Show that if $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible












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If $B in M_{n times n}(mathbb{R})$ and $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible.




I get that $$mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated.










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    7












    $begingroup$



    If $B in M_{n times n}(mathbb{R})$ and $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible.




    I get that $$mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated.










    share|cite|improve this question











    $endgroup$















      7












      7








      7


      1



      $begingroup$



      If $B in M_{n times n}(mathbb{R})$ and $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible.




      I get that $$mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated.










      share|cite|improve this question











      $endgroup$





      If $B in M_{n times n}(mathbb{R})$ and $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible.




      I get that $$mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated.







      linear-algebra matrices






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      edited Aug 1 '17 at 11:32









      Rodrigo de Azevedo

      13.1k41960




      13.1k41960










      asked Aug 1 '17 at 1:18









      MigosMigos

      1014




      1014






















          3 Answers
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          21












          $begingroup$

          A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.



          If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.



          Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.



          Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.



          Hence, either way it follows that $B$ is not injective.



          VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.






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            If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.



            But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.






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              $begingroup$

              Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.






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                3 Answers
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                3 Answers
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                21












                $begingroup$

                A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.



                If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.



                Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.



                Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.



                Hence, either way it follows that $B$ is not injective.



                VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.






                share|cite|improve this answer











                $endgroup$


















                  21












                  $begingroup$

                  A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.



                  If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.



                  Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.



                  Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.



                  Hence, either way it follows that $B$ is not injective.



                  VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.






                  share|cite|improve this answer











                  $endgroup$
















                    21












                    21








                    21





                    $begingroup$

                    A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.



                    If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.



                    Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.



                    Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.



                    Hence, either way it follows that $B$ is not injective.



                    VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.






                    share|cite|improve this answer











                    $endgroup$



                    A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.



                    If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.



                    Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.



                    Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.



                    Hence, either way it follows that $B$ is not injective.



                    VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 8 at 16:50

























                    answered Aug 1 '17 at 1:26









                    астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

                    40.6k33678




                    40.6k33678























                        13












                        $begingroup$

                        If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.



                        But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.






                        share|cite|improve this answer









                        $endgroup$


















                          13












                          $begingroup$

                          If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.



                          But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.






                          share|cite|improve this answer









                          $endgroup$
















                            13












                            13








                            13





                            $begingroup$

                            If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.



                            But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.






                            share|cite|improve this answer









                            $endgroup$



                            If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.



                            But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 1 '17 at 1:26









                            carmichael561carmichael561

                            47.2k54583




                            47.2k54583























                                9












                                $begingroup$

                                Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.






                                share|cite|improve this answer









                                $endgroup$


















                                  9












                                  $begingroup$

                                  Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    9












                                    9








                                    9





                                    $begingroup$

                                    Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Aug 1 '17 at 4:09









                                    Bob KruegerBob Krueger

                                    4,2002822




                                    4,2002822






























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