There exists a direct sum decomposition V=W⊕Z into two subspaces, with T the projection from V onto W along...











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Let V be a vector space, and T∈L(V). Show that the following statements (i), (ii) are equivalent:



(i) There exists a direct sum decomposition V=W⊕Z into two subspace, with T the projection from V onto W along Z.



(ii) T∘T=T.



any one can help with it, i have no idea with this question










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  • 1




    One direction is easy. For the other: what is the kernel of $T$? How about the image?
    – user3482749
    Nov 16 at 22:15















up vote
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Let V be a vector space, and T∈L(V). Show that the following statements (i), (ii) are equivalent:



(i) There exists a direct sum decomposition V=W⊕Z into two subspace, with T the projection from V onto W along Z.



(ii) T∘T=T.



any one can help with it, i have no idea with this question










share|cite|improve this question


















  • 1




    One direction is easy. For the other: what is the kernel of $T$? How about the image?
    – user3482749
    Nov 16 at 22:15













up vote
0
down vote

favorite
2









up vote
0
down vote

favorite
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2





Let V be a vector space, and T∈L(V). Show that the following statements (i), (ii) are equivalent:



(i) There exists a direct sum decomposition V=W⊕Z into two subspace, with T the projection from V onto W along Z.



(ii) T∘T=T.



any one can help with it, i have no idea with this question










share|cite|improve this question













Let V be a vector space, and T∈L(V). Show that the following statements (i), (ii) are equivalent:



(i) There exists a direct sum decomposition V=W⊕Z into two subspace, with T the projection from V onto W along Z.



(ii) T∘T=T.



any one can help with it, i have no idea with this question







linear-algebra






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asked Nov 16 at 22:12









DORCT

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406








  • 1




    One direction is easy. For the other: what is the kernel of $T$? How about the image?
    – user3482749
    Nov 16 at 22:15














  • 1




    One direction is easy. For the other: what is the kernel of $T$? How about the image?
    – user3482749
    Nov 16 at 22:15








1




1




One direction is easy. For the other: what is the kernel of $T$? How about the image?
– user3482749
Nov 16 at 22:15




One direction is easy. For the other: what is the kernel of $T$? How about the image?
– user3482749
Nov 16 at 22:15










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










(i) $Longrightarrow$ (ii):



$V = W oplus Z; tag 1$



by definition means that



$V = W + Z, ; W cap Z = {0}; tag 2$



we note that the decomposition of any $v in V$ into



$v = w + z, ; w in W, ; z in Z, tag 3$



is unique, for if



$w_1 + z_1 = w_2 + z_2, tag 4$



then



$W ni w_1 - w_2 = z_2 - z_1 in Z; tag 5$



thus,



$w_1 - w_2 = 0 = z_2 - z_1 Longrightarrow w_1 = w_2, ; z_1 = z_2 tag 6$



as claimed; therefore, since $w$ is unambiguously determined by $v$, we may define a function



$T:V to W, ; T(v) = T(w + z) = w; tag 7$



we investigate the linearity of $T$: if



$v = av_1 + v_2, tag 8$



we may uniquely write



$v_1 = w_1 + z_1, ; v_2 = w_2 + z_2, tag 9$



whence



$v = a(w_1 + z_1) + w_2 + z_2 = (aw_1 + w_2) + (az_1 + z_2) in W + Z, tag{10}$



uniquely; it follows that



$Tv = T(av_1 + v_2) = aw_1 + w_2 = aTv_1 + Tv_2, tag{11}$



establishing the linearity of $T$.



We compute



$T^2(w + z) = T(T(w + z)) = Tw = T(w + z), tag{12}$



whence



$T^2 = T. tag{13}$



(ii) $Longrightarrow$ (i):



$T^2 = T Longrightarrow T(T - I) = T^2 - T = 0; tag{14}$



set



$W = T(V); tag{15}$



then, via (14):



$w in W Longrightarrow exists v in V, ; w = Tv; ; Tw = T^2v = Tv = w; tag{16}$



we see that $T$ fixes $w$ pointwise; it acts as the identity on the subspace $W$. We may also set



$Z = (I - T)V; tag{17}$



then, again by (14),



$z in Z Longrightarrow exists v in V, z = (I - T)v; ; Tz = T(I -T)v = 0 Longrightarrow z in ker T; tag{18}$



likewise,



$z in ker T Longrightarrow Tz = 0 Longrightarrow z = Iz - Tz = (I - T)z Longrightarrow z in (I - T)V = Z; tag{19}$



thus,



$Z = ker T; tag{20}$



now if



$y in Z cap W, tag{21}$



we have



$y = Tv, ; v in V; tag{22}$



$Ty = 0; tag{23}$



therefore, again invoking (14),



$y = Tv = T^2v = T(Tv) = Ty = 0; tag{24}$



we have then shown that



$Z cap W = {0}; tag{25}$



finally, for $v in V$,



$v = Iv - Tv + Tv = (I - T)v + Tv in Z + W; tag{26}$



(25) and (26) show that



$V = W oplus Z; tag{27}$



(16) and (20) show that $T$ is a projection onto $W$ "along $Z$". Thus we have (ii) $Longrightarrow$ (i).






share|cite|improve this answer






























    up vote
    1
    down vote













    The implication (i) $Rightarrow$ (ii) follows from the definition.



    The implication (ii) $Rightarrow$ (i) is not hard to verify once you know that $W=operatorname{im}T$ and $Z=operatorname{ker}T$.






    share|cite|improve this answer





















    • what is im T?..
      – DORCT
      Nov 16 at 23:31










    • The image of $T$. You could also write it as $T(V)$.
      – Servaes
      Nov 16 at 23:31











    Your Answer





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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    (i) $Longrightarrow$ (ii):



    $V = W oplus Z; tag 1$



    by definition means that



    $V = W + Z, ; W cap Z = {0}; tag 2$



    we note that the decomposition of any $v in V$ into



    $v = w + z, ; w in W, ; z in Z, tag 3$



    is unique, for if



    $w_1 + z_1 = w_2 + z_2, tag 4$



    then



    $W ni w_1 - w_2 = z_2 - z_1 in Z; tag 5$



    thus,



    $w_1 - w_2 = 0 = z_2 - z_1 Longrightarrow w_1 = w_2, ; z_1 = z_2 tag 6$



    as claimed; therefore, since $w$ is unambiguously determined by $v$, we may define a function



    $T:V to W, ; T(v) = T(w + z) = w; tag 7$



    we investigate the linearity of $T$: if



    $v = av_1 + v_2, tag 8$



    we may uniquely write



    $v_1 = w_1 + z_1, ; v_2 = w_2 + z_2, tag 9$



    whence



    $v = a(w_1 + z_1) + w_2 + z_2 = (aw_1 + w_2) + (az_1 + z_2) in W + Z, tag{10}$



    uniquely; it follows that



    $Tv = T(av_1 + v_2) = aw_1 + w_2 = aTv_1 + Tv_2, tag{11}$



    establishing the linearity of $T$.



    We compute



    $T^2(w + z) = T(T(w + z)) = Tw = T(w + z), tag{12}$



    whence



    $T^2 = T. tag{13}$



    (ii) $Longrightarrow$ (i):



    $T^2 = T Longrightarrow T(T - I) = T^2 - T = 0; tag{14}$



    set



    $W = T(V); tag{15}$



    then, via (14):



    $w in W Longrightarrow exists v in V, ; w = Tv; ; Tw = T^2v = Tv = w; tag{16}$



    we see that $T$ fixes $w$ pointwise; it acts as the identity on the subspace $W$. We may also set



    $Z = (I - T)V; tag{17}$



    then, again by (14),



    $z in Z Longrightarrow exists v in V, z = (I - T)v; ; Tz = T(I -T)v = 0 Longrightarrow z in ker T; tag{18}$



    likewise,



    $z in ker T Longrightarrow Tz = 0 Longrightarrow z = Iz - Tz = (I - T)z Longrightarrow z in (I - T)V = Z; tag{19}$



    thus,



    $Z = ker T; tag{20}$



    now if



    $y in Z cap W, tag{21}$



    we have



    $y = Tv, ; v in V; tag{22}$



    $Ty = 0; tag{23}$



    therefore, again invoking (14),



    $y = Tv = T^2v = T(Tv) = Ty = 0; tag{24}$



    we have then shown that



    $Z cap W = {0}; tag{25}$



    finally, for $v in V$,



    $v = Iv - Tv + Tv = (I - T)v + Tv in Z + W; tag{26}$



    (25) and (26) show that



    $V = W oplus Z; tag{27}$



    (16) and (20) show that $T$ is a projection onto $W$ "along $Z$". Thus we have (ii) $Longrightarrow$ (i).






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      (i) $Longrightarrow$ (ii):



      $V = W oplus Z; tag 1$



      by definition means that



      $V = W + Z, ; W cap Z = {0}; tag 2$



      we note that the decomposition of any $v in V$ into



      $v = w + z, ; w in W, ; z in Z, tag 3$



      is unique, for if



      $w_1 + z_1 = w_2 + z_2, tag 4$



      then



      $W ni w_1 - w_2 = z_2 - z_1 in Z; tag 5$



      thus,



      $w_1 - w_2 = 0 = z_2 - z_1 Longrightarrow w_1 = w_2, ; z_1 = z_2 tag 6$



      as claimed; therefore, since $w$ is unambiguously determined by $v$, we may define a function



      $T:V to W, ; T(v) = T(w + z) = w; tag 7$



      we investigate the linearity of $T$: if



      $v = av_1 + v_2, tag 8$



      we may uniquely write



      $v_1 = w_1 + z_1, ; v_2 = w_2 + z_2, tag 9$



      whence



      $v = a(w_1 + z_1) + w_2 + z_2 = (aw_1 + w_2) + (az_1 + z_2) in W + Z, tag{10}$



      uniquely; it follows that



      $Tv = T(av_1 + v_2) = aw_1 + w_2 = aTv_1 + Tv_2, tag{11}$



      establishing the linearity of $T$.



      We compute



      $T^2(w + z) = T(T(w + z)) = Tw = T(w + z), tag{12}$



      whence



      $T^2 = T. tag{13}$



      (ii) $Longrightarrow$ (i):



      $T^2 = T Longrightarrow T(T - I) = T^2 - T = 0; tag{14}$



      set



      $W = T(V); tag{15}$



      then, via (14):



      $w in W Longrightarrow exists v in V, ; w = Tv; ; Tw = T^2v = Tv = w; tag{16}$



      we see that $T$ fixes $w$ pointwise; it acts as the identity on the subspace $W$. We may also set



      $Z = (I - T)V; tag{17}$



      then, again by (14),



      $z in Z Longrightarrow exists v in V, z = (I - T)v; ; Tz = T(I -T)v = 0 Longrightarrow z in ker T; tag{18}$



      likewise,



      $z in ker T Longrightarrow Tz = 0 Longrightarrow z = Iz - Tz = (I - T)z Longrightarrow z in (I - T)V = Z; tag{19}$



      thus,



      $Z = ker T; tag{20}$



      now if



      $y in Z cap W, tag{21}$



      we have



      $y = Tv, ; v in V; tag{22}$



      $Ty = 0; tag{23}$



      therefore, again invoking (14),



      $y = Tv = T^2v = T(Tv) = Ty = 0; tag{24}$



      we have then shown that



      $Z cap W = {0}; tag{25}$



      finally, for $v in V$,



      $v = Iv - Tv + Tv = (I - T)v + Tv in Z + W; tag{26}$



      (25) and (26) show that



      $V = W oplus Z; tag{27}$



      (16) and (20) show that $T$ is a projection onto $W$ "along $Z$". Thus we have (ii) $Longrightarrow$ (i).






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        (i) $Longrightarrow$ (ii):



        $V = W oplus Z; tag 1$



        by definition means that



        $V = W + Z, ; W cap Z = {0}; tag 2$



        we note that the decomposition of any $v in V$ into



        $v = w + z, ; w in W, ; z in Z, tag 3$



        is unique, for if



        $w_1 + z_1 = w_2 + z_2, tag 4$



        then



        $W ni w_1 - w_2 = z_2 - z_1 in Z; tag 5$



        thus,



        $w_1 - w_2 = 0 = z_2 - z_1 Longrightarrow w_1 = w_2, ; z_1 = z_2 tag 6$



        as claimed; therefore, since $w$ is unambiguously determined by $v$, we may define a function



        $T:V to W, ; T(v) = T(w + z) = w; tag 7$



        we investigate the linearity of $T$: if



        $v = av_1 + v_2, tag 8$



        we may uniquely write



        $v_1 = w_1 + z_1, ; v_2 = w_2 + z_2, tag 9$



        whence



        $v = a(w_1 + z_1) + w_2 + z_2 = (aw_1 + w_2) + (az_1 + z_2) in W + Z, tag{10}$



        uniquely; it follows that



        $Tv = T(av_1 + v_2) = aw_1 + w_2 = aTv_1 + Tv_2, tag{11}$



        establishing the linearity of $T$.



        We compute



        $T^2(w + z) = T(T(w + z)) = Tw = T(w + z), tag{12}$



        whence



        $T^2 = T. tag{13}$



        (ii) $Longrightarrow$ (i):



        $T^2 = T Longrightarrow T(T - I) = T^2 - T = 0; tag{14}$



        set



        $W = T(V); tag{15}$



        then, via (14):



        $w in W Longrightarrow exists v in V, ; w = Tv; ; Tw = T^2v = Tv = w; tag{16}$



        we see that $T$ fixes $w$ pointwise; it acts as the identity on the subspace $W$. We may also set



        $Z = (I - T)V; tag{17}$



        then, again by (14),



        $z in Z Longrightarrow exists v in V, z = (I - T)v; ; Tz = T(I -T)v = 0 Longrightarrow z in ker T; tag{18}$



        likewise,



        $z in ker T Longrightarrow Tz = 0 Longrightarrow z = Iz - Tz = (I - T)z Longrightarrow z in (I - T)V = Z; tag{19}$



        thus,



        $Z = ker T; tag{20}$



        now if



        $y in Z cap W, tag{21}$



        we have



        $y = Tv, ; v in V; tag{22}$



        $Ty = 0; tag{23}$



        therefore, again invoking (14),



        $y = Tv = T^2v = T(Tv) = Ty = 0; tag{24}$



        we have then shown that



        $Z cap W = {0}; tag{25}$



        finally, for $v in V$,



        $v = Iv - Tv + Tv = (I - T)v + Tv in Z + W; tag{26}$



        (25) and (26) show that



        $V = W oplus Z; tag{27}$



        (16) and (20) show that $T$ is a projection onto $W$ "along $Z$". Thus we have (ii) $Longrightarrow$ (i).






        share|cite|improve this answer














        (i) $Longrightarrow$ (ii):



        $V = W oplus Z; tag 1$



        by definition means that



        $V = W + Z, ; W cap Z = {0}; tag 2$



        we note that the decomposition of any $v in V$ into



        $v = w + z, ; w in W, ; z in Z, tag 3$



        is unique, for if



        $w_1 + z_1 = w_2 + z_2, tag 4$



        then



        $W ni w_1 - w_2 = z_2 - z_1 in Z; tag 5$



        thus,



        $w_1 - w_2 = 0 = z_2 - z_1 Longrightarrow w_1 = w_2, ; z_1 = z_2 tag 6$



        as claimed; therefore, since $w$ is unambiguously determined by $v$, we may define a function



        $T:V to W, ; T(v) = T(w + z) = w; tag 7$



        we investigate the linearity of $T$: if



        $v = av_1 + v_2, tag 8$



        we may uniquely write



        $v_1 = w_1 + z_1, ; v_2 = w_2 + z_2, tag 9$



        whence



        $v = a(w_1 + z_1) + w_2 + z_2 = (aw_1 + w_2) + (az_1 + z_2) in W + Z, tag{10}$



        uniquely; it follows that



        $Tv = T(av_1 + v_2) = aw_1 + w_2 = aTv_1 + Tv_2, tag{11}$



        establishing the linearity of $T$.



        We compute



        $T^2(w + z) = T(T(w + z)) = Tw = T(w + z), tag{12}$



        whence



        $T^2 = T. tag{13}$



        (ii) $Longrightarrow$ (i):



        $T^2 = T Longrightarrow T(T - I) = T^2 - T = 0; tag{14}$



        set



        $W = T(V); tag{15}$



        then, via (14):



        $w in W Longrightarrow exists v in V, ; w = Tv; ; Tw = T^2v = Tv = w; tag{16}$



        we see that $T$ fixes $w$ pointwise; it acts as the identity on the subspace $W$. We may also set



        $Z = (I - T)V; tag{17}$



        then, again by (14),



        $z in Z Longrightarrow exists v in V, z = (I - T)v; ; Tz = T(I -T)v = 0 Longrightarrow z in ker T; tag{18}$



        likewise,



        $z in ker T Longrightarrow Tz = 0 Longrightarrow z = Iz - Tz = (I - T)z Longrightarrow z in (I - T)V = Z; tag{19}$



        thus,



        $Z = ker T; tag{20}$



        now if



        $y in Z cap W, tag{21}$



        we have



        $y = Tv, ; v in V; tag{22}$



        $Ty = 0; tag{23}$



        therefore, again invoking (14),



        $y = Tv = T^2v = T(Tv) = Ty = 0; tag{24}$



        we have then shown that



        $Z cap W = {0}; tag{25}$



        finally, for $v in V$,



        $v = Iv - Tv + Tv = (I - T)v + Tv in Z + W; tag{26}$



        (25) and (26) show that



        $V = W oplus Z; tag{27}$



        (16) and (20) show that $T$ is a projection onto $W$ "along $Z$". Thus we have (ii) $Longrightarrow$ (i).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 17 at 23:10

























        answered Nov 17 at 20:40









        Robert Lewis

        42k22760




        42k22760






















            up vote
            1
            down vote













            The implication (i) $Rightarrow$ (ii) follows from the definition.



            The implication (ii) $Rightarrow$ (i) is not hard to verify once you know that $W=operatorname{im}T$ and $Z=operatorname{ker}T$.






            share|cite|improve this answer





















            • what is im T?..
              – DORCT
              Nov 16 at 23:31










            • The image of $T$. You could also write it as $T(V)$.
              – Servaes
              Nov 16 at 23:31















            up vote
            1
            down vote













            The implication (i) $Rightarrow$ (ii) follows from the definition.



            The implication (ii) $Rightarrow$ (i) is not hard to verify once you know that $W=operatorname{im}T$ and $Z=operatorname{ker}T$.






            share|cite|improve this answer





















            • what is im T?..
              – DORCT
              Nov 16 at 23:31










            • The image of $T$. You could also write it as $T(V)$.
              – Servaes
              Nov 16 at 23:31













            up vote
            1
            down vote










            up vote
            1
            down vote









            The implication (i) $Rightarrow$ (ii) follows from the definition.



            The implication (ii) $Rightarrow$ (i) is not hard to verify once you know that $W=operatorname{im}T$ and $Z=operatorname{ker}T$.






            share|cite|improve this answer












            The implication (i) $Rightarrow$ (ii) follows from the definition.



            The implication (ii) $Rightarrow$ (i) is not hard to verify once you know that $W=operatorname{im}T$ and $Z=operatorname{ker}T$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 16 at 23:17









            Servaes

            21.5k33792




            21.5k33792












            • what is im T?..
              – DORCT
              Nov 16 at 23:31










            • The image of $T$. You could also write it as $T(V)$.
              – Servaes
              Nov 16 at 23:31


















            • what is im T?..
              – DORCT
              Nov 16 at 23:31










            • The image of $T$. You could also write it as $T(V)$.
              – Servaes
              Nov 16 at 23:31
















            what is im T?..
            – DORCT
            Nov 16 at 23:31




            what is im T?..
            – DORCT
            Nov 16 at 23:31












            The image of $T$. You could also write it as $T(V)$.
            – Servaes
            Nov 16 at 23:31




            The image of $T$. You could also write it as $T(V)$.
            – Servaes
            Nov 16 at 23:31


















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