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Unformatted text preview: (i) = i for all i = j , which
only occurs when σ is the identity. When we plug 0 into the equation, we get that f (0) = tr(A). 1.4 Similarity (cont’d) 1.4.27 Theorem. A matrix A ∈ Mn is diagonalizable if and only if the geometric and algebraic
multiplicities of each eigenvalue are equal.
Proof. We begin by noting that if λj and λk are eigenvalues of A with λj = λk , then their
eigenspaces intersect trivially; that is Eλj ∩ Eλk = {0}. To show this, let x ∈ Eλj ∩ Eλk . Then
λj x = Ax = λk x, which implies x = 0 because λj = λk . Thus if {v1 , v2 , . . . , vmj } is a basis
for Eλj and {u1 , u2 , . . . , umk } is a basis for Eλk , then {v1 , v2 , . . . , vmj } ∪ {u1 , u2 , . . . , umk } is a
linearly independent set, and forms a basis for Eλj + Eλk . Inductively iterating this, we obtain a
basis for Eλ1 + · · · + Eλr , forming a subspace of dimension r=1 dim(Eλj ). If all algebraic and
j
geometric multiplicities are equal, then this space has dimension
r
dim(Eλj ) = j =1 r
mj = n. j =1 In this case, the subspace equals the entire space, and we have a basis of eigenvectors of A.
Thus, A is diagonalizable.
If not all algebraic and geometric multiplicities are equal, then ∃λk such that dim(Eλk...
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 Fall '12
 BernhardBodmann
 Algebra, Matrices

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