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
linear-algebra
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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
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 |
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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
42k22760
42k22760
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
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
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 |
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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