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Prove or disprove: If $A$ is $ntimes n$ and $exists;min BbbN:;A^m=I_n$, then $A$ is invertible.



The Next CEO of Stack OverflowDeterminants of matrices; Does the equation $det M=det(I_n-M)$ have a solution?If $A^2+2A+I_n=O_n$ then $A$ is invertibleDeterminant of a $2 times 2$ block matrixprove that if a square matrix $A$ is invertible then $AA^T$ is invertible.Matrices such that $M^2+M^T=I_n$ are invertibleIf a matrix is row equivalent to some invertible matrix then it is invertibleProve that $A+I_n$ is invertible, where $leftlVert ArightrVert<1$Is it true that $m=nimplies A$ is invertible, for an $mtimes n$ matrix satisfying $(AA^T)^r=I$Help with proof or counterexample: $A^3=0 implies I_n+A$ is invertibleThe matrix $I_n - v^t x$ is invertible when $langle v, x rangle neq 1$










3












$begingroup$


Is this statement true? If $A$ is an $ntimes n$ matrix and $A^m=I_n$ for some $min BbbN$, then $A$ is invertible.



My trial



Let $nin BbbN$ be fixed. Then, $$[det(A)]^m=det(A^m)=I_n=1.$$
Hence, $$det(A)=1neq 0.$$
Thus, $A$ is invertible since $det(A)neq 0.$. I'm I right or is there a counter-example?






matrices algebra-precalculus matrix-equations







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
    $endgroup$
    – Chrystomath
    Mar 18 at 11:32
















3












$begingroup$


Is this statement true? If $A$ is an $ntimes n$ matrix and $A^m=I_n$ for some $min BbbN$, then $A$ is invertible.



My trial



Let $nin BbbN$ be fixed. Then, $$[det(A)]^m=det(A^m)=I_n=1.$$
Hence, $$det(A)=1neq 0.$$
Thus, $A$ is invertible since $det(A)neq 0.$. I'm I right or is there a counter-example?






matrices algebra-precalculus matrix-equations







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
    $endgroup$
    – Chrystomath
    Mar 18 at 11:32














3












3








3


1



$begingroup$


Is this statement true? If $A$ is an $ntimes n$ matrix and $A^m=I_n$ for some $min BbbN$, then $A$ is invertible.



My trial



Let $nin BbbN$ be fixed. Then, $$[det(A)]^m=det(A^m)=I_n=1.$$
Hence, $$det(A)=1neq 0.$$
Thus, $A$ is invertible since $det(A)neq 0.$. I'm I right or is there a counter-example?






matrices algebra-precalculus matrix-equations







share|cite|improve this question











$endgroup$




Is this statement true? If $A$ is an $ntimes n$ matrix and $A^m=I_n$ for some $min BbbN$, then $A$ is invertible.



My trial



Let $nin BbbN$ be fixed. Then, $$[det(A)]^m=det(A^m)=I_n=1.$$
Hence, $$det(A)=1neq 0.$$
Thus, $A$ is invertible since $det(A)neq 0.$. I'm I right or is there a counter-example?






matrices algebra-precalculus matrix-equations




matrices algebra-precalculus matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 18 at 11:34







Omojola Micheal

















asked Mar 18 at 11:29









Omojola MichealOmojola Micheal

1,999424




1,999424







  • 2




    $begingroup$
    Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
    $endgroup$
    – Chrystomath
    Mar 18 at 11:32













  • 2




    $begingroup$
    Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
    $endgroup$
    – Chrystomath
    Mar 18 at 11:32








2




2




$begingroup$
Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
$endgroup$
– Chrystomath
Mar 18 at 11:32





$begingroup$
Looks good. Alternatively, $A(A^m-1)=I=(A^m-1)A$ so $A^-1=A^m-1$.
$endgroup$
– Chrystomath
Mar 18 at 11:32











3 Answers
3






active

oldest

votes


















4












$begingroup$

The right conclusion is $det(A) ne 0$, hence it is invertible.



Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Sorry, I'll edit!
    $endgroup$
    – Omojola Micheal
    Mar 18 at 11:33










  • $begingroup$
    Thanks for the prompt reply.
    $endgroup$
    – Omojola Micheal
    Mar 18 at 11:34


















6












$begingroup$

I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      The right conclusion is $det(A) ne 0$, hence it is invertible.



      Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Sorry, I'll edit!
        $endgroup$
        – Omojola Micheal
        Mar 18 at 11:33










      • $begingroup$
        Thanks for the prompt reply.
        $endgroup$
        – Omojola Micheal
        Mar 18 at 11:34















      4












      $begingroup$

      The right conclusion is $det(A) ne 0$, hence it is invertible.



      Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Sorry, I'll edit!
        $endgroup$
        – Omojola Micheal
        Mar 18 at 11:33










      • $begingroup$
        Thanks for the prompt reply.
        $endgroup$
        – Omojola Micheal
        Mar 18 at 11:34













      4












      4








      4





      $begingroup$

      The right conclusion is $det(A) ne 0$, hence it is invertible.



      Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.






      share|cite|improve this answer









      $endgroup$



      The right conclusion is $det(A) ne 0$, hence it is invertible.



      Notice that we can't conclude that $det(A)=1$. After all, it can take value $-1$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Mar 18 at 11:32









      Siong Thye GohSiong Thye Goh

      103k1468119




      103k1468119











      • $begingroup$
        Sorry, I'll edit!
        $endgroup$
        – Omojola Micheal
        Mar 18 at 11:33










      • $begingroup$
        Thanks for the prompt reply.
        $endgroup$
        – Omojola Micheal
        Mar 18 at 11:34
















      • $begingroup$
        Sorry, I'll edit!
        $endgroup$
        – Omojola Micheal
        Mar 18 at 11:33










      • $begingroup$
        Thanks for the prompt reply.
        $endgroup$
        – Omojola Micheal
        Mar 18 at 11:34















      $begingroup$
      Sorry, I'll edit!
      $endgroup$
      – Omojola Micheal
      Mar 18 at 11:33




      $begingroup$
      Sorry, I'll edit!
      $endgroup$
      – Omojola Micheal
      Mar 18 at 11:33












      $begingroup$
      Thanks for the prompt reply.
      $endgroup$
      – Omojola Micheal
      Mar 18 at 11:34




      $begingroup$
      Thanks for the prompt reply.
      $endgroup$
      – Omojola Micheal
      Mar 18 at 11:34











      6












      $begingroup$

      I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?






      share|cite|improve this answer









      $endgroup$

















        6












        $begingroup$

        I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?






        share|cite|improve this answer









        $endgroup$















          6












          6








          6





          $begingroup$

          I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?






          share|cite|improve this answer









          $endgroup$



          I think it is easier to see it this way: what is $AA^m-1 = A^m-1A$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 18 at 11:33









          MariahMariah

          2,1431718




          2,1431718





















              2












              $begingroup$

              It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.






                  share|cite|improve this answer









                  $endgroup$



                  It's invertible since it has a inverse, $A^m-1$ that is. By the way, $det(A)$ could also be -1.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 18 at 11:46









                  Sander KortewegSander Korteweg

                  566




                  566



























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