-a)^r$, then we can find a good basis of E when the matrix of $f$ has the Jordan form.
But, this is my problem : how to find the good basis ? Cause on $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$, we know the basis on which the matrix of the multiplication by $X$ have the Jordan form, but actually, to determine it on $E$, we should know, in the explicit way, the isomorphism between $E$ and $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$, and it's seems very complicated, in a practical way...
When we want to apply this principle in a concrete way (for example for the Linear difference equation), we make a change of basis to transform the matrix in order to obtain a matrix which have the jordan form, but we should also know the change of basis if we want to solve the problem, finally...
So, how to do ? And, are we able, from this application of invariant factors, to find the basis adapted to the jordan form ?
linear-algebra abstract-algebra jordan-normal-form
asked Jan 3 at 12:40
ChocoSavourChocoSavour
33118
$endgroup$
add a comment |
$begingroup$
I have a question about the Jordan reduction using the module theory and especially the invariant factors. If we have a vector space $E$ of dimension $n$ over a field $k$, and $f in End(E)$ then it's also a $k[X]$-module for the operation : $P.v = P(f)(v)$. And of course it's a finitely generated module, and then :
$E cong frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$
as a $k[X]$-module, and $f$ is the multiplication by $X$ for $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$. Then, $E=E_1 bigoplus ... bigoplus E_n$ and on a good basis of E, the matrix of $f$ is the compagnon matrix of $P_i$ on each subspace $E_i$ which is stable by $f$.
($P_i$ can be find as the invariant factors of $det(Mat(f) - XId)$)
And then, if $P_i$ have the form $(X-a)^r$, then we can find a good basis of E when the matrix of $f$ has the Jordan form.
But, this is my problem : how to find the good basis ? Cause on $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$, we know the basis on which the matrix of the multiplication by $X$ have the Jordan form, but actually, to determine it on $E$, we should know, in the explicit way, the isomorphism between $E$ and $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$, and it's seems very complicated, in a practical way...
When we want to apply this principle in a concrete way (for example for the Linear difference equation), we make a change of basis to transform the matrix in order to obtain a matrix which have the jordan form, but we should also know the change of basis if we want to solve the problem, finally...
So, how to do ? And, are we able, from this application of invariant factors, to find the basis adapted to the jordan form ?
linear-algebra abstract-algebra jordan-normal-form
asked Jan 3 at 12:40
ChocoSavourChocoSavour
33118
$endgroup$
I have a question about the Jordan reduction using the module theory and especially the invariant factors. If we have a vector space $E$ of dimension $n$ over a field $k$, and $f in End(E)$ then it's also a $k[X]$-module for the operation : $P.v = P(f)(v)$. And of course it's a finitely generated module, and then :
$E cong frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$
as a $k[X]$-module, and $f$ is the multiplication by $X$ for $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$. Then, $E=E_1 bigoplus ... bigoplus E_n$ and on a good basis of E, the matrix of $f$ is the compagnon matrix of $P_i$ on each subspace $E_i$ which is stable by $f$.
($P_i$ can be find as the invariant factors of $det(Mat(f) - XId)$)
And then, if $P_i$ have the form $(X-a)^r$, then we can find a good basis of E when the matrix of $f$ has the Jordan form.
But, this is my problem : how to find the good basis ? Cause on $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$, we know the basis on which the matrix of the multiplication by $X$ have the Jordan form, but actually, to determine it on $E$, we should know, in the explicit way, the isomorphism between $E$ and $frac{k[X]}{(P_1)} times ... times frac{k[X]}{(P_s)}$, and it's seems very complicated, in a practical way...
When we want to apply this principle in a concrete way (for example for the Linear difference equation), we make a change of basis to transform the matrix in order to obtain a matrix which have the jordan form, but we should also know the change of basis if we want to solve the problem, finally...
So, how to do ? And, are we able, from this application of invariant factors, to find the basis adapted to the jordan form ?
linear-algebra abstract-algebra jordan-normal-form
linear-algebra abstract-algebra jordan-normal-form
asked Jan 3 at 12:40
ChocoSavourChocoSavour
33118
asked Jan 3 at 12:40
ChocoSavourChocoSavour
33118
asked Jan 3 at 12:40
ChocoSavourChocoSavour
33118
asked Jan 3 at 12:40
ChocoSavourChocoSavour
33118
asked Jan 3 at 12:40
ChocoSavourChocoSavour
33118
33118
add a comment |
add a comment |
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