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# qr

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```Projeção ortogonal de um vetor v
sobre o plano definido pelas colunas da matriz A.

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5
5,0
1
v

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12
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12
A

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3.6154
2.3462
1.4615
Pv

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0.653850.461540.11538
0.461540.384620.15385-
0.115380.15385-0.96154
AAAAP TT
1

P = QQ
T
, se Q é obtida da decomposição A = QR.

P é idempotente (P = P
2
) e simétrica (P = P
T
).

Assim, (I \u2013 P) é ortogonal a P, ou seja, (I \u2013 P)TP = 0.

span(P) + span(I \u2013 P) = IR3.
-2
0
2
-3-2
-10
12
3
-5
-4
-3
-2
-1
0
1
2
3
4
5
x
Projeçao ortogonal de um vetor no plano
y
v
Pv
a1
a2
(I-P)v

\uf0b7 Transformando o vetor x no vetor \uf0b1 ||x||2e1.
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1
x \uf0fa
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0
1
1012ex

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3
110
12
xexv

\uf0b7 A projeção ortogonal de x sobre H é dada por:
\uf028 \uf029\uf028 \uf029 v
vv
xv
xx
vv
vv
IxvvvvIPx
T
T
T
T
TT
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\uf0f8
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\uf0f8
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\uf0e7\uf0e7
\uf0e8
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\uf02d\uf03d\uf02d\uf03d
\uf02d1.
\uf0b7 ||x||2e1 é o reflexo de x com relação a H:
Qxx
vv
vv
Iex
T
T
\uf03d\uf0f7\uf0f7
\uf0f8
\uf0f6
\uf0e7\uf0e7
\uf0e8
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.
\uf0b7 A matriz Q é ortogonal (Q
T
Q = I).
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31623,094868,0
94868,031623,0
Q
\uf0b7 Poderíamos ter transformado x em \u2013 ||x||2e1.
Neste caso, teríamos v = x + ||x||2e1.
Px
H
x
||x||2e1
x \u2013 ||x||2e1 = v
DECOMPOSIÇÃO DE HOUSEHOLDER

\uf0b7 Algoritmo da Decomposição de Householder.

1. Para i = 1 até n,
1.1. x = Ai:m,i;
1.2. vi = x + sign(x1)||x||2e1;
1.3. vi = vi / ||vi||2;
1.4. Ai:m,i:n = (I \u2013 2vivi
T
)Ai:m,i:n;

Custo computacional: 32
3
2
2 nmn \uf02d
operações.

\uf0b7 Resolução do sistema Ax = b:

Se A = QR, temos QRx = b, ou seja, Rx = Q
T
b.

Fazendo Qi = (I \u2013 2vivi
T
), temos Q
T
= QnQn-1 ... Q1.

\uf0b7 Algoritmo da resolução do sistema Ax = b:

1. y = b;
2. Para i = 1 até n,
2.1. yi:m = Qi yi:m;
3. Resolver o sistema Rx = y;

\uf0b7 Decomposição de Householder de
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10
12
21
A :

1
o
. passo:

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2
1
x ,
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0
0.52573
0.85065
v1 ,
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100
00.447210.89443-
00.89443-0.44721-
Q1

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1-0
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1.7889-2.2361-
AQA 11

2
o
. passo:

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1-
1.3416-
x ,
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0.31481-
0.94915-
0
v2

(v2 já está normalizado e tem primeiro elemento nulo)

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0.801780.59761-0
0.59761-0.80178-0
001
Q2

R
00
1.67330
1.7889-2.2361-
AQQAQA \uf03d
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\uf03d\uf03d\uf03d 12122
\uf0b7 Resolução de um sistema Ax = b.

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A ,
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1
1
1
b

(observe que
)(ARb\uf0ce
)

Calculando y = Q
T
b = Q2Q1b.

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bQy 1 ,

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0
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yQy 2 .

(como y3 = 0, o sistema é compatível)

Resolvendo o sistema Rx = y.

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0
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x
x
00
1.67330
1.7889-2.2361-
yRx
2
1

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1
1
x```