Is it possible to find a $nxn$-matrix $M$ such that $M^2 =- I_n$, where $-I_n$ is the identity matrix?











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Is it possible to find a $nxn$-matrix $M$ such that $M^2 = -I_n$, where $I_n$ is the identity matrix?



For odd $n$, this cannot hold, since $(-1)^n = det(-I_n) = det (M^2) = det(M)^2$ but what can one say about even $n$?






linear-algebra matrices







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  • I think you're looking for $M^2=-I_n$, right?
    – Carl Schildkraut
    Nov 28 at 21:47










  • For the case where $n$ is even, you can start by solving the case for $n=2$. By Block matrix multiplication, it's not difficult to find an example in generals cases. en.wikipedia.org/wiki/Block_matrix
    – 曾靖國
    Nov 28 at 21:53












  • @CarlSchildkraut yeah right.
    – MPB94
    Nov 28 at 22:04






  • 1




    Take any $mathbb{R}$-basis for $mathbb{C}^n$, and then consider what multiplication by $i$ in $mathbb{C}^n$ does to the basis...
    – Joppy
    Nov 28 at 22:18















up vote
1
down vote

favorite
1












Is it possible to find a $nxn$-matrix $M$ such that $M^2 = -I_n$, where $I_n$ is the identity matrix?



For odd $n$, this cannot hold, since $(-1)^n = det(-I_n) = det (M^2) = det(M)^2$ but what can one say about even $n$?






linear-algebra matrices







share|cite|improve this question
























  • I think you're looking for $M^2=-I_n$, right?
    – Carl Schildkraut
    Nov 28 at 21:47










  • For the case where $n$ is even, you can start by solving the case for $n=2$. By Block matrix multiplication, it's not difficult to find an example in generals cases. en.wikipedia.org/wiki/Block_matrix
    – 曾靖國
    Nov 28 at 21:53












  • @CarlSchildkraut yeah right.
    – MPB94
    Nov 28 at 22:04






  • 1




    Take any $mathbb{R}$-basis for $mathbb{C}^n$, and then consider what multiplication by $i$ in $mathbb{C}^n$ does to the basis...
    – Joppy
    Nov 28 at 22:18













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Is it possible to find a $nxn$-matrix $M$ such that $M^2 = -I_n$, where $I_n$ is the identity matrix?



For odd $n$, this cannot hold, since $(-1)^n = det(-I_n) = det (M^2) = det(M)^2$ but what can one say about even $n$?






linear-algebra matrices







share|cite|improve this question















Is it possible to find a $nxn$-matrix $M$ such that $M^2 = -I_n$, where $I_n$ is the identity matrix?



For odd $n$, this cannot hold, since $(-1)^n = det(-I_n) = det (M^2) = det(M)^2$ but what can one say about even $n$?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













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edited Nov 28 at 22:04

























asked Nov 28 at 21:46









MPB94

23816




23816












  • I think you're looking for $M^2=-I_n$, right?
    – Carl Schildkraut
    Nov 28 at 21:47










  • For the case where $n$ is even, you can start by solving the case for $n=2$. By Block matrix multiplication, it's not difficult to find an example in generals cases. en.wikipedia.org/wiki/Block_matrix
    – 曾靖國
    Nov 28 at 21:53












  • @CarlSchildkraut yeah right.
    – MPB94
    Nov 28 at 22:04






  • 1




    Take any $mathbb{R}$-basis for $mathbb{C}^n$, and then consider what multiplication by $i$ in $mathbb{C}^n$ does to the basis...
    – Joppy
    Nov 28 at 22:18


















  • I think you're looking for $M^2=-I_n$, right?
    – Carl Schildkraut
    Nov 28 at 21:47










  • For the case where $n$ is even, you can start by solving the case for $n=2$. By Block matrix multiplication, it's not difficult to find an example in generals cases. en.wikipedia.org/wiki/Block_matrix
    – 曾靖國
    Nov 28 at 21:53












  • @CarlSchildkraut yeah right.
    – MPB94
    Nov 28 at 22:04






  • 1




    Take any $mathbb{R}$-basis for $mathbb{C}^n$, and then consider what multiplication by $i$ in $mathbb{C}^n$ does to the basis...
    – Joppy
    Nov 28 at 22:18
















I think you're looking for $M^2=-I_n$, right?
– Carl Schildkraut
Nov 28 at 21:47




I think you're looking for $M^2=-I_n$, right?
– Carl Schildkraut
Nov 28 at 21:47












For the case where $n$ is even, you can start by solving the case for $n=2$. By Block matrix multiplication, it's not difficult to find an example in generals cases. en.wikipedia.org/wiki/Block_matrix
– 曾靖國
Nov 28 at 21:53






For the case where $n$ is even, you can start by solving the case for $n=2$. By Block matrix multiplication, it's not difficult to find an example in generals cases. en.wikipedia.org/wiki/Block_matrix
– 曾靖國
Nov 28 at 21:53














@CarlSchildkraut yeah right.
– MPB94
Nov 28 at 22:04




@CarlSchildkraut yeah right.
– MPB94
Nov 28 at 22:04




1




1




Take any $mathbb{R}$-basis for $mathbb{C}^n$, and then consider what multiplication by $i$ in $mathbb{C}^n$ does to the basis...
– Joppy
Nov 28 at 22:18




Take any $mathbb{R}$-basis for $mathbb{C}^n$, and then consider what multiplication by $i$ in $mathbb{C}^n$ does to the basis...
– Joppy
Nov 28 at 22:18










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










Consider the $2 times 2$ matrix



$J = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix}; tag 1$



we have



$J^2 = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} = -I_2, tag 2$



as may be readily verified by an easy calculation. We may then construct a $2n times 2n$ matrix $M_0$ by placing $n$ copies of $J$ along the (block) diagonal:



$M_0 = begin{bmatrix} J & 0 & ldots & 0 \ 0 & J & ldots & 0 \ dots & vdots & J & 0 \ 0 & 0 & ldots & Jend{bmatrix}; tag 3$



it is equally easy to see that



$M_0^2 = -I_{2n}; tag 4$



this establishes the existence of a desired $2n times 2n$ matrix. An entire family of such matrices may be had by taking matrices $M$ similar to $M_0$:



$M = SM_0S^{-1}; tag 5$



then



$M^2 = SM_0S^{-1} SM_0S^{-1} = SM_0 M_0S^{-1} = S(-I_{2n}) S^{-1} = -I_{2n}. tag 6$



We see there are many matrices of the requisite form.






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    up vote
    1
    down vote













    $begin{pmatrix}
    1 & 2\
    -1 & -1
    end{pmatrix}$






    share|cite|improve this answer




























      up vote
      0
      down vote













      A classical example is
      $$
      M=left(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)
      $$



      This is a classical example as it allows to mimic (or represent) complex number computations using real arithmetic.
      $$
      z=a+ib leftrightarrow aleft(begin{array}{cc}1 & 0 \ 0 & 1end{array}right) + bleft(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)=left(begin{array}{cc}a & -b \ b & aend{array}right)
      $$






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        Consider the $2 times 2$ matrix



        $J = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix}; tag 1$



        we have



        $J^2 = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} = -I_2, tag 2$



        as may be readily verified by an easy calculation. We may then construct a $2n times 2n$ matrix $M_0$ by placing $n$ copies of $J$ along the (block) diagonal:



        $M_0 = begin{bmatrix} J & 0 & ldots & 0 \ 0 & J & ldots & 0 \ dots & vdots & J & 0 \ 0 & 0 & ldots & Jend{bmatrix}; tag 3$



        it is equally easy to see that



        $M_0^2 = -I_{2n}; tag 4$



        this establishes the existence of a desired $2n times 2n$ matrix. An entire family of such matrices may be had by taking matrices $M$ similar to $M_0$:



        $M = SM_0S^{-1}; tag 5$



        then



        $M^2 = SM_0S^{-1} SM_0S^{-1} = SM_0 M_0S^{-1} = S(-I_{2n}) S^{-1} = -I_{2n}. tag 6$



        We see there are many matrices of the requisite form.






        share|cite|improve this answer

























          up vote
          2
          down vote



          accepted










          Consider the $2 times 2$ matrix



          $J = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix}; tag 1$



          we have



          $J^2 = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} = -I_2, tag 2$



          as may be readily verified by an easy calculation. We may then construct a $2n times 2n$ matrix $M_0$ by placing $n$ copies of $J$ along the (block) diagonal:



          $M_0 = begin{bmatrix} J & 0 & ldots & 0 \ 0 & J & ldots & 0 \ dots & vdots & J & 0 \ 0 & 0 & ldots & Jend{bmatrix}; tag 3$



          it is equally easy to see that



          $M_0^2 = -I_{2n}; tag 4$



          this establishes the existence of a desired $2n times 2n$ matrix. An entire family of such matrices may be had by taking matrices $M$ similar to $M_0$:



          $M = SM_0S^{-1}; tag 5$



          then



          $M^2 = SM_0S^{-1} SM_0S^{-1} = SM_0 M_0S^{-1} = S(-I_{2n}) S^{-1} = -I_{2n}. tag 6$



          We see there are many matrices of the requisite form.






          share|cite|improve this answer























            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Consider the $2 times 2$ matrix



            $J = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix}; tag 1$



            we have



            $J^2 = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} = -I_2, tag 2$



            as may be readily verified by an easy calculation. We may then construct a $2n times 2n$ matrix $M_0$ by placing $n$ copies of $J$ along the (block) diagonal:



            $M_0 = begin{bmatrix} J & 0 & ldots & 0 \ 0 & J & ldots & 0 \ dots & vdots & J & 0 \ 0 & 0 & ldots & Jend{bmatrix}; tag 3$



            it is equally easy to see that



            $M_0^2 = -I_{2n}; tag 4$



            this establishes the existence of a desired $2n times 2n$ matrix. An entire family of such matrices may be had by taking matrices $M$ similar to $M_0$:



            $M = SM_0S^{-1}; tag 5$



            then



            $M^2 = SM_0S^{-1} SM_0S^{-1} = SM_0 M_0S^{-1} = S(-I_{2n}) S^{-1} = -I_{2n}. tag 6$



            We see there are many matrices of the requisite form.






            share|cite|improve this answer












            Consider the $2 times 2$ matrix



            $J = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix}; tag 1$



            we have



            $J^2 = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} = -I_2, tag 2$



            as may be readily verified by an easy calculation. We may then construct a $2n times 2n$ matrix $M_0$ by placing $n$ copies of $J$ along the (block) diagonal:



            $M_0 = begin{bmatrix} J & 0 & ldots & 0 \ 0 & J & ldots & 0 \ dots & vdots & J & 0 \ 0 & 0 & ldots & Jend{bmatrix}; tag 3$



            it is equally easy to see that



            $M_0^2 = -I_{2n}; tag 4$



            this establishes the existence of a desired $2n times 2n$ matrix. An entire family of such matrices may be had by taking matrices $M$ similar to $M_0$:



            $M = SM_0S^{-1}; tag 5$



            then



            $M^2 = SM_0S^{-1} SM_0S^{-1} = SM_0 M_0S^{-1} = S(-I_{2n}) S^{-1} = -I_{2n}. tag 6$



            We see there are many matrices of the requisite form.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 28 at 22:26









            Robert Lewis

            43k22863




            43k22863






















                up vote
                1
                down vote













                $begin{pmatrix}
                1 & 2\
                -1 & -1
                end{pmatrix}$






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  $begin{pmatrix}
                  1 & 2\
                  -1 & -1
                  end{pmatrix}$






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    $begin{pmatrix}
                    1 & 2\
                    -1 & -1
                    end{pmatrix}$






                    share|cite|improve this answer












                    $begin{pmatrix}
                    1 & 2\
                    -1 & -1
                    end{pmatrix}$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 28 at 22:23









                    Fat ninja

                    634




                    634






















                        up vote
                        0
                        down vote













                        A classical example is
                        $$
                        M=left(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)
                        $$



                        This is a classical example as it allows to mimic (or represent) complex number computations using real arithmetic.
                        $$
                        z=a+ib leftrightarrow aleft(begin{array}{cc}1 & 0 \ 0 & 1end{array}right) + bleft(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)=left(begin{array}{cc}a & -b \ b & aend{array}right)
                        $$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          A classical example is
                          $$
                          M=left(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)
                          $$



                          This is a classical example as it allows to mimic (or represent) complex number computations using real arithmetic.
                          $$
                          z=a+ib leftrightarrow aleft(begin{array}{cc}1 & 0 \ 0 & 1end{array}right) + bleft(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)=left(begin{array}{cc}a & -b \ b & aend{array}right)
                          $$






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            A classical example is
                            $$
                            M=left(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)
                            $$



                            This is a classical example as it allows to mimic (or represent) complex number computations using real arithmetic.
                            $$
                            z=a+ib leftrightarrow aleft(begin{array}{cc}1 & 0 \ 0 & 1end{array}right) + bleft(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)=left(begin{array}{cc}a & -b \ b & aend{array}right)
                            $$






                            share|cite|improve this answer












                            A classical example is
                            $$
                            M=left(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)
                            $$



                            This is a classical example as it allows to mimic (or represent) complex number computations using real arithmetic.
                            $$
                            z=a+ib leftrightarrow aleft(begin{array}{cc}1 & 0 \ 0 & 1end{array}right) + bleft(begin{array}{cc}0 & -1 \ 1 & 0end{array}right)=left(begin{array}{cc}a & -b \ b & aend{array}right)
                            $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 28 at 22:25









                            Picaud Vincent

                            1,10036




                            1,10036






























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