# Math 2270 Spring 2004 Homework 13 Solutions Section 5.1

advertisement ```Math 2270 Spring 2004
Homework 13 Solutions
Section 5.1) 6, 7, 12, 17, 28, 29, 33
(6)




~u &middot; ~v = 
1
−1
2
−2
 
 
 
&middot;
 
2
3
4
5





= 2 − 3 + 8 − 10 = −3
√
√
||~u|| = 1 + 1 + 4 + 4 = 10
√
√
√
||~v || = 4 + 9 + 16 + 25 = 54 = 3 6
−1
−3
cos (α) = √ = √
3 60
2 15
!
−1
α = cos−1 √
≈ 97o ≈ 1.7radians
2 15
(7)
~u &middot; ~v =
&quot;
2
−3
# &quot;
&middot;
5
4
#
= 10 − 12 = −2 &lt; 0
The cosine of the angle is negative, so the angle must be obtuse.
(12) First, by the definition of length
||~v + w||
~ 2 = (~v + w)
~ &middot; (~v + w)
~
Then, by properties of dot products
= ~v &middot; ~v + 2~v &middot; w
~ +w
~ &middot;w
~
Again, by the definition of length
= ||~v ||2 + ||w||
~ 2 + 2 (~v &middot; w)
~
By the Cauchy-Schwarz inequality
≤ ||~v ||2 + ||w||
~ 2 + 2 ||~v || ||w||
~ = (||~v|| + ||w||)
~ 2
In summary, we have
||~v + w||
~ 2 ≤ (||~v || + ||w||)
~ 2
||~v + w||
~ ≤ ||~v || + ||w||
~
1
Math 2270 Spring 2004
Homework 13 Solutions
(17) We must find the kernel of the matrix
A=
&quot;
1 2 3 4
5 6 7 8
#
→
&quot;
1 2 3
4
0 −4 −8 −12
#
→
&quot;
1 0 −1 −2
0 1 2 3
#
Elements of the kernel are of the form




~x = r 
1
−2
1
0






 + s


2
−3
0
1





where r and s are scalars. Therefore, a basis for the orthogonal complement of W is












(28)





1
0
0
0
 
 
 
&middot;
 
1/2
1/2
1/2
1/2

1
0
0
0





projV 








=
=
1
2
1


2
 
 
 
,
 
2
−3
0
1

1
0
0
0
 
1/2
1/2
−1/2
−1/2
1/2
1/2
1/2
1/2


1/2
1/2
−1/2
−1/2





1
−2
1
0
 
 
&middot;
 


+

1


2
















=



+


1
2





1


2
1
0
0
0
1/2
−1/2
−1/2
1/2
 
 
 
&middot;
 









=
1/2
−1/2
−1/2
1/2

3/4
1/4
−1/4
1/4





=
1
2




(29) We are given a set of orthonormal vectors. Therefore
~vi &middot; ~vj = 0 if i 6= j
~vi &middot; ~vi = 1
||~x||2 = ||7~v1 − 3~v2 + 2~v3 + ~v4 − ~v5 ||2 = (7~v1 − 3~v2 + 2~v3 + ~v4 − ~v5 ) &middot; (7~v1 − 3~v2 + 2~v3 + ~v4 − ~v5 )
= 49~v1 &middot; ~v1 + 9~v2 &middot; ~v2 + 4~v3 &middot; ~v3 + ~v4 &middot; ~v4 + ~v5 &middot; ~v5 = 49 + 9 + 4 + 1 + 1 = 64
2
Math 2270 Spring 2004
Homework 13 Solutions
√
Therefore ||~x|| = 64 = 8. Notice that the majority of the terms in the dot product
disappear (are equal to zero) since the vectors are orthogonal to one another. Since the
vectors are orthogonal to one another, we could have also used the pythagorean theorem.
(33) First, let’s look at the case where n = 2. We have a vector ~x with x1 + x2 = 1. Consider
the vector with one’s as both components. Then, by the Cauchy-Schwarz inequality we have
&quot;
x1
x2
# &quot;
&middot;
1
1
#
≤
&quot;
||~x|| 1
1
√
|x1 + x2 | ≤ ||~x|| 1 + 1
1≤
√1
2
√
#
2 ||~x||
≤ ||~x||
√
Therefore, any vector with the given property has length greater than or equal
to
1/
2.
q
Consider the vector with x1 = x2 = 1/2. This vector has magnitude of 1/4 + 1/4 =
q
√
1/2 = 1/ 2. Therefore, the vector with x1 = x2 = 1/2 has the minimal length.
In general, we have





x1
x2
&middot;&middot;&middot;
xn
 
 
 
&middot;
 
1
1
&middot;&middot;&middot;
1





≤



||~x|| 

1
1
&middot;&middot;&middot;
1





√
|x1 + x2 + &middot; &middot; &middot; + xn | ≤ ||~x|| 1 + 1 + &middot; &middot; &middot; + 1
1≤
√1
n
√
n ||~x||
≤ ||~x||
Again we see that the vector with this minimal length is the one with all components equal
to one another, x1 = x2 = . . . = xn = 1/n. We also note that the Cauchy-Schwarz inequality
is an equality if and only if the two vectors involved are parallel to one another. Therefore,
the vector ~x has its minimal length only when it is parallel to the vector of all ones, or its
components are equal to one another.
3
Math 2270 Spring 2004
Homework 13 Solutions
Section 5.2) 4, 13, 18, 27, 32, 34
(4, 18) We are given the basis






0


~v3 =  −2 
0
25


~v2 =  0  ,
25
4


~v1 =  0  ,
3
We use the Gram-Schmidt process:
||~v1 || =
√


4/5


w
~1 =  0 
3/5
16 + 9 = 5
w
~ 1 &middot; ~v2 = 54 (25) + 35 (25) = 35








−3
25 − 28
4/5
25








0
~v2 − projV1 ~v2 =  0  − 35  0  = 
= 0 
4
25 − 21
3/5
25
v2
~
− projV1 ~v2 √
= 9 + 16 = 5
w
~ 1 &middot; ~v3 = 0
~
v3
− projV2 ~v3 =
~
v3
− ~0


−3/5
w
~2 = 
0 


4/5
w
~ 2 &middot; ~v3 = 0
=
√


0


w
~ 3 =  −1 
0
4=2
M = QR





5 35 0
4/5 −3/5 0
4 25 0




0 −1   0 5 0 
0 −2  =  0

 0
0 0 2
3/5 4/5 0
3 −25 0
4
Math 2270 Spring 2004
Homework 13 Solutions
(13, 27) We are given the basis
1
1
1
1



~v1 = 






,


~v2 = 

1
0
0
1





,


~v3 = 

0
2
1
−1





We use the Gram-Schmidt process:
||~v1 || =
√
w
~1 = 






~v2 − projV1 ~v2 = 
1
2
1
0
0
1
+0+0+


 
 
−
 
1 1
−
2 2

0
 2

~v3 − projV2 ~v3 = 
 1
−1
− projV2 ~v3 =
s

1/2
1/2
1/2
1/2









=

w
~2 = 
=1
1/2
1/2
1/2
1/2





1
=1
2



− projV1 ~v2 = 1
w
~ 1 &middot; ~v3 = 0 + 1 +
~
v3

1+1+1+1=2
w
~ 1 &middot; ~v2 =
~
v2

1/2
−1/2
−1/2
1/2
1/2
−1/2
−1/2
1/2










1 1
− = −2
2 2



1/2

 1/2 




=

 −1/2 
−1/2
w
~ 2 &middot; ~v3 = 0 − 1 −

 
 
−
 
1/2
1/2
1/2
1/2






+2


1 1 1 1
+ + + =1
4 4 4 4
5
1/2
−1/2
−1/2
1/2




w
~3 = 
1/2
1/2
−1/2
−1/2





Math 2270 Spring 2004
Homework 13 Solutions
M = QR





1
1
1
1
1 0
0 2
0 1
1 −1





=
1/2 1/2 1/2
1/2 −1/2 1/2
1/2 −1/2 −1/2
1/2 1/2 −1/2












2 1 1
0 1 −2 

0 0 1
(32) We start by finding a basis for the plane x1 + x2 + x3 = 0 or, x1 = −x2 − x3 . Vectors
on the plane have the general form




−1
−1




~x = r  1  + s  0 
1
0
Let


−1


~v2 =  0 
1
We use the Gram-Schmidt process.
||~v1 || =
√ 
−1/√ 2


w
~ 1 =  1/ 2 
0

√
√
1+1+0= 2
w
~ 1 &middot; ~v2 =



−1


~v1 =  1 
0
√1
2
+0+0=

√1
2




−1/2
−1/2
−1



 

~v2 − projV1 ~v2 =  0  −  1/2  =  −1/2 
1
0
1
v2
~
− projV1 ~v2 =
q
1
4
+ 41 + 1 =
6
√
6
2
√ 
−1/√6


w
~ 2 =  −1/√ 6 
2/ 6

Math 2270 Spring 2004
Homework 13 Solutions
(34) We first find a basis for the kernel of the given matrix.
A=
&quot;
1 1 1 1
1 2 3 4
#
→
&quot;
1 0 −1 −2
0 1 2 3
#
The general form for vectors in the kernel is
1
−2
1
0




~x = r 
Let




~v1 = 
1
−2
1
0
We use the Gram-Schmidt process.






 + s



2
−3
0
1








~v2 − projV1 ~v2 = 
v2
~
− projV1 ~v2 =
2
−3
0
1
q
4
9
√2
6


 
 
−
 
+ 19 +
16
9








~v2 = 
√
√
||~v1 || = 1 + 4 + 1 = 6
w
~ 1 &middot; ~v2 =

+
√6
6
2
−3
0
1
w
~1 =
+0+0=
8/6
−16/6
8/6
0
+1=





√




=
30
3





√
1/ √6
−2/√ 6
1/ 6
0











√8
6
4/6
−2/6
−8/6
1









=



w
~2 = 


7

2/3
−1/3
−4/3
1





√
2/ √30
−1/√30
−4/√ 30
3/ 30






```