Resolvent matrix












2












$begingroup$


Suppose $A$ is a triangular matrix. What is the most efficient known algorithm to compute the polynomial (in $x$) matrix $(xI-A)^{-1}$?



Of course, $(xI-A)^{-1}= N(x)/p_A(x)$, where $p_A$ is the characteristic polynomial of $A$, which is easy to compute once we know an eigendecomposition of $A$. But what about $N(x)$?



I am aware of the Leverrier-Fadeev algorithm, which requires $O(n^4)$ operations if $A$ is $ntimes n$. Moreover, it makes use of power iteration, which can lead to numerical instability.






linear-algebra dynamical-systems







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the Leverrier-Fadeev algorithm you mentioned?
    $endgroup$
    – DisintegratingByParts
    Nov 27 '14 at 11:27










  • $begingroup$
    @Dis, this one.
    $endgroup$
    – J. M. is not a mathematician
    Aug 28 '17 at 4:32
















2












$begingroup$


Suppose $A$ is a triangular matrix. What is the most efficient known algorithm to compute the polynomial (in $x$) matrix $(xI-A)^{-1}$?



Of course, $(xI-A)^{-1}= N(x)/p_A(x)$, where $p_A$ is the characteristic polynomial of $A$, which is easy to compute once we know an eigendecomposition of $A$. But what about $N(x)$?



I am aware of the Leverrier-Fadeev algorithm, which requires $O(n^4)$ operations if $A$ is $ntimes n$. Moreover, it makes use of power iteration, which can lead to numerical instability.






linear-algebra dynamical-systems







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the Leverrier-Fadeev algorithm you mentioned?
    $endgroup$
    – DisintegratingByParts
    Nov 27 '14 at 11:27










  • $begingroup$
    @Dis, this one.
    $endgroup$
    – J. M. is not a mathematician
    Aug 28 '17 at 4:32














2












2








2


1



$begingroup$


Suppose $A$ is a triangular matrix. What is the most efficient known algorithm to compute the polynomial (in $x$) matrix $(xI-A)^{-1}$?



Of course, $(xI-A)^{-1}= N(x)/p_A(x)$, where $p_A$ is the characteristic polynomial of $A$, which is easy to compute once we know an eigendecomposition of $A$. But what about $N(x)$?



I am aware of the Leverrier-Fadeev algorithm, which requires $O(n^4)$ operations if $A$ is $ntimes n$. Moreover, it makes use of power iteration, which can lead to numerical instability.






linear-algebra dynamical-systems







share|cite|improve this question









$endgroup$




Suppose $A$ is a triangular matrix. What is the most efficient known algorithm to compute the polynomial (in $x$) matrix $(xI-A)^{-1}$?



Of course, $(xI-A)^{-1}= N(x)/p_A(x)$, where $p_A$ is the characteristic polynomial of $A$, which is easy to compute once we know an eigendecomposition of $A$. But what about $N(x)$?



I am aware of the Leverrier-Fadeev algorithm, which requires $O(n^4)$ operations if $A$ is $ntimes n$. Moreover, it makes use of power iteration, which can lead to numerical instability.






linear-algebra dynamical-systems




linear-algebra dynamical-systems






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 '14 at 10:12









MicheleMichele

112




112












  • $begingroup$
    What is the Leverrier-Fadeev algorithm you mentioned?
    $endgroup$
    – DisintegratingByParts
    Nov 27 '14 at 11:27










  • $begingroup$
    @Dis, this one.
    $endgroup$
    – J. M. is not a mathematician
    Aug 28 '17 at 4:32


















  • $begingroup$
    What is the Leverrier-Fadeev algorithm you mentioned?
    $endgroup$
    – DisintegratingByParts
    Nov 27 '14 at 11:27










  • $begingroup$
    @Dis, this one.
    $endgroup$
    – J. M. is not a mathematician
    Aug 28 '17 at 4:32
















$begingroup$
What is the Leverrier-Fadeev algorithm you mentioned?
$endgroup$
– DisintegratingByParts
Nov 27 '14 at 11:27




$begingroup$
What is the Leverrier-Fadeev algorithm you mentioned?
$endgroup$
– DisintegratingByParts
Nov 27 '14 at 11:27












$begingroup$
@Dis, this one.
$endgroup$
– J. M. is not a mathematician
Aug 28 '17 at 4:32




$begingroup$
@Dis, this one.
$endgroup$
– J. M. is not a mathematician
Aug 28 '17 at 4:32










2 Answers
2






active

oldest

votes


















0












$begingroup$

Since $A$ is triangular, you may try to first diagonalize if, $A=PDP^{-1}$. You already know what the eigenvalues of $A$ are. Then, $$(xI-A)^{-1} = (P(xI-D)P^{-1})^{-1} = P(xI-D)^{-1}P^{-1}$$ and $(xI-D)^{-1}$ is trivial to calculate. Does this help?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Not much unfortunately, because $A$ may not be diagonalizable.
    $endgroup$
    – Michele
    Nov 26 '14 at 12:15










  • $begingroup$
    @Michele Then you can at least find the Jordan normal form of the matrix. In any case, even simply calculating $(xI-A)^{-1}$ is not a difficult operation to perform if $A$ is triangular.
    $endgroup$
    – 5xum
    Nov 26 '14 at 12:21










  • $begingroup$
    Thanks. What is the computational cost of computing the Jordan Normal Form for a triangular $A$? In any case, note that the difficulty here is that we have to carry out a symbolic computation, because of the indeterminate $x$ (I know of course that for specific values of $x$ this is easy).
    $endgroup$
    – Michele
    Nov 26 '14 at 13:00





















0












$begingroup$

$(xI-A)^{-1}=Adj(xI-A)/p(x)$ where $p(x)$ is the (known) characteristic polynomial of $A$. On the other hand, $Adj(sI-A)=Delta p(s,A)$ where $Delta p(s,t)=dfrac{p(s)-p(t)}{s-t}$ (a symmetric polynomial).



For example, let $n=3,p(x)=x^3-2x^2-3x+2$; then



$Delta p(s,t)=dfrac{s^3-t^3-2s^2+2t^2-3s+3t}{s-t}=s^2+st+t^2-2s-2t-3$.



Finally $Adj(xI-A)=(x^2-2x-3)I+(x-2)A+A^2$.



EDIT. The resolvent is essentially used to calculate $f(A)$ when $f$ is an holomorphic function.



Here $(xI-A)^{-1}=sum_j a_j(x)/p(x) A^j$ where $a_j$ is a polynomial.



Then $f(A)=1/2ipisum_j (int_{gamma} f(x)a_j(x)/p(x)dx) A^j$ where $gamma$ is a ad hoc curve. Consequently, we must only calculate $n$ numerical integrals.



In particular, when $lambda_1,cdots,lambda_n$, the roots of $p$, are simple:



$f(A)=sum_j(sum_k f(lambda_k)a_j(lambda_k)/p'(lambda_k)) A^j$.



Anyway, the complexity of the calculation is that of the computation of the $(A^j)_j$, that is here $sim n^4/6$ (of course, if the calculation of the Jordan form of $A$ is stable, then we can do the job with complexity about $30 n³$).






share|cite|improve this answer











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    2 Answers
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    2 Answers
    2






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    active

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    0












    $begingroup$

    Since $A$ is triangular, you may try to first diagonalize if, $A=PDP^{-1}$. You already know what the eigenvalues of $A$ are. Then, $$(xI-A)^{-1} = (P(xI-D)P^{-1})^{-1} = P(xI-D)^{-1}P^{-1}$$ and $(xI-D)^{-1}$ is trivial to calculate. Does this help?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Not much unfortunately, because $A$ may not be diagonalizable.
      $endgroup$
      – Michele
      Nov 26 '14 at 12:15










    • $begingroup$
      @Michele Then you can at least find the Jordan normal form of the matrix. In any case, even simply calculating $(xI-A)^{-1}$ is not a difficult operation to perform if $A$ is triangular.
      $endgroup$
      – 5xum
      Nov 26 '14 at 12:21










    • $begingroup$
      Thanks. What is the computational cost of computing the Jordan Normal Form for a triangular $A$? In any case, note that the difficulty here is that we have to carry out a symbolic computation, because of the indeterminate $x$ (I know of course that for specific values of $x$ this is easy).
      $endgroup$
      – Michele
      Nov 26 '14 at 13:00


















    0












    $begingroup$

    Since $A$ is triangular, you may try to first diagonalize if, $A=PDP^{-1}$. You already know what the eigenvalues of $A$ are. Then, $$(xI-A)^{-1} = (P(xI-D)P^{-1})^{-1} = P(xI-D)^{-1}P^{-1}$$ and $(xI-D)^{-1}$ is trivial to calculate. Does this help?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Not much unfortunately, because $A$ may not be diagonalizable.
      $endgroup$
      – Michele
      Nov 26 '14 at 12:15










    • $begingroup$
      @Michele Then you can at least find the Jordan normal form of the matrix. In any case, even simply calculating $(xI-A)^{-1}$ is not a difficult operation to perform if $A$ is triangular.
      $endgroup$
      – 5xum
      Nov 26 '14 at 12:21










    • $begingroup$
      Thanks. What is the computational cost of computing the Jordan Normal Form for a triangular $A$? In any case, note that the difficulty here is that we have to carry out a symbolic computation, because of the indeterminate $x$ (I know of course that for specific values of $x$ this is easy).
      $endgroup$
      – Michele
      Nov 26 '14 at 13:00
















    0












    0








    0





    $begingroup$

    Since $A$ is triangular, you may try to first diagonalize if, $A=PDP^{-1}$. You already know what the eigenvalues of $A$ are. Then, $$(xI-A)^{-1} = (P(xI-D)P^{-1})^{-1} = P(xI-D)^{-1}P^{-1}$$ and $(xI-D)^{-1}$ is trivial to calculate. Does this help?






    share|cite|improve this answer









    $endgroup$



    Since $A$ is triangular, you may try to first diagonalize if, $A=PDP^{-1}$. You already know what the eigenvalues of $A$ are. Then, $$(xI-A)^{-1} = (P(xI-D)P^{-1})^{-1} = P(xI-D)^{-1}P^{-1}$$ and $(xI-D)^{-1}$ is trivial to calculate. Does this help?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 26 '14 at 10:31









    5xum5xum

    91.4k394161




    91.4k394161












    • $begingroup$
      Not much unfortunately, because $A$ may not be diagonalizable.
      $endgroup$
      – Michele
      Nov 26 '14 at 12:15










    • $begingroup$
      @Michele Then you can at least find the Jordan normal form of the matrix. In any case, even simply calculating $(xI-A)^{-1}$ is not a difficult operation to perform if $A$ is triangular.
      $endgroup$
      – 5xum
      Nov 26 '14 at 12:21










    • $begingroup$
      Thanks. What is the computational cost of computing the Jordan Normal Form for a triangular $A$? In any case, note that the difficulty here is that we have to carry out a symbolic computation, because of the indeterminate $x$ (I know of course that for specific values of $x$ this is easy).
      $endgroup$
      – Michele
      Nov 26 '14 at 13:00




















    • $begingroup$
      Not much unfortunately, because $A$ may not be diagonalizable.
      $endgroup$
      – Michele
      Nov 26 '14 at 12:15










    • $begingroup$
      @Michele Then you can at least find the Jordan normal form of the matrix. In any case, even simply calculating $(xI-A)^{-1}$ is not a difficult operation to perform if $A$ is triangular.
      $endgroup$
      – 5xum
      Nov 26 '14 at 12:21










    • $begingroup$
      Thanks. What is the computational cost of computing the Jordan Normal Form for a triangular $A$? In any case, note that the difficulty here is that we have to carry out a symbolic computation, because of the indeterminate $x$ (I know of course that for specific values of $x$ this is easy).
      $endgroup$
      – Michele
      Nov 26 '14 at 13:00


















    $begingroup$
    Not much unfortunately, because $A$ may not be diagonalizable.
    $endgroup$
    – Michele
    Nov 26 '14 at 12:15




    $begingroup$
    Not much unfortunately, because $A$ may not be diagonalizable.
    $endgroup$
    – Michele
    Nov 26 '14 at 12:15












    $begingroup$
    @Michele Then you can at least find the Jordan normal form of the matrix. In any case, even simply calculating $(xI-A)^{-1}$ is not a difficult operation to perform if $A$ is triangular.
    $endgroup$
    – 5xum
    Nov 26 '14 at 12:21




    $begingroup$
    @Michele Then you can at least find the Jordan normal form of the matrix. In any case, even simply calculating $(xI-A)^{-1}$ is not a difficult operation to perform if $A$ is triangular.
    $endgroup$
    – 5xum
    Nov 26 '14 at 12:21












    $begingroup$
    Thanks. What is the computational cost of computing the Jordan Normal Form for a triangular $A$? In any case, note that the difficulty here is that we have to carry out a symbolic computation, because of the indeterminate $x$ (I know of course that for specific values of $x$ this is easy).
    $endgroup$
    – Michele
    Nov 26 '14 at 13:00






    $begingroup$
    Thanks. What is the computational cost of computing the Jordan Normal Form for a triangular $A$? In any case, note that the difficulty here is that we have to carry out a symbolic computation, because of the indeterminate $x$ (I know of course that for specific values of $x$ this is easy).
    $endgroup$
    – Michele
    Nov 26 '14 at 13:00













    0












    $begingroup$

    $(xI-A)^{-1}=Adj(xI-A)/p(x)$ where $p(x)$ is the (known) characteristic polynomial of $A$. On the other hand, $Adj(sI-A)=Delta p(s,A)$ where $Delta p(s,t)=dfrac{p(s)-p(t)}{s-t}$ (a symmetric polynomial).



    For example, let $n=3,p(x)=x^3-2x^2-3x+2$; then



    $Delta p(s,t)=dfrac{s^3-t^3-2s^2+2t^2-3s+3t}{s-t}=s^2+st+t^2-2s-2t-3$.



    Finally $Adj(xI-A)=(x^2-2x-3)I+(x-2)A+A^2$.



    EDIT. The resolvent is essentially used to calculate $f(A)$ when $f$ is an holomorphic function.



    Here $(xI-A)^{-1}=sum_j a_j(x)/p(x) A^j$ where $a_j$ is a polynomial.



    Then $f(A)=1/2ipisum_j (int_{gamma} f(x)a_j(x)/p(x)dx) A^j$ where $gamma$ is a ad hoc curve. Consequently, we must only calculate $n$ numerical integrals.



    In particular, when $lambda_1,cdots,lambda_n$, the roots of $p$, are simple:



    $f(A)=sum_j(sum_k f(lambda_k)a_j(lambda_k)/p'(lambda_k)) A^j$.



    Anyway, the complexity of the calculation is that of the computation of the $(A^j)_j$, that is here $sim n^4/6$ (of course, if the calculation of the Jordan form of $A$ is stable, then we can do the job with complexity about $30 n³$).






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      $(xI-A)^{-1}=Adj(xI-A)/p(x)$ where $p(x)$ is the (known) characteristic polynomial of $A$. On the other hand, $Adj(sI-A)=Delta p(s,A)$ where $Delta p(s,t)=dfrac{p(s)-p(t)}{s-t}$ (a symmetric polynomial).



      For example, let $n=3,p(x)=x^3-2x^2-3x+2$; then



      $Delta p(s,t)=dfrac{s^3-t^3-2s^2+2t^2-3s+3t}{s-t}=s^2+st+t^2-2s-2t-3$.



      Finally $Adj(xI-A)=(x^2-2x-3)I+(x-2)A+A^2$.



      EDIT. The resolvent is essentially used to calculate $f(A)$ when $f$ is an holomorphic function.



      Here $(xI-A)^{-1}=sum_j a_j(x)/p(x) A^j$ where $a_j$ is a polynomial.



      Then $f(A)=1/2ipisum_j (int_{gamma} f(x)a_j(x)/p(x)dx) A^j$ where $gamma$ is a ad hoc curve. Consequently, we must only calculate $n$ numerical integrals.



      In particular, when $lambda_1,cdots,lambda_n$, the roots of $p$, are simple:



      $f(A)=sum_j(sum_k f(lambda_k)a_j(lambda_k)/p'(lambda_k)) A^j$.



      Anyway, the complexity of the calculation is that of the computation of the $(A^j)_j$, that is here $sim n^4/6$ (of course, if the calculation of the Jordan form of $A$ is stable, then we can do the job with complexity about $30 n³$).






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        $(xI-A)^{-1}=Adj(xI-A)/p(x)$ where $p(x)$ is the (known) characteristic polynomial of $A$. On the other hand, $Adj(sI-A)=Delta p(s,A)$ where $Delta p(s,t)=dfrac{p(s)-p(t)}{s-t}$ (a symmetric polynomial).



        For example, let $n=3,p(x)=x^3-2x^2-3x+2$; then



        $Delta p(s,t)=dfrac{s^3-t^3-2s^2+2t^2-3s+3t}{s-t}=s^2+st+t^2-2s-2t-3$.



        Finally $Adj(xI-A)=(x^2-2x-3)I+(x-2)A+A^2$.



        EDIT. The resolvent is essentially used to calculate $f(A)$ when $f$ is an holomorphic function.



        Here $(xI-A)^{-1}=sum_j a_j(x)/p(x) A^j$ where $a_j$ is a polynomial.



        Then $f(A)=1/2ipisum_j (int_{gamma} f(x)a_j(x)/p(x)dx) A^j$ where $gamma$ is a ad hoc curve. Consequently, we must only calculate $n$ numerical integrals.



        In particular, when $lambda_1,cdots,lambda_n$, the roots of $p$, are simple:



        $f(A)=sum_j(sum_k f(lambda_k)a_j(lambda_k)/p'(lambda_k)) A^j$.



        Anyway, the complexity of the calculation is that of the computation of the $(A^j)_j$, that is here $sim n^4/6$ (of course, if the calculation of the Jordan form of $A$ is stable, then we can do the job with complexity about $30 n³$).






        share|cite|improve this answer











        $endgroup$



        $(xI-A)^{-1}=Adj(xI-A)/p(x)$ where $p(x)$ is the (known) characteristic polynomial of $A$. On the other hand, $Adj(sI-A)=Delta p(s,A)$ where $Delta p(s,t)=dfrac{p(s)-p(t)}{s-t}$ (a symmetric polynomial).



        For example, let $n=3,p(x)=x^3-2x^2-3x+2$; then



        $Delta p(s,t)=dfrac{s^3-t^3-2s^2+2t^2-3s+3t}{s-t}=s^2+st+t^2-2s-2t-3$.



        Finally $Adj(xI-A)=(x^2-2x-3)I+(x-2)A+A^2$.



        EDIT. The resolvent is essentially used to calculate $f(A)$ when $f$ is an holomorphic function.



        Here $(xI-A)^{-1}=sum_j a_j(x)/p(x) A^j$ where $a_j$ is a polynomial.



        Then $f(A)=1/2ipisum_j (int_{gamma} f(x)a_j(x)/p(x)dx) A^j$ where $gamma$ is a ad hoc curve. Consequently, we must only calculate $n$ numerical integrals.



        In particular, when $lambda_1,cdots,lambda_n$, the roots of $p$, are simple:



        $f(A)=sum_j(sum_k f(lambda_k)a_j(lambda_k)/p'(lambda_k)) A^j$.



        Anyway, the complexity of the calculation is that of the computation of the $(A^j)_j$, that is here $sim n^4/6$ (of course, if the calculation of the Jordan form of $A$ is stable, then we can do the job with complexity about $30 n³$).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 4 '17 at 10:24

























        answered Sep 2 '17 at 15:54









        loup blancloup blanc

        23.7k21851




        23.7k21851






























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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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