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

33.2k33293




33.2k33293










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

        33.2k33293




        33.2k33293























            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

            63.7k1385165




            63.7k1385165








            • 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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more