Prove or disprove: if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible












2












$begingroup$



if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible




I really can't figure it out. I know it's true but don't know how to prove it






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Prove is the verb, proof is the noun.
    $endgroup$
    – Pedro Tamaroff♦
    Nov 11 '13 at 4:17










  • $begingroup$
    I changed the quadratic equation, didn't check the grammar, but you are right though.
    $endgroup$
    – imranfat
    Nov 11 '13 at 4:20






  • 1




    $begingroup$
    "I know it's true..." how do you know this? You may want to be less certain of that.
    $endgroup$
    – alex.jordan
    Nov 11 '13 at 8:20


















2












$begingroup$



if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible




I really can't figure it out. I know it's true but don't know how to prove it






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Prove is the verb, proof is the noun.
    $endgroup$
    – Pedro Tamaroff♦
    Nov 11 '13 at 4:17










  • $begingroup$
    I changed the quadratic equation, didn't check the grammar, but you are right though.
    $endgroup$
    – imranfat
    Nov 11 '13 at 4:20






  • 1




    $begingroup$
    "I know it's true..." how do you know this? You may want to be less certain of that.
    $endgroup$
    – alex.jordan
    Nov 11 '13 at 8:20
















2












2








2





$begingroup$



if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible




I really can't figure it out. I know it's true but don't know how to prove it






linear-algebra matrices







share|cite|improve this question











$endgroup$





if $A$ is nonzero $2 times 2$ matrix such that $A^2+A=0$, then A is invertible




I really can't figure it out. I know it's true but don't know how to prove it






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 11 '13 at 4:22







user61527

















asked Nov 11 '13 at 4:10









12321232

111




111












  • $begingroup$
    Prove is the verb, proof is the noun.
    $endgroup$
    – Pedro Tamaroff♦
    Nov 11 '13 at 4:17










  • $begingroup$
    I changed the quadratic equation, didn't check the grammar, but you are right though.
    $endgroup$
    – imranfat
    Nov 11 '13 at 4:20






  • 1




    $begingroup$
    "I know it's true..." how do you know this? You may want to be less certain of that.
    $endgroup$
    – alex.jordan
    Nov 11 '13 at 8:20




















  • $begingroup$
    Prove is the verb, proof is the noun.
    $endgroup$
    – Pedro Tamaroff♦
    Nov 11 '13 at 4:17










  • $begingroup$
    I changed the quadratic equation, didn't check the grammar, but you are right though.
    $endgroup$
    – imranfat
    Nov 11 '13 at 4:20






  • 1




    $begingroup$
    "I know it's true..." how do you know this? You may want to be less certain of that.
    $endgroup$
    – alex.jordan
    Nov 11 '13 at 8:20


















$begingroup$
Prove is the verb, proof is the noun.
$endgroup$
– Pedro Tamaroff♦
Nov 11 '13 at 4:17




$begingroup$
Prove is the verb, proof is the noun.
$endgroup$
– Pedro Tamaroff♦
Nov 11 '13 at 4:17












$begingroup$
I changed the quadratic equation, didn't check the grammar, but you are right though.
$endgroup$
– imranfat
Nov 11 '13 at 4:20




$begingroup$
I changed the quadratic equation, didn't check the grammar, but you are right though.
$endgroup$
– imranfat
Nov 11 '13 at 4:20




1




1




$begingroup$
"I know it's true..." how do you know this? You may want to be less certain of that.
$endgroup$
– alex.jordan
Nov 11 '13 at 8:20






$begingroup$
"I know it's true..." how do you know this? You may want to be less certain of that.
$endgroup$
– alex.jordan
Nov 11 '13 at 8:20












4 Answers
4






active

oldest

votes


















2












$begingroup$

If $A$ had an inverse, then



$$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$



So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yea just the inverse of the identity, can you clarify a little more plz
    $endgroup$
    – 1232
    Nov 11 '13 at 4:19










  • $begingroup$
    @1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
    $endgroup$
    – user61527
    Nov 11 '13 at 4:21










  • $begingroup$
    no i cannot find one
    $endgroup$
    – 1232
    Nov 11 '13 at 4:26






  • 4




    $begingroup$
    @1232 Look harder.
    $endgroup$
    – Will Nelson
    Nov 11 '13 at 7:51



















2












$begingroup$

Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    This is not true. For example
    $$
    A=begin{pmatrix}
    -1&0\
    0&0
    end{pmatrix}
    $$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      The answer becomes evident, when you view it in terms of the minimal polynomial.



      Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0



      Since A is not 0,
      A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
      or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
      Hence its not invertible.
      So either A = -I or A is not invertible.






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        If $A$ had an inverse, then



        $$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$



        So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          yea just the inverse of the identity, can you clarify a little more plz
          $endgroup$
          – 1232
          Nov 11 '13 at 4:19










        • $begingroup$
          @1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
          $endgroup$
          – user61527
          Nov 11 '13 at 4:21










        • $begingroup$
          no i cannot find one
          $endgroup$
          – 1232
          Nov 11 '13 at 4:26






        • 4




          $begingroup$
          @1232 Look harder.
          $endgroup$
          – Will Nelson
          Nov 11 '13 at 7:51
















        2












        $begingroup$

        If $A$ had an inverse, then



        $$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$



        So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          yea just the inverse of the identity, can you clarify a little more plz
          $endgroup$
          – 1232
          Nov 11 '13 at 4:19










        • $begingroup$
          @1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
          $endgroup$
          – user61527
          Nov 11 '13 at 4:21










        • $begingroup$
          no i cannot find one
          $endgroup$
          – 1232
          Nov 11 '13 at 4:26






        • 4




          $begingroup$
          @1232 Look harder.
          $endgroup$
          – Will Nelson
          Nov 11 '13 at 7:51














        2












        2








        2





        $begingroup$

        If $A$ had an inverse, then



        $$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$



        So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?






        share|cite|improve this answer









        $endgroup$



        If $A$ had an inverse, then



        $$I = A A^{-1} = -A^2 A^{-1} = -A^{-1}$$



        So we conclude that $A^{-1} = -I$, implying that $A = -I$. Can you think of a matrix $A$ satisfying the given condition that is not $-I$?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 11 '13 at 4:13







        user61527



















        • $begingroup$
          yea just the inverse of the identity, can you clarify a little more plz
          $endgroup$
          – 1232
          Nov 11 '13 at 4:19










        • $begingroup$
          @1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
          $endgroup$
          – user61527
          Nov 11 '13 at 4:21










        • $begingroup$
          no i cannot find one
          $endgroup$
          – 1232
          Nov 11 '13 at 4:26






        • 4




          $begingroup$
          @1232 Look harder.
          $endgroup$
          – Will Nelson
          Nov 11 '13 at 7:51


















        • $begingroup$
          yea just the inverse of the identity, can you clarify a little more plz
          $endgroup$
          – 1232
          Nov 11 '13 at 4:19










        • $begingroup$
          @1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
          $endgroup$
          – user61527
          Nov 11 '13 at 4:21










        • $begingroup$
          no i cannot find one
          $endgroup$
          – 1232
          Nov 11 '13 at 4:26






        • 4




          $begingroup$
          @1232 Look harder.
          $endgroup$
          – Will Nelson
          Nov 11 '13 at 7:51
















        $begingroup$
        yea just the inverse of the identity, can you clarify a little more plz
        $endgroup$
        – 1232
        Nov 11 '13 at 4:19




        $begingroup$
        yea just the inverse of the identity, can you clarify a little more plz
        $endgroup$
        – 1232
        Nov 11 '13 at 4:19












        $begingroup$
        @1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
        $endgroup$
        – user61527
        Nov 11 '13 at 4:21




        $begingroup$
        @1232 The problem is reduced to asking whether there is a matrix not equal to $-I$ such that $A^2 + A = 0$. Can you find such a matrix?
        $endgroup$
        – user61527
        Nov 11 '13 at 4:21












        $begingroup$
        no i cannot find one
        $endgroup$
        – 1232
        Nov 11 '13 at 4:26




        $begingroup$
        no i cannot find one
        $endgroup$
        – 1232
        Nov 11 '13 at 4:26




        4




        4




        $begingroup$
        @1232 Look harder.
        $endgroup$
        – Will Nelson
        Nov 11 '13 at 7:51




        $begingroup$
        @1232 Look harder.
        $endgroup$
        – Will Nelson
        Nov 11 '13 at 7:51











        2












        $begingroup$

        Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.






            share|cite|improve this answer









            $endgroup$



            Try a diagonal matrix $A$ that's not invertible and not zero, but for which $A^2+A=0$. It might be very easy to find one.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 11 '13 at 7:05









            Will NelsonWill Nelson

            4,4541023




            4,4541023























                2












                $begingroup$

                This is not true. For example
                $$
                A=begin{pmatrix}
                -1&0\
                0&0
                end{pmatrix}
                $$






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  This is not true. For example
                  $$
                  A=begin{pmatrix}
                  -1&0\
                  0&0
                  end{pmatrix}
                  $$






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    This is not true. For example
                    $$
                    A=begin{pmatrix}
                    -1&0\
                    0&0
                    end{pmatrix}
                    $$






                    share|cite|improve this answer











                    $endgroup$



                    This is not true. For example
                    $$
                    A=begin{pmatrix}
                    -1&0\
                    0&0
                    end{pmatrix}
                    $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 11 '17 at 16:18

























                    answered Jan 11 '17 at 16:05









                    boazboaz

                    2,405614




                    2,405614























                        0












                        $begingroup$

                        The answer becomes evident, when you view it in terms of the minimal polynomial.



                        Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0



                        Since A is not 0,
                        A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
                        or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
                        Hence its not invertible.
                        So either A = -I or A is not invertible.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The answer becomes evident, when you view it in terms of the minimal polynomial.



                          Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0



                          Since A is not 0,
                          A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
                          or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
                          Hence its not invertible.
                          So either A = -I or A is not invertible.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The answer becomes evident, when you view it in terms of the minimal polynomial.



                            Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0



                            Since A is not 0,
                            A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
                            or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
                            Hence its not invertible.
                            So either A = -I or A is not invertible.






                            share|cite|improve this answer









                            $endgroup$



                            The answer becomes evident, when you view it in terms of the minimal polynomial.



                            Now given the above equation A^2 + A = 0, this can also be written as A(A+I) = 0



                            Since A is not 0,
                            A either has A+I = 0, i.e, A = -I. (when A+I is the minimal polynomial of A)
                            or has A(A+I) = 0 as its minimal polynomial, in this case the eigenvalues of A are 0 and 1, which implies the determinant of A = 0.
                            Hence its not invertible.
                            So either A = -I or A is not invertible.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 8 '18 at 19:37









                            manav gaddammanav gaddam

                            11




                            11






























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