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Unformatted text preview: sequently, Schur’s triangularization (over R) gives
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 ∗ · · · ∗
0 ∗ · · · ∗
x1 A x1 ∗ = . . . . . ,
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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