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Ans 4 - v1 - C LASSROOM ] Rajendra B hatia I ndian...

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CLASSROOM Rajendra Bhatia Indian Statistical Institute New Delhi 110 016, India. Email: rbh@isid.ac.in Keywords Eigenvalues, idempotent, projec- tion operator, spectrum, Hilbert space. ] Eigenvalues of AB and BA / Let A, B be n x n matrices with complex entries. Given below are several proofs of the fact that AB have the same eigenvalues. Each proof brings out a dif- ferent viewpoint and may be presented at the appropri- ate time in a linear algebra course. Let tr(T) stand for the trace of T, and det(T) for the determinant of T. The relations tr(AB) = tr(BA) det(AB) = det(BA). (1) are usually proved early in linear algebra courses. The first is easy to verify; the second takes more work to prove. Let ,~n-Cl(T)~n-1 + + (-1)'cn(T) (2) be the characteristic polynomial of T, and let AI(T), A2(T),. .., An(T) be its n roots, counted with multiplic- ities and in any order. These are the eigenvalues of T. We know that ck(T) is the k, th elementary symmetric polynomial in these numbers. Thus n l(r) = y:At(T) = tr (T) j=l c2(T) = i<j n c.,(T) = I] Aj(T) = det(T). To say that have the same eigenvalues amounts to saying that ck(AB)=ck(BA) for 1_< k_<n. (3) 88 RESONANCE J January 2002
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CLASSROOM We know that this is true when k = 1, or n ; and want to prove it for other values of k. Proof 1. It suffices to prove that, for 1 < m < n, A~'(AB)+. ..+A'~(AB) = A~'(BA)+".+Am(BA) (4) (Recall Newton's identities by which the n elementary symmetric polynomials in n variables are expressed in terms of the n sums of powers.) Note that the eigenval- ues of T m are the ruth powers of the eigenvalues of T. So, ~A~n(T) = ~/~3(T m) = tr (Tin). Thus the state- ment (4) is equivalent to [(AB) m] = tr [(BA)m]. But this follows fr(~m (1) tr [(AB) m]= tr (ABAB. ..AB) = tr(BABA. ..BA) = [(BA)m]. Proof 2. One can prove the relations (3) directly. The coefficient ck(T) is the sum of all the k • k principal minors of T. A direct computation (the Binet-Cauchy formula) leads to (3). A more sophisticated version of this argument involves the antisymmetric tensor prod- uct Ak(T). This is a matrix of order (~.) whose entries are the k x k minors of T. So ck(T)= tr A k(T),l_< k<n. Among the pleasant properties of A k is multiplicativity: Ak(AB) = Ak(A) A k (B). So ck(AB) = [Ak(AB)] = tr [Ak(A) A k (B)] = tr[Ak(B) Ak(A)]= Ak(BA)=ck(BA). Proof 3. This proof invokes a continuity argument that is useful in many contexts. Suppose A is invertible (non- singular). Then AB = A(BA)A -1. So AB and BA are Recall Newton's
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Ans 4 - v1 - C LASSROOM ] Rajendra B hatia I ndian...

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