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Lecture #4 Notes - Matrix Theory Math6304 Lecture Notes...

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Matrix Theory, Math6304 Lecture Notes from September 6, 2012 taken by Nathaniel Hammen Last Time (9/4/12) Diagonalization: conditions for diagonalization Eigenvalue Multiplicity: algebraic and geometric multiplicity 1 Further Review 1.1 Warm-up questions 1.1.1 Question. If A M n has only one eigenvalue λ (of multiplicity n ) and is diagonalizable, what is A ? Answer. Because A is diagonalizable, there exists an invertible S M n such that S 1 AS is diagonal. In fact, because all eigenvalues are λ , we have A = S ( S 1 AS ) S 1 = S λ IS 1 = λ SS 1 = λ I 1.1.2 Question. Let A M n and f ( t ) = det ( I + tA ) . What is f ( t ) in terms of A ? Answer. The i, j th entry of ( I + tA ) is δ i,j + ta i,j , so f ( t ) = det ( I + tA ) = σ S n sgn ( σ ) n j =1 ( δ σ ( j ) ,j + ta σ ( j ) ,j ) = σ S n sgn ( σ ) n j =1 δ σ ( j ) ,j + t n j =1 a σ ( j ) ,j i = j δ σ ( i ) ,i + o ( t 2 ) where S n is the set of permutations of n elements and sgn ( σ ) is +1 if σ is an even permutation and -1 if σ is an odd permutation. Di ff erentiating f ( t ) gives f ( t ) = σ S n sgn ( σ ) n j =1 a σ ( j ) ,j i = j δ σ ( i ) ,i + o ( t ) = n j =1 a j,j + o ( t ) = tr ( A ) + o ( t ) 1
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This last equality holds because i = j δ σ ( i ) ,i is only nonzero when σ ( 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 M n is diagonalizable if and only if the geometric and algebraic multiplicities of each eigenvalue are equal. Proof. We begin by noting that if λ j and λ k are eigenvalues of A with λ j = λ k , then their eigenspaces intersect trivially; that is E λ j E λ k = { 0 } . To show this, let x E λ j E λ k . Then λ j x = Ax = λ k x , which implies x = 0 because λ j = λ k . Thus if { v 1 , v 2 , . . . , v m j } is a basis for E λ j and { u 1 , u 2 , . . . , u m k } is a basis for E λ k , then { v 1 , v 2 , . . . , v m j } { u 1 , u 2 , . . . , u m k } is a linearly independent set, and forms a basis for E λ j + E λ k . Inductively iterating this, we obtain a basis for E λ 1 + · · · + E λ r
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