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18.06 Problem Set 4 Solution
Due Wednesday, 11 March 2009 at 4 pm in 2106.
Total: 175 points.
Problem 1:
A
is an
m
×
n
matrix of rank
r
. Suppose there are righthandsides
~
b
for which
A~x
=
~
b
has no solution.
(a) What are all the inequalities (
<
or
≤
) that must be true between
m
,
n
, and
r
?
(b)
A
T
~
y
=
~
0 has solutions other than
~
y
=
~
0. Why must this be true?
Solution
(15 points = 10+5)
(a) First of all, the rank
r
of a matrix is the number of column (row) pivots, it
must be less than equal to
m
and
n
. If the matrix were of full row rank, i.e.,
r
=
m
,
it would imply that
A~x
=
~
b
always has a solution; we know that this is not the case,
and hence
r
6
=
m
. To sum up, the inequalities among
m,n,r
are
r
≤
n,r < m
.
(b) Since
A
T
is an
n
×
m
matrix, the null space
N
(
A
T
) has dimension
m

r
,
which is positive by (a). Hence,
A
T
~
y
=
~
0 has solutions other than
~
y
=
~
0.
Problem 2:
A
is an
m
×
n
matrix of rank
r
. Which of the four fundamental
subspaces are the same for:
(a)
A
and
±
A
A
²
(b)
±
A
A
²
and
±
A A
A A
²
Explain why all three matrices
A
,
±
A
A
²
, and
±
A A
A A
²
must have the same rank
r
.
Solution
(20 points = 10+10)
(a) Note that if we do invertible row operations on the matrix
±
A
A
²
, we may
kill the bottom
A
and get
±
A
0
²
.
1
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View Full Document Nullspace: Since the nullspace is invariant under row operations, the two matri
ces have the same nullspace.
Column space: Since the two matrices do not have the same number of rows,
their column space must not be the same.
Row space: Since the row space is also invariant under row operations, the two
matrices have the same row space.
Left nullspace: The transpose of
±
A
A
²
is
(
A
T
A
T
)
. Hence, the left nullspace is
a subspace of
R
2
m
. However, the left nullspace is a subspace of
R
m
. They cannot
be the same.
(b) Using invertible column operations, we can turn
±
A A
A A
²
into
±
A
0
A
0
²
.
The nullspace and the row space of
±
A A
A A
²
are subspaces of
R
2
n
, whereas the
nullspace and the row space of
±
A
A
²
are subspaces of
R
n
. They cannot be the same.
Since the column space and the left nullspace are invariant under column oper
ations, the two matrices have the same column space and left nullspace.
REMARK: For an
m
×
n
matrix of rank
r
, we have
Fundamental space
subspace of
dimension
Nullspace
R
n
n

r
Column space
R
m
r
Row space
R
n
r
Left nullspace
R
m
m

r
Problem 3:
Find a basis for each of the four subspaces for
A
=
0 1 2 3 4
0 1 2 4 6
0 0 0 1 2
Solution
(25 points = 5(echelon form)+5+5+5+5)
We ﬁrst write
A
as in rowreduced echelon form.
0 1 2 3 4
0 1 2 4 6
0 0 0 1 2
;
0 1 2 3 4
0 0 0 1 2
0 0 0 1 2
;
0 1 2 3 4
0 0 0 1 2
0 0 0 0 0
;
0 1 2 0

2
0 0 0 1
2
0 0 0 0
0
2
The second and the fourth variables are the pivots. The ﬁrst, third, and the ﬁfth
variables are free variables.
The row operation matrix
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This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Spring '08 term at MIT.
 Spring '08
 Strang

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