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Unformatted text preview: projections onto orthogonal vectors. This is important.
The projections of .1; 1/ onto .1; 0/ and .1; 2/ are p1 D .1; 0/ and p2 D .0:6; 1:2/.
Then p1 C p2 ¤ b.
Since A is invertible, P D A.AT A/ 1 AT D AA 1 .AT / 1 AT D I : project on all of R2 .
0:2 0:4
0:2
10
0:2 This is not a1 D .1; 0/
P2 D
, P2 a1 D
, P1 D
, P1 P2 a1 D
.
0:4 0:8
0:4
00
0
No, P1 P2 ¤ .P1 P2 /2 . 11 (a) p D A.AT A/ " 1T
A b D .2; 3; 0/, e D .0; 0; 4/, AT e D 0 (b) p D .4; 4; 6/, e D 0.
# 100
0 1 0 D projection matrix onto the column space of A (the xy plane)
000
"
#
0:5 0:5 0
Projection matrix onto the second column space.
P2 D 0:5 0:5 0 =
Certainly .P2 /2 D P2 .
0
0
1
2
3
2
3
23 23
100
1000
1
1
60 1 07
60 1 0 07
627 627
13 A D 4
, P D square matrix D 4
, p D P 4 5 D 4 5.
0 0 15
0 0 1 05
3
3
000
0000
4
0
14 The projection of this b onto the column space of A is b itself when b is in that space.
"
#
"#
58
4
0
1
8 17
2 and b D P b D p D 2 .
But P is not necessarily I . P D
21
4 2 20
4
12 P1 D 15 2A has the same column space as A. b for 2A is half of b for A.
x
x
16 1
.1; 2;
2
2 1/ C 3 .1; 0; 1/ D .2; 1; 1/. So b is in the plane. Projection shows P b D b.
2 P /2 D .I P /.I P / D I P I I P C P 2 D I P . When
P projects onto the column space, I P projects onto the left nullspace.
(a) I P is the projection matrix onto .1; 1/ in the perpendicular direction to .1; 1/
(b) I P projects onto the plane x C y C z D 0 perpendicular to .1; 1; 1/.
"
#
5=6
1=6
1=3
For any basis vectors in the plane x y 2z D 0,
1=6
5=6
1=3 .
say .1; 1; 0/ and .2; 0; 1/, the matrix P is
1=3
1=3
1=3
"#
"
#
"
#
1
1=6
1=6
1=3
5=6
1=6
1=3
ee T D
1 , Q D eTe
1=6
1=6
1=3 , I Q D 1=6
5=6
1=3 .
eD
2
1=3
1=3 2=3
1=3
1=3
1=3
2
A.AT A/ 1 AT D A.AT A/ 1 .AT A/.AT A/ 1 AT D A.AT A/ 1 AT . So P 2 D P .
P b is in the column space (where P projects). Then its projection P .P b/ is P b.
P T D .A.AT A/ 1 AT /T D A..AT A/ 1 /T AT D A.AT A/ 1 AT D P . (AT A is symmetric!)
If A is invertible then its column space is all of Rn . So P D I and e D 0.
The nullspace of AT is orthogonal to the column space C .A/. So if AT b D 0, the projection of b onto C .A/ should be p D 0. Check P b D A.AT A/ 1 AT b D A.AT A/ 1 0. 17 If P D P then .I
18 19 20
21
22
23
24 Solutions to Exercises 46 25 The column space of P will be S . Then r D dimension of S D n. 26 A 1 exists since the rank is r D m. Multiply A2 D A by A T T 1 to get A D I . 27 If A Ax D 0 then Ax is in the nullspace of A . But Ax is always in the column space of A. To be in both of those perpendicular spaces, Ax must be zero. So A and AT A
have the same nullspace. 28 P 2 D P D P T give P T P D P . Then the .2; 2/ entry of P equals the .2; 2/ entry of P T P which is the length squared of column 2. 29 A D B T has independent columns, so AT A (which is BB T ) must be invertible.
aaT
1 9 12
3
so PC D T D
.
4
aa
25 12 25
(b) The row space is the line through v D .1; 2; 2/ and PR D vvT =vT v. Always
PC A D A (columns of A project to themselves) and APR D A. Then PC APR D A ! 30 (a)...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

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