No. of distinct matrices satisfying a certain equation [duplicate]

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  • Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?

    3 answers




[TIFR-GS 2011, Part B, problem no. 1] Let $A$ be a $2×2$ matrix with complex entries. The no. of $2×2$ matrices A with complex entries satisfying the equation $A^3=A$ is infinite. They asked if the statement is true or false.




From the given equation, I've found that the eigenvalues of $A$ can be $0$, $1$, and $-1$. Now I took a trial matrix and tried to satisfy the equation, but it doesn't help in any way. Any help would be appreciated.




linear-algebra eigenvalues-eigenvectors




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marked as duplicate by Community♦ 9 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • The identity doesn't have trace $1$ or determinant $0$.
    – Arnaud Mortier
    12 hours ago










  • Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
    – Anik Bhowmick
    12 hours ago










  • Plus, there is the possible eigenvalue $-1$.
    – Arnaud Mortier
    12 hours ago











  • Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
    – Anik Bhowmick
    12 hours ago














up vote
2
down vote

favorite













This question already has an answer here:



  • Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?

    3 answers




[TIFR-GS 2011, Part B, problem no. 1] Let $A$ be a $2×2$ matrix with complex entries. The no. of $2×2$ matrices A with complex entries satisfying the equation $A^3=A$ is infinite. They asked if the statement is true or false.




From the given equation, I've found that the eigenvalues of $A$ can be $0$, $1$, and $-1$. Now I took a trial matrix and tried to satisfy the equation, but it doesn't help in any way. Any help would be appreciated.




linear-algebra eigenvalues-eigenvectors




share|cite|improve this question













marked as duplicate by Community♦ 9 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • The identity doesn't have trace $1$ or determinant $0$.
    – Arnaud Mortier
    12 hours ago










  • Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
    – Anik Bhowmick
    12 hours ago










  • Plus, there is the possible eigenvalue $-1$.
    – Arnaud Mortier
    12 hours ago











  • Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
    – Anik Bhowmick
    12 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite












This question already has an answer here:



  • Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?

    3 answers




[TIFR-GS 2011, Part B, problem no. 1] Let $A$ be a $2×2$ matrix with complex entries. The no. of $2×2$ matrices A with complex entries satisfying the equation $A^3=A$ is infinite. They asked if the statement is true or false.




From the given equation, I've found that the eigenvalues of $A$ can be $0$, $1$, and $-1$. Now I took a trial matrix and tried to satisfy the equation, but it doesn't help in any way. Any help would be appreciated.




linear-algebra eigenvalues-eigenvectors




share|cite|improve this question














This question already has an answer here:



  • Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?

    3 answers




[TIFR-GS 2011, Part B, problem no. 1] Let $A$ be a $2×2$ matrix with complex entries. The no. of $2×2$ matrices A with complex entries satisfying the equation $A^3=A$ is infinite. They asked if the statement is true or false.




From the given equation, I've found that the eigenvalues of $A$ can be $0$, $1$, and $-1$. Now I took a trial matrix and tried to satisfy the equation, but it doesn't help in any way. Any help would be appreciated.





This question already has an answer here:



  • Are there infinitely many $Ain mathbbC^2 times 2$ satisfying $A^3 = A$?

    3 answers





linear-algebra eigenvalues-eigenvectors





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 10 hours ago
























asked 12 hours ago









Anik Bhowmick

849




849




marked as duplicate by Community♦ 9 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Community♦ 9 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • The identity doesn't have trace $1$ or determinant $0$.
    – Arnaud Mortier
    12 hours ago










  • Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
    – Anik Bhowmick
    12 hours ago










  • Plus, there is the possible eigenvalue $-1$.
    – Arnaud Mortier
    12 hours ago











  • Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
    – Anik Bhowmick
    12 hours ago
















  • The identity doesn't have trace $1$ or determinant $0$.
    – Arnaud Mortier
    12 hours ago










  • Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
    – Anik Bhowmick
    12 hours ago










  • Plus, there is the possible eigenvalue $-1$.
    – Arnaud Mortier
    12 hours ago











  • Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
    – Anik Bhowmick
    12 hours ago















The identity doesn't have trace $1$ or determinant $0$.
– Arnaud Mortier
12 hours ago




The identity doesn't have trace $1$ or determinant $0$.
– Arnaud Mortier
12 hours ago












Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
– Anik Bhowmick
12 hours ago




Okay, identity matrix just have the eigenvalue $1$. Thanks for correcting my route.
– Anik Bhowmick
12 hours ago












Plus, there is the possible eigenvalue $-1$.
– Arnaud Mortier
12 hours ago





Plus, there is the possible eigenvalue $-1$.
– Arnaud Mortier
12 hours ago













Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
– Anik Bhowmick
12 hours ago




Yes, sorry, I computer it wrong. $0$, $1$, $-1$ are the roots of characteristic equation.
– Anik Bhowmick
12 hours ago










2 Answers
2






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up vote
4
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For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.






share|cite|improve this answer





















  • (+1) Nice point of view.
    – Arnaud Mortier
    12 hours ago










  • Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
    – Anik Bhowmick
    12 hours ago

















up vote
2
down vote













Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?



For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
is infinite.






share|cite|improve this answer




























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote













    For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.






    share|cite|improve this answer





















    • (+1) Nice point of view.
      – Arnaud Mortier
      12 hours ago










    • Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
      – Anik Bhowmick
      12 hours ago














    up vote
    4
    down vote













    For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.






    share|cite|improve this answer





















    • (+1) Nice point of view.
      – Arnaud Mortier
      12 hours ago










    • Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
      – Anik Bhowmick
      12 hours ago












    up vote
    4
    down vote










    up vote
    4
    down vote









    For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.






    share|cite|improve this answer













    For any basis $x_1,x_2$ of $mathbbC^2$, any linear transformation $A:mathbbC^2tomathbbC^2$ with eigenvectors $x_1$ and $x_2$ and eigenvalues $-1, 0,$ or $1$ satisfies the equation.







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered 12 hours ago









    alphacapture

    1,682319




    1,682319











    • (+1) Nice point of view.
      – Arnaud Mortier
      12 hours ago










    • Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
      – Anik Bhowmick
      12 hours ago
















    • (+1) Nice point of view.
      – Arnaud Mortier
      12 hours ago










    • Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
      – Anik Bhowmick
      12 hours ago















    (+1) Nice point of view.
    – Arnaud Mortier
    12 hours ago




    (+1) Nice point of view.
    – Arnaud Mortier
    12 hours ago












    Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
    – Anik Bhowmick
    12 hours ago




    Got it !! Since any matrix is some sort of linear transformation, it'll happen for all L.T having eigenvalues $0$, $1$, $-1$ !!
    – Anik Bhowmick
    12 hours ago










    up vote
    2
    down vote













    Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
    The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?



    For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
    is infinite.






    share|cite|improve this answer

























      up vote
      2
      down vote













      Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
      The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?



      For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
      is infinite.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
        The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?



        For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
        is infinite.






        share|cite|improve this answer













        Since $X^3-X$ has only roots with multiplicity $1$, such a matrix $A$ has to be diagonalizable. So you are looking at the set $$leftA=Pleft(matrixa&0\0&bright)P^-1mid Ptext is invertible and a,bin-1,0,1right$$
        The three subsets corresponding to $a=b$ have one element each, now what happens when $aneq b$?



        For instance, $$leftleft(matrix1&0\z&1right)left(matrix1&0\0&-1right)left(matrix1&0\-z&1right)mid zinBbb Cright$$
        is infinite.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 12 hours ago









        Arnaud Mortier

        17.6k21757




        17.6k21757












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