How can I prove that a linear map $T:Vto V$ such that $T^2 = I$ is equal to the Identity map?












2












$begingroup$


Suppose $T in mathcal{L}(V)$, $T^2=I$, and $-1$ is not an eigenvalue of $T$. Prove that $T=I$.



I'm not sure where to start, though I know I somehow have to show that $T circ T$ yields the identity matrix, but there isn't much information given.






linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations







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  • $begingroup$
    I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 3 '18 at 23:07
















2












$begingroup$


Suppose $T in mathcal{L}(V)$, $T^2=I$, and $-1$ is not an eigenvalue of $T$. Prove that $T=I$.



I'm not sure where to start, though I know I somehow have to show that $T circ T$ yields the identity matrix, but there isn't much information given.






linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 3 '18 at 23:07














2












2








2


1



$begingroup$


Suppose $T in mathcal{L}(V)$, $T^2=I$, and $-1$ is not an eigenvalue of $T$. Prove that $T=I$.



I'm not sure where to start, though I know I somehow have to show that $T circ T$ yields the identity matrix, but there isn't much information given.






linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations







share|cite|improve this question











$endgroup$




Suppose $T in mathcal{L}(V)$, $T^2=I$, and $-1$ is not an eigenvalue of $T$. Prove that $T=I$.



I'm not sure where to start, though I know I somehow have to show that $T circ T$ yields the identity matrix, but there isn't much information given.






linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations




linear-algebra matrices vector-spaces eigenvalues-eigenvectors linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 23:46









Batominovski

1




1










asked Dec 3 '18 at 22:34









JaigusJaigus

2218




2218












  • $begingroup$
    I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 3 '18 at 23:07


















  • $begingroup$
    I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
    $endgroup$
    – Robert Lewis
    Dec 3 '18 at 23:07
















$begingroup$
I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
$endgroup$
– Robert Lewis
Dec 3 '18 at 23:07




$begingroup$
I eidited the title of your post to make it better conform to the problem presented in the body. Cheers!
$endgroup$
– Robert Lewis
Dec 3 '18 at 23:07










3 Answers
3






active

oldest

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4












$begingroup$

Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).



... or this:



Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
    $endgroup$
    – Jaigus
    Dec 3 '18 at 23:04










  • $begingroup$
    @Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
    $endgroup$
    – Omnomnomnom
    Dec 4 '18 at 1:05



















2












$begingroup$

Note that



$(T + I)(T - I) = T^2 - I; tag 1$



since



$T^2 = I, tag 2$



$T^2 - I = 0; tag 3$



now suppose



$T ne I; tag 4$



then



$exists ; x in v, ; Tx ne Ix = x; tag 5$



therefore,



$w = (T - I) x = Tx - x ne 0; tag 6$



thus, by (1) and (3),



$(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$



but



$(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$



which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence



$forall x in v, ; Tx = Ix = x, tag 9$



as was to be proved.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.




    Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
    From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.







    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









      4












      $begingroup$

      Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).



      ... or this:



      Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
        $endgroup$
        – Jaigus
        Dec 3 '18 at 23:04










      • $begingroup$
        @Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
        $endgroup$
        – Omnomnomnom
        Dec 4 '18 at 1:05
















      4












      $begingroup$

      Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).



      ... or this:



      Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
        $endgroup$
        – Jaigus
        Dec 3 '18 at 23:04










      • $begingroup$
        @Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
        $endgroup$
        – Omnomnomnom
        Dec 4 '18 at 1:05














      4












      4








      4





      $begingroup$

      Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).



      ... or this:



      Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).






      share|cite|improve this answer











      $endgroup$



      Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).



      ... or this:



      Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 4 '18 at 10:11

























      answered Dec 3 '18 at 22:36









      lhflhf

      163k10168392




      163k10168392












      • $begingroup$
        I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
        $endgroup$
        – Jaigus
        Dec 3 '18 at 23:04










      • $begingroup$
        @Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
        $endgroup$
        – Omnomnomnom
        Dec 4 '18 at 1:05


















      • $begingroup$
        I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
        $endgroup$
        – Jaigus
        Dec 3 '18 at 23:04










      • $begingroup$
        @Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
        $endgroup$
        – Omnomnomnom
        Dec 4 '18 at 1:05
















      $begingroup$
      I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
      $endgroup$
      – Jaigus
      Dec 3 '18 at 23:04




      $begingroup$
      I understand the expansion but, I'm not sure how to reason that $T+I$ is invertible. Since it's given that -1 is not an eigenvalue for T, is $T=I$ the only solution for $(T-I)(T+I)$ because of that?
      $endgroup$
      – Jaigus
      Dec 3 '18 at 23:04












      $begingroup$
      @Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
      $endgroup$
      – Omnomnomnom
      Dec 4 '18 at 1:05




      $begingroup$
      @Jaigus not quite. Because $-1$ is not an eigenvalue, the only solution for $(T+I)x = 0$ is $x = 0$. That is, the matrix $T+I$ has a trivial kernel, which is to say that $T+I$ is invertible.
      $endgroup$
      – Omnomnomnom
      Dec 4 '18 at 1:05











      2












      $begingroup$

      Note that



      $(T + I)(T - I) = T^2 - I; tag 1$



      since



      $T^2 = I, tag 2$



      $T^2 - I = 0; tag 3$



      now suppose



      $T ne I; tag 4$



      then



      $exists ; x in v, ; Tx ne Ix = x; tag 5$



      therefore,



      $w = (T - I) x = Tx - x ne 0; tag 6$



      thus, by (1) and (3),



      $(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$



      but



      $(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$



      which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence



      $forall x in v, ; Tx = Ix = x, tag 9$



      as was to be proved.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Note that



        $(T + I)(T - I) = T^2 - I; tag 1$



        since



        $T^2 = I, tag 2$



        $T^2 - I = 0; tag 3$



        now suppose



        $T ne I; tag 4$



        then



        $exists ; x in v, ; Tx ne Ix = x; tag 5$



        therefore,



        $w = (T - I) x = Tx - x ne 0; tag 6$



        thus, by (1) and (3),



        $(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$



        but



        $(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$



        which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence



        $forall x in v, ; Tx = Ix = x, tag 9$



        as was to be proved.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Note that



          $(T + I)(T - I) = T^2 - I; tag 1$



          since



          $T^2 = I, tag 2$



          $T^2 - I = 0; tag 3$



          now suppose



          $T ne I; tag 4$



          then



          $exists ; x in v, ; Tx ne Ix = x; tag 5$



          therefore,



          $w = (T - I) x = Tx - x ne 0; tag 6$



          thus, by (1) and (3),



          $(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$



          but



          $(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$



          which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence



          $forall x in v, ; Tx = Ix = x, tag 9$



          as was to be proved.






          share|cite|improve this answer











          $endgroup$



          Note that



          $(T + I)(T - I) = T^2 - I; tag 1$



          since



          $T^2 = I, tag 2$



          $T^2 - I = 0; tag 3$



          now suppose



          $T ne I; tag 4$



          then



          $exists ; x in v, ; Tx ne Ix = x; tag 5$



          therefore,



          $w = (T - I) x = Tx - x ne 0; tag 6$



          thus, by (1) and (3),



          $(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; tag 7$



          but



          $(T + I)w = 0 Longrightarrow Tw + w = 0 Longrightarrow Tw = -w, tag 8$



          which asserts that $w ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence



          $forall x in v, ; Tx = Ix = x, tag 9$



          as was to be proved.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 23:16

























          answered Dec 3 '18 at 23:01









          Robert LewisRobert Lewis

          44.7k22964




          44.7k22964























              1












              $begingroup$

              Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.




              Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
              From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.







              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.




                Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
                From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.







                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.




                  Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
                  From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.







                  share|cite|improve this answer











                  $endgroup$



                  Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=text{im}(T+I)oplus ker(T+I),.$$ Then, prove that $text{im}(T-I)=ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $ker(T+I)=0$. Therefore, $text{im}(T-I)=0$, proving that $T=I$.




                  Observe that, for any $vin V$, $$v=frac{1}{2}(T+I)(v)-frac{1}{2}(T-I)(v),,tag{*}$$ with $frac{1}{2}(T+I)(v)in text{im}(T+I)$ and $-frac12(T-I)(v)in text{im}(T-I)subseteq ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=text{im}(T+I)+ker(T+I),.tag{#}$$ If $vin text{im}(T+I)cap ker(T+I)$, then $v=(T+I)(u)$ for some $uin V$, so $$(T+I)^2(u)=(T+I)(v)=0,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=frac{1}{2}big((T+I)^2-(T^2-I)big)(u)=0,.$$ Thus, the sum (#) is direct.
                  From (*), we can also see that $V=text{im}(T-I)+text{im}(T+I)$ and $text{im}(T-I)subseteq ker(T+I)$. This implies $text{im}(T-I)=ker(T+I)$.








                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 4 '18 at 0:03

























                  answered Dec 3 '18 at 23:37









                  BatominovskiBatominovski

                  1




                  1






























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