There exists a direct sum decomposition V=W⊕Z into two subspaces, with T the projection from V onto W along...
up vote
0
down vote
favorite
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
add a comment |
up vote
0
down vote
favorite
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
1
One direction is easy. For the other: what is the kernel of $T$? How about the image?
– user3482749
Nov 16 at 22:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
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
linear-algebra
asked Nov 16 at 22:12
DORCT
406
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
add a comment |
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
add a comment |
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).
add a comment |
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 $Z=operatorname{ker}T$.
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
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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).
add a comment |
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).
add a comment |
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).
(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).
edited Nov 17 at 23:10
answered Nov 17 at 20:40
Robert Lewis
41.9k22760
41.9k22760
add a comment |
add a comment |
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 $Z=operatorname{ker}T$.
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
add a comment |
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 $Z=operatorname{ker}T$.
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
add a comment |
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 $Z=operatorname{ker}T$.
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 $Z=operatorname{ker}T$.
answered Nov 16 at 23:17
Servaes
20.9k33792
20.9k33792
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
add a comment |
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
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001696%2fthere-exists-a-direct-sum-decomposition-v-w%25e2%258a%2595z-into-two-subspaces-with-t-the-pro%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
One direction is easy. For the other: what is the kernel of $T$? How about the image?
– user3482749
Nov 16 at 22:15