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












3












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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







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  • $begingroup$
    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




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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

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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, $$text{adj}^n(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{n}-1}{n}},A&text{if }ntext{ is even},, \ big(det(A)big)^{frac{(n-1)^{n}-(n-1)}{n}},text{adj}(A)&text{if }ntext{ is odd},. end{cases}$$







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$endgroup$





















    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














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    2 Answers
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    2 Answers
    2






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    active

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    active

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    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, $$text{adj}^n(A)=begin{cases}big(det(A)big)^{frac{(n-1)^{n}-1}{n}},A&text{if }ntext{ is even},, \ big(det(A)big)^{frac{(n-1)^{n}-(n-1)}{n}},text{adj}(A)&text{if }ntext{ is odd},. end{cases}$$








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        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




















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