正規直交化法 問題 1 解答
・ 正規直交化法 用 , 次 基底 正規直交基底 作 .
(1) a1 =
1 0 0
, a2 =
1 2
−1
, a3 =
−1 1 2
[解]: 与 基底 直交基底 .
b1 =a1 =
1 0 0
,
b2 =a2− (a2, b1) (b1, b1)b1 =
1 2
−1
−
1 0 0
=
0 2
−1
,
b3 =a3− (a3, b1)
(b1, b1)b1− (a3, b2) (b2, b2)b2 =
−1 1 2
−
−1 0 0
−
0 0 0
=
0 1 2
.
直交基底 b1, b2, b3 正規化 行 正規直交基底 求 . ,求 正規直交基底c1, c2, c3 ,
c1 = 1
1 0 0
=
1 0 0
, c2 =
√5 5
0 2
−1
=
0
2√ 5
−5√55
, c3 =
√5 5
0 1 2
=
√0
5 5 2√
5 5
.
(2) a1 =
0 1 1
, a2 =
1 0 1
, a3 =
1 1 0
[解]: 与 基底 直交基底 .
b1 =a1 =
0 1 1
,
b2 =a2−(a2, b1) (b1, b1)b1 =
1 0 1
−
0
1 2 1 2
=
1
−12
1 2
,
b3 =a3−(a3, b1)
(b1, b1)b1− (a3, b2) (b2, b2)b2 =
1 1 0
−
0
1 2 1 2
−
1
−316
1 6
=
2 3 2
−323
.
直交基底 b1, b2, b3 正規化 行 正規直交基底 求 . ,求 正規直交基底c1, c2, c3 ,
c1 =
√2 2
0 1 1
=
√0
2
√2 2 2
, c2 =
√6 3
1
−12
1 2
=
√6 3
−√√66
6 6
, c3 =
√3 2
2 3 2
−323
=
√3
√3 3
−3√33
.
(3) a1 =
1 1 1
, a2 =
2 0 1
, a3 =
1
−2
−1
[解]: 与 基底 直交基底 .
b1 =a1 =
1 1 1
,
b2 =a2−(a2, b1) (b1, b1)b1 =
2 0 1
−
1 1 1
=
1
−1 0
,
b3 =a3−(a3, b1)
(b1, b1)b1− (a3, b2) (b2, b2)b2 =
1
−2
−1
−
−23
−23
−23
−
3
−232
0
=
1 6 1
−613
.
直交基底 b1, b2, b3 正規化 行 正規直交基底 求 . ,求 正規直交基底c1, c2, c3 ,
c1 =
√3 3
1 1 1
=
√3
√3 3
√3 3 3
, c2 =
√2 2
1
−1 0
=
√2 2
−√22 0
, c3 =√ 6
1 6 1
−613
=
√6
√6 6 6
−√36
.
(4) a1 =
−1 2 2
, a2 =
0 1 2
, a3 =
1 1 1
[解]: 与 基底 直交基底 .
b1 =a1 =
−1 2 2
,
b2 =a2−(a2, b1) (b1, b1)b1 =
0 1 2
−
−23
4 3 4 3
=
2
−313
2 3
,
b3 =a3−(a3, b1)
(b1, b1)b1− (a3, b2) (b2, b2)b2 =
1 1 1
−
−13
2 3 2 3
−
2
−313
2 3
=
2 3 2
−313
.
直交基底 b1, b2, b3 正規化 行 正規直交基底 求 . ,求 正規直交基底c1, c2, c3 ,
c1 = 1 3
−1 2 2
=
−13
2 3 2 3
, c2 = 1
2
−313
2 3
=
2
−313
2 3
, c3 = 1
2 3 2
−313
=
2 3 2
−313
.
1
2
3
[解]: 与 基底 直交基底 .
b1 =a1 =
1 2 3
,
b2 =a2− (a2, b1) (b1, b1)b1 =
2 3 4
−
10 7 20
7 30
7
=
4 7 1
−727
,
b3 =a3− (a3, b1)
(b1, b1)b1−(a3, b2) (b2, b2)b2 =
3 4 1
−
1 2 3
−
8 3 2
−343
=
−23
4
−323
.
直交基底 b1, b2, b3 正規化 行 正規直交基底 求 . ,求 正規直交基底c1, c2, c3 ,
c1 =
√14 14
1 2 3
=
√14
√14 14 7 3√ 14 14
, c2 =
√21 3
4 7 1
−727
=
4√ 21
√21 21 21
−2√2121
, c3 =
√6 4
−23
4
−323
=
−√√66
6 3
−√66
.
(6) a1 =
1
−1 1
−1
, a2 =
1 1 0 0
, a3 =
0 0 1 1
, a4 =
0 1 1 0
[解]: 与 基底 直交基底 .
b1 =a1 =
1
−1 1
−1
,
b2 =a2− (a2, b1) (b1, b1)b1 =
1 1 0 0
−
0 0 0 0
=
1 1 0 0
,
b3 =a3− (a3, b1)
(b1, b1)b1− (a3, b2) (b2, b2)b2 =
0 0 1 1
−
0 0 0 0
−
0 0 0 0
=
0 0 1 1
,
b4 =a4− (a4, b1)
(b1, b1)b1− (a4, b2)
(b2, b2)b2− (a4, b3) (b3, b3)b3 =
0 1 1 0
−
0 0 0 0
−
1 2 1 2
0 0
−
0 0
1 2 1 2
=
−12
1 2 1
−212
.
直交基底 b1, b2, b3, b4 正規化 行 正規直交基底 求 . , 求
正規直交基底 c1, c2, c3, c4 ,
c1 = 1 2
1
−1 1
−1
=
1
−212
1
−212
, c2 =
√2 2
1 1 0 0
=
√2
√2 2 2
0 0
,
c3 =
√2 2
0 0 1 1
=
0
√0
2
√2 2 2
, c4 = 1
−12
1 2 1
−212
=
−12
1 2 1
−212
.
(7) a1 =
1 3 0 0
, a2 =
−1 1 0 0
, a3 =
0 0 1 1
, a4 =
0 0 3 4
[解]: 与 基底 直交基底 .
b1 =a1 =
1 3 0 0
,
b2 =a2− (a2, b1) (b1, b1)b1 =
−1 1 0 0
−
1 5 3 5
0 0
=
−65
2 5
0 0
,
b3 =a3− (a3, b1)
(b1, b1)b1− (a3, b2) (b2, b2)b2 =
0 0 1 1
−
0 0 0 0
−
0 0 0 0
=
0 0 1 1
,
b4 =a4− (a4, b1)
(b1, b1)b1− (a4, b2)
(b2, b2)b2− (a4, b3) (b3, b3)b3 =
0 0 3 4
−
0 0 0 0
−
0 0 0 0
−
0 0
7 2 7 2
=
0 0
−12
1 2
.
直交基底 b1, b2, b3, b4 正規化 行 正規直交基底 求 . , 求 正規直交基底 c1, c2, c3, c4 ,
c1 =
√10 10
1 3 0 0
=
√10 10 3√
10 10
0 0
, c2 =
√10 4
−65
2 5
0 0
=
−√3√1010
10 10
0 0
,
c3 =
√2
0 0
=
0
√0
, c4 =√ 2
0 0
−
=
0 0
−√
.
(8) a1 =
1 1 1 1
, a2 =
2 1 2 1
, a3 =
0
−2 1
−1
, a4 =
1
−3 1
−1
[解]: 与 基底 直交基底 .
b1 =a1 =
1 1 1 1
,
b2 =a2− (a2, b1) (b1, b1)b1 =
2 1 2 1
−
3 2 3 2 3 2 3 2
=
1
−212
1
−212
,
b3 =a3− (a3, b1)
(b1, b1)b1− (a3, b2) (b2, b2)b2 =
0
−2 1
−1
−
−12
−12
−12
−12
−
1
−1 1
−1
=
−12
−12
1 2 1 2
,
b4 =a4− (a4, b1)
(b1, b1)b1− (a4, b2)
(b2, b2)b2− (a4, b3) (b3, b3)b3 =
1
−3 1
−1
−
−12
−12
−12
−12
−
3
−232
3
−232
−
−12
−12
1 2 1 2
=
1
−212
−12
1 2
.
直交基底 b1, b2, b3, b4 正規化 行 正規直交基底 求 . , 求 正規直交基底 c1, c2, c3, c4 ,
c1 = 1 2
1 1 1 1
=
1 2 1 2 1 2 1 2
, c2 = 1
1
−212
1
−212
=
1
−212
1
−212
,
c3 = 1
−12
−12
1 2 1 2
=
−12
−12
1 2 1 2
, c4 = 1
1
−212
−12
1 2
=
1
−212
−12
1 2
.
(9) a1 =
1 1
−1
−1
, a2 =
1 1 1
−1
, a3 =
1
−1 1 1
, a4 =
1 1 1 1
[解]: 与 基底 直交基底 .
b1 =a1 =
1 1
−1
−1
,
b2 =a2− (a2, b1) (b1, b1)b1 =
1 1 1
−1
−
1 2 1
−212
−12
=
1 2 1 2 3
−212
,
b3 =a3− (a3, b1)
(b1, b1)b1− (a3, b2) (b2, b2)b2 =
1
−1 1 1
−
−12
−12
1 2 1 2
−
1 6 1 6 1
−216
=
4
−323
0
2 3
,
b4 =a4− (a4, b1)
(b1, b1)b1− (a4, b2)
(b2, b2)b2− (a4, b3) (b3, b3)b3 =
1 1 1 1
−
0 0 0 0
−
1 3 1 3
1
−13
−
2
−313
0
1 3
=
0 1 0 1
.
直交基底 b1, b2, b3, b4 正規化 行 正規直交基底 求 . , 求 正規直交基底 c1, c2, c3, c4 ,
c1 = 1 2
1 1
−1
−1
=
1 2 1
−212
−12
, c2 =
√3 3
1 2 1 2 3
−212
=
√3
√6 3
√6 3 2
−√63
,
c3 =
√6 4
4
−323
0
2 3
=
√6 3
−√66
√0
6 6
, c4 =
√2 2
0 1 0 1
=
√0
2 2
√0
2 2
.