Page 163 - Linear Algebra Done Right

Isometries
2
If θ ∈ R, then the operator on R of counterclockwise rotation (cen-
tered at the origin) by an angle of θ has matrix 7.39 with respect to
the standard basis, as you should verify. The next result states that ev- 151
ery isometry on a real inner-product space is composed of pieces that
look like rotations on two-dimensional subspaces, pieces that equal the
identity operator, and pieces that equal multiplication by −1.
7.38 Theorem: Suppose that V is a real inner-product space and This theorem implies
S ∈L(V). Then S is an isometry if and only if there is an orthonormal that an isometry on an
basis of V with respect to which S has a block diagonal matrix where odd-dimensional real
each block on the diagonal is a 1-by-1 matrix containing 1 or −1 or a inner-product space
2-by-2 matrix of the form must have 1 or −1 as
an eigenvalue.

cos θ − sin θ
7.39 ,
sin θ cos θ
with θ ∈ (0,π).

Proof: First suppose that S is an isometry. Because S is normal,
there is an orthonormal basis of V such that with respect to this basis
S has a block diagonal matrix, where each block is a 1-by-1 matrix or a
2-by-2 matrix of the form

a −b
7.40 ,
b a
with b> 0 (see 7.25).
If λ is an entry in a 1-by-1 along the diagonal of the matrix of S (with
respect to the basis mentioned above), then there is a basis vector e j
such that Se j = λe j . Because S is an isometry, this implies that |λ|= 1.
Thus λ = 1or λ =−1 because these are the only real numbers with
absolute value 1.
Now consider a 2-by-2 matrix of the form 7.40 along the diagonal of
the matrix of S. There are basis vectors e j ,e j+1 such that

Se j = ae j + be j+1 .

Thus
2
2
2
2
1 = e j = Se j = a + b .
The equation above, along with the condition b> 0, implies that there
exists a number θ ∈ (0,π) such that a = cos θ and b = sin θ. Thus the