7Transformation groups

IA Vectors and Matrices

7.2 Length preserving matrices

Theorem. Let P ∈ O(n). Then the following are equivalent:

(i) P is orthogonal

(ii) |P x| = |x|

(iii) (P x)

T

(P y) = x

T

y, i.e. (P x) · (Py) = x ·y.

(iv) If (v

1

, v

2

, ··· , v

n

) are orthonormal, so are (P v

1

, P v

2

, ··· , P v

n

)

(v) The columns of P are orthonormal.

Proof. We do them one by one:

(i) ⇒ (ii): |P x|

2

= (P x)

T

(P x) = x

T

P

T

P x = x

T

x = |x|

2

(ii) ⇒ (iii): |P (x + y)|

2

= |x + y|

2

. The right hand side is

(x

T

+ y

T

)(x + y) = x

T

x + y

T

y + y

T

x + x

T

y = |x|

2

+ |y|

2

+ 2x

T

y.

Similarly, the left hand side is

|P x + P y|

2

= |P x|

2

+ |P y| + 2(P x)

T

P y = |x|

2

+ |y|

2

+ 2(P x)

T

P y.

So (P x)

T

P y = x

T

y.

(iii) ⇒ (iv): (P v

i

)

T

P v

j

= v

T

i

v

j

= δ

ij

. So P v

i

’s are also orthonormal.

(iv) ⇒

(v): Take the

v

i

’s to be the standard basis. So the columns of

P

, being

P e

i

, are orthonormal.

(v) ⇒

(i): The columns of

P

are orthonormal. Then (

P P

T

)

ij

=

P

ik

P

jk

=

(P

i

) · (P

j

) = δ

ij

, viewing P

i

as the ith column of P . So P P

T

= I.

Therefore the set of length-preserving matrices is precisely O(n).