1
Normpreserving (or lengthkeeping) linear operator
T
on
R
n
:

T
(
v
)

=

v
 ∀
v
∈
R
n
.
Example:
T
: linear operator on
R
2
that rotates a vector by
θ
.
⇒
T
is normpreserving.
Example:
U
: linear operator on
R
n
that has an eigenvalue
λ≠
±
1.
⇒
U
is
not
normpreserving, since for the corresponding
eigenvector
v
,

U
(
v
)

=
λ
v

=
λ
·

v
 ≠ 
v

.
Necessary conditions
for a linear operator to be normpreserving:
Let
Q
= [
q
1
q
2
"
q
n
] be the standard matrix of the linear operator.
Then (1)

q
j

=

Q
e
j

=

e
j

= 1, and
(2)

q
i
+
q
j

2
=

Q
e
i
+
Q
e
j

2
=

Q
(
e
i
+
e
j
)

2
=

e
i
+
e
j

2
= 2
=

q
i

2
+

q
j

2
, i.e.,
q
i
and
q
j
are orthogonal.
Example:
is an orthogonal matrix.
Proof
(a)
⇒
(b) with
Q
= [
q
1
q
2
"
q
n
],
q
i
•
q
i
= 1 = [
Q
T
Q
]
ii
∀
i
,
and
q
i
•
q
j
= 0 = [
Q
T
Q
]
ij
∀
i
≠
j
.
⇒
Q
T
Q
=
I
n
, and
Q
is invertible with
Q
−
1
=
Q
T
.
(b)
⇒
(c)
∀
u
,
v
∈
R
n
,
Q
u
•
Q
v
=
u
•
Q
T
Q
v
=
u
•
Q
−
1
Q
v
=
u
•
v
.
(
⇔
Q
T
Q
=
I
n
.)
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(c)
⇒
(d)
∀
u
∈
R
n
,

Q
u

= (
Q
u
•
Q
u
)
1/2
= (
u
•
u
)
1/2
=

u

.
(d)
⇒
(a) The above necessary conditions.
Corollary: (a)
Q
is orthogonal if and only if the rows
Q
of form an
orthonormal basis of
R
n
, or equivalently,
QQ
T
=
I
n
.
(b)
Q
is orthogonal if and only if
Q
T
is orthogonal.
Proof
Q
T
Q
=
I
n
⇔
QQ
T
=
I
n
⇔
Q
T
=
Q
−
1
⇔
(
Q
T
)
T
Q
T
=
I
n
.
Proof
(a)
QQ
T
=
I
n
⇒
1 = det(
I
n
) = det(
QQ
T
) = det(
Q
)det(
Q
T
)
= det(
Q
)
2
.
⇒
det(
Q
) =
±
1.
(b) (
PQ
)
T
=
Q
T
P
T
=
Q
−
1
P
−
1
= (
PQ
)
−
1
.
(c) (
Q
−
1
)
T
= (
Q
T
)
−
1
= (
Q
−
1
)
−
1
.
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 Spring '09
 Fong
 Linear Algebra, Matrices, Orthogonal matrix, linear operator, ∈ Rn

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