Math
2940
Solutions, Fall
2011
Section
4
.
1
4
)I
f
AB
= 0 then the columns of
B
are in the
nullspace
of
A
. The rows of
A
are in the
left nullspace
of
B
.
Say
A
and
B
are both 3
×
3 of rank 2. Then the nullspace of
A
has dimension 3
−
2=1
.
But this nullspace contains the columns of
B
, whose span has dimension 2, a contradiction.
11
)F
o
r
A
=
°
12
36
±
the columns are visibly dependent (as are the rows) so the vector
°
1
3
±
is a basis for the column space and (1
,
2) is a basis for the row space. The row
reduced echelon form of
A
is
°
00
±
so a basis for the nullspace is
°
−
2
1
±
. Finally,
A
T
=
°
13
26
±
→
°
±
so the nullspace of
A
T
has basis
°
−
3
1
±
. Note the orthogonality
between the column space of
A
and the nullspace of
A
T
. And between the nullspace of
A
and the rowspace of
A
.
For
B
=
°
10
30
±
the ±rst column forms a basis for the column space and the rows are
visibly dependent so the vector
°
1
3
±
is a basis for the column space and (1
,
0) is a basis for
the row space. The row reduced echelon form of
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '05
 HUI
 Math, Linear Algebra, Algebra, Dot Product, column space

Click to edit the document details