t 0 and we proceed with a lower right

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Unformatted text preview: sequently, Schur’s triangularization (over R) gives orthogonal O, Ot AO = T, with T triangular. But T is normal and triangular, thus T is diagonal. 1.5.7 Question. What if A is not symmetric? 1.6 Block Triangularization 1.6.8 Theorem. If A ∈ Mn (R), then there exits a real orthogonal matrix O ∈ Mn (R) such that A1 ∗ ∗∗ 0 A2 ∗ ∗ t O AO = . . . . . , . . . . ∗ 0 · · · 0 Ar with diagonal blocks Aj ∈ M1 (R) or M2 (R). Proof. Repeat Schur’s triangularization procedure. To begin with, if λ1 is real, then there is a (real) x1 ∈ Rn and Ax1 = λ1 x1 . So normalizing x1 , complementing to orthonormal basis of Rn viewed as columns gives ￿ ￿￿ ￿ A x1 ∗ = λ1 x1 ∗ , and λ1 ∗ · · · ∗ ￿ t￿ ￿ ￿ 0 ∗ · · · ∗ x1 A x1 ∗ = . . . . . , ∗t . . . . .. . 0 ∗ ··· ∗ and we proceed with a lower right block as before. If λ1 = α + iβ , α, β ∈ R, β ￿= 0, then there is an eigenvalue λ2 such that λ2 = λ1 = α − iβ . Why? Take an eigenvector x ∈ Cn belonging to eigenvalue λ, take real and imaginary parts: x = u + iv where u, v ∈ Rn . Then Ax = A(u + iv ) = Au + iAv = λ(u + iv ) = (α + iβ )(u + iv ) = αu −...
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