Let $A= begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}$. Find the entry in the first row and second column...












4












$begingroup$


I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19
















4












$begingroup$


I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19














4












4








4


2



$begingroup$


I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$




I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.

Where:
$P=begin{pmatrix} 1&1 \ -4&-1 end{pmatrix}$
$D=begin{pmatrix} 0&0 \ 0&6 end{pmatrix}$
$P^{-1}=frac{1}{3}begin{pmatrix} -1&-1 \ 4&1 end{pmatrix}$.



So that $A^{2014}=PD^{2014}P^{-1}$

But i just want to know whether there is an alternate method.






linear-algebra eigenvalues-eigenvectors diagonalization




linear-algebra eigenvalues-eigenvectors diagonalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 0:18









vadim123

76.5k897191




76.5k897191










asked Jan 6 at 0:16









DD90DD90

2598




2598








  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19














  • 1




    $begingroup$
    This is the standard elegant way to do it.I don't know another method other than brute force.
    $endgroup$
    – Ethan Bolker
    Jan 6 at 0:19






  • 1




    $begingroup$
    I think that what you did is very well, other ways will come to the decomposition anyways
    $endgroup$
    – José Alejandro Aburto Araneda
    Jan 6 at 0:19








1




1




$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19




$begingroup$
This is the standard elegant way to do it.I don't know another method other than brute force.
$endgroup$
– Ethan Bolker
Jan 6 at 0:19




1




1




$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19




$begingroup$
I think that what you did is very well, other ways will come to the decomposition anyways
$endgroup$
– José Alejandro Aburto Araneda
Jan 6 at 0:19










3 Answers
3






active

oldest

votes


















5












$begingroup$

Hint Show that
$$A^2=6A$$



Note: You can show the stronger statement:
$$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
but this is overkill.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
    $endgroup$
    – Mindlack
    Jan 6 at 0:25










  • $begingroup$
    Very nice indeed!
    $endgroup$
    – Mike
    Jan 6 at 0:33










  • $begingroup$
    Can you please explain it a little further how to proceed with this
    $endgroup$
    – DD90
    Jan 6 at 0:35






  • 1




    $begingroup$
    If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
    $endgroup$
    – Mindlack
    Jan 6 at 0:52










  • $begingroup$
    @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
    $endgroup$
    – N. S.
    Jan 6 at 3:34



















2












$begingroup$

By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
    $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



    Which now immediately gives you
    $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






    share|cite|improve this answer









    $endgroup$














      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34
















      5












      $begingroup$

      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34














      5












      5








      5





      $begingroup$

      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.






      share|cite|improve this answer









      $endgroup$



      Hint Show that
      $$A^2=6A$$



      Note: You can show the stronger statement:
      $$begin{pmatrix} 8&2 \ -8&-2 end{pmatrix}begin{pmatrix} a&b \ -a&-b end{pmatrix}=6begin{pmatrix} a&b \ -a&-b end{pmatrix}$$
      but this is overkill.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 6 at 0:23









      N. S.N. S.

      105k7115210




      105k7115210












      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34


















      • $begingroup$
        This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
        $endgroup$
        – Mindlack
        Jan 6 at 0:25










      • $begingroup$
        Very nice indeed!
        $endgroup$
        – Mike
        Jan 6 at 0:33










      • $begingroup$
        Can you please explain it a little further how to proceed with this
        $endgroup$
        – DD90
        Jan 6 at 0:35






      • 1




        $begingroup$
        If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
        $endgroup$
        – Mindlack
        Jan 6 at 0:52










      • $begingroup$
        @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
        $endgroup$
        – N. S.
        Jan 6 at 3:34
















      $begingroup$
      This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
      $endgroup$
      – Mindlack
      Jan 6 at 0:25




      $begingroup$
      This is a cleaner and easier argument. Of course, it relies on the fact that the matrix $A$ has rank $1$.
      $endgroup$
      – Mindlack
      Jan 6 at 0:25












      $begingroup$
      Very nice indeed!
      $endgroup$
      – Mike
      Jan 6 at 0:33




      $begingroup$
      Very nice indeed!
      $endgroup$
      – Mike
      Jan 6 at 0:33












      $begingroup$
      Can you please explain it a little further how to proceed with this
      $endgroup$
      – DD90
      Jan 6 at 0:35




      $begingroup$
      Can you please explain it a little further how to proceed with this
      $endgroup$
      – DD90
      Jan 6 at 0:35




      1




      1




      $begingroup$
      If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
      $endgroup$
      – Mindlack
      Jan 6 at 0:52




      $begingroup$
      If $A^2=6A$, then $A^3=A^2A=6AA=6*6A=6^2A$. Moreover, $A^4=A^3A=6^2AA=6^2*6A=6^3A$, and so on...
      $endgroup$
      – Mindlack
      Jan 6 at 0:52












      $begingroup$
      @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
      $endgroup$
      – N. S.
      Jan 6 at 3:34




      $begingroup$
      @DD90 $$A^3=A^2 cdot A=6A cdot A =6 A^2 = 6^2A \A^4=A^3 cdot A=6^2A cdot A =6^2 A^2 = 6^3A \...$$
      $endgroup$
      – N. S.
      Jan 6 at 3:34











      2












      $begingroup$

      By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



      Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



        Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



          Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.






          share|cite|improve this answer











          $endgroup$



          By the Cayley-Hamilton theorem, $A^{2014}=aI+bA$ for some unknown coefficients $a$ and $b$. This equation is also satisfied by $A$’s eigenvalues, which generates the system of equations $a=0$, $a+6b=6^{2014}$, from which $b=6^{2013}$. The required entry of $A^{2014}$ is therefore $2cdot6^{2013}$.



          Note, too, that the eigenvalues of $A$ can be found by inspection. Its rows are obviously linearly dependent, so one of its eigenvalues is $0$. The other eigenvalue is then equal to $A$’s trace.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 1:08

























          answered Jan 6 at 1:02









          amdamd

          31.6k21052




          31.6k21052























              0












              $begingroup$

              Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
              $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



              Which now immediately gives you
              $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
                $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



                Which now immediately gives you
                $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
                  $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



                  Which now immediately gives you
                  $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$






                  share|cite|improve this answer









                  $endgroup$



                  Since $A$ is clearly rank-1 (the two columns are multiples), you can directly decompose it into the outer product of two vectors:
                  $$A=begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}$$



                  Which now immediately gives you
                  $$A^{2014}=left(begin{bmatrix}1\-1end{bmatrix}begin{bmatrix}8&2end{bmatrix}right)^{2014}=begin{bmatrix}1\-1end{bmatrix}left(begin{bmatrix}8&2end{bmatrix}begin{bmatrix}1\-1end{bmatrix}right)^{2013}begin{bmatrix}8&2end{bmatrix}=begin{bmatrix}1\-1end{bmatrix}6^{2013}begin{bmatrix}8&2end{bmatrix}=6^{2013}A$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 3:26









                  obscuransobscurans

                  1,152311




                  1,152311






























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                      Wiesbaden

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                      Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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                      Marschland

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                      Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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                      Dieringhausen

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                      Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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