What is the $n$-time iterated adjugate of an $ntimes n$ matrix $A$?












3












$begingroup$



What is $underbrace{text{adj}Big(text{adj}big(ldots(text{adj}}_{ntext{ adj}} A)ldotsbig)Big)$, where $text{adj}$ is written $n$ times, and the order of the matrix $A$ is $ntimes n$?




Can you show the proof for each $n$ (I mean by induction)!!






linear-algebra matrices induction matrix-equations function-and-relation-composition







share|cite|improve this question











$endgroup$












  • $begingroup$
    i have till n =2 don't know how do further !!
    $endgroup$
    – user416571
    Dec 13 '18 at 9:46
















3












$begingroup$



What is $underbrace{text{adj}Big(text{adj}big(ldots(text{adj}}_{ntext{ adj}} A)ldotsbig)Big)$, where $text{adj}$ is written $n$ times, and the order of the matrix $A$ is $ntimes n$?




Can you show the proof for each $n$ (I mean by induction)!!






linear-algebra matrices induction matrix-equations function-and-relation-composition







share|cite|improve this question











$endgroup$












  • $begingroup$
    i have till n =2 don't know how do further !!
    $endgroup$
    – user416571
    Dec 13 '18 at 9:46














3












3








3


1



$begingroup$



What is $underbrace{text{adj}Big(text{adj}big(ldots(text{adj}}_{ntext{ adj}} A)ldotsbig)Big)$, where $text{adj}$ is written $n$ times, and the order of the matrix $A$ is $ntimes n$?




Can you show the proof for each $n$ (I mean by induction)!!






linear-algebra matrices induction matrix-equations function-and-relation-composition







share|cite|improve this question











$endgroup$





What is $underbrace{text{adj}Big(text{adj}big(ldots(text{adj}}_{ntext{ adj}} A)ldotsbig)Big)$, where $text{adj}$ is written $n$ times, and the order of the matrix $A$ is $ntimes n$?




Can you show the proof for each $n$ (I mean by induction)!!






linear-algebra matrices induction matrix-equations function-and-relation-composition




linear-algebra matrices induction matrix-equations function-and-relation-composition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 11:37









Batominovski

33.1k33293




33.1k33293










asked Dec 13 '18 at 9:44









user416571user416571

414




414












  • $begingroup$
    i have till n =2 don't know how do further !!
    $endgroup$
    – user416571
    Dec 13 '18 at 9:46


















  • $begingroup$
    i have till n =2 don't know how do further !!
    $endgroup$
    – user416571
    Dec 13 '18 at 9:46
















$begingroup$
i have till n =2 don't know how do further !!
$endgroup$
– user416571
Dec 13 '18 at 9:46




$begingroup$
i have till n =2 don't know how do further !!
$endgroup$
– user416571
Dec 13 '18 at 9:46










2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint. For any integer $ngeq 2$, for any $n$-by-$n$ matrix $X$, and for each scalar $k$, show that
$$text{adj}(k,X)=k^{n-1},text{adj}(X),,$$ $$detbig(text{adj}(X)big)=big(det(X)big)^{n-1},,$$ and $$text{adj}big(text{adj}(X)big)=big(det(X)big)^{n-2} X,.$$ Here, $0^0:=1$. (For $n=1$, $text{adj}(X)=1$ always.)




For each $minmathbb{Z}_{geq 0}$, let $text{adj}^m$ be the $m$-time iteration of $text{adj}$. Then, for any $n$-by-$n$ matrix $A$, $$text{adj}^m(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{m}-1}{n}},A&text{if }mtext{ is even},, \ big(det(A)big)^{frac{(n-1)^{m}-(n-1)}{n}},text{adj}(A)&text{if }mtext{ is odd},. end{cases}tag{*}$$ For $n=1$, $$text{adj}^0(A)=A,,text{ and }text{adj}^m(A)=1text{ for any }minmathbb{Z}_{>0},.$$ For $n=2$, $$text{adj}^m(A)=begin{cases}A&text{if }mtext{ is even},, \ text{adj}(A)&text{if }mtext{ is odd},. end{cases}$$ Clearly, if $ngeq 3$ and $det(A)=0$, then $$text{adj}^0(A)=A,,,, text{adj}^1(A)=text{adj}(A),,text{ and }text{adj}^m(A)=0text{ for all }mgeq 2,.$$ If $det(A)neq 0$, then you may simplify (*) to $$text{adj}^m(A)=big(det(A)big)^{frac{(n-1)^{m}-(-1)^m}{n}},A^{(-1)^m}text{ for every }minmathbb{Z}_{geq 0},.$$ In particular, $$text{adj}^n(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{n}-1}{n}},A&text{if }ntext{ is even},, \ big(det(A)big)^{frac{(n-1)^{n}-(n-1)}{n}},text{adj}(A)&text{if }ntext{ is odd},. end{cases}$$







share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Hint. We know that
    $$
    mathrm{adj},Acdot A=det Acdot I, tag{1}
    $$
    and hence, if $det Ane 0$,
    $$
    mathrm{adj},A=det A cdot A^{-1},
    $$
    and
    $$
    det(mathrm{adj},A)cdot det A=(det A)^n.
    $$
    Hence $det(mathrm{adj},A)=(det A)^{n-1}$.



    Now $(1)$ implies that
    $$
    mathrm{adj}(mathrm{adj},A)cdot mathrm{adj},A=det (mathrm{adj},A)cdot I
    =(det A)^{n-1}I,
    $$
    and hence
    $$
    mathrm{adj}(mathrm{adj},A)=(det A)^{n-1}cdot(mathrm{adj},A)^{-1}=(det A)^ncdot A
    $$
    and
    $$
    detbig(mathrm{adj}(mathrm{adj},A)big)=big(det(mathrm{adj},A)big)^{n-1}=(det A)^{(n-1)^2}, quad det(mathrm{adj}^k,A)=(det A)^{(n-1)^k}
    tag{2}$$
    and
    $$
    mathrm{adj}^{k+1},Acdot mathrm{adj}^{k},A=det (mathrm{adj}^k,A)cdot I
    =(det A)^{(n-1)^k}I, tag{3}
    $$
    Finally, we obtain that:
    $$
    mathrm{adj}^{2k}A=c_{k}A, quad mathrm{adj}^{2k+1}A=d_{k}A^{-1}
    $$
    where $c_k,d_k$ can be obtained from (2) and (3).






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      What about $det A=0$?
      $endgroup$
      – Christoph
      Dec 13 '18 at 10:40






    • 1




      $begingroup$
      The $mathrm{adj}^k A$, are still defined in the same exactly way, using a density argument, since the invertible matrices are dense in the set of all square matrices.
      $endgroup$
      – Yiorgos S. Smyrlis
      Dec 13 '18 at 10:50






    • 5




      $begingroup$
      I think you made a mistake at the beginning. $operatorname{adj}(A) = det(A),A^{-1}$, not $frac{1}{det(A)},A^{-1}$.
      $endgroup$
      – Christoph
      Dec 13 '18 at 10:59













    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%2f3037812%2fwhat-is-the-n-time-iterated-adjugate-of-an-n-times-n-matrix-a%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Hint. For any integer $ngeq 2$, for any $n$-by-$n$ matrix $X$, and for each scalar $k$, show that
    $$text{adj}(k,X)=k^{n-1},text{adj}(X),,$$ $$detbig(text{adj}(X)big)=big(det(X)big)^{n-1},,$$ and $$text{adj}big(text{adj}(X)big)=big(det(X)big)^{n-2} X,.$$ Here, $0^0:=1$. (For $n=1$, $text{adj}(X)=1$ always.)




    For each $minmathbb{Z}_{geq 0}$, let $text{adj}^m$ be the $m$-time iteration of $text{adj}$. Then, for any $n$-by-$n$ matrix $A$, $$text{adj}^m(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{m}-1}{n}},A&text{if }mtext{ is even},, \ big(det(A)big)^{frac{(n-1)^{m}-(n-1)}{n}},text{adj}(A)&text{if }mtext{ is odd},. end{cases}tag{*}$$ For $n=1$, $$text{adj}^0(A)=A,,text{ and }text{adj}^m(A)=1text{ for any }minmathbb{Z}_{>0},.$$ For $n=2$, $$text{adj}^m(A)=begin{cases}A&text{if }mtext{ is even},, \ text{adj}(A)&text{if }mtext{ is odd},. end{cases}$$ Clearly, if $ngeq 3$ and $det(A)=0$, then $$text{adj}^0(A)=A,,,, text{adj}^1(A)=text{adj}(A),,text{ and }text{adj}^m(A)=0text{ for all }mgeq 2,.$$ If $det(A)neq 0$, then you may simplify (*) to $$text{adj}^m(A)=big(det(A)big)^{frac{(n-1)^{m}-(-1)^m}{n}},A^{(-1)^m}text{ for every }minmathbb{Z}_{geq 0},.$$ In particular, $$text{adj}^n(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{n}-1}{n}},A&text{if }ntext{ is even},, \ big(det(A)big)^{frac{(n-1)^{n}-(n-1)}{n}},text{adj}(A)&text{if }ntext{ is odd},. end{cases}$$







    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Hint. For any integer $ngeq 2$, for any $n$-by-$n$ matrix $X$, and for each scalar $k$, show that
      $$text{adj}(k,X)=k^{n-1},text{adj}(X),,$$ $$detbig(text{adj}(X)big)=big(det(X)big)^{n-1},,$$ and $$text{adj}big(text{adj}(X)big)=big(det(X)big)^{n-2} X,.$$ Here, $0^0:=1$. (For $n=1$, $text{adj}(X)=1$ always.)




      For each $minmathbb{Z}_{geq 0}$, let $text{adj}^m$ be the $m$-time iteration of $text{adj}$. Then, for any $n$-by-$n$ matrix $A$, $$text{adj}^m(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{m}-1}{n}},A&text{if }mtext{ is even},, \ big(det(A)big)^{frac{(n-1)^{m}-(n-1)}{n}},text{adj}(A)&text{if }mtext{ is odd},. end{cases}tag{*}$$ For $n=1$, $$text{adj}^0(A)=A,,text{ and }text{adj}^m(A)=1text{ for any }minmathbb{Z}_{>0},.$$ For $n=2$, $$text{adj}^m(A)=begin{cases}A&text{if }mtext{ is even},, \ text{adj}(A)&text{if }mtext{ is odd},. end{cases}$$ Clearly, if $ngeq 3$ and $det(A)=0$, then $$text{adj}^0(A)=A,,,, text{adj}^1(A)=text{adj}(A),,text{ and }text{adj}^m(A)=0text{ for all }mgeq 2,.$$ If $det(A)neq 0$, then you may simplify (*) to $$text{adj}^m(A)=big(det(A)big)^{frac{(n-1)^{m}-(-1)^m}{n}},A^{(-1)^m}text{ for every }minmathbb{Z}_{geq 0},.$$ In particular, $$text{adj}^n(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{n}-1}{n}},A&text{if }ntext{ is even},, \ big(det(A)big)^{frac{(n-1)^{n}-(n-1)}{n}},text{adj}(A)&text{if }ntext{ is odd},. end{cases}$$







      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint. For any integer $ngeq 2$, for any $n$-by-$n$ matrix $X$, and for each scalar $k$, show that
        $$text{adj}(k,X)=k^{n-1},text{adj}(X),,$$ $$detbig(text{adj}(X)big)=big(det(X)big)^{n-1},,$$ and $$text{adj}big(text{adj}(X)big)=big(det(X)big)^{n-2} X,.$$ Here, $0^0:=1$. (For $n=1$, $text{adj}(X)=1$ always.)




        For each $minmathbb{Z}_{geq 0}$, let $text{adj}^m$ be the $m$-time iteration of $text{adj}$. Then, for any $n$-by-$n$ matrix $A$, $$text{adj}^m(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{m}-1}{n}},A&text{if }mtext{ is even},, \ big(det(A)big)^{frac{(n-1)^{m}-(n-1)}{n}},text{adj}(A)&text{if }mtext{ is odd},. end{cases}tag{*}$$ For $n=1$, $$text{adj}^0(A)=A,,text{ and }text{adj}^m(A)=1text{ for any }minmathbb{Z}_{>0},.$$ For $n=2$, $$text{adj}^m(A)=begin{cases}A&text{if }mtext{ is even},, \ text{adj}(A)&text{if }mtext{ is odd},. end{cases}$$ Clearly, if $ngeq 3$ and $det(A)=0$, then $$text{adj}^0(A)=A,,,, text{adj}^1(A)=text{adj}(A),,text{ and }text{adj}^m(A)=0text{ for all }mgeq 2,.$$ If $det(A)neq 0$, then you may simplify (*) to $$text{adj}^m(A)=big(det(A)big)^{frac{(n-1)^{m}-(-1)^m}{n}},A^{(-1)^m}text{ for every }minmathbb{Z}_{geq 0},.$$ In particular, $$text{adj}^n(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{n}-1}{n}},A&text{if }ntext{ is even},, \ big(det(A)big)^{frac{(n-1)^{n}-(n-1)}{n}},text{adj}(A)&text{if }ntext{ is odd},. end{cases}$$







        share|cite|improve this answer











        $endgroup$



        Hint. For any integer $ngeq 2$, for any $n$-by-$n$ matrix $X$, and for each scalar $k$, show that
        $$text{adj}(k,X)=k^{n-1},text{adj}(X),,$$ $$detbig(text{adj}(X)big)=big(det(X)big)^{n-1},,$$ and $$text{adj}big(text{adj}(X)big)=big(det(X)big)^{n-2} X,.$$ Here, $0^0:=1$. (For $n=1$, $text{adj}(X)=1$ always.)




        For each $minmathbb{Z}_{geq 0}$, let $text{adj}^m$ be the $m$-time iteration of $text{adj}$. Then, for any $n$-by-$n$ matrix $A$, $$text{adj}^m(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{m}-1}{n}},A&text{if }mtext{ is even},, \ big(det(A)big)^{frac{(n-1)^{m}-(n-1)}{n}},text{adj}(A)&text{if }mtext{ is odd},. end{cases}tag{*}$$ For $n=1$, $$text{adj}^0(A)=A,,text{ and }text{adj}^m(A)=1text{ for any }minmathbb{Z}_{>0},.$$ For $n=2$, $$text{adj}^m(A)=begin{cases}A&text{if }mtext{ is even},, \ text{adj}(A)&text{if }mtext{ is odd},. end{cases}$$ Clearly, if $ngeq 3$ and $det(A)=0$, then $$text{adj}^0(A)=A,,,, text{adj}^1(A)=text{adj}(A),,text{ and }text{adj}^m(A)=0text{ for all }mgeq 2,.$$ If $det(A)neq 0$, then you may simplify (*) to $$text{adj}^m(A)=big(det(A)big)^{frac{(n-1)^{m}-(-1)^m}{n}},A^{(-1)^m}text{ for every }minmathbb{Z}_{geq 0},.$$ In particular, $$text{adj}^n(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{n}-1}{n}},A&text{if }ntext{ is even},, \ big(det(A)big)^{frac{(n-1)^{n}-(n-1)}{n}},text{adj}(A)&text{if }ntext{ is odd},. end{cases}$$








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 13 '18 at 11:05

























        answered Dec 13 '18 at 10:31









        BatominovskiBatominovski

        33.1k33293




        33.1k33293























            0












            $begingroup$

            Hint. We know that
            $$
            mathrm{adj},Acdot A=det Acdot I, tag{1}
            $$
            and hence, if $det Ane 0$,
            $$
            mathrm{adj},A=det A cdot A^{-1},
            $$
            and
            $$
            det(mathrm{adj},A)cdot det A=(det A)^n.
            $$
            Hence $det(mathrm{adj},A)=(det A)^{n-1}$.



            Now $(1)$ implies that
            $$
            mathrm{adj}(mathrm{adj},A)cdot mathrm{adj},A=det (mathrm{adj},A)cdot I
            =(det A)^{n-1}I,
            $$
            and hence
            $$
            mathrm{adj}(mathrm{adj},A)=(det A)^{n-1}cdot(mathrm{adj},A)^{-1}=(det A)^ncdot A
            $$
            and
            $$
            detbig(mathrm{adj}(mathrm{adj},A)big)=big(det(mathrm{adj},A)big)^{n-1}=(det A)^{(n-1)^2}, quad det(mathrm{adj}^k,A)=(det A)^{(n-1)^k}
            tag{2}$$
            and
            $$
            mathrm{adj}^{k+1},Acdot mathrm{adj}^{k},A=det (mathrm{adj}^k,A)cdot I
            =(det A)^{(n-1)^k}I, tag{3}
            $$
            Finally, we obtain that:
            $$
            mathrm{adj}^{2k}A=c_{k}A, quad mathrm{adj}^{2k+1}A=d_{k}A^{-1}
            $$
            where $c_k,d_k$ can be obtained from (2) and (3).






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              What about $det A=0$?
              $endgroup$
              – Christoph
              Dec 13 '18 at 10:40






            • 1




              $begingroup$
              The $mathrm{adj}^k A$, are still defined in the same exactly way, using a density argument, since the invertible matrices are dense in the set of all square matrices.
              $endgroup$
              – Yiorgos S. Smyrlis
              Dec 13 '18 at 10:50






            • 5




              $begingroup$
              I think you made a mistake at the beginning. $operatorname{adj}(A) = det(A),A^{-1}$, not $frac{1}{det(A)},A^{-1}$.
              $endgroup$
              – Christoph
              Dec 13 '18 at 10:59


















            0












            $begingroup$

            Hint. We know that
            $$
            mathrm{adj},Acdot A=det Acdot I, tag{1}
            $$
            and hence, if $det Ane 0$,
            $$
            mathrm{adj},A=det A cdot A^{-1},
            $$
            and
            $$
            det(mathrm{adj},A)cdot det A=(det A)^n.
            $$
            Hence $det(mathrm{adj},A)=(det A)^{n-1}$.



            Now $(1)$ implies that
            $$
            mathrm{adj}(mathrm{adj},A)cdot mathrm{adj},A=det (mathrm{adj},A)cdot I
            =(det A)^{n-1}I,
            $$
            and hence
            $$
            mathrm{adj}(mathrm{adj},A)=(det A)^{n-1}cdot(mathrm{adj},A)^{-1}=(det A)^ncdot A
            $$
            and
            $$
            detbig(mathrm{adj}(mathrm{adj},A)big)=big(det(mathrm{adj},A)big)^{n-1}=(det A)^{(n-1)^2}, quad det(mathrm{adj}^k,A)=(det A)^{(n-1)^k}
            tag{2}$$
            and
            $$
            mathrm{adj}^{k+1},Acdot mathrm{adj}^{k},A=det (mathrm{adj}^k,A)cdot I
            =(det A)^{(n-1)^k}I, tag{3}
            $$
            Finally, we obtain that:
            $$
            mathrm{adj}^{2k}A=c_{k}A, quad mathrm{adj}^{2k+1}A=d_{k}A^{-1}
            $$
            where $c_k,d_k$ can be obtained from (2) and (3).






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              What about $det A=0$?
              $endgroup$
              – Christoph
              Dec 13 '18 at 10:40






            • 1




              $begingroup$
              The $mathrm{adj}^k A$, are still defined in the same exactly way, using a density argument, since the invertible matrices are dense in the set of all square matrices.
              $endgroup$
              – Yiorgos S. Smyrlis
              Dec 13 '18 at 10:50






            • 5




              $begingroup$
              I think you made a mistake at the beginning. $operatorname{adj}(A) = det(A),A^{-1}$, not $frac{1}{det(A)},A^{-1}$.
              $endgroup$
              – Christoph
              Dec 13 '18 at 10:59
















            0












            0








            0





            $begingroup$

            Hint. We know that
            $$
            mathrm{adj},Acdot A=det Acdot I, tag{1}
            $$
            and hence, if $det Ane 0$,
            $$
            mathrm{adj},A=det A cdot A^{-1},
            $$
            and
            $$
            det(mathrm{adj},A)cdot det A=(det A)^n.
            $$
            Hence $det(mathrm{adj},A)=(det A)^{n-1}$.



            Now $(1)$ implies that
            $$
            mathrm{adj}(mathrm{adj},A)cdot mathrm{adj},A=det (mathrm{adj},A)cdot I
            =(det A)^{n-1}I,
            $$
            and hence
            $$
            mathrm{adj}(mathrm{adj},A)=(det A)^{n-1}cdot(mathrm{adj},A)^{-1}=(det A)^ncdot A
            $$
            and
            $$
            detbig(mathrm{adj}(mathrm{adj},A)big)=big(det(mathrm{adj},A)big)^{n-1}=(det A)^{(n-1)^2}, quad det(mathrm{adj}^k,A)=(det A)^{(n-1)^k}
            tag{2}$$
            and
            $$
            mathrm{adj}^{k+1},Acdot mathrm{adj}^{k},A=det (mathrm{adj}^k,A)cdot I
            =(det A)^{(n-1)^k}I, tag{3}
            $$
            Finally, we obtain that:
            $$
            mathrm{adj}^{2k}A=c_{k}A, quad mathrm{adj}^{2k+1}A=d_{k}A^{-1}
            $$
            where $c_k,d_k$ can be obtained from (2) and (3).






            share|cite|improve this answer











            $endgroup$



            Hint. We know that
            $$
            mathrm{adj},Acdot A=det Acdot I, tag{1}
            $$
            and hence, if $det Ane 0$,
            $$
            mathrm{adj},A=det A cdot A^{-1},
            $$
            and
            $$
            det(mathrm{adj},A)cdot det A=(det A)^n.
            $$
            Hence $det(mathrm{adj},A)=(det A)^{n-1}$.



            Now $(1)$ implies that
            $$
            mathrm{adj}(mathrm{adj},A)cdot mathrm{adj},A=det (mathrm{adj},A)cdot I
            =(det A)^{n-1}I,
            $$
            and hence
            $$
            mathrm{adj}(mathrm{adj},A)=(det A)^{n-1}cdot(mathrm{adj},A)^{-1}=(det A)^ncdot A
            $$
            and
            $$
            detbig(mathrm{adj}(mathrm{adj},A)big)=big(det(mathrm{adj},A)big)^{n-1}=(det A)^{(n-1)^2}, quad det(mathrm{adj}^k,A)=(det A)^{(n-1)^k}
            tag{2}$$
            and
            $$
            mathrm{adj}^{k+1},Acdot mathrm{adj}^{k},A=det (mathrm{adj}^k,A)cdot I
            =(det A)^{(n-1)^k}I, tag{3}
            $$
            Finally, we obtain that:
            $$
            mathrm{adj}^{2k}A=c_{k}A, quad mathrm{adj}^{2k+1}A=d_{k}A^{-1}
            $$
            where $c_k,d_k$ can be obtained from (2) and (3).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 15 '18 at 10:35









            user416571

            414




            414










            answered Dec 13 '18 at 10:31









            Yiorgos S. SmyrlisYiorgos S. Smyrlis

            63.4k1385163




            63.4k1385163








            • 1




              $begingroup$
              What about $det A=0$?
              $endgroup$
              – Christoph
              Dec 13 '18 at 10:40






            • 1




              $begingroup$
              The $mathrm{adj}^k A$, are still defined in the same exactly way, using a density argument, since the invertible matrices are dense in the set of all square matrices.
              $endgroup$
              – Yiorgos S. Smyrlis
              Dec 13 '18 at 10:50






            • 5




              $begingroup$
              I think you made a mistake at the beginning. $operatorname{adj}(A) = det(A),A^{-1}$, not $frac{1}{det(A)},A^{-1}$.
              $endgroup$
              – Christoph
              Dec 13 '18 at 10:59
















            • 1




              $begingroup$
              What about $det A=0$?
              $endgroup$
              – Christoph
              Dec 13 '18 at 10:40






            • 1




              $begingroup$
              The $mathrm{adj}^k A$, are still defined in the same exactly way, using a density argument, since the invertible matrices are dense in the set of all square matrices.
              $endgroup$
              – Yiorgos S. Smyrlis
              Dec 13 '18 at 10:50






            • 5




              $begingroup$
              I think you made a mistake at the beginning. $operatorname{adj}(A) = det(A),A^{-1}$, not $frac{1}{det(A)},A^{-1}$.
              $endgroup$
              – Christoph
              Dec 13 '18 at 10:59










            1




            1




            $begingroup$
            What about $det A=0$?
            $endgroup$
            – Christoph
            Dec 13 '18 at 10:40




            $begingroup$
            What about $det A=0$?
            $endgroup$
            – Christoph
            Dec 13 '18 at 10:40




            1




            1




            $begingroup$
            The $mathrm{adj}^k A$, are still defined in the same exactly way, using a density argument, since the invertible matrices are dense in the set of all square matrices.
            $endgroup$
            – Yiorgos S. Smyrlis
            Dec 13 '18 at 10:50




            $begingroup$
            The $mathrm{adj}^k A$, are still defined in the same exactly way, using a density argument, since the invertible matrices are dense in the set of all square matrices.
            $endgroup$
            – Yiorgos S. Smyrlis
            Dec 13 '18 at 10:50




            5




            5




            $begingroup$
            I think you made a mistake at the beginning. $operatorname{adj}(A) = det(A),A^{-1}$, not $frac{1}{det(A)},A^{-1}$.
            $endgroup$
            – Christoph
            Dec 13 '18 at 10:59






            $begingroup$
            I think you made a mistake at the beginning. $operatorname{adj}(A) = det(A),A^{-1}$, not $frac{1}{det(A)},A^{-1}$.
            $endgroup$
            – Christoph
            Dec 13 '18 at 10:59




















            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%2f3037812%2fwhat-is-the-n-time-iterated-adjugate-of-an-n-times-n-matrix-a%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            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...
            Read more

            Musa

            Image
            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...
            Read more