Solutions to Review Problems for Exam Two
1. Find the dimension and a basis for the set
S
of symmetric 2
×
2 matrices.
The basis is
{
1
0
0
0
,
0
0
0
1
,
0
1
1
0
}
and tThe dimension is 3.
2. Find the dimension and a basis for the set
S
of polynomials in
P
4
p
(
x
) such
that
p
(2) = 0.
The basis is
{
x

2
, x
2

4
, x
3

8
}
and the dimension is 3.
3. Suppose that the
m
×
n
matrices A and B are row equivalent. Which of the
following are True/False: (a) A and B have the same rank
(b) A and B have
the same rowspace
(c) A and B have the same column space
(d) A and B have the same nullspace
(a), (b) and (d) are true–see (19).
(c) is false, by the following counterexample:
A
=
1
0
1
0
and
B
=
1
0
0
0
.
4.
Suppose the vector space
V
has dimension
n
and
r < n < m
.
Which of
the following are True/False:
(a) Any set of
m
vectors must span
V
.
(b)
Any set of
m
vectors must be dependent.
(c) Any set of
r
vectors must be
independent.
(a) is false, by the counterexample
{
(1 0)
T
,
(2 0)
T
,
(3 0)
T
}
(b) is true –see (21).
(c) is false, by the counterexample
{
(1 0 0)
T
,
(2 0 0)
T
}
in
R
3
.
5. Suppose that a 3
×
5 matrix A has rank 3. Which of (a) (d) are True/False:
(a) The rows of A linearly independent
(b) The rows of A span
R
5
(c) The columns of A linearly independent
(d) The columns of A span
R
3
(e) Find the dimension of the nullspace of A.
Recall that the rank of
A
is the dimension of the rowspace, which equals the
dimension of the column space by (19).
(a) is true, since the rows are 3 vectors in the 3dimensional rowspace.
(b) is false since 3 rows cannot span a 5dimensional space.
(c) is false since 5 vectors in a 3dimensional space must be dependent.
(d) is true, since the column space is a 3dimensional subspace of
R
3
.
(e)
Rk
(
A
) +
Null
(
A
) = 5, so that
dim
(
nullspace
(
A
) =
Null
(
A
) = 5

3 = 2.
6. Let A be an
m
×
n
matrix, so that
T
A
:
F
n
→
F
m
. Which of the following
statements is equivalent to the statement that
T
A
is onto?
(a) rk(A) = m
(b) rk(A) = n
(c) the columns of A span
F
m
(d) the columns of A are independent
(e) for each
→
b
∈
F
m
, there is at most one solution
→
x
such that
A
→
x
=
→
b
(f) for each
→
b
∈
F
m
, there is at least one solution
→
x
such that
A
→
x
=
→
b
(g) A has a left inverse
(h) A has a right inverse
By definition,
T
A
is onto if and only if every
v
∈
F
m
is equal to
T
A
(
x
) =
Ax
for some
x
∈
R
n
. Since
A
(
x
1
, x
2
, . . . , x
n
)
T
=
x
1
A
1
+
· · ·
+
x
n
A
n
, this is equivalent
to saying that the columnspace of
A
is
R
m
.
Recall also that the rank(A) is the
dimension of the column space by (22).
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(a), (c), (f) and (h) are true.
T
a
onto
⇐⇒
(f) and (f)
⇐⇒
(c) by the discussion
above. (c)
⇐⇒
(a), since the columnspace is a subspace of
F
m
and therefore has
dimension
m
if and only of it equals
F
m
, which is if and only if the columns span
F
m
.
(f)
→
(h), since we can find a right inverse by simultaneously solving
Ax
=
e
i
for all
i
.
(h)
→
(f), since once we have
AR
=
I
, we can solve
Ax
=
b
by letting
x
=
Rb
.
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 Spring '09
 RUDYAK
 Linear Algebra, Vectors, basis, TA, Rowspace

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