[2b.] The system (
A
T
A
)
x
=
A
T
b
is
•
3
6
6
18
‚
x
=
•
18
30
‚
with (unique) solution
•
8

1
‚
. This is the least square solution to the system
Ax
=
b
.
[2c.] The vector
p
=
1
4
4
is in the column space of
A
. Moreover, the vector
0

1
1
spans the nullspace of
P
. Hence there are infinitely many choices for the vector
q
and they are
q
=
p
+
t
0

1
1
,
t
∈
R
.
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3.
[15 marks]
3a.
[4 marks]
Give a 3
×
3matrix
A
with the following properties:
i.
A
T
=
A

1
.
ii. det(
A
) = 1. (
A
is not allowed to be a diagonal matrix)
3b.
[4 marks]
Give a 3
×
3matrix with the following properties:
i.
A
T
=
A
.
ii.
A
2
=
A
.
iii. rk(
A
) = 1. (
A
is not allowed to be a diagonal matrix)
3c.
[4 marks]
Suppose
A
is a 5
×
3matrix with orthonormal columns. Evaluate
the following determinants:
i det(
A
T
A
)
ii det(
AA
T
)
iii det(
A
(
A
T
A
)

1
A
T
).
3d.
[3 marks]
Which value(s) of
α
∈
R
give det(
A
) = 0, if
A
=
α
2
3

α
α
0
3
2
5
?
Sol.
[3a.] We consider, for example an orthogonal matrix
A
, with det(
A
) = 1. Permutation
matrices such as
0
1
0
0
0
1
1
0
0
or
0
0
1
1
0
0
0
1
0
have this property.
[3b.] We may consider a projection matrix of rank 1: for example (cfr. question 2c.)
A
=
v
(
v
T
v
)

1
v
T
, where
v
=
1
2
3
. Namely the matrix
A
=
1
14
1
2
3
2
4
6
3
6
9
.
[3c.]
det(
A
T
A
) = det(
I
) = 1
det(
AA
T
) = 0
det(
A
(
A
T
A
)

1
A
T
) = 0
AA
T
must have dependent columns and determinant zero because
A
(
A
T
x
) = 0
for
any nonzero vector
x
in the nullspace of
A
T
. The 3
×
5matrix
A
T
has 3 linearly
independent (orthonormal!) rows and a nontrivial nullspace of dimension 5

3 = 2.
Notice that det(
A
(
A
T
A
)

1
A
T
) = det(
AA
T
) = 0 as
A
T
A
=
I
.
[3d.] det(
A
) = 5
α
(
α

1) = 0. Therefore,
α
= 0 or
α
= 1.
4.
[15 marks]
Suppose the following information is known about a matrix
A
:
i.
A
2
0
0
=
2
4

6
ii.
A
0

1
0
=

2

4
6
iii.
A
is symmetric.
The following questions refer to any
matrix
A
with the above properties