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Lecture 36

# MATH125 Lecture 36: 36

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University of Alberta

Mathematics

MATH125

Nikita

Winter

Description

Lecture 36, April 10, 2017.
Example. Show that the following matrix is orthogonal and ▯nd its
inverse:
[ ]
B = cos▯ ▯sin▯
sin▯ cos▯
Solution.
[ 2 2 ] [ ]
B B = cos ▯ + sin ▯ ▯cos▯ sin▯ + sin▯ cos▯ = 1 0
▯sin▯ cos▯ + cos▯ sin▯ sin ▯ + cos ▯ 0 1
Therefore B is orthogonal and
[ ]
▯1 T cos▯ sin▯
B = B = ▯sin▯ cos▯
Remark. Matrix B is the matrix of a rotation through the angle ▯
in R . Any rotation is a length-preserving transformation, also called
isometry. The next theorem shows that every orthogonal matrix trans-
formation is an isometry. Orthogonal matrices also preserve dot prod-
ucts. In fact, orthogonal matrices are characterized by either one of
these two properties.
Theorem. Let Q be an n ▯ n matrix. The following statements are
equivalent:
a.Q is orthogonal.
b. ▯▯Qx▯▯ = ▯▯x▯▯ for every x in R .
n
c. Qx ▯ Qy = x ▯ y for every x and y in R .
T
Proof. Note that x ▯ y = x y.
(a) ▯ (c). Assume that Q is orthogonal. Then Q Q = I, and we have
Qx ▯ Qy = (Qx) Qy = x Q Qy = x Iy = x y = x ▯ y:
(c) ▯ (b). If Qx▯Qy = x▯y for every x;y, then, taking y = x, we have
Qx ▯ Qx = x ▯ x, so
√ ▯
▯▯Qx▯▯ = Qx ▯ Qx = x ▯ x = ▯▯x▯▯:
(b) ▯ (c). Assume that (b) holds and let q ienote the ith column of
Q. We have
1 1
x ▯ y = (▯▯x + y▯▯ ▯ ▯▯x ▯ y▯▯ ) (▯▯Q(x +

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