4
)
A
=

1
1
0

1
0
1
0

1
1
. Note the sum of the columns is the zero vector so the rank of
A
is
at most 2. It is in fact 2 as the first two columns are visibly dependent.
Taking
b
to be the first column of
A
, we can solve
Ax
=
b
.
A vector perpendicular to the first two columns will not be in the column space of
A
.
Playing with numbers a bit (or taking the cross product of the first two columns) gives
c
=
1

1
1
as a vector for which there is no solution to
Ax
=
c
.
For a given
b
, if we want a solution,
b
must lie in the plane spanned by the first two
columns. this plane is perpendicular to
c
, so
b
must be perpendicular to
c
.
7
)
A
T
CA
=

1

1
0
1
0

1
0
1
1
1
0
0
0
2
0
0
0
2

1
1
0

1
0
1
0

1
1
=

1

1
0
1
0

1
0
1
1

1
1
0

2
0
2
0

2
2
=
3

1

2

1
3

2

2

2
4
.
The sum of the columns is zero so
1
1
1
is in the nullspace.
The first two columns are
visibly independent so the rank is 2. In
A
we have the loo along edge 1, then edge 3, then
edge 2 backwards. This corresponds to Row 1 + Row 3  Row 2 being the 0 Row.
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 '05
 HUI
 Linear Algebra, Algebra, −1, column space

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