CAAM 336
·
DIFFERENTIAL EQUATIONS
Problem Set 4
·
Solutions
Posted Wednesday 15 September 2010. Due Wednesday 22 September 2010, 5pm.
1. [20 points]
The equation
x
1
+
x
2
+
x
3
= 0 deﬁnes a plane in
R
3
that passes through the origin.
(a) Find two linearly independent vectors in
R
3
whose span is this plane.
(b) Find the point in this plane closest (in the standard Euclidean norm,
k
z
k
=
√
z
T
z
) to the vector
v
=
1
0
1
by formulating this as a best approximation problem. (You may use MATLAB to invert a matrix.)
Solution.
(a) Since two linearly independent vectors determine a plane, we simply need to ﬁnd two linearly
independent vectors that satisfy
x
1
+
x
2
+
x
3
= 0. One can do this by inspection, for example,
and ﬁnd
1

1
0
,
0
1

1
.
However, it would be nice to have an orthogonal basis for this space. To do that, pick one vector,
say the ﬁrst vector given above; set the second vector to be (
α,β,γ
)
T
. We would like the this
vector to be in the plane:
α
+
β
+
γ
= 0
and to be orthogonal to the ﬁrst vector:
1

1
0
T
α
β
γ
=
α

β
+ 0 = 0
.
This gives two equations in three unknowns, which will be satisﬁed if
β
=
α
and
γ
=

2
α
for any
α
, i.e., we have the vector
α
α

2
α
.
With
α
= 1, we have two orthogonal vectors whose span is the desired plane:
x
=
1

1
0
,
y
=
1
1

2
.
(b) The closest point in the plane to the vector
v
is found solving the usual bestapproximation
problem matrix equation:
±
x
T
x x
T
y
y
T
x y
T
y
²±
c
1
c
2
²
=
±
x
T
v
y
T
v
²
,
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View Full Documentthat is,
±
2 0
0 6
²±
c
1
c
2
²
=
±
1

1
²
.
The orthogonality of the vectors
x
and
y
make this an easy problem to solve:
c
1
= 1
/
2
,
c
2
=

1
/
6
.
Thus, the best approximation to
v
= (1
,
0
,
1)
T
is the vector
b
v
=
c
1
x
+
c
2
y
=
1
/
2

1
/
6

1
/
2

1
/
6
1
/
3
=
1
/
3

2
/
3
1
/
3
.
We can verify this answer by checking (1) that
b
v
is in the desired plane: 1
/
3

2
/
3+1
/
3 = 0, and
(2) verifying that the error
v

b
v
=
2
/
3

2
/
3
2
/
3
is orthogonal to the two basis vectors
x
and
y
for the plane, (
v

b
v
)
T
x
= (
v

b
v
)
T
y
= 0.
2. [25 points]
Recall that a linear operator
P
is a projection from the vector space
V
to the vector space
V
provided
P
2
=
P
, that is,
P
(
Pf
) =
Pf
for all
f
∈
V
. Consider
V
=
C
[

1
,
1] with the usual inner product
(
u,v
) =
Z
1

1
u
(
x
)
v
(
x
)
dx,
and the two linear operators
P
e
and
P
o
the project a function onto their even and odd parts. That is,
(
P
e
f
)(
x
) =
f
(
x
) +
f
(

x
)
2
,
(
P
o
f
)(
x
) =
f
(
x
)

f
(

x
)
2
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 Fall '09
 Tompson
 Linear Algebra

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