17.11 Which Functions Respect the Order
Chapter 13 §05 ended on two facts that sit awkwardly
together. If
Throughout,
17.11.1. Functions of a Hermitian matrix
Chapter 12 §08 already applies a function to a
matrix, and nothing needs rebuilding. Let
- Polynomials give the expected
answer (part (a)):
gives , and gives . - Sums and products go through (part
(b)):
, , and . - Real-valued
gives a Hermitian (part (c)). This is what lets be compared with anything in the Loewner order. - The eigenvectors do not change
(part (d)):
has the same orthonormal eigenbasis as .
Two instances have names already. If
17.11.2. Operator monotone and operator convex functions
Chapter 13 §05 introduced the Loewner order
The obvious guess is “the increasing ones”, since
that is the answer for numbers. It is wrong. The
function
A function is operator monotone when applying it to Hermitian matrices never breaks the Loewner order, at every size at once.
Let
- (a)
is operator monotone on if, for every size and every pair of Hermitian matrices whose spectra lie in , - (b)
is operator convex on if, for every size , every pair of Hermitian matrices whose spectra lie in , and every , It is operator concave on if is operator convex on .
Here
In words: the first condition takes any two Hermitian
matrices of the same size with spectra inside the domain
of
Everything in the definition makes
sense. Since
The smallest size already says something, and the proof is three lines.
If
Proof. Let
The converse is false, and the failure is the reason this section exists.
Warning. Increasing is not
enough. “Increasing” is exactly the case
Quick check. Is
Solution. For
17.11.3. The examples we can prove
Four families are within reach, and the first is the simplest. It includes a degenerate case, the constant functions, which respect every order because they flatten everything to one matrix.
Let
Proof. By Theorem 12.8.3
(a),
The second example is the one that behaves like the
square root:
The function
Idea. A congruence removes the
noncommutativity. Conjugating by
Proof. Let
Two things make the whitening work, and they are
worth separating. First, once one of the two matrices is
To build more examples out of these, one needs to know which ways of combining functions preserve the property.
Let
- (a) For every
and every , the functions and are operator monotone on . - (b) For every
, the translate is operator monotone on the interval .
Proof.
(a) Let
be Hermitian of the same size with spectra in . By Theorem 12.8.3 (b), . Both brackets are by hypothesis, and for every the quadratic form of the sum is the sum of the two nonnegative quadratic forms. So the sum is by Definition 13.1.1. Likewise, using (b) and (a) of the same theorem, , whose quadratic form is times a nonnegative one.(b) Let
on , and let be Hermitian with spectrum in , with spectral resolution . By Theorem 12.8.1 (c) and (d), and the numbers are distinct. The are nonzero self-adjoint idempotents with for and (Theorem 12.8.1 (b), (c)). So by Theorem 12.8.2 this is the spectral resolution of , whose spectrum lies in . By Definition 12.8.2, applied to each side, Now if with spectra in , then , and both shifted matrices have spectra in . Hence .
Combining the last two propositions produces a whole one-parameter family. It will reappear in the final part of the section as the building block of every operator monotone function.
For every
Proof. By Proposition 17.11.3
and Proposition 17.11.4
(b),
The fourth example is the one that started the
section, and it is already proved. By Proposition 13.5.4,
A nonexample by minimal change. The
function
Why “for every
17.11.4. Eigenvalues move monotonically for every increasing function
Operator monotonicity is a strong requirement, and a
weaker one comes for free. §02 proved that the whole
ordered list of eigenvalues respects the Loewner order:
by Corollary 17.2.4,
Let
Proof. The matrix
The lemma needs
Let
Proof. By Corollary 17.2.4,
So the squaring map, which is not operator monotone, is nevertheless monotone on eigenvalues. The Chapter 13 witness shows both at once.
Let
Solution. From Example (Squaring does not preserve the Loewner order),
So for every increasing
Warning. Eigenvalues in
order do not make matrices in order. It is
tempting to read Proposition 17.11.7 as
“almost” operator monotonicity. It is not, because
17.11.5. Operator convexity, and where the two notions part
Operator convexity is the second notion of Definition 17.11.1, and one might expect it to behave like operator monotonicity. It does not: the function that fails to be operator monotone turns out to be operator convex.
The function
Idea. For numbers,
Proof. By Theorem 12.8.3
(a), the function
The contrast is now complete, and it runs both ways.
is operator convex but not operator monotone on : Proposition 17.11.8 and Example (Squaring does not preserve the Loewner order). is operator monotone but not operator convex on : Proposition 17.11.3, and at the scalar function is not even convex, since with , , and the value at the midpoint is , above the chord’s .
So neither notion implies the other. Each implies its
scalar version (increasing, convex) by taking
17.11.6. What is not proved here
The question the section opened with has a complete answer, and this chapter does not prove it. Here is the statement, so that the reader knows what the answer looks like and which of the facts above are pieces of it.
Let
- (i)
is operator monotone on . - (ii) There are
, , and a positive measure on with , such that for every - (iii)
is the restriction to of a function that is analytic on and sends every point of the upper half-plane to a point with nonnegative imaginary part.
We do not prove this theorem here. Its proof needs complex analysis and some measure theory, which lie outside Chapter 16’s list (A1)–(A6); Chapter 21 proves it. Nothing in this chapter depends on it.
Read (ii) against what was proved above. The
functions it superposes are
Condition (iii) answers the opening question in one
line. For
Quick check. Find a point
Solution. Take
The same test settles the powers
17.11.7. Summary and transfer
The chapter began with a list of eigenvalues that the book had been able to order since Chapter 13 without being able to describe. It ends with each eigenvalue described as an optimization, and with a set of comparison theorems that follow from that description. Five moves did the work, and each transfers beyond this chapter.
- The dimension count. Two subspaces
of an
-dimensional space whose dimensions add to more than share a nonzero vector. This one line drives Courant–Fischer (Theorem 17.2.3) and Weyl’s inequalities (§03) directly, and through Courant–Fischer everything after them. Transfer: whenever an extremal problem over vectors has to be compared with a fixed subspace, count dimensions before computing anything. - Min–max. Describe a quantity as a
minimum over subspaces of a maximum over vectors, and
prove it with two inequalities of different characters:
one exhibits a witness subspace and computes, and the
other takes an arbitrary subspace and intersects it with
a span of eigenvectors. Applying the result to
gives the dual form for free. Transfer: a description that names no eigenvector is what makes two different matrices comparable. The same shape describes singular values (§09) and inertia (§10). - Compress and interlace. Restricting
a Hermitian form to a subspace gives a smaller Hermitian
matrix whose eigenvalues are trapped by the original’s:
compressing to a
-dimensional subspace puts of the piece between and of the whole, which are consecutive when . Deleting a row and column (§04), compressing to an orthonormal frame (§06) and deleting a row or a column of a rectangular matrix (§09) are all this move; adding a rank-one term (§05) reaches the same interlaced picture by a different route, the Loewner order together with Weyl’s rank bound. Transfer: to bound the spectrum of a piece, look for the whole of which it is a compression. - Perturb in order. Compare
with index by index, never “each eigenvalue with its nearest neighbor”. For Hermitian perturbations the Lipschitz constant is in the spectral norm, independent of , of the gaps, and of the eigenvectors (§03). For singular values it needs no hypothesis on at all (§09). Transfer: where the Jordan block of Chapter 16 §07 showed that eigenvalues of general matrices can move like , the Hermitian and singular-value problems are perfectly conditioned. Reduce a question to one of them whenever possible. - Majorize. When the right comparison
is between sums of the largest entries rather
than individual ones, compare partial sums of decreasing
rearrangements. The diagonal of a Hermitian matrix is
majorized by its spectrum, and by the Schur–Horn theorem
majorization describes the possible diagonals
exactly (§08); and the Ky Fan and Lidskii sums
(§§06–07) are of the same type. Transfer: a
family of partial-sum inequalities indexed by
, together with equality of the totals, is a majorization in disguise, and writing it as one lets the whole family be used at once.
Two supporting moves recur under these five:
apply the result to
The chapter also paid a series of promises made in earlier chapters, some of them long ago.
| The promise | Made in | Paid by |
|---|---|---|
| The maximization route to eigenvalues “returns in Chapter 17, where the Rayleigh quotient and the Courant–Fischer theorem describe every eigenvalue of a self-adjoint operator by optimization” | Chapter 12 §02 | §01 (The Rayleigh quotient) and §02 (The min–max theorem), Theorem 17.2.3 |
| “Chapter 17 characterizes the eigenvalues of a
Hermitian matrix as maxima and minima of |
Chapter 13 §13 | §01 and §02, Theorem 17.2.3 |
| Chapter 13 §05 | §02, Corollary 17.2.4 | |
| “The eigenvalues of a Hermitian matrix move by at
most |
Chapter 16 §07 | §03 (How far eigenvalues move), Weyl’s perturbation bound |
| Lemma 13.11.2 is “the extreme case of a family of statements about sums of the largest eigenvalues” | Chapter 13 §11 | §06 (Subspaces and Sums of Eigenvalues), Ky Fan’s theorem, of which it is a special case |
| The second half of the approximation theorem, in the spectral norm | Chapter 13 §11 and Chapter 16 §08 | §09 (The same theory for singular values), Eckart–Young in the spectral norm |
| Interlacing of the zeros of consecutive orthogonal polynomials, “Chapter 17’s business” | Chapter 11 §10 | §04 (Deleting a row and a column): strict interlacing, through the bordered-matrix criterion |
| The operator monotone functions named in Chapter 13 §05 | Chapter 13 §05 | §11: definition and first examples; |
Four earlier theorems also received second proofs from the variational side, each of which shows something the first proof did not. The positivity of principal submatrices of a positive semidefinite matrix now follows from interlacing (§04, Deleting a row and a column). Hadamard’s inequality Theorem 13.7.3 follows from AM–GM after a diagonal scaling, with Schur’s theorem supplying the equal totals (§08, The diagonal and the spectrum). Sylvester’s law of inertia Theorem 14.5.1 is now a statement about the dimensions of subspaces, with no basis in it (§10, Inertia, again and additively). And the block test Theorem 13.6.1 is now a consequence of inertia additivity (§10).
Some things the chapter did not do. Bounds for the
eigenvalues of non-Hermitian matrices under
perturbation, beginning with Bauer–Fike, are Chapter
20’s. Chapter 21 has Loewner’s theorem with the powers
17.11.8. Exercises
A. Check your understanding
- (a) Define “
is operator monotone on ”, and say what the clause “for every ” ranges over. - (b) Explain why an operator monotone function is increasing, and name an increasing function that is not operator monotone.
- (c) Decide whether
the following is correct, and justify your answer: if
is increasing and are Hermitian with spectra in , then for every . - (d) Decide whether
the following is correct, and justify your answer: if
are Hermitian of the same size and for every , then . - (e) Which of
and is operator monotone on , and which is operator convex there? - (f) What does this section prove about Loewner’s theorem?
Solution.
(a) For every
and all Hermitian with spectra in , implies , where is formed by Definition 12.8.2 (Definition 17.11.1 (a)). The clause ranges over all matrix sizes: the implication must hold for , , , … matrices simultaneously.(b) Taking
gives whenever in (Proposition 17.11.1). The function on is increasing and is not operator monotone (Example (Squaring does not preserve the Loewner order)).(c) Correct. This is Proposition 17.11.7: Corollary 17.2.4 orders the eigenvalues of
and index by index, preserves each inequality, and Lemma 17.11.6 identifies with .(d) Incorrect.
and have the same eigenvalues, but their difference is indefinite, so neither dominates the other.(e)
is operator monotone on (Proposition 17.11.3) and is not operator convex, since already the scalar function is not convex: at , , and their midpoint , . is operator convex (Proposition 17.11.8) and not operator monotone on . The witness of Example (Squaring does not preserve the Loewner order) has in the spectrum of , but the pair works: has determinant .(f) Only the statement (Theorem 17.11.9) and the fact that its building blocks
are operator monotone (Corollary 17.11.5). The theorem itself is proved in Chapter 21.
B. Practice
Determine which of the following functions are operator monotone on the interval given. Justify your answer.
- (i)
on . - (ii)
on . - (iii)
on . - (iv)
on . - (v)
on .
Solution.
(i) Operator monotone, by Proposition 17.11.2 with
.(ii) Not operator monotone. It is not increasing, since
, so it fails at with ; compare Proposition 17.11.1.(iii) Operator monotone. By Corollary 17.11.5 with
, is operator monotone on , and with , so Proposition 17.11.4 (a) applies.(iv) Operator monotone, by Corollary 17.11.5 with
.(v) Not operator monotone: Example (Squaring does not preserve the Loewner order) gives
, with spectra and in , and .
Let
- (a) Show that
. - (b) Compute
and , and verify directly that , as Proposition 17.11.3 predicts. - (c) Show that
by exhibiting a vector with . - (d) Compute the
eigenvalues of
and , and verify that for . Hence explain which result of this section (c) and (d) illustrate together.
Solution.
(a)
, a diagonal matrix with nonnegative entries, so and . The leading principal minors of are and , both positive, so by Sylvester’s criterion (Theorem 13.1.5 (d)).(b)
and , so For every , , so .(c)
and , so , with determinant . With ,(d)
has trace and determinant , so its eigenvalues are , approximately and . has trace and determinant , so its eigenvalues are , approximately and . For , because . For , is . Both sides are positive, and squaring gives , that is , which holds because . Hence (c) and (d) together illustrate Proposition 17.11.7 and the warning after it: the increasing function orders the eigenvalues but not the matrices.
With
Solution. From B2,
C. Going deeper
- (a) Let
. Prove that , where is for on . - (b) Deduce that
is operator monotone on for every integer . - (c) Explain why the
argument of (b) cannot reach
, and say where the operator monotonicity of for every is proved.
Hint for (a): the positive square root is unique.
Solution.
(a) Let
on . Since , its spectrum lies in (Theorem 13.1.4 (b)), and is Hermitian with nonnegative eigenvalues by Theorem 12.8.3 (c) and (d), so . By Theorem 12.8.3 (b), , and , so . By the uniqueness in Theorem 13.1.4 (d), applied to , is the positive square root of : . The same holds for . Now Proposition 13.5.4 applied to gives , and by Theorem 13.2.1. Applying Proposition 13.5.4 again, to , gives , that is .(b) Induction on
. For the function is , operator monotone by Proposition 17.11.2. Suppose whenever . The argument of (a), with in place of and in place of , shows that , and similarly for . Both and are , so Proposition 13.5.4 applied to gives .(c) Each step of (b) takes a square root, which halves the exponent. Starting from
, the exponents reached are exactly , and is not among them. The operator monotonicity of for every needs a new idea, an integral representation of the kind in Theorem 17.11.9, and it is proved in Chapter 21.
- (a) Let
with . Prove that the matrix is Hermitian and . - (b) Let
be of the same size and . Prove that that is, is operator convex on . - (c) Deduce that
is both operator monotone and operator concave on .
Hint: Schur complements.
Solution.
(a)
is Hermitian, and so is , since . The off-diagonal blocks are and , so the block matrix is Hermitian. Its corner is , and its Schur complement is By Theorem 13.6.1 (b), .(b) Put
. For , . Both inner products are , and and are and not both zero, so the sum is . Thus . By (a), and , so for every the quadratic form of is , and this block matrix is . Its corner is , so Theorem 13.6.1 (b) gives which is the claim. Since sends a positive definite matrix to its inverse (Definition 12.8.2), and a Hermitian matrix has spectrum in exactly when it is (Theorem 13.1.5 (b)), this is Definition 17.11.1 (b) for .(c) Operator monotone by Proposition 17.11.3. Operator concave by definition, because its negative
is operator convex by (b).
- (a) With
and as in B2, show that . Conclude that is not operator monotone on . - (b) One might hope
to deduce (a) from the failure of
. Suppose were operator monotone on . Applying it and (Proposition 13.5.4) to both sides of , repeatedly and in any order, yields inequalities for certain exponents . Show that every such has the form with integers , and that never has this form. Explain why this means (a) needed a witness of its own.
Solution.
(a) From B2,
, with spectra in . Multiplying the squares of B2 once more, The determinant is , so the difference is indefinite. Explicitly, with , By Theorem 12.8.3 (a), sends to , so it is not operator monotone on .(b) Start from
. Taking the square root of gives , and cubing it gives , by Theorem 12.8.3 (b) and the uniqueness of the positive square root, as in C1 (a); likewise for . If , the results are and , of the same form. By induction on the number of steps, every exponent reached has the form . If , then , and the left side is odd while the right side is even, since . So no exponent reached equals . The hypothetical operator monotonicity of the cube therefore cannot be converted, by these means, into that of the square. The failure of the square does not formally imply the failure of the cube, and (a) had to find a pair of matrices of its own.