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.