1
◦
S =
(a)
(x,y,z) ∈
R
4
x + 2y − 2z + w = 0 2x + 7y + 2z − 2w = 0 x − y − 8z − 3w = 0
S
(b)
U
1
◦
U ∪ S =
R4
3
R
S = {v1 , v2 , v3}
T = { w1 , w2 , w3}
S := { (1, 0, 1);(1, 2, 0);(1, 1, 1)} T := { (6, 3, 3);(4, −1, 3);(5, 5, 2)} [I ]S T
(a)
v = { 4, −9, 5}
(b) (c)
[v ]S
1
◦
[v]T = { 1, 2, −2}
3
R
2
T
R
→
3
R
T (x, y) = (3x − 2y, x + 2y, 2y − x) [T ]αβ
(a)
α
2
β
R
3
R
T (3, 1/2) (b) (c)
α = { (2, 1);(0, 3)}
1
◦
y
β = { (1, 1, 1);(2, 0, 1);(0, 0, 2)}
S =
(a) (b)
(x,y,z) ∈
R
4
x + 2y − 2z + w = 0 2x + 7y + 2z − 2w = 0 x − y − 8z − 3w = 0
S U ∪ S =
U
2 −2 1 0 2 7 2 −2 0 1 −1 −8 −3 0 2 −2 1 0 0 6 −4 0 0 −3 −6 −4 0 2 −2 6 0 0
0 0
1 0 −4 0 0
4
R
F 2 − 2F 1 → F 2 F 1 → F 3
F 2 + F 3 → F 3
w = 0 y = − 2z
x = 6z
z ∈
R 4
R
(x,y,z,w) = (6z, −2z,z, 0) = z · (6, −2, 1, 0)
S = (6, −2, 1, 0)
;
dim(S ) = 1 4
U
R
U ∪ S =
U =
1
◦
1 0 0 0
;
0 1 0 0
4
R
;
0 0 0 1
F 3 −
3
S = {v1 , v2 , v3}
R
3
T = { w1 , w2 , w3}
R
S := { (1, 0, 1);(1, 2, 0);(1, 1, 1)} T := { (6, 3, 3);(4, −1, 3);(5, 5, 2)} [I ]S T
(a) (b) (c)
v = { 4, −9, 5}
[v]T = { 1, 2, −2}
[v ]S
1 1 6 4 5 0 2 1 3 −1 5 1 0 1 3 3 2
1 1 6 4 5 0 2 1 3 −1 5 0 −1 0 −3 −1 −3
0 0
1 1 6 4 5 0 −3 −1 −3 2 1 3 −1 5
0 0
0 1 3 3 2 0 −3 −1 −3 0 1 −3 −3 −1 0
0 0
0
0 0
6 6 3 3 1 3 −3 −3 −1
F 3 − F 1 → F 3
[T ]S T =
F 3 F 2
F 3 + 2F 2 → F 3 F 2 → F 1
−1
6 6 3 3 1 3 −3 −3 −1
F 1 − F 3 → F 1
(4, −9, 5) = a 11(6, 3, 3) + a2,1 (4, −1, 3) + a3,1 (5, 5, 2)
1
◦
6 4 5 4 3 −1 5 −9 3 3 2 5
F 1 +
→
6 4 0 14 0 −6 0 −12 0 0 1 −2
→
1 0 1 1 0 1 0 2 0 0 1 −2
[v]T =
1 2 −2
[v]S = [I ]S T [v]T
6 6 3 3 1 3 −3 −3 −1
[v]S →
12 −1 −7
→
[v]S = 2
T
R
→
3
1 2 −2
12 −1 −7
R
T (x, y) = (3x − 2y, x + 2y, 2y − x) [T ]αβ
(a)
α
2
β
R
3
R
T (3, 1/2) (b) (c)
α = { (2, 1);(0, 3)}
T (1, 0) = (3, 1, −1) [T ]βα =
y
β = { (1, 1, 1);(2, 0, 1);(0, 0, 2)}
T (0, 1) = (−2, 2, 2)
3 −2 1 2 2 −1
[T ]βα =
∗
3 1/2
8 4 −2
T (x, y) = (3x − 2y, x + 2y, 2y − x) = (0, 0) 1
◦
ker (T ) = 0
dim(ker(T )) + dim(Img(T )) = dim(R2 ) dim(Img(T )) = 2 T (x, y) = (3x − 2y, x + 2y, 2y − x) Img(T ) = (3, 1, −1);(−2, 2, 2)
α = { (2, 1);(0, 3)}
[I ]αα =
[T ]βα
β = { (1, 1, 1);(2, 0, 1);(0, 0, 2)}
y
2 0 1 3
[I ]ββ =
1 2 0 1 0 0 1 1 2
= [I ]ββ [T ]βα [I ]αα
=
1 2 0 1 0 0 1 1 2
−1
3 −2 1 2 2 −1
∗
∗
2 0 1 3
[I ]ββ = ([I ]ββ )
[T ]βα =
[I ]ββ
[T ]βα
=
=
−
=
0
1
1 2 1 4
1 2 1 4
− −
1 2 1 −2 − 32 1 [T ]βα
1
◦
−
3 3 0 −3 − 32 3 0
1
1 2 1 4
1 2 1 4
− −
0 0 1 2
0 0
1 2
∗
=
∗
3 −2 1 2 −1 2
2 0 1 3
4 6 0 −6 −2 3
∗
2 0 1 3
−1
