MAT 211, Spring 2012
Solutions to Homework Assignment 9
Maximal grade for HW9: 100 points
Section 5.1. 16. (10 points) Consider the vectors
u1 = (1/2, 1/2, 1/2, 1/2), u2 = (1/2, 1/2, −1/2, −1/2), u3 = (1/2, −1/2, 1/2, −1/2)
in R4 . Find all vectors u4 such that u1 , u2 , u3 , u4 for an orthonormal basis in
R4 .
Answer: (1/2,-1/2,-1/2,1/2), (-1/2,1/2,1/2,-1/2).
Solution: Let u4 = (x1 , x2 , x3 , x4 ). One can check that
u1 · u1 = 1/4 + 1/4 + 1/4 = 1/4 = 1, u1 · u2 = 1/4 + 1/4 − 1/4 − 1/4 = 0,
and similarly
u2 · u2 = u3 · u3 = 1, u1 · u3 = u2 · u3 = 0.
Let us impose the condition that u4 is orthogonal to u1 , u2 , u3 :
1/2x1 + 1/2x2 + 1/2x3 + 1/2x4
=0
1/2x1 + 1/2x2 − 1/2x3 − 1/2x4 =0
1/2x1 − 1/2x2 + 1/2x3 − 1/2x4 =0
Let us write this in matrix notation:
1/2 1/2 1/2 1/2 0
1/2 1/2 −1/2 −1/2 0
1/2 −1/2 1/2 −1/2 0
1
Multiply the whole matrix by 2:
1 1 1 1 0
1 1 −1 −1 0
1 −1 1 −1 0
Subtract the first row from the second and third:
1 1 1 1 0
0 0 −2 −2 0
0 −2 0 −2 0
Swap second and third row and divide them by (-2):
1 1 1 1 0
0 1 0 1 0
0 0 1 1 0
Subtract the second and third row from the first:
1 0 0 −1 0
0 1 0 1 0
0 0 1 1 0
Therefore u4 = (x4 , −x4 , −x4 , x4 ).
The remaining condition is that u4 should have unit length. Since u4 ·u4 =
2
4x4 = 1, we have x4 = ±1/2.
17. (15 points) Find a basis for W ⊥ , where W = Span((1, 2, 3, 4), (5, 6, 7, 8)).
Solution: The space W ⊥ is defined by the system of two linear equations:
(
x1 + 2x2 + 3x3 + 4x4 =0
5x1 + 6x2 + 7x3 + 8x4 = 0
In matrix notation:
1 2 3 4 0
5 6 7 8 0
Subtract the first row (multiplied by 5) from the second:
1 2 3 4 0
0 −4 −8 −12 0
2
Divide second row by (-4):
1 2 3 4 0
0 1 2 3 0
Subtract the second row, multiplied by 2, from the first:
1 0 −1 −2 0
0 1 2 3 0
Therefore x3 , x4 are free parameters, x1 = x3 + 2x4 , x2 = −2x3 − 3x4 . The
basis in the space of solutions: (1, −2, 1, 0), (2, −4, 0, 1).
25. a) (5 points) Consider a vector v and a scalar k. Show that ||k · v|| =
|k|||v||.
b) (5 points) Show that if v is a nonzero vector in Rn , then u = v/||v|| is
a unit vector.
Solution: a)
||k · v||2 = (k · v) · (k · v) = k 2 (v · v) = k 2 ||v||2 .
Therefore p
||k · v|| = k 2 ||v||2 = |k|||v||.
b) Let us apply (a) to k = 1/||v||:
||u|| = ||k · v|| = |k|||v|| = ||v||/||v|| = 1.
Therefore u is a unit vector.
26. (20 points) Find the orthogonal projection of v = (49, 49, 49) onto
the subspace of R3 spanned by v1 = (2, 3, 6) and v2 = (3, −6, 2).
Answer: (19, 39, 64).
Solution: Remark that
v1 ·v1 = 22 +32 +62 = 49, , v2 ·v2 = 32 +(−6)2 +22 = 49.
v1 ·v2 = 2·3−3·6+2·6 = 0,
√
Therefore v1 and v2 are perpendicular, and both have lengths 49 = 7.
Therefore the vectors
1 1
u1 = v1 = (2/7, 3/7, 6/7), u2 = v2 = (3/7, −6/7, 2/7)
7 7
3
form an orthonormal basis of the subspace.
We have
v · u1 = 14 + 21 + 42 = 77, v · u2 = 21 − 42 + 14 = −7,
so the projection of v can be found by a formula
v|| = (v·u1 )u1 +(v·u2 )u2 = 77u1 −7u2 = (22, 33, 66)−(3, −6, 2) = (19, 39, 64).
28. (20 points) Find the orthogonal projection of v = (1, 0, 0, 0) onto
the subspace in R4 spanned by v1 = (1, 1, 1, 1), v2 = (1, 1, −1, −1), v3 =
(1, −1, −1, 1).
Answer: (3/4, 1/4, −1/4, 1/4).
Solution: Similarly to the previous problem one can √ check that v1 , v2 , v3
are pairwise orthogonal and ||v1 || = ||v2 || = ||v3 || = 4 = 2. Therefore one
can choose an orthonormal basis in the subspace:
1 1
u1 = v1 = (1/2, 1/2, 1/2, 1/2), u2 = v2 = (1/2, 1/2, −1/2, −1/2),
2 2
1
u3 = v3 = (1/2, −1/2, −1/2, 1/2).
2
We have:
v · u1 = 1/2, v · u2 = 1/2, v · u3 = 1/2,
so
v|| = (v · u1 )u1 + (v · u2 )u2 + (v · u3 )u3 =
= 1/2(u1 + u2 + u3 ) = 1/2(3/2, 1/2, −1/2, 1/2) = (3/4, 1/4, −1/4, 1/4).
29. (15 points) Consider the orthonormal vectors u1 , u2 , u3 , u4 , u5 in R10 .
Find the length of the vector
x = 7u1 − 3u2 + 2u3 + u4 − u5 .
Solution: Since ui are orthonormal, we have
||x||2 = 72 + (−3)2 + 22 + 12 + (−1)2 = 64, ||x|| = 8.
4
33. (10 points) Among all the vectors in Rn whose components add up
to 1, find the vector of minimal length.
Answer: (1/n, . . . , 1/n).
Solution: We have to project the origin onto the hyperplane consisting
of all vectors with sum of coordinates 1. It is clear that the projection does
not change if we permute the coordinates, so all its coordinates should be
equal to each other. Since their sum is 1, the desired vector is (1/n, . . . , 1/n).
