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Lecture 37

# MATH125 Lecture 37: 37

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University of Alberta

Mathematics

MATH125

Nikita

Winter

Description

Final Lecture 37, April 12, 2017.
Theorem. Let Q be an orthogonal matrix.
a. The matrix Q ▯1 is orthogonal.
b. detQ = ▯1.
c. If ▯ is an eigenvalue of Q, then ▯ = ▯1.
d. If Q1and Q ar2 orthogonal n ▯ n matrices, then so is the
product Q Q .
1 2
▯1 T
Proof. (a) Q = Q is orthogonal.
(b) 1 = detI = det(Q Q) = detQ detQ = (detQ) , so that detQ =
▯1.
(c) If v is an eigenvector with eigenvalue ▯, then jjvjj = jjQvjj = jj▯vjj =
j▯jjjvjj so that j▯j = 1.
(d) (Q1Q 2 Q Q1=2Q Q Q2Q 1 Q1I2 = I.2 2 ▯
x5.2: Orthogonal Complements
n
De▯nition. Let W be a subspace in R . We say that a vector v in
R is orthogonal to W, if v is orthogonal to every vector in W. The
set of all vectors that are orthogonal to W is called the orthogonal
complement of W, denoted W . That is,
W ? = fv in Rn : v ▯ w = 0 for all w in Wg:
Theorem. Let W be a subspace of R . n
a. W ? is a subspace of R .
? ?
b. (W ) = W.
c. W \ W ? = f0g.
?
d. If W is spanned by vectors w1;:::;wk, then v is in W is and
only if v ▯iw = 0 for all i = 1;:::;k.
Proof. (a) 0 is in W because 0▯w = 0 for any w in W. If u and v are
?
in W , then for any w in W we have
(u + v) ▯ w = u ▯ w + v ▯ w = 0 + 0 = 0
?
so that u+v is in W, too. Finally, if u is W , then for any scalar c we
have

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