Math 310, Lesieutre Problem set #13 November 25, 2015 Problems

Math 310, Lesieutre Problem set #13 November 25, 2015 Problems
Math 310, Lesieutre
Problem set #13
November 25, 2015
Problems for M 11/16:
6.2.1 Determine whether the following vectors are an orthogonal set:
 
 
 
−1
5
3
 4  , 2 , −4 .
−3
1
−7
We need to check each pair and see if they dot to 0:
   
−1
5
 4  · 2 = −5 + 8 − 3 = 0
−3
1
   
−1
3
 4  · −4 = −3 − 16 + 21 = 2
−3
−7
   
5
3
2 · −4 = 15 − 8 − 7 = 0.
1
−7
They’re not orthogonal, since the first and third are not perpendicular to each other.
6.2.8 Show that the following vectors are an orthogonal basis for R2 , and express x as a
linear combination of the u’s.
2
6
9
u1 =
, u2 =
, x=
.
−3
4
−7
We have
u1 · u2 = 12 − 12 = 0.
So they’re orthogonal. A set of orthogonal vectors is always linearly independent, so
we don’t need to check that by row reduction like we usually would (though there’s no
harm in doing so).
We then have
u2 · x
u1 · x
u1 +
u2
u1 · u1
u2 · u2
39
26
1
= u1 + u2 = 3u1 + u2 .
13
52
2
x=
6.2.10 Show that the following vectors are an orthogonal basis for R3 , and express x as a
linear combination of the u’s.
 
 
 
 
3
2
1
5
u1 = −3 , u2 =  2  , u3 = 1 , x = −3 .
0
−1
4
1
We compute the dot products of all these things to show that they’re an orthogonal
set.
   
3
2
−3 ·  2  = 0,
0
−1
   
3
1
−3 · 1 = 0,
0
4
   
2
1
 2  · 1 = 0.
−1
4
Looks good. Then
u2 · x
u3 · x
u1 · x
u1 +
u2 +
u3
u1 · u1
u2 · u2
u3 · u3
24
3
6
4
1
1
= u1 + u2 + u3 = u1 + u2 + u3 .
18
9
18
3
3
3
1
−1
6.2.12 Compute the orthogonal projection of
onto the line through
and the ori−1
3
gin.
x=
This is just going to be
1
−1
·
−1
3
−4 −1
−1
2/5
=
=
.
3
−6/5
10 3
−1
−1
·
3
3
2
4
6.2.13 Let y =
and u =
. Write y as a sum of two orthogonal vectors, one in he
3
−7
span of u and one orthogonal to u. (We didn’t do one quite like this in lecture; take a
look at Example 3 in the book.)
We can do this like the above one: take the component that’s parallel to u. The result
when we subtract is going to be perpendicular to u (this is how Gram–Schmidt works).
We get
y·u
−13
1
−4/5
projy u =
u=
u=− u=
.
7/5
u·u
65
5
Then the other part is
2
−4/5
14/5
y − projy u =
−
=
.
3
7/5
8/5
We know that this should work, but let’s do a sanity check here. Is this vector actually
orthogonal to u, like it’s supposed to be?
56 56
14/5
4
−
= 0,
·
=
8/5
−7
5
5
as it must.
Problems for W 11/20:
6.3.3 Verify that the given vectors are an orthogonal set, and then find the projection of y
onto W = span(u1 , u2 ).
 
 
 
−1
1
−1





y = 4 , u1 = 1 , u2 = 1  .
3
0
0
We have
   
1
−1
u1 · u2 = 1 ·  1  = (1)(−1) + (1)(1) + (0)(0) = 0.
0
0
So they’re orthogonal.
The orthogonal projection is
y · u2
y · u1
u1 +
u2
u ·u
u ·u
1 1   2 2    
−1
−1
−1
1
 4  · 1    4  ·  1   
1
−1
3
3
0
0



=     1 +    1
1
1
−1
−1
0
0
1 · 1
 1 · 1 
0
0
0
0
 
   
1
−1
−1
3  5   
1 +
1 = 4 .
=
2
2
0
0
0
proju1 ,u2 y =
6.3.9 Let W be the subspace spanned by u1 , u2 and u3 . Write y as the sum of a vector in
W and a vector orthogonal to W .
 
 
 
 
4
1
−1
−1
3
1
3
0

 
 
 
y=
 3  , u1 = 0 , u2 =  1  , u3 =  1  ,
−1
1
−2
1
This time we want
y · u1
y · u2
y · u3
u1 +
u2 +
u3
u1 · u1
u2 · u2
u3 · u3
   
   
4
1
4
−1
 3  1
3 3
 ·    ·  
 3  0 1
 3   1  −1


−1
1
−1
−2  3 
1



 +
=     
+
1
1 0
−1
−1  1 
1 1 1
 3   3  −2
 · 
 · 
0 0
1 1
1
1
−2
−2
 
 
   34 
1
−1
−1
15

 3  0  0   58 
8
6
1







=  +
+
= 15  .
3 0
15  1  3  1   25 
28
1
−2
1
15
proju1 ,u2 ,u3 y =

  
4
−1
3 0
 ·  
 3   1  −1
−1
1 0

   
−1
−1  1 
0 0 1
 · 
1 1
1
1
6.3.12 Find the closest point to y in the subspace spanned by v1 and v2 .
 
 
 
3
1
−4
−1
−2
1

 
 
y=
 1  , v1 = −1 , v2 =  0  .
13
2
3
This is the same general deal as the last one.
y · v2
y · v1
v1 +
v2
projv1 ,v2 y =
v ·v
v ·v
1 1  2 2
   
3
3
1
−4
−1 −2
−1  1 
 ·    ·  
 1  −1 1
 1   0  −4
13
2 −2
13
3  
+     1 
=    
1
1 −1
−4
−4  0 
−2 −2 2
1 1 3
 · 
 · 
−1 −1
0 0
2
2
3
3
 
   
1
−4
−1
 26  1  −5
30 
−2
 +
  =  .
=
10 −1 26  0  −3
2
3
9
6.3.15 Let


5
y = −9 ,
5
 
−3
u1 = −5 ,
1
 
−3
u2 =  2  .
1
Find the distance from y to the plane in R3 spanned by u1 and u2 . (Hint: what point
in that plane is closest to y?)
We’re supposed to find the distance from y to the plane. To do that, we figure out
what the closest point on the plane is (which we know how to do), and then we figure
out the distance to that point.
Here we go again:
y · u2
y · u1
u1 +
u2
u1 · u1
u2 · u2
   
   
5
−3
5
−3
−9 · −5   −9 ·  2   
−3
−3
5
1
5
1



=     −5 +     2 
−3
−3
−3
−3
1
1
−5 · −5
 2 · 2 
1
1
1
1
 
   
−3
−3
3
35   −28    
−5 +
2 = −9 .
=
35
14
1
1
−1
proju1 ,u2 y =
The distance from y to the plane is the length of
     
5
−3
8
−9 − −9 = 0 .
5
−1
6
The length (which is our final answer) is then
√
82 + 02 + 62 = 10.
Problems for F 11/22:
6.4.1 The given set is a basis for a subspace W . Use the Gram–Schmidt process to produce
an orthogonal basis for W .
 
 
3
8
 0 ,  5 .
−1
−6
We want


3
v1 = x1 =  0 
−1

  
8
3
   5 · 0    
8
3
−1
−6
−1
x2 · v1
v1 =  5  −      0  =  5  .
v2 = x2 −
v1 · v1
3
3
−6
−1
−3
 0 · 0 
−1
−1
6.4.9 Find an orthogonal basis for the column space of the matrix


3 −5 1
1
1
1

A=
−1 5 −2 .
3 −7 8


3
1

v1 = x1 = 
−1
3

  
−5
3
1 1
  
  
 5  · −1  3   1 
−5
1
−7
3    
x2 · v1
−     1 = 3 
v1 = 
v2 = x2 −
5
v1 · v1
3
3 −1  3 
1 1 3
−7
−1
 · 
−1 −1
3
3
x3 · v2
x3 · v1
v1 −
v2
v3 = x3 −
v1 · v1
v2 · v2
   
   
1
1
3
1
1 1
1 3
  
   
  
−2 · −1  3  −2 ·  3   1  −3
1
1
−1  3   1 
8
3 1
8

 
   
=
−2 −  3   3  −1 −  1   1   3  =  1  .
1 1 3
 3   3  −1
3
8
 · 
 · 
−1 −1
3 3
3
3
−1
−1
These are all orthogonal, so looks like we got it right.
6.4.10 Find an orthogonal basis for the column space of the matrix below.


−1 6
6
 3 −8 3 

A=
 1 −2 6 
1 −4 −3
This is identical to the one above.
 
−1
3

v1 = x1 = 
1
1
x2 · v1
v1
v2 = x2 −
v1 · v1
   
6
−1
−8  3 
  
  
   
−2 ·  1  −1  6 
6
−1
3
−8
−4
1  3  −8 −36  3   1 
   
   

=
−2 − −1 −1  1  = −2 − 12  1  =  1 
3 3 1
−4
1
−1
−4
 · 
1 1
1
1
x3 · v2
x3 · v1
v1 −
v2
v3 = x3 −
v1 · v1
v2 · v2
   
   
6
−1
6
3
3 3
3 1
  
   
  
 6  ·  1  −1  6  ·  1   3 
6
3
−3
1  
−3
−1  
−     3 −     1 
=
6
−1
−1  1 
3
3 1
3 3 1
 1   1  −1
−3
 · 
 · 
1 1
1 1
1
1
−1
−1
 
   
 
−1
3
−1
6
3
 3  30  1  −1
6

 
   
=
 6  − 12  1  − 12  1  =  3  .
−3
1
−1
−1
6.4.13 The columns of Q below were obtained by running Gram–Schmidt orthonormalization
on the columns of A. Find an upper-triangular R with A = QR.




5
9
5/6 −1/6
1

7
5/6 
 , Q =  1/6

A=
−3 −5
−3/6 1/6 
1
5
1/6
3/6
The rule is R = QT A. This gives

5
9
5/6 1/6 −3/6 1/6 
7
T
1

R=Q A=

−1/6 5/6 1/6 3/6 −3 −5
1
5
6 12
=
.
0 6

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