b.
A
=
QQ
T
, for some matrix
Q
with orthonormal columns
u
1
, . . . , u
m
. Note that
Q
T
Q
=
I
m
, since the
ij
th entry of
Q
T
Q
is
u
i
·
u
j
.
Then
A
2
=
QQ
T
QQ
T
=
Q
(
Q
T
Q
)
Q
T
=
QI
m
Q
T
=
QQ
T
=
A
.
43. Examine how
A
acts on
u
, and on a vector
v
orthogonal to
u
:
Au
= (2
uu
T

I
3
)
u
= 2
uu
T
u

u
=
u,
since
u
T
u
=
u
·
u
=
u
2
= 1
.
Av
= (2
uu
T

I
3
)
v
= 2
uu
T
v

v
=

v,
since
u
T
v
=
u
·
v
= 0
.
Since
A
leaves the vectors in
L
= span(
u
) unchanged and reverses the vectors in
V
=
L
⊥
,
it represents the
reflection about L
.
Note that
B
=

A
, so that
B
reverses the vectors in
L
and leaves the vectors in
V
unchanged; that is,
B
represents the reflection about
V
.
44. Note that
A
T
is an
m
×
n
matrix. By Facts 3.3.7 and 5.3.9c we have
dim(ker(
A
T
)) =
n

rank(
A
T
) =
n

rank(
A
).
By Fact 3.3.6, dim(im(
A
)) = rank(
A
), so that dim(im(
A
)) + dim(ker(
A
T
)) =
n
.
45. Note that
A
T
is an
m
×
n
matrix. By Facts 3.3.7 and 5.3.9c, we have
dim(ker(
A
)) =
m

rank(
A
) and dim(ker(
A
T
)) =
n

rank(
A
T
) =
n

rank(
A
),
so that dim(ker(
A
)) = dim(ker(
A
T
)) if (and only if)
A
is a square matrix.
260
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ISM:
Linear Algebra
Section 5.3
46. By Fact 5.2.2, the columns
u
1
, . . . , u
m
of
Q
are orthonormal. Therefore,
Q
T
Q
=
I
m
, since
the
ij
th entry of
Q
T
Q
is
u
i
·
u
j
.
If we multiply the equation
M
=
QR
by
Q
T
from the left then
Q
T
M
=
Q
T
QR
=
R
, as
claimed.
47. By Fact 5.2.2, the columns
u
1
, . . . , u
m
of
Q
are orthonormal. Therefore,
Q
T
Q
=
I
m
, since
the
ij
th entry of
Q
T
Q
is
u
i
·
u
j
.
By Fact 5.3.9a, we now have
A
T
A
= (
QR
)
T
QR
=
R
T
Q
T
QR
=
R
T
R
.
48. As suggested, we consider the
QR
factorization
A
T
=
PR
of
A
T
, where
P
is orthogonal and
R
is upper triangular with positive diagonal entries.
By Fact 5.3.9a,
A
= (
PR
)
T
=
R
T
P
T
.
Note that
L
=
R
T
is lower triangular and
Q
=
P
T
is orthogonal.
49. Yes! By Exercise 5.2.45, we can write
A
T
=
PL
, where
P
is orthogonal and
L
is lower
triangular.
By Fact 5.3.9a,
A
= (
PL
)
T
=
L
T
P
T
.
Note that
R
=
L
T
is upper triangular, and
Q
=
P
T
is orthogonal (by Exercise 11).
50. a. If an
n
×
n
matrix
A
is orthogonal and upper triangular, then
A

1
is both lower
triangular (since
A

1
=
A
T
) and upper triangular (being the inverse of an upper
triangular matrix; compare with Exercise 2.3.35c).
Therefore,
A

1
=
A
T
is a diagonal matrix, and so is
A
itself. Since
A
is orthogonal
with positive diagonal entries, all the diagonal entries must be 1, so that
A
=
I
n
.
b. Using the terminology suggested in the hint, we observe that
51. a. Using the terminology suggested in the hint, we observe that
I
m
=
Q
T
1
Q
1
= (
Q
2
S
)
T
Q
2
S
=
S
T
Q
T
2
Q
2
S
=
S
T
S
, so that
S
is orthogonal, by Fact 5.3.7.
b. Using the terminology suggested in the hint, we observe that
R
2
R

1
1
is both orthogonal
(let
S
=
R
2
R

1
1
in part a) and upper triangular, with positive diagonal entries. By
Exercise 50a, we have
R
2
R

1
1
=
I
m
, so that
R
1
=
R
2
. Then
Q
1
=
Q
2
R
2
R

1
1
=
Q
2
, as
claimed.