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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 GramSchmidt 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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 Fall '12
 BernhardBodmann
 Addition, Matrices

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