Solution.
This space is just Col
A
, where
A
=
1
0
2
2
3
0
1

2

1

1

2
2

8
10

6
3
3
0
3
9
. So we just
need to find a basis for Col
A
. This part is left to you.
Ex. 4.4.27 Determine whether
{
1 + 2
t
3
,
2 +
t

3
t
2
,

t
+ 2
t
2

t
3
}
is linearly independent or not.
Solution.
Say
E
=
{
1
, t, t
2
, t
3
}
.
Then
{
[1 + 2
t
3
]
E
,
[2 +
t

3
t
2
]
E
,
[

t
+ 2
t
2

t
3
]
E
}
=
1
0
0
2
,
2
1

3
0
,
0

1
2

1
is linearly independent iff
{
1 + 2
t
3
,
2 +
t

3
t
2
,

t
+ 2
t
2

t
3
}
is linearly independent.
As
1
2
0
0
1

1
0

3
2
2
0

1
∼
1
2
0
0
1

1
0
0
1
0
0
0
, every column is a
pivot column, the columns are linearly independent. (The matrix is not onto, hence
the columns do not span
R
4
.)
Ex. 4.6.15 If
A
is a 3
×
7 matrix, find the smallest and the largest possible dimension of Nul
A
,
Col
A
, Row
A
. Also find the minimum and the maximum rank of
A
.
Solution.
For a 3
×
7 matrix, at most it has 3 pivots and at least it has 0 pivot, hence
max rank
A
= max dim Col
A
= max dim Row
A
= 3 (for example,
A
=
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
)
and min rank
A
= min dim Col
A
= min dim Row
A
= 0 (for example,
A
is the zero ma
trix). As we have rank theorem, saying rank
A
+dim Nul
A
= 7, hence max dim Nul
A
=
7 and min dim Nul
A
= 4.
7
Ex. 4.7.9 Find
P
C←B
, where
B
=
4
4
,
8
4
and
C
=
2
2
,

2
2
. Find
C
coordinate
of
4
4
+
8
4
by using
P
C←B
.
Solution.
2

2
4
8
2
2
4
4
∼
1
0
2
3
0
1
0

1
. Hence
P
C←B
=
2
3
0

1
. And
[
4
4
+
8
4
]
C
=
P
C←B
[
4
4
+
8
4
]
B
=
2
3
0

1
1
1
=
5

1
Ex. 4.7.14 Find
P
C←B
, where
B
=
{
1
,
1 +
t,
1 +
t
+
t
2
}
and
C
=
{
t
2
, t,
1
}
.
Solution.
It is sufficient to consider
1
0
0
,
1
1
0
,
1
1
1
and
0
0
1
,
0
1
0
,
1
0
0
.
Then
0
0
1
1
1
1
0
1
0
0
1
1
1
0
0
0
0
1
∼
1
0
0
0
0
1
0
1
0
0
1
1
0
0
1
1
1
1
. Hence
P
C←B
=
0
0
1
0
1
1
1
1
1
.
Eigenvalues and Eigenvectors.
(1)
~x
is an eigenvector of
A
∈
R
n
×
n
corresponding to eigenvalue
λ
iff
~x
6
=
~
0 and
A~x
=
λ~x
.
(2) Eigenspace corresponding to
m
is
E
m
=
{
eigenvectors corresponding to
m
} ∪
n
~
0
o
.
(3) Eigenvalues are roots of the characteristic equation det(
A

λI
) = 0.
(4) Eigenspace corresponding to
m
is
E
m
= Nul(
A

mI
).
(5) Algebraic multiplicity of
m
is
a
if det(
A

λI
) = (
m

λ
)
a
· · ·
.
(6) Geometric multiplicity of
m
is dim
E
m
.
Properties.
(1) The eigenvalues of a triangular matrix are the entries on its diagonal.
(2) Eigenvectors corresponding to distinct eigenvalues are linearly independent.
(3) If
A
is similar to
B
, that is,
A
=
PBP

1
for some invertible
P
, then
A
and
B
have the
same characteristic polynomial, same eigenvalues with the same algebraic multiplicities
and the same geometric multiplicities.
8
Matrix for a linear transformation.
(1) The matrix for
T
:
V
→
W
relative to the bases
B
and
C
is
A
=
h
[
T
~
b
1
]
C
· · ·
[
T
~
b
n
]
C
i
,
where
B
,
C
are bases for
V
,
W
, respectively. It will satisfy [
T~x
]
C
=
A
[
~x
]
B
and
A
=
P

1
C
TP
B
. In particular
T
:
V
→
V
is the identity map, then
A
=
P
C←B
.
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 Spring '08
 Chorin
 Differential Equations, Linear Algebra, Algebra, Determinant, Equations, basis, ~vP