1.3. Eigenvalues and sums 43
Remark 1.3.3. By homogeneity, one can replace the restriction |v| = 1
with v = 0 provided that one replaces the quadratic form
v∗Av
with the
Rayleigh quotient
v∗Av/v∗v.
A closely related formula is as follows. Given an n×n Hermitian matrix
A and an m-dimensional subspace V of
Cn,
we define the partial trace
tr(A
V
) to be the expression
tr(A
V
) :=
m
i=1
vi
∗Avi
where v1,...,vm is any orthonormal basis of V . It is easy to see that this
expression is independent of the choice of orthonormal basis, and so the
partial trace is well-defined.
Proposition 1.3.4 (Extremal partial trace). Let A be an n × n Hermitian
matrix. Then for any 1 k n, one has
λ1(A) + · · · + λk(A) = sup
dim(V )=k
tr(A
V
)
and
λn−k+1(A) + · · · + λn(A) = inf
dim(V )=k
tr(A
V
).
As a corollary, we see that A λ1(A)+···+λk(A) is a convex function,
and A λn−k+1(A) + · · · + λn(A) is a concave function.
Proof. Again, by symmetry it suffices to prove the first formula. As before,
we may assume, without loss of generality, that A has the standard eigenba-
sis e1,...,en corresponding to λ1(A),...,λn(A), respectively. By selecting
V to be the span of e1,...,ek we have the inequality
λ1(A) + · · · + λk(A) sup
dim(V )=k
tr(A
V
),
so it suffices to prove the reverse inequality. For this we induct on the dimen-
sion n. If V has dimension k, then it has a k−1-dimensional subspace V that
is contained in the span of e2,...,en. By the induction hypothesis applied
to the restriction of A to this span (which has eigenvalues λ2(A),...,λn(A)),
we have
λ2(A) + · · · + λk(A) tr(A
V
).
On the other hand, if v is a unit vector in the orthogonal complement of V
in V , we see from (1.53) that
λ1(A)
v∗Av.
Adding the two inequalities we obtain the claim.
Previous Page Next Page