Math 2270, Fall 2015
Instructor: Thomas Goller
24 November 2015
Quiz #8 Solutions
2
3
2 3
2 3
1
4
3
6 27
617
6 17
7
6 7
6 7
(1) Let W = Span{~u1 , ~u2 }, where ~u1 = 6
4 15 and ~u2 = 4 0 5. Let ~y = 4 1 5.
2
3
13
(a) Show that ~u1 and ~u2 are orthogonal. (1 point)
Solution: ~u1 · ~u2 =
4
2 + 0 + 6 = 0.
(b) Compute the closest point to ~y in W . (3 points)
Solution:
2
3
2 3 2
1
4
7 26 6 1 7 6
~y · ~u1
~y · ~u2
30 6
2
6 7+
6 7=6
ŷ =
~u1 +
~u2 =
~u1 · ~u1
~u2 · ~u2
10 4 15 26 4 0 5 4
2
3
3
1
57
7
35
9
(c) Write ~y as the sum of a vector in W and a vector orthogonal to W . (2 points)
Solution:
~z = ~y
2
3
3
6 17
7
ŷ = 6
415
13
2
3 2 3
1
4
6 57 647
6 7 = 6 7,
4 35 445
9
4
2
3 2
3
6 17 6
6 7=6
415 4
13
(d) Compute the distance from ~y to W . (1 point)
Solution:
k~z k =
p
42 + 4 2 + 42 + 42 =
p
64 = 8.
3 2 3
1
4
647
57
7 + 6 7.
35 445
9
4
Math 2270, Fall 2015
Instructor: Thomas Goller
24 November 2015
2 3
2 3
1
0
(2) Bonus problem: Let W be the plane in R3 spanned by ~u1 = 405 and ~u2 = 415. (1
0
1
bonus point)
(a) Compute projW ~e1 , projW ~e2 , and projW ~e3 .
Solution: projW ~e1 = ~e1 since ~e1 is already in W .
2 3
2 3
2 3
2 3
2 3
2 3
1
0
0
1
0
0
0
1
1
0
1
1
projW ~e2 = 405 + 415 = 415 ;
projW ~e3 = 405 + 415 = 415 .
1
2
2
1
2
2
0
1
1
0
1
1
(b) Write down the matrix A for the linear transformation projW .
Solution:
2
3
1 0 0
A = 40 12 12 5
0 12 12
(c) Compute a basis for the range of projW .
Solution: The range of projW is W (and also Col A), which has basis {~u1 , ~u2 }.
(d) Compute a basis for the kernel of projW .
Solution: The kernel of projW is W ?
8 2 39
< 0 =
and also Nul A, which has basis 4 15 .
:
;
1