Show that the associated matrix is $PAP^{-1}$












0












$begingroup$


Let $eta in operatorname{End}_{R}(R^n)$ and let $A$ be the associated matrix of $eta$ with respect to the basis $(e_1, dots e_n)$. Let $f_i = sum p_{i,j}e_j$ where the matrix $P=(p_{i,j}) in GL_n(R)$. Show that $PAP^{-1}$ is the associated matrix of the basis $(f_1 dots f_n)$.



$underline{My space attempt:}$



$$eta(f_i) = etaleft(sum_{j=1}^n p_{i,j}e_j right) = sum_{j=1}^n eta(p_{i,j} e_j) = sum_{j=1}^n p_{i,j} eta(e_j)$$



$$=sum_{j=1}^n p_{i,j} left( sum_{k=1}^n a_{i,k} e_k right) = sum_{j=1}^n sum_{k=1}^n(p_{i,j}a_{i,k})e_k$$



$$=sum_{j=1}^nsum_{k=1}^n (p_{i,j}a_{i,k}p_{i,j}^{-1})p_{i,j}e_k$$



From here I see that I am very close but I'm not sure how to turn that $p_{i,j}e_k$ into the $f_{k}$ because the sum includes all the other terms. Any help is appreciated, thanks!






linear-algebra abstract-algebra change-of-basis free-modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:14










  • $begingroup$
    Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
    $endgroup$
    – Bernard
    Dec 29 '18 at 21:15










  • $begingroup$
    Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
    $endgroup$
    – Justin Stevenson
    Dec 29 '18 at 21:28






  • 1




    $begingroup$
    I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
    $endgroup$
    – Omnomnomnom
    Dec 29 '18 at 22:17
















0












$begingroup$


Let $eta in operatorname{End}_{R}(R^n)$ and let $A$ be the associated matrix of $eta$ with respect to the basis $(e_1, dots e_n)$. Let $f_i = sum p_{i,j}e_j$ where the matrix $P=(p_{i,j}) in GL_n(R)$. Show that $PAP^{-1}$ is the associated matrix of the basis $(f_1 dots f_n)$.



$underline{My space attempt:}$



$$eta(f_i) = etaleft(sum_{j=1}^n p_{i,j}e_j right) = sum_{j=1}^n eta(p_{i,j} e_j) = sum_{j=1}^n p_{i,j} eta(e_j)$$



$$=sum_{j=1}^n p_{i,j} left( sum_{k=1}^n a_{i,k} e_k right) = sum_{j=1}^n sum_{k=1}^n(p_{i,j}a_{i,k})e_k$$



$$=sum_{j=1}^nsum_{k=1}^n (p_{i,j}a_{i,k}p_{i,j}^{-1})p_{i,j}e_k$$



From here I see that I am very close but I'm not sure how to turn that $p_{i,j}e_k$ into the $f_{k}$ because the sum includes all the other terms. Any help is appreciated, thanks!






linear-algebra abstract-algebra change-of-basis free-modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:14










  • $begingroup$
    Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
    $endgroup$
    – Bernard
    Dec 29 '18 at 21:15










  • $begingroup$
    Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
    $endgroup$
    – Justin Stevenson
    Dec 29 '18 at 21:28






  • 1




    $begingroup$
    I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
    $endgroup$
    – Omnomnomnom
    Dec 29 '18 at 22:17














0












0








0





$begingroup$


Let $eta in operatorname{End}_{R}(R^n)$ and let $A$ be the associated matrix of $eta$ with respect to the basis $(e_1, dots e_n)$. Let $f_i = sum p_{i,j}e_j$ where the matrix $P=(p_{i,j}) in GL_n(R)$. Show that $PAP^{-1}$ is the associated matrix of the basis $(f_1 dots f_n)$.



$underline{My space attempt:}$



$$eta(f_i) = etaleft(sum_{j=1}^n p_{i,j}e_j right) = sum_{j=1}^n eta(p_{i,j} e_j) = sum_{j=1}^n p_{i,j} eta(e_j)$$



$$=sum_{j=1}^n p_{i,j} left( sum_{k=1}^n a_{i,k} e_k right) = sum_{j=1}^n sum_{k=1}^n(p_{i,j}a_{i,k})e_k$$



$$=sum_{j=1}^nsum_{k=1}^n (p_{i,j}a_{i,k}p_{i,j}^{-1})p_{i,j}e_k$$



From here I see that I am very close but I'm not sure how to turn that $p_{i,j}e_k$ into the $f_{k}$ because the sum includes all the other terms. Any help is appreciated, thanks!






linear-algebra abstract-algebra change-of-basis free-modules







share|cite|improve this question











$endgroup$




Let $eta in operatorname{End}_{R}(R^n)$ and let $A$ be the associated matrix of $eta$ with respect to the basis $(e_1, dots e_n)$. Let $f_i = sum p_{i,j}e_j$ where the matrix $P=(p_{i,j}) in GL_n(R)$. Show that $PAP^{-1}$ is the associated matrix of the basis $(f_1 dots f_n)$.



$underline{My space attempt:}$



$$eta(f_i) = etaleft(sum_{j=1}^n p_{i,j}e_j right) = sum_{j=1}^n eta(p_{i,j} e_j) = sum_{j=1}^n p_{i,j} eta(e_j)$$



$$=sum_{j=1}^n p_{i,j} left( sum_{k=1}^n a_{i,k} e_k right) = sum_{j=1}^n sum_{k=1}^n(p_{i,j}a_{i,k})e_k$$



$$=sum_{j=1}^nsum_{k=1}^n (p_{i,j}a_{i,k}p_{i,j}^{-1})p_{i,j}e_k$$



From here I see that I am very close but I'm not sure how to turn that $p_{i,j}e_k$ into the $f_{k}$ because the sum includes all the other terms. Any help is appreciated, thanks!






linear-algebra abstract-algebra change-of-basis free-modules




linear-algebra abstract-algebra change-of-basis free-modules






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 21:26







Justin Stevenson

















asked Dec 29 '18 at 21:02









Justin StevensonJustin Stevenson

957519




957519












  • $begingroup$
    The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:14










  • $begingroup$
    Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
    $endgroup$
    – Bernard
    Dec 29 '18 at 21:15










  • $begingroup$
    Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
    $endgroup$
    – Justin Stevenson
    Dec 29 '18 at 21:28






  • 1




    $begingroup$
    I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
    $endgroup$
    – Omnomnomnom
    Dec 29 '18 at 22:17


















  • $begingroup$
    The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
    $endgroup$
    – angryavian
    Dec 29 '18 at 21:14










  • $begingroup$
    Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
    $endgroup$
    – Bernard
    Dec 29 '18 at 21:15










  • $begingroup$
    Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
    $endgroup$
    – Justin Stevenson
    Dec 29 '18 at 21:28






  • 1




    $begingroup$
    I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
    $endgroup$
    – Omnomnomnom
    Dec 29 '18 at 22:17
















$begingroup$
The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
$endgroup$
– angryavian
Dec 29 '18 at 21:14




$begingroup$
The $i,j$ element of $P^{-1}$ is not the inverse of $p_{i,j}$.
$endgroup$
– angryavian
Dec 29 '18 at 21:14












$begingroup$
Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
$endgroup$
– Bernard
Dec 29 '18 at 21:15




$begingroup$
Are you sure of the formulæ in this question? Normally the coordinates of the $f-i$s in basis $(e_1,dots,e_n)$ should be column vectors in $P$, not row vectors.
$endgroup$
– Bernard
Dec 29 '18 at 21:15












$begingroup$
Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
$endgroup$
– Justin Stevenson
Dec 29 '18 at 21:28




$begingroup$
Yeah I double checked what I have written is exactly what is in the book. Perhaps its a typo on the books part?
$endgroup$
– Justin Stevenson
Dec 29 '18 at 21:28




1




1




$begingroup$
I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
$endgroup$
– Omnomnomnom
Dec 29 '18 at 22:17




$begingroup$
I think $p^{-1}_{i,j}$ is supposed to mean the $i,j$ element of $P^{-1}$, rather than $(p_{ij})^{-1}$
$endgroup$
– Omnomnomnom
Dec 29 '18 at 22:17










1 Answer
1






active

oldest

votes


















0












$begingroup$

I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



We could finish your proof as follows:
$$
eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
= sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
\= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
= sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
$$
Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056261%2fshow-that-the-associated-matrix-is-pap-1%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



    We could finish your proof as follows:
    $$
    eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
    = sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
    sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
    \= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
    = sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
    $$
    Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



      We could finish your proof as follows:
      $$
      eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
      = sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
      sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
      \= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
      = sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
      $$
      Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



        We could finish your proof as follows:
        $$
        eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
        = sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
        sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
        \= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
        = sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
        $$
        Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.






        share|cite|improve this answer









        $endgroup$



        I'll use the notation $p^{-1}_{ij}$ to refer to the $i,j$ entry of $P^{-1}$, which is what I suspect your book did.



        We could finish your proof as follows:
        $$
        eta(f_i) = eta left( sum_{j=1}^n p_{ij} e_{j}right) = sum_{j=1}^n p_{ij} eta(e_{j})\
        = sum_{j=1}^n p_{ij} left(sum_{k=1}^n a_{jk}e_{k}right) =
        sum_{k=1}^n sum_{k=1}^n p_{ij}a_{jk} e_k
        \= sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk} left(sum_{ell=1}^n p^{-1}_{kell}f_{ell}right)
        = sum_{j=1}^n sum_{k=1}^n sum_{ell=1}^n p_{ij}a_{jk}p^{-1}_{kell} f_ell
        $$
        Now, it suffices to note that $sum_{j=1}^n sum_{k=1}^n p_{ij}a_{jk}p^{-1}_{kell}$ is indeed the $(ell,i)$ of the matrix $PAP^{-1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 10:33









        OmnomnomnomOmnomnomnom

        129k792185




        129k792185






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056261%2fshow-that-the-associated-matrix-is-pap-1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Wiesbaden

            Image
            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
            Read more

            Marschland

            Image
            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
            Read more

            Dieringhausen

            Image
            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
            Read more