Unformatted text preview: nput basis. Solutions to Exercises 78 4 The third derivative matrix has 6 in the .1; 4/ position; since the third derivative of x 3 is 6. This matrix also comes from AB . The fourth derivative of a cubic is zero, and B 2
is the zero matrix. 5 T .v1 C v2 C v3 / D 2w1 C w2 C 2w3 ; A times .1; 1; 1/ gives .2; 1; 2/.
6 v D c.v2 v 3 / gives T .v/ D 0; nullspace is .0; c; c /; solutions .1; 0; 0/ C .0; c; c /. 7 .1; 0; 0/ is not in the column space of the matrix A, and w1 is not in the range of the linear transformation T . Key point: Column space of matrix matches range of
transformation.
8 We don’t know T .w/ unless the w’s are the same as the v’s. In that case the matrix is A2 . 9 Rank of A D 2 D dimension of the range of T . The outputs Av (column space) match the outputs T .v/ (the range of T ). The “output space” W is like Rm : it contains all
outputs but may not be ﬁlled up.
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#
"#
"#
100
1
1
1D
10 The matrix for T is A D 1 1 0 . For the output 0 choose input v D
111
0
0
"#
1
A 1 0 . This means: For the output w1 choose the input v1 v2 .
0
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#
1
00
1
1
1 0 so T 1 .w1 / D v1 v2 ; T 1 .w2 / D v2 v3 ; T 1 .w3 / D v3 .
11 A D
0
11
The columns of A 1 describe T 1 from W back to V . The only solution to T .v/ D 0
is v D 0. 12 (c) T 1 .T .w1 // D w1 is wrong because w1 is not generally in the input space. 13 (a) T .v1 / D v2 ; T .v2 / D v1 is its own inverse (b) T .v1 / D v1 ; T .v2 / D 0 has
T 2 D T (c) If T 2 D I for part (a) and T 2 D T for part (b), then T must be I .
1
2
3
1
21
.
must be 2A
D inverse of (a)
(c) A
(b)
14 (a)
6
3
5
2
53
0
r
s
1
rs
to
and
; this is the “easy” direcand
15 (a) M D
transforms
u
0
1
t
tu
1
ab
tion. (b) N D
transforms in the inverse direction, back to the standard
cd
basis vectors. (c) ad D bc will make the forward matrix singular and the inverse
impossible.
1
3
1
10 21
D
.
16 M W D
7
3
12 53
17 Recording basis vectors is done by a Permutation matrix. Changing lengths is done by a positive diagonal matrix.
18 .a; b/ D .cos ; sin /. Minus sign from Q 1 D QT . Solutions to Exercises
11
a
5
19 M D
;
D
D ﬁrst column of M
45
b
4
11
.
45 79
1
1
D coordinates of
in basis
0 1
x 2 ; w3 .x/ D 2 .x 2 x /; y D 4w1 C 5w2 C 6w3 .
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"
#
0
1
0
1
11
0
:5 : v’s to w’s: inverse matrix D 1
0 0 . The key
w’s to v’s: :5
:5
1
:5
1
11
idea: The matrix multiplies the coordinates in the v basis to give the coordinates in the
w basis.
2
32 3 2 3
A
4
1 a a2
4 1 b b 2 5 4 B 5 D 4 5 5. This
The 3 equations to match 4; 5; 6 at x D a; b; c are
1 c c2
C
6
Vandermonde determinant equals .b a/.c a/.c b /. So a; b; c must be distinct to
have det ¤ 0 and one solution A; B; C .
The matrix M with these nine entries must be invertible.
Start from A D QR. Column 2 is a2 D r12 q 1 C r22 q 2 . This gives a2 as a combination
of the q ’s. So the change of basis matrix is R .
Start from A D LU . Row 2 of A is `21 (row 1 of U / C `22 (row 2 of U ). The change
of basis matr...
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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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