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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
,~nCl(T)~n1
+ + (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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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 BinetCauchy
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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