resumen factorizacion LU

1. FACTORIZACIÓN LU

2. Introducción
𝑆𝑒 𝑝𝑢𝑒𝑑𝑒 𝑑𝑒𝑐𝑖𝑟 𝑞𝑢𝑒 𝑒𝑙 𝑀é𝑡𝑜𝑑𝑜 𝑑𝑒 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑐𝑖ó𝑛 𝑳𝑼 𝑒𝑠 𝑏á𝑠𝑖𝑐𝑎𝑚𝑒𝑛𝑡𝑒
𝑒𝑙 𝑀é𝑡𝑜𝑑𝑜 𝑑𝑒 𝐺𝑎𝑢𝑠𝑠 𝑒𝑛 𝑓𝑜𝑟𝑚𝑎 𝑚𝑎𝑡𝑟𝑖𝑐𝑖𝑎𝑙.
𝐶𝑜𝑛𝑠𝑖𝑠𝑡𝑒 𝑒𝑛 𝑑𝑒𝑠𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑟 𝑢𝑛𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑑𝑒 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑨 𝑒𝑛 𝑑𝑜𝑠
𝑠𝑢𝑏𝑚𝑎𝑡𝑟𝑖𝑐𝑒𝑠 𝑳 𝑦 𝑼, 𝑑𝑜𝑛𝑑𝑒 𝑳 𝑒𝑠 𝑢𝑛𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑓𝑒𝑟𝑖𝑜𝑟 𝑦 𝑼 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑢𝑝𝑒𝑟𝑖𝑜𝑟.
𝐸𝑠𝑡𝑒 𝑝𝑟𝑜𝑐𝑒𝑑𝑖𝑚𝑖𝑒𝑛𝑡𝑜 𝑝𝑢𝑒𝑑𝑒 𝑠𝑒𝑟 𝑠𝑖𝑒𝑚𝑝𝑟𝑒 𝑟𝑒𝑎𝑙𝑖𝑧𝑎𝑑𝑜 𝑠𝑖 𝑨 𝑡𝑖𝑒𝑛𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑎.

3. Procedimiento
𝑃𝑎𝑟𝑎 𝑒𝑙𝑙𝑜, 𝑠𝑒 𝑑𝑒𝑓𝑖𝑛𝑒 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑇𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝐼𝑛𝑓𝑒𝑟𝑖𝑜𝑟 𝑈𝑛𝑖𝑡𝑎𝑟𝑖𝑎 𝑑𝑒𝑙 𝑝𝑎𝑠𝑜 𝑘:
𝐿𝑘
=
1 0
0 1
⋯ 0 0
0 0 0
⋯ 0
0 0
0 0
⋮ ⋮
⋱ 0 0
0 1 0
0 0
0 0
0 0
0
0
0
0
0 −𝑙𝑘+1,𝑘 1
0
0
⋮
−𝑙𝑛,𝑘
0
0
⋱ 0
⋱
⋯
0
1
(1)
𝐷𝑜𝑛𝑑𝑒 𝑙𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑑𝑒 𝑙𝑎 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑠𝑜𝑛 "𝑢𝑛𝑜𝑠" 𝑦 𝑙𝑜𝑠 𝑜𝑡𝑟𝑜𝑠 "𝑐𝑒𝑟𝑜𝑠" 𝑒𝑥𝑐𝑒𝑝𝑡𝑜
𝑎𝑞𝑢𝑒𝑙𝑙𝑜𝑠 𝑒𝑛 𝑙𝑎 𝑐𝑜𝑙𝑢𝑚𝑛𝑎 𝑘 − é𝑠𝑖𝑚𝑎 𝑝𝑜𝑟 𝑑𝑒𝑏𝑎𝑗𝑜 𝑑𝑒 𝑙𝑎 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙, 𝑑𝑜𝑛𝑑𝑒 𝑒𝑠𝑡á𝑛 𝑙𝑜𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑜𝑠
𝑑𝑒 𝑙𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑣𝑜𝑠 − 𝑙𝑖𝑘.
𝑇𝑎𝑙 𝑞𝑢𝑒:
𝑙𝑖𝑘 =
𝑎𝑖𝑘
(𝑘)
𝑎𝑘𝑘
(𝑘) (2)

4. Procedimiento
𝐸𝑙 𝑝𝑟𝑜𝑐𝑒𝑠𝑜 𝑠𝑒 𝑖𝑛𝑖𝑐𝑖𝑎 𝑝𝑎𝑟𝑎 𝐾 = 1, ℎ𝑎𝑠𝑡𝑎 𝑘 = 𝑛 − 1, 𝑡𝑎𝑙 𝑞𝑢𝑒:
𝑨(𝑲+𝟏)
= 𝑳(𝒌)
∙ 𝑨(𝒌)
(3)
𝑆𝑒 𝑐𝑜𝑛𝑣𝑒𝑛𝑐𝑖𝑜𝑛𝑎 𝑞𝑢𝑒 𝑨(𝟏)
= 𝑨
𝐸𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑝𝑎𝑟𝑎 𝑘 = 𝑛 − 1, 𝑨 𝒏
= 𝑼 (4)

5. Procedimiento
𝐷𝑜𝑛𝑑𝑒 𝑼 𝑒𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑢𝑝𝑒𝑟𝑖𝑜𝑟 𝑦 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑓𝑒𝑟𝑖𝑜𝑟 𝑢𝑛𝑖𝑡𝑎𝑟𝑖𝑎 𝑳 𝑒𝑠:
𝑳 =
1 0 ⋯ ⋯ 0
𝑙21
⋮
⋮
𝑙𝑛1
1
𝑙32
⋮
𝑙𝑛2
⋱
⋱
⋱
⋯
⋯ 0
0 0
1 0
𝑙𝑛,𝑛−1 1
(5)

6. Ejemplo.
𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑟 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧:
𝑨 =
2 1 1
1 0 1
1 1 1
𝑷𝒂𝒓𝒂 𝒌 = 𝟏:
𝑙21 =
𝑎21
(1)
𝑎11
(1)
=
1
2
, 𝑙31 =
𝑎31
(1)
𝑎11
(1)
=
1
2
𝐴(2)
= 𝐿(1)
∙ 𝐴 1
=
1 0 0
Τ
−1 2 1 0
Τ
−1 2 0 1
∙
2 1 1
1 0 1
1 1 1
=
2 1 1
0 −1/2 1/2
0 1/2 1/2
𝑷𝒂𝒓𝒂 𝒌 = 𝟐:
𝑙32 =
𝑎32
(2)
𝑎22
(2)
=
1/2
−1/2
= −1
𝐴(3)
= 𝐿(2)
∙ 𝐴 2
=
1 0 0
0 1 0
0 1 1
∙
2 1 1
0 −1/2 1/2
0 1/2 1/2
=
2 1 1
0 Τ
−1 2 Τ
1 2
0 0 1
= 𝑼
𝑦 𝑳 =
1 0 0
Τ
1 2 1 0
Τ
1 2 −1 1

7. Ejemplo.
𝐹𝑖𝑎𝑛𝑙𝑚𝑒𝑛𝑡𝑒:
𝑨 =
2 1 1
1 0 1
1 1 1
=
1 0 0
Τ
1 2 1 0
Τ
1 2 −1 1
∙
2 1 1
0 Τ
−1 2 Τ
1 2
0 0 1
𝑳 𝑼

8. NOTA.
𝐿𝑎 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑐𝑖ó𝑛 𝑳𝑼 𝑠𝑒 𝑝𝑢𝑒𝑑𝑒 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑒𝑛 𝑒𝑙 𝑓𝑜𝑟𝑚𝑎𝑡𝑜 𝑑𝑒 𝑇𝑎𝑏𝑙𝑎, 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑛𝑑𝑜
𝑒𝑙 𝑎𝑙𝑔𝑜𝑟𝑖𝑡𝑚𝑜 𝑑𝑒 𝑬𝒍𝒊𝒎𝒊𝒏𝒂𝒄𝒊ó𝒏 𝒅𝒆 𝑮𝒂𝒖𝒔𝒔 𝑦 𝑎𝑝𝑟𝑜𝑣𝑒𝑐ℎ𝑎𝑛𝑑𝑜 𝑙𝑜𝑠 𝑒𝑠𝑝𝑎𝑐𝑖𝑜𝑠, 𝑑𝑜𝑛𝑑𝑒 𝑠𝑒
𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑛 𝑙𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑑𝑒 𝑨, 𝑝𝑎𝑟𝑎 𝑎𝑙𝑚𝑎𝑐𝑒𝑛𝑎𝑟 𝑙𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑑𝑒 𝑳.

9. Ejemplo.
𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑒𝑙 𝑒𝑗𝑒𝑚𝑝𝑙𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑒𝑛 𝑓𝑜𝑟𝑚𝑎 𝑑𝑒 𝑇𝑎𝑏𝑙𝑎.
𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑟 𝑙𝑎 𝑚𝑎𝑡𝑟𝑖𝑧: 𝑨 =
2 1 1
1 0 1
1 1 1

10. Ejemplo, cont. …
𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜 𝑞𝑢𝑒:
𝑳 ∙ 𝑼 =
1 0 0
0.5 1 0
0.5 −1 1
2 1 1
0 −0.5 0.5
0 0 1
=
2 1 1
1 0 1
1 1 1
= 𝑨

11. 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑐𝑖ó𝑛 𝐿𝑈 𝑒𝑛 𝑙𝑎 𝑆𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒
𝑆𝑖𝑠𝑡𝑒𝑚𝑎 𝑑𝑒 𝐸𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝐿𝑖𝑛𝑒𝑎𝑙𝑒𝑠
𝑆𝑒𝑎 𝑑𝑎𝑑𝑜 𝑒𝑙 𝑆𝑖𝑠𝑡𝑒𝑚𝑎 𝑑𝑒 𝐸𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠: 𝑨 ∙ 𝒙 = 𝒃
𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑨 𝑝𝑜𝑟 𝑳𝑼 𝑠𝑒 𝑡𝑖𝑒𝑛𝑒 𝑳𝑼 ∙ 𝒙 = 𝒃
𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑛𝑑𝑜 𝑙𝑎 𝑝𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑 𝑎𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑣𝑎 𝑳 𝑼 ∙ 𝒙 = 𝒃
ℎ𝑎𝑐𝑖𝑒𝑛𝑑𝑜 𝑼 ∙ 𝒙 = 𝒄 , 𝑠𝑒 𝑜𝑏𝑡𝑖𝑒𝑛𝑒𝑛 𝑑𝑜𝑠 𝑠𝑖𝑠𝑡𝑒𝑚𝑠, 𝑡𝑎𝑙 𝑞𝑢𝑒:
1) 𝑳 ∙ 𝒄 = 𝒃 𝑇𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝐼𝑛𝑓𝑒𝑟𝑖𝑜𝑟, 𝑠𝑒 𝑟𝑒𝑠𝑢𝑒𝑙𝑣𝑒 𝑝𝑜𝑟 𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖ó𝑛 𝐹𝑟𝑜𝑛𝑡𝑎𝑙
𝑦 2) 𝑼 ∙ 𝑿 = 𝒄 (𝑇𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑆𝑢𝑝𝑒𝑟𝑖𝑜𝑟, 𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑝𝑜𝑟 𝑠𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖ó𝑛 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑣𝑎)

12. 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑐𝑖ó𝑛 𝐿𝑈 𝑒𝑛 𝑙𝑎 𝑆𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑑𝑒
𝑆𝑖𝑠𝑡𝑒𝑚𝑎 𝑑𝑒 𝐸𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝐿𝑖𝑛𝑒𝑎𝑙𝑒𝑠
Observación. 𝐿𝑎 𝑃𝑖𝑣𝑜𝑡𝑎𝑐𝑖ó𝑛 𝑡𝑎𝑚𝑏𝑖é𝑛 𝑒𝑠 𝑜𝑏𝑙𝑖𝑔𝑎𝑡𝑜𝑟𝑖𝑎 𝑒𝑛 𝑙𝑎 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑐𝑖ó𝑛 𝑳𝑼, 𝑑𝑒 𝑚𝑎𝑛𝑒𝑟𝑎 𝑞𝑢𝑒
𝑙𝑎𝑠 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑑𝑒 𝑓𝑖𝑙𝑎 𝑎𝑓𝑒𝑐𝑡𝑎𝑛 𝑡𝑎𝑚𝑏í𝑒𝑛 𝑎 𝑙𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑑𝑒 𝑳 𝑦 𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝒃.
𝑇𝑎𝑙 𝑞𝑢𝑒:
1) 𝑳 ∙ 𝒄 = 𝒃∗
(𝒃∗
= 𝒃 𝑠𝑖 𝑛𝑜 𝑠𝑒 𝑟𝑒𝑎𝑙𝑖𝑧𝑎 𝑛𝑖𝑛𝑔𝑢𝑛𝑎 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑐𝑖ó𝑛)
2) 𝑼 ∙ 𝒙 = 𝒄

13. EJEMPLO.
𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑒𝑙 𝑆𝑖𝑠𝑡𝑒𝑚𝑎 𝑢𝑡𝑖𝑙𝑖𝑧𝑎𝑛𝑑𝑜 𝑙𝑎 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑐𝑖ó𝑛 𝑳𝑼
2 1 1
1 0 1
1 1 1
∙
𝑥1
𝑥2
𝑥3
=
1
2
1
𝐴𝑝𝑟𝑜𝑣𝑒𝑐ℎ𝑎𝑛𝑑𝑜 𝑙𝑜𝑠 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜𝑠 𝑑𝑒𝑙 𝑒𝑗𝑒𝑚𝑝𝑙𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟, 𝑠𝑒 𝑡𝑖𝑒𝑛𝑒 𝑞𝑢𝑒:
𝑳 =
1 0 0
0.5 1 0
0.5 −1 1
, 𝑼 =
2 1 1
0 −0.5 0.5
0 0 1
𝑅𝑒𝑠𝑜𝑙𝑣𝑖𝑒𝑛𝑑𝑜 𝑒𝑙 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 1): 𝑳 ∙ 𝒄 = 𝒃∗ (𝑛°𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑐𝑖𝑜𝑛𝑒𝑠 = 0, → 𝒃∗ = 𝒃)
1 0 0
0.5 1 0
0.5 −1 1
𝑐1
𝑐2
𝑐3
=
1
2
1
→ 𝑐1 = 1
→ 𝑐2 = 2 − 0.5 ∗ 1 = 1.5
→ 𝑐3 = 1 − 0.5 ∗ 1 + 1 ∗ 1.5 = 2

14. EJEMPLO, Cont. ….
𝐿𝑢𝑒𝑔𝑜, 𝑟𝑒𝑠𝑜𝑙𝑣𝑖𝑒𝑛𝑑𝑜 𝑒𝑙 𝑆𝑖𝑠𝑡𝑒𝑚𝑎 2): 𝑼 ∙ 𝒙 = 𝒄
2 1 1
0 −0.5 0.5
0 0 1
∙
𝑥1
𝑥2
𝑥3
=
1
1.5
2
→ 𝑥1 = ( Τ
1 − 1 ∗ 2 − 1 ∗ (−1)) 2 = 0
→ 𝑥2 = Τ
(1.5 − 0.5 ∗ 2) (−0.5) = −1
→ 𝑥3 = 2
𝐿𝑎 𝑠𝑜𝑙𝑢𝑐𝑖ó𝑛 𝑒𝑠:
𝑥1
𝑥2
𝑥3
=
0
−1
2

15. Ejercicio.
𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑝𝑜𝑟 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑎𝑐𝑖ó𝑛 𝐿𝑈:
1 7 3
5 2 2
3 1 2
𝑥1
𝑥2
𝑥3
=
24
15
11
𝐸𝑙𝑖𝑚𝑖𝑛𝑎𝑐𝑖ó𝑛 𝑑𝑒 𝐺𝑎𝑢𝑠𝑠:

16. Ejercicio, Cont. …
𝑆𝑖𝑠𝑡𝑒𝑚𝑎 1) 𝑳 ∙ 𝒄 = 𝒃∗
(𝑛°𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑐𝑖𝑜𝑛𝑒𝑠 = 1, → 𝒃∗
=
15
24
11
1 0 0
0.2 1 0
0.6 −0.0303 1
𝑐1
𝑐2
𝑐3
=
15
24
11
→ 𝑐1 = 15
→ 𝑐2 = 24 − 0.2 ∗ 15 = 21
→ 𝑐3 = 11 − 0.6 ∗ 15 + 0.0303 ∗ 21 = 2.64
𝑆𝑖𝑠𝑡𝑒𝑚𝑎 2) 𝑼 ∙ 𝒙 = 𝒄
5 2 2
0 6.6 2.6
0 0 0.879
𝑥1
𝑥2
𝑥3
=
15
21
2.64
→ 𝑥1 = Τ
(15 − 2 ∗ 3 − 2 ∗ 2) 5 = 1.00
→ 𝑥2 = Τ
(21 − 2.6 ∗ 3) 6.6 = 2.00
→ 𝑥3 = Τ
2.64 0.879 = 3.00
𝑆𝑜𝑙.
𝑥1
𝑥2
𝑥3
=
1.00
2.00
3.00