Problem 22
AT A must be a diagonal matrix since it represents every column of A times every row of A.
When the two columns are different the result is zero. When they are the same the norm
(squared) of that column results.
Problem 23
The lines 3x + y = b1 and 6x + 2y = b2 are parallel. They are the same line if 2b1 = b2. Then
[b1, b2]T is perpendicular to the left nullspace of
3 1
A =
6 2
−2
or
. Note we can check that this vector is an element of the left nullspace by computing
1
b
−2 1
1
= −2b
b
1 + b2 = −2b1 + 2b2 = 0
2
The nullspace of the matrix is the line 3x + y = 0. One vector in this nullspace is [−1, 3]T .
Problem 24
Part (a): As discussed in the book if two subspaces are orthogonal then they can only meet
at the origin. But for the two planes given we have many intersections. To find them we
solve the system given by
x
1 1
1
y
1 1 −1
= 0 ,
z
then the point (x, y, z) will be on both planes. Performing row reduction we obtain
x
1 1 0
y
0 0 1
= 0
z
so we see that z = 0 and x + y = 0, giving the fact that any vector that is a multiple of
1
−1 is in both planes and these two spaces cannot be orthogonal.
0
Part (b): The two lines specified are described as the spans of the two vectors
2
1
4
and
−3
5
2
respectively. For their subspaces to be orthogonal, the subspace generating vectors must be
1
orthogonal. In this case 2 4 5 T −3 = 2 − 12 + 10 = 0 and they are orthogonal.
2
We still need to show that they are not orthogonal components. To do so it suffices to find
a vector orthogonal to one space that is not in the other space. Consider 2 4 5 , which as a nullspace given by setting the free variables equal to a basis and solving for the pivot
variables. Since the free variables are x2 and x3 we have a first vector in the nullspace given
by setting x2 = 1,x3 = 0, which implies that x1 = −2. Also setting x2 = 0, x3 = 1, we have
that x1 = −5, giving two vector of
2
−2
−5/2
1
and
0
0
1
−2
2
Now consider the vector 1 it is orthogonal to 4 and thus is in its orthogonal
0
5
1
complement. This vector however is not in the span of −3 . Thus the two spaces are
2
not the orthogonal complement of each other.
0
1
Part (c): Consider the subspaces spanned by the vectors
, and
, respectively.
1
1
They meet only at the origin but are not orthogonal.
Problem 25
Let
1 2 3
A = 2 4 5 ,
3 6 7
then A has [1 , 2 , 3]T in both its row space and its nullspace. Let B be defined by
1 1 −1
B = 2 2 −2 ,
3 3 −3
then B has [1 , 2 , 3]T in the column space of B and
1
0
B 2 = 0 .
3
0
Now v could not be both in the row space of A and in the nullspace of A. Also v could not
both be in the column space of A and in the left nullspace of A. It could however be in the
row space and the left nullspace or in the nullspace and the left nullspace.
Problem 26
A basis for the left nullspace of A.
Section 4.2 (Projections)
Problem 1 (simple projections)
Part (a): The coefficient of projection ˆ
x is given by
aT b
1 + 2 + 2
5
ˆ
x =
=
=
aT a
1 + 1 + 1
3
so the projection is then
1
aT b
5
p = a
=
1
aT a
3
1
and the error e is given by
1
1
1
−2
e = b − p = 2 − 1 =
1
3
.
2
1
1
To check that e is perpendicular to a we compute eT a = 1 (−2 + 1 + 1) = 0.
3
Part (b): The projection coefficient is given by
aT b
−1 − 9 − 1
ˆ
x =
=
= −1 .
aT a
1 + 9 + 1
so the projection p is then
1
p = ˆ
xa = −a = 3 .
1
The error e = b − p = 0 is certainly orthogonal to a.
Problem 2 (drawing projections)
Part (a): Our projection is given by
aT b
1
cos(θ)
p = ˆ
xa =
a = cos(θ)
=
aT a
0
0
Part (b): From Figure XXX of b onto a is zero. Algebraically we have
aT b
1 − 1 1
0
p = ˆ
xa =
a =
=
aT a
2
−1
0
Problem 3 (computing a projection matrix)
Part (a): The projection matrix P equals P = aaT , which in this case is
aT a
1
1
1 1 1
1
1 1 1
1
P =
=
1 1 1
3
3
.
1 1 1
For this projection matrix note that
3 3 3
1 1 1
1
1
P 2 =
3 3 3
1 1 1
9
= 3
= P .
3 3 3
1 1 1
The requested product P b is
1 1 1 1
5
1
1
P b =
1 1 1
2
5
3
= 3
.
1 1 1
2
5
Part (b): The projection matrix P equals P = aaT , which in this case is
aT a
−1
−3
−1 −3 −1
−1
1 3 1
1
P =
=
3 9 3
1 + 9 + 1
11
.
1 3 1
For this projection matrix note that P 2 is given by
1 3 1 1 3 1
11 33 11
1 3 1
1
1
1
P 2 =
3 9 3
3 9 3
33 99 33
3 9 3
112
= 112
= 11
= P .
1 3 1
1 3 1
11 33 11
1 3 1
The requested product P b is then given by
1 3 1 1
11
1
1
1
P b =
3 9 3
3
33
3
11
= 11
=
.
1 3 1
1
11
1
Problem 4 (more calculations with projection matrices)
Part (a): Our first projection matrix is given by P1 = aaT which in this case is
aT a
1
1 0
P
1 =
1 0 =
0
0 0
Calculating P 21 we have that
1 0
P 21 =
= P
0 0
1 ,
as required.
Part (b): Our second projection matrix is given by P2 = aaT which in this case is
aT a
1 1
1 1
−1
P
2 =
1 −1 =
2
−1
2
−1
1
Calculating P 2 we have that
2
1 2
−2
1 1
−1
P 2 =
=
= P
2
4
−2
2
2
−1
1
2 ,
as required. In each case, P 2 should equal P because the action of the second application of
our projection will not change the vector produced by the action of the first application of
our projection matrix.
Problem 5 (more calculations with projection matrices)