Solutions: Assignment 4
3.3.20
Find the redundant column vectors of the given matrix
A
“by inspection”.
Then find a basis of the image of
A
and a basis of the kernel of
A
.
A
=
1
0
5
3

3
0
0
0
1
3
0
0
0
0
0
0
0
0
0
0
The second and third columns are mutliples of the first.
And the fifth
column is 3 times the third column minus 12 times the first. So the second,
third, and fifth columns are redundant. And a basis for the image is just
1
0
0
0
and
3
1
0
0
. To get a basis for the kernel we look at
A
x
=
0. This
tells us that relationship between entries of
x
are just
x
1
=

5
x
3

3
x
4
+3
x
5
and
x
4
=

3
x
5
. So this gives us a basis of 3 elements:
x
2
=
1
x
3
=
0
x
5
=
0
x
2
=
0
x
3
=
1
x
5
=
0
x
2
=
0
x
3
=
0
x
5
=
1
0
1
0
0
0

5
0
1
0
0
3
0
0

3
1
3.3.22
Find the reduced rowechelon form of the given matrix
A
.
Then find a
basis of the image of
A
and a basis for the kernel of
A
.
A
=
2
4
8
4
5
1
7
9
3
2
4
8
4
5
1
7
9
3
→
1
2
4
4
5
1
7
9
3
→
1
2
4
0

3

15
0

5

25
→
1
2
4
0
1
5
0

5

25
→
1
0

6
0
1
5
0
0
0
There are leading ones in the first two columns of
rref
(
A
), So a basis for
the image is
2
4
7
and
4
5
9
. We know that
x
is in the kernel is equiv
alent to
A
x
=
0. That’s equivalent to
rref
(
A
)
x
=
0. Which is equivalent
1
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to
x
1
= 6
x
3
and
x
2
=

5
x
3
. Which is equivalent to
x
=
x
3
6

5
1
. So
6

5
1
is a basis for the kernel.
3.3.33
A subspace
V
of
R
n
is called a hyperplane if
V
is defined by the homoge
neous equation
c
1
x
1
+
c
2
x
2
+
. . .
+
c
n
x
n
= 0
where at least one of the coefficients
c
i
is nonzero.
What is the dimen
sion of a hyperplane in
R
n
?
Justify your answer carefully.
What is a
hyperplane in
R
3
? What is it in
R
2
.
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 Spring '08
 MOVSHEV
 Linear Algebra, Vectors, Vector Space, basis, linear transformation, row space

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