Lecture #10 Notes

If a mn r has n real eigenvalues counting multiplicity

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Unformatted text preview: rem. If A ∈ Mn (R) has n distinct (real) eigenvalues, then it is diagonalizable. Proof. As before, we use the fact that eigenvectors belonging to distinct eigenvalues are linearly independent. 1.5.4 Theorem. A matrix A ∈ Mn (R) is diagonalizable if and only if it has n eigenvalues (with multiplicities counted) and the geometric and algebraic eigenvalues are equal. Proof. Extends preceding theorem, same strategy as before. The following theorem is the real case of Schur’s triangularization theorem. 1.5.5 Theorem. If A ∈ Mn (R) has n (real) eigenvalues (counting multiplicity), then there exists an orthogonal matrix O ∈ Mn (R) such that Ot AO = T, where T is triangular, and the eigenvalues of A are the diagonal entries of T . Proof. This proof is identical to the proof of Schur’s theorem in the complex case. We prove it by induction on the dimension n. For n = 1, it is trivially true. Now suppose the theorem holds for all matrices in Mn−1 (R). Let A ∈ Mn (R) has real eigenvalues λ1 , λ2 , · · · , λn . Choose an eigenvector x ∈ Rn , ||x|| = 1. Now x may be extended to a basis {x, y2 , · · · , yn } of Rn . Apply Gram-Schmidt orthonormalization to this basis to produce an orthonormal basis {x, z2 , · · · , zn } of Rn . Define the matrix O1 = [x, z2 , · · · , zn ]. Note that O1 is orthogonal, since its columns are orthonormal. Thus, ￿ ￿ t t O1 AO1 = O1 λ1 x1 ∗ · · ·...
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