Lecture #10 Notes

Then ot ao o1 o2 t ao1 o2 t t o2 o1 ao1 o2 1 0 t

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Unformatted text preview: ∗ λ1 ∗ 0 = . , . B . 0 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 = . . ￿ . O 0 Note that O2 is orthogonal (its columns are orthonormal). Let O = O1 O2 . O is orthogonal since O t = ( O1 O2 ) t tt = O2 O1 − − = O2 1 O1 1 = ( O1 O2 ) −1 = O −1 . Then OT AO = (O1 O2 )t AO1 O2 t t = O2 (O1 AO1 )O2 λ1 ∗ · · · ∗ 0 t = O2 . O2 . . B 0 λ1 ∗ · · · ∗ λ1 ∗ · · · ∗ 0 0 = . . Ot B O = . ￿ ￿ . ￿ . . T 0 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) such that 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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