(Partial) Solutions to Homework 7
Problem 20.3:
Let
P
n
(
t
) be the vector space of degree
n
polynomials, and
d
dt
:
P
n
(
t
)
7→
P
n

1
(
t
) be the derivative operator. Find the matrix of
d
dt
in the
bases
{
1
, t, . . . , t
n
}
for
P
n
(
t
) and
{
1
, t, . . . , t
n

1
}
for
P
n

1
(
t
).
Solution:
In order to find the matrix of a given linear transformation with
respect to some bases, we calculate the linear transformation on each element
of the basis of the domain, and express the answer in terms of the basis of the
range.
d
dt
(1) = 0 = 0
·
1 + 0
·
t
+ 0
·
t
2
+
· · ·
d
dt
(
t
) = 1 = 1
·
1 + 0
·
t
+ 0
·
t
2
+
· · ·
d
dt
(
t
2
) = 2
t
= 0
·
1 + 2
·
t
+ 0
·
t
2
+
· · ·
d
dt
(
t
3
) = 3
t
2
= 0
·
1 + 0
·
t
+ 3
·
t
2
+
· · ·
.
.
.
Then each list of coefficients becomes a column vector in the matrix of the linear
transformation with respect to those two bases:
M
=
0
1
0
0
· · ·
0
0
2
0
· · ·
0
0
0
3
· · ·
.
.
.
.
.
.
Not part of the problem, but note that this matrix is not invertible (its deter
minant is zero). This makes sense, since differentiation is not invertible (if I tell
you
d
dt
(
f
(
t
)) = 3, you don’t know if
f
(
t
) = 3
t
, or
f
(
t
) = 3
t
+ 2, etc.).
Problem 21.3:
Let
u, v
be independent vectors in
R
3
, and
P
= span
{
u, v
}
be the plane spanned by
u
and
v
.
i.
Is the vector
v
⊥
=
v

u
·
v
u
·
u
u
in the plane
P
?
ii.
What is the angle between
v
⊥
and
u
?
iii.
Given your solution to the above, how can you find a third vector perpen
dicular to both
u
and
v
⊥
?
iv.
Construct an orthonormal basis for
R
3
from
u
and
v
.
v.
Test your abstract formulae starting with
u
=
(
1
2
0
)
and
v
=
(
0
1
1
)
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Solution:
i. Yes, this vector is in the plane
P
because it is a linear combination of
u
and
v
.
ii. The angle between
v
⊥
and
u
is
π/
2 because we can calculate the dot
product:
v
⊥
·
u
= (
v

u
·
v
u
·
u
u
)
·
u
=
v
·
u

(
u
·
v
)(
u
·
u
)
u
·
u
= 0
(remember, the dot product is symmetric, i.e.,
u
·
v
=
v
·
u
.)
iii. We could get a third vector orthogonal to the both
u
and
v
⊥
by taking
the cross product of the two.
iv. An orthogonal basis is given by
{
u, v
⊥
, u
×
v
⊥
}
. To get an orthonormal
basis, we divide each vector by its magnitude:
{
u

u

,
v
⊥

v
⊥

,
u
×
v
⊥

u
×
v
⊥

}
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Pon
 Linear Algebra, Polynomials, Derivative, Vector Space, basis

Click to edit the document details