Unformatted text preview: ∗ λ1 ∗
0 = .
, . B
3 where B ∈ Mn−1 (R). Now the characteristic polynomial of A factors as
p A ( t ) = ( t − λ1 ) p B ( t ) ,
where pB (t) is the characteristic polynomial of B . This means that λ2 , · · · , λn are the eigenvalues
of B . Now, by the induction assumption, there exists an orthogonal O ∈ Mn−1 (R) such that
Ot B T ,
where T is upper triangular with λ2 , · · · , λn on the diagonal. Deﬁne 1 0···0
0 O2 = . .
0 Note that O2 is orthogonal (its columns are orthonormal). Let O = O1 O2 . O is orthogonal since
O t = ( O1 O2 ) t
= O2 O1
= O2 1 O1 1
= ( O1 O2 ) −1 = O −1 .
OT AO = (O1 O2 )t AO1 O2
= O2 (O1 AO1 )O2 λ1 ∗ · · · ∗
= O2 . O2
0 λ1 ∗ · · · ∗
λ1 ∗ · · · ∗
0 0 = .
. Ot B O = .
0 = T. Thus, we have triangularized A to the matrix T which has the eigenvalues of A in its diagonal.
1.5.6 Theorem. If A ∈ Mn (R) is symmetric, then there exists an orthogonal matrix O ∈ Mn (R)
Ot AO = D,
where D is a diagonal matrix containing the eigenvalues of A as its diagonal entries. 4 Proof. By symmetry, all eigenvalues of A are real (previously proven). This would also apply to
all (complex) eigenvalues obtained from factoring the characteristic polynomial pA of A over C.
This emplies A has n real eigenvalues. Con...
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