Let be a Hermitian matrix. By the spectral theorem for Hermitian matrices (which, for sake of completeness, we prove below), one can diagonalise using a sequence

of real eigenvalues, together with an orthonormal basis of eigenvectors . (The eigenvalues are uniquely determined by , but the eigenvectors have a little ambiguity to them, particularly if there are repeated eigenvalues; for instance, one could multiply each eigenvector by a complex phase . In these notes we are arranging eigenvalues in descending order; of course, one can also arrange eigenvalues in increasing order, which causes some slight notational changes in the results below.) The set is known as the spectrum of .

A basic question in linear algebra asks the extent to which the eigenvalues and of two Hermitian matrices constrains the eigenvalues of the sum. For instance, the linearity of trace

when expressed in terms of eigenvalues, gives the trace constraint

(together with the counterparts for and ) gives the inequality

The complete answer to this problem is a fascinating one, requiring a strangely recursive description (once known as *Horn’s conjecture*, which is now solved), and connected to a large number of other fields of mathematics, such as geometric invariant theory, intersection theory, and the combinatorics of a certain gadget known as a “honeycomb”. See for instance my survey with Allen Knutson on this topic some years ago.

In typical applications to random matrices, one of the matrices (say, ) is “small” in some sense, so that is a perturbation of . In this case, one does not need the full strength of the above theory, and instead rely on a simple aspect of it pointed out by Helmke and Rosenthal and by Totaro, which generates several of the *eigenvalue inequalities* relating , , and , of which (1) and (3) are examples. (Actually, this method eventually generates *all* of the eigenvalue inequalities, but this is a non-trivial fact to prove.) These eigenvalue inequalities can mostly be deduced from a number of *minimax* characterisations of eigenvalues (of which (2) is a typical example), together with some basic facts about intersections of subspaces. Examples include the Weyl inequalities

valid whenever and , and the *Ky Fan inequality*

One consequence of these inequalities is that the spectrum of a Hermitian matrix is *stable* with respect to small perturbations.

We will also establish some closely related inequalities concerning the relationships between the eigenvalues of a matrix, and the eigenvalues of its minors.

Many of the inequalities here have analogues for the singular values of non-Hermitian matrices (which is consistent with the discussion near Exercise 16 of Notes 3). However, the situation is markedly different when dealing with *eigenvalues* of *non-Hermitian* matrices; here, the spectrum can be far more unstable, if pseudospectrum is present. Because of this, the theory of the eigenvalues of a random non-Hermitian matrix requires an additional ingredient, namely upper bounds on the prevalence of pseudospectrum, which after recentering the matrix is basically equivalent to establishing lower bounds on least singular values. We will discuss this point in more detail in later notes.

We will work primarily here with Hermitian matrices, which can be viewed as self-adjoint transformations on complex vector spaces such as . One can of course specialise the discussion to real symmetric matrices, in which case one can restrict these complex vector spaces to their real counterparts . The specialisation of the complex theory below to the real case is straightforward and is left to the interested reader.

** — 1. Proof of spectral theorem — **

To prove the spectral theorem, it is convenient to work more abstractly, in the context of self-adjoint operators on finite-dimensional Hilbert spaces:

Theorem 1 (Spectral theorem)Let be a finite-dimensional complex Hilbert space of some dimension , and let be a self-adjoint operator. Then there exists an orthonormal basis of and eigenvalues such that for all .

The spectral theorem as stated in the introduction then follows by specialising to the case and ordering the eigenvalues.

*Proof:* We induct on the dimension . The claim is vacuous for , so suppose that and that the claim has already been proven for .

Let be a unit vector in (thus ) that maximises the form ; this maximum exists by compactness. By the method of Lagrange multipliers, is a critical point of for some . Differentiating in an arbitrary direction , we conclude that

this simplifies using self-adjointness to

Since was arbitrary, we conclude that , thus is a unit eigenvector of . By self-adjointness, this implies that the orthogonal complement of is preserved by . Restricting to this lower-dimensional subspace and applying the induction hypothesis, we can find an orthonormal basis of eigenvectors of on . Adjoining the new unit vector to the orthonormal basis, we obtain the claim.

Suppose we have a self-adjoint transformation , which of course can be identified with a Hermitian matrix. Using the orthogonal eigenbasis provided by the spectral theorem, we can perform an orthonormal change of variables to set that eigenbasis to be the standard basis , so that the matrix of becomes diagonal. This is very useful when dealing with just a single matrix – for instance, it makes the task of computing functions of , such as or , much easier. However, when one has *several* Hermitian matrices in play (e.g. ), then it is usually not possible to standardise all the eigenbases simultaneously (i.e. to simultaneously diagonalise all the matrices), except when the matrices all commute. Nevertheless one can still normalise *one* of the eigenbases to be the standard basis, and this is still useful for several applications, as we shall soon see.

Exercise 1Suppose that the eigenvalues of an Hermitian matrix are distinct. Show that the associated eigenbasis is unique up to rotating each individual eigenvector by a complex phase . In particular, thespectral projectionsare unique. What happens when there is eigenvalue multiplicity?

** — 2. Minimax formulae — **

The eigenvalue functional is not a linear functional (except in dimension one). It is not even a convex functional (except when ) or a concave functional (except when ). However, it is the next best thing, namely it is a *minimax* expression of linear functionals. (Note that a convex functional is the same thing as a max of linear functionals, while a concave functional is the same thing as a min of linear functionals.) More precisely, we have

Theorem 2 (Courant-Fischer min-max theorem)Let be an Hermitian matrix. Then we havefor all , where ranges over all subspaces of with the indicated dimension.

*Proof:* It suffices to prove (6), as (7) follows by replacing by (noting that ).

We first verify the case, i.e. (2). By the spectral theorem, we can assume that has the standard eigenbasis , in which case we have

whenever . The claim (2) is then easily verified.

To prove the general case, we may again assume has the standard eigenbasis. By considering the space spanned by , we easily see the inequality

so we only need to prove the reverse inequality. In other words, for every -dimensional subspace of , we have to show that contains a unit vector such that

Let be the space spanned by . This space has codimension , so it must have non-trivial intersection with . If we let be a unit vector in , the claim then follows from (8).

Remark 1By homogeneity, one can replace the restriction with provided that one replaces the quadratic form with the Rayleigh quotient .

A closely related formula is as follows. Given an Hermitian matrix and an -dimensional subspace of , we define the *partial trace* to be the expression

where is any orthonormal basis of . 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 3 (Extremal partial trace)Let be an Hermitian matrix. Then for any , one hasand

As a corollary, we see that is a convex function, and 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 has the standard eigenbasis. By selecting to be the span of we have the inequality

so it suffices to prove the reverse inequality. For this we induct on dimension. If has dimension , then it has a -dimensional subspace that is contained in the span of . By the induction hypothesis, we have

On the other hand, if is a unit vector in the orthogonal complement of in , we see from (2) that

Adding the two inequalities we obtain the claim.

Specialising Proposition 3 to the case when is a coordinate subspace (i.e. the span of of the basis vectors ), we conclude the *Schur-Horn inequalities*

for any , where are the diagonal entries of .

Exercise 2Show that the inequalities (9) are equivalent to the assertion that the diagonal entries lies in thepermutahedronof , defined as the convex hull of the permutations of in .

Remark 2It is a theorem of Schur and Horn that these are the complete set of inequalities connecting the diagonal entries of a Hermitian matrix to its spectrum. To put it another way, the image of anycoadjoint orbitof a matrix with a given spectrum under the diagonal map is the permutahedron of . Note that the vertices of this permutahedron can be attained by considering the diagonal matrices inside this coadjoint orbit, whose entries are then a permutation of the eigenvalues. One can interpret this diagonal map as the moment map associated with the conjugation action of the standard maximal torus of (i.e. the diagonal unitary matrices) on the coadjoint orbit. When viewed in this fashion, the Schur-Horn theorem can be viewed as the special case of the more general Atiyah convexity theorem (also proven independently by Guillemin and Sternberg) in symplectic geometry. Indeed, the topic of eigenvalues of Hermitian matrices turns out to be quite profitably viewed as a question in symplectic geometry (and also in algebraic geometry, particularly when viewed through the machinery of geometric invariant theory).

There is a simultaneous generalisation of Theorem 2 and Proposition 3:

Exercise 3 (Wielandt minimax formula)Let be integers. Define a partial flag to be a nested collection of subspaces of such that for all . Define the associated Schubert variety to be the collection of all -dimensional subspaces such that . Show that for any matrix ,

** — 3. Eigenvalue inequalities — **

Using the above minimax formulae, we can now quickly prove a variety of eigenvalue inequalities. The basic idea is to exploit the linearity relationship

for any unit vector , and more generally

For instance, as mentioned before, the inequality (3) follows immediately from (2) and (10). Similarly, for the Ky Fan inequality (5), one observes from (11) and Proposition 3 that

for any -dimensional subspace . Substituting this into Proposition 3 gives the claim. If one uses Exercise 3 instead of Proposition 3, one obtains the more general *Lidskii inequality*

In a similar spirit, using the inequality

for unit vectors , combined with (10) and (6), we obtain the eigenvalue stability inequality

thus the spectrum of is close to that of if is small in operator norm. In particular, we see that the map is Lipschitz continuous on the space of Hermitian matrices, for fixed .

More generally, suppose one wants to establish the Weyl inequality (4). From (6) that it suffices to show that every -dimensional subspace contains a unit vector such that

But from (6), one can find a subspace of codimension such that for all unit vectors in , and a subspace of codimension such that for all unit vectors in . The intersection has codimension at most and so has a nontrivial intersection with ; and the claim follows.

Remark 3More generally, one can generate an eigenvalue inequality whenever the intersection numbers of three Schubert varieties of compatible dimensions is non-zero; see the paper of Helmke and Rosenthal. In fact, this generates a complete set of inequalities; this is a result of Klyachko. One can in fact restrict attention to those varieties whose intersection number is exactly one; this is a result of Knutson, Woodward, and myself. Finally, in those cases, the fact that the intersection is one can be proven by entirely elementary means (based on the standard inequalities relating the dimension of two subspaces to their intersection and sum ); this is a result of Bercovici, Collins, Dykema, Li, and Timotin. As a consequence, the methods in this section can, in principle, be used to derive all possible eigenvalue inequalities for sums of Hermitian matrices.

Exercise 4Verify the inequalities (12) and (4) by hand in the case when and commute (and are thus simultaneously diagonalisable), without the use of minimax formulae.

Exercise 5Establish the dual Lidskii inequalityfor any and the dual Weyl inequality

whenever .

Exercise 6Use the Lidskii inequality to establish the more general inequalitywhenever , and is the decreasing rearrangement of . (

Hint:express as the integral of as runs from to infinity. For each fixed , apply (12).) Combine this with Hölder’s inequality to conclude the-Weilandt-Hoffman inequalityis the usual norm (with the usual convention that ), and

is the -Schatten norm of .

Exercise 7Show that the -Schatten norms are indeed a norm on the space of Hermitian matrices for every .

Exercise 8Show that for any and any Hermitian matrix , one has

Exercise 9Establish the Hölder inequalitywhenever with , and are Hermitian matrices. (

Hint:Diagonalise one of the matrices and use the preceding exercise.)

The most important -Schatten norms are the *-Schatten norm* , which is just the operator norm, and the -Schatten norm , which is also the Frobenius norm (or Hilbert-Schmidt norm)

where are the coeffiicents of . (The -Schatten norm , also known as the *nuclear norm* or trace class norm, is important in a number of applications, such as matrix completion, but will not be used often in this course.) Thus we see that the case of the Weilandt-Hoffman inequality can be written as

We will give an alternate proof of this inequality, based on eigenvalue deformation, in the next section.

** — 4. Eigenvalue deformation — **

From the Weyl inequality (13), we know that the eigenvalue maps are Lipschitz continuous on Hermitian matrices (and thus also on real symmetric matrices). It turns out that we can obtain better regularity, provided that we avoid repeated eigenvalues. Fortunately, repeated eigenvalues are rare:

Exercise 10 (Dimension count)Suppose that . Show that the space of Hermitian matrices with at least one repeated eigenvalue has codimension in the space of all Hermitian matrices, and the space of real symmetric matrices with at least one repeated eigenvalue has codimension in the space of all real symmetric matrices. (When , repeated eigenvalues of course do not occur.)

Let us say that a Hermitian matrix has *simple spectrum* if it has no repeated eigenvalues. We thus see from the above exercise and (13) that the set of Hermitian matrices with simple spectrum forms an open dense set in the space of all Hermitian matrices, and similarly for real symmetric matrices; thus simple spectrum is the *generic* behaviour of such matrices. Indeed, the unexpectedly high codimension of the non-simple matrices (naively, one would expect a codimension set for a collision between, say, and ) suggests a *repulsion* phenomenon: because it is unexpectedly rare for eigenvalues to be equal, there must be some “force” that “repels” eigenvalues of Hermitian (and to a lesser extent, real symmetric) matrices from getting too close to each other. We now develop some machinery to make this more precise.

We first observe that when has simple spectrum, the zeroes of the characteristic polynomial are simple (i.e. the polynomial has nonzero derivartive at those zeroes). From this and the inverse function theorem, we see that each of the eigenvalue maps are smooth on the region where has simple spectrum. Because the eigenvectors are determined (up to phase) by the equations and , another application of the inverse function theorem tells us that we can (locally) select the maps to also be smooth. (There may be topological obstructions to smoothly selecting these vectors globally, but this will not concern us here as we will be performing a local analysis only. In later notes, we will in fact not work with the at all due to their phase ambiguity, and work instead with the *spectral projections* , which do not have this ambiguity.)

Now suppose that depends smoothly on a time variable , so that (when has simple spectrum) the eigenvalues and eigenvectors also depend smoothly on . We can then differentiate the equations

to obtain various equations of motion for and in terms of the derivatives of .

Let’s see how this works. Taking first derivatives of (18), (19) using the product rule, we obtain

The equation (21) simplifies to , thus is orthogonal to . Taking inner products of (20) with , we conclude the *Hadamard first variation formula*

This can already be used to give alternate proofs of various eigenvalue identities. For instance, If we apply this to , we see that

whenever has simple spectrum. The right-hand side can be bounded in magnitude by , and so we see that the map is Lipschitz with norm whenever has simple spectrum, which happens for generic (and all ) by Exercise 10. By the fundamental theorem of calculus, we thus conclude (13).

Exercise 11Use a similar argument to the one above to establish (17) without using minimax formulae or Lidskii’s inequality.

Exercise 12Use a similar argument to the one above to deduce Lidskii’s inequality (12) from Proposition 3 rather than Exercise 3.

One can also compute the second derivative of eigenvalues:

Exercise 13Suppose that depends smoothly on . By differentiating (20), (21) twice, establish theHadamard second variation formulawhenever has simple spectrum and .

If one interprets the second derivative of the eigenvalues as being proportional to a “force” on those eigenvalues (in analogy with Newton’s second law), (23) is asserting that each eigenvalue “repels” the other eigenvalues by exerting a force that is inversely proportional to their separation (and also proportional to the square of the matrix coefficient of in the eigenbasis). See this earlier blog post for more discussion.

Remark 4In the proof of the four moment theorem of Van Vu and myself, which we will discuss in a subsequent lecture, we will also need the variation formulae for the third, fourth, and fifth derivatives of the eigenvalues (the first four derivatives match up with the four moments mentioned in the theorem, and the fifth derivative is needed to control error terms). Fortunately, we do not need the precise formulae for these derivatives (which, as one can imagine, are quite complicated), but only their general form, and in particular an upper bound for these derivatives in terms of more easily computable quantities.

** — 5. Minors — **

In the previous sections, we perturbed Hermitian matrices by adding a (small) Hermitian correction matrix to them to form a new Hermitian matrix . Another important way to perturb a matrix is to pass to a principal minor, for instance to the top left minor of . There is an important relationship between the eigenvalues of the two matrices:

Exercise 14 (Cauchy interlacing inequalities)For any Hermitian matrix with top left minor , thenfor all . (

Hint:use the Courant-Fischer min-max theorem, Theorem 2.) Show furthermore that the space of for which equality holds in one of the inequalities in (24) has codimension (for Hermitian matrices) or (for real symmetric matrices).

Remark 5If one takes successive minors of an Hermitian matrix , and computes their spectra, then (24) shows that this triangular array of numbers forms a pattern known as aGelfand-Tsetlin pattern. These patterns are discussed a little more in this blog post.

One can obtain a more precise formula for the eigenvalues of in terms of those for :

Exercise 15 (Eigenvalue equation)Let be an Hermitian matrix with top left minor . Suppose that is an eigenvalue of distinct from all the eigenvalues of (and thus simple, by (24)). Show thatwhere is the bottom right entry of , and is the right column of (minus the bottom entry). (

Hint:Expand out the eigenvalue equation into the and components.) Note the similarities between (25) and (23).

Observe that the function is a rational function of which is increasing away from the eigenvalues of , where it has a pole (except in the rare case when the inner product vanishes, in which case it can have a removable singularity). By graphing this function one can see that the interlacing formula (24) can also be interpreted as a manifestation of the intermediate value theorem.

The identity (25) suggests that under typical circumstances, an eigenvalue of can only get close to an eigenvalue if the associated inner product is small. This type of observation is useful to achieve *eigenvalue repulsion* – to show that it is unlikely that the gap between two adjacent eigenvalues is small. We shall see examples of this in later notes.

** — 6. Singular values — **

The theory of eigenvalues of Hermitian matrices has an analogue in the theory of singular values of non-Hermitian matrices. We first begin with the counterpart to the spectral theorem, namely the singular value decomposition.

Theorem 4 (Singular value decomposition)Let , and let be a linear transformation from an -dimensional complex Hilbert space to a -dimensional complex Hilbert space . (In particular, could be an matrix with complex entries, viewed as a linear transformation from to .) Then there exist non-negative real numbers(known as the

singular valuesof ) and orthonormal sets and (known assingular vectorsof ), such thatfor all , where we abbreviate , etc.

Furthermore, whenever is orthogonal to all of the .

We adopt the convention that for . The above theorem only applies to matrices with at least as many rows as columns, but one can also extend the definition to matrices with more columns than rows by adopting the convention (it is easy to check that this extension is consistent on square matrices). All of the results below extend (with minor modifications) to the case when there are more columns than rows, but we have not displayed those extensions here in order to simplify the notation.

*Proof:* We induct on . The claim is vacuous for , so suppose that and that the claim has already been proven for .

We follow a similar strategy to the proof of Theorem 1. We may assume that is not identically zero, as the claim is obvious otherwise. The function is continuous on the unit sphere of , so there exists a unit vector which maximises this quantity. If we set , one easily verifies that is a critical point of the map , which then implies that . Thus, if we set , then and . This implies that maps the orthogonal complement of in to the orthogonal complement of in . By induction hypothesis, the restriction of to (and ) then admits a singular value decomposition with singular values and singular vectors , with the stated properties. By construction we see that are less than or equal to . If we now adjoin to the other singular values and vectors we obtain the claim.

Exercise 16Show that the singular values of a matrix are unique. If we have , show that the singular vectors are unique up to rotation by a complex phase.

By construction (and the above uniqueness claim) we see that whenever is a matrix, is a unitary matrix, and is a unitary matrix. Thus the singular spectrum of a matrix is invariant under left and right unitary transformations.

Exercise 17If is a complex matrix for some , show that the augmented matrixis a Hermitian matrix whose eigenvalues consist of , together with copies of the eigenvalue zero. (This generalises Exercise 16 from Notes 3.) What is the relationship between the singular vectors of and the eigenvectors of ?

Exercise 18If is an Hermitian matrix, show that the singular values of are simply the absolute values of , arranged in descending order. Show that the same claim also holds when is a normal matrix. What is the relationship between the singular vectors and eigenvectors of ?

Remark 6When is not normal, the relationship between eigenvalues and singular values is more subtle. We will discuss this point in later notes.

Exercise 19If is a complex matrix for some , show that has eigenvalues , and has eigenvalues together with copies of the eigenvalue zero. Based on this observation, give an alternate proof of the singular value decomposition theorem using the spectral theorem for (positive semi-definite) Hermitian matrices.

Exercise 20Show that the rank of a matrix is equal to the number of non-zero singular values.

Exercise 21Let be a complex matrix for some . Establish the Courant-Fischer min-max formulafor all , where the supremum ranges over all subspaces of of dimension .

One can use the above exercises to deduce many inequalities about singular values from analogous ones about eigenvalues. We give some examples below.

Exercise 22Let be complex matrices for some .

- (i) Establish the Weyl inequality whenever .
- (ii) Establish the Lidskii inequality
whenever .

- (iii) Show that for any , the map defines a norm on the space of complex matrices (this norm is known as the
Ky Fan norm).- (iv) Establish the Weyl inequality for all .
- (v) More generally, establish the -Weilandt-Hoffman inequality for any , where is the -Schatten norm of . (Note that this is consistent with the previous definition of the Schatten norms.)
- (vi) Show that the -Schatten norm is indeed a norm on for any .
- (vii) If is formed by removing one row from , show that for all .
- (viii) If and is formed by removing one column from , show that for all and . What changes when ?

Exercise 23Let be a complex matrix for some . Observe that the linear transformation naturally induces a linear transformation from -forms on to -forms on . We give the structure of a Hilbert space by declaring the basic forms for to be orthonormal.For any , show that the operator norm of is equal to .

Exercise 24Let be a matrix for some , let be a matrix, and let be a matrix for some .Show that and for any .

Exercise 25Let be a matrix for some , let be distinct, and let be distinct. Show thatUsing this, show that if are distinct, then

for every .

Exercise 26Establish the Hölder inequalitywhenever are complex matrices and are such that .

*Acknowledgments:* Thanks to Allen Knutson for corrections.

## 53 comments

Comments feed for this article

13 January, 2010 at 6:10 am

MariusHi Terry,

Linear algebra is indeed fascinating. Maybe the readers of this post may find

interesting some connections between this subject and quasiconvexity in the

calculus of variations and nonlinear elasticity:

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V0R-4SPSHN6-3&_user=3797462&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=1164865628&_rerunOrigin=google&_acct=C000061416&_version=1&_urlVersion=0&_userid=3797462&md5=d8361adf4dc60e9323fc42216695b1cd

or

http://www.imar.ro/~mbuliga/major_stem.pdf

14 January, 2010 at 1:50 pm

AnonymousA potential typo: in eq. 19 (and elsewhere), I thought the norm should be 1, not 0.

[Corrected, thanks – T.]15 January, 2010 at 7:59 am

kashifA small typo start of section 4: “From the Weyl inequality (13) we now that…” should be “From the Weyl inequality (13) we know that…”

[Corrected, thanks – T.]18 January, 2010 at 6:29 pm

254A, Notes 3b: Brownian motion and Dyson Brownian motion « What’s new[…] from the first and second Hadamard variation formulae (see Section 4 of Notes 3a) we […]

19 January, 2010 at 4:15 pm

David SpeyerPossible typo: Should equation (19) read u_i^* u_i =1 (rather than 0)?

[Ack, I thought I corrected that typo already. Corrected again, thanks – T.]2 February, 2010 at 1:34 pm

254A, Notes 4: The semi-circular law « What’s new[…] first observation is that the Cauchy interlacing law (Exercise 14 from Notes 3a) shows that the ESD of is very stable in . Indeed, we see from the interlacing law […]

17 February, 2010 at 8:30 am

Steven HeilmanSome pretty pedantic potential corrections:

1. A basic question in linear algebra …\lambda_{1}(B),\ldots,\lambda_{n}(B)

2. Eqs. (5),(9),(12): numerical label outside viewing window?

3. Section 3: Similarly, [for] the Ky Fan…

4. Section 4: From the Weyl inequality… we [k]now

5. After Exercise 10: … from the above exercise and [ ] (13)

6(a). Statement of Theorem 4: A^* v_j=\sigma_j u_j

6(b). Proof of Theorem 4: \sigma_1 :=\|Au_1 \|>0 ?

7. Exercise 23: induces a linear transformation … \bigwedge^k {\mathbb C}^p …

——————-

[and a few for Notes 3b]

1. After Remark 6: Applying [Lemma] (5)

2. … with disjoint increments being jointly indepen[d]ent

3. Proof of Theorem 6: …G continues [to] have the GUE distribution…

[Corrected, thanks – T.]23 February, 2010 at 10:02 pm

254A, Notes 6: Gaussian ensembles « What’s new[…] is simple. Since almost all Hermitian matrices have simple spectrum (see Exercise 10 of Notes 3a), this gives the full spectral distribution of GUE, except for the issue of the unspecified […]

14 March, 2010 at 11:33 am

254A, Notes 8: The circular law « What’s new[…] case, where eigenvalue inequalities such as the Weyl inequalities or Hoffman-Wielandt inequalities (Notes 3a) ensure stability of the […]

8 May, 2010 at 1:30 am

AnonymousDear Prof. Tao,

I had some problems on Ex. 8 and Ex. 13:

Ex. 8: Could you give some hint on how to prove it? This is most probably very trivial, but I tried it for several hours.

Ex. 13: I tried to do it according to your hint, but it is hard for me how to make the other eigenvalues involved in the equations of .

Thank you very much in advance!

8 May, 2010 at 1:12 pm

Terence TaoFor Ex 8, apply the method of Ex6 but with (12) replaced by (9).

For Ex13, a solution can be found at the link provided.

21 May, 2010 at 10:11 am

BenHi,

I’m a bit confused about your use of the inverse/implicit function theorem to get smoothness of the eigenvectors in a neighbourhood of . Is there not an issue that the matrix is singular? Is this where the normalization comes in? Thanks for your help.

21 May, 2010 at 12:07 pm

Terence TaoYes, with the normalisation (and working in the real symmetric setting for simplicity), the map is a map from an n-dimensional space to an n-dimensoinal space , and one can check by hand that the Jacobian is nondegenerate when the eigenvalues are simple, so the inverse theorem lets one select both and smoothly in this case. In the complex Hermitian setting it is a bit more complicated, one has to also fix the phase of one of the coordinates of (e.g. insisting that the first coordinate is a positive real) to make the dimensions match.

There is a theorem of Rellich that shows that eigenvectors and eigenvalues can in fact be selected in an analytic fashion even when there are repeated eigenvalues, but only if one abandons the idea of keeping the order of the eigenvalues fixed (i.e. one allows one eigenvalue to overtake another). This is a nontrivial theorem to prove, though.

2 October, 2014 at 10:02 am

hmobahiThank you Terrence. This was an extremely helpful comment. Does your comment imply that the function \sum_{k=1}^n (\frac{d}{d t} \lambda_k(t)) f_k(\lambda_k(t)) is well-defined even when eigenvalues are not simple? Here f_k(x) is any functional that analytically depends on x. I think about this, because the sum kills the dependency of a specific ordering of the eigenvaules.

30 May, 2010 at 5:12 pm

AnonymousQuestion and a correction.

In Exercise 5, the second inequality should be A->A+B

n->A, n->B.

Question: could a hint be given on how to prove the p-Weilandt-Hoffman inequality using Holder’s inequality? I do not have any clue.

30 May, 2010 at 11:05 pm

Terence TaoThanks for the correction!

By Holder’s inequality, the norm of a sequence is equal to the largest inner product between that sequence and a sequence of unit norm. Now use the fact that the norm is invariant with respect to rearrangement.

6 February, 2012 at 9:03 pm

AnonymousDear Prof. Tao, to get to get the first inequality in Exercise 6, do you multiply the Lidskii’s inequality by the integral given in the hint, and then how can one get or have an rearrangement of in front of ? Thanks for your help!

31 May, 2010 at 11:48 am

AnonymousThanks for the reply. But I am still confused about how to get (14) using ONLY the first inequality given in Exercise 6 and the Holder’s inequality.

TO generate the l_p norm for (\lamda_{i}(A+B)-\lamda_{i}(A)) using Holder’s inequality, we need to use a c_i that is negative if (\lamda_{i}(A+B)-\lamda_{i}(A)) is negative.

But this inequality is only about c_i that is nonnegative.

For example, consider the scalar case when A,B are both scalars, \lamda_{1}(A+B)-\lamda_{1}(A) <\lamda_1(B), does not necessarily mean the absolute value of the former is smaller than the absolute value of the latter.

Should we use some other conditions?

31 May, 2010 at 6:56 pm

Terence TaoBecause the trace of A+B is the sum of the trace of A and the trace of B, one can extend the original inequality to the case when the c_i are arbitrary reals.

6 February, 2012 at 9:06 pm

AnonymousDear Prof. Tao, to get the first inequality in Exercise 6, do you multiply the Lidskii’s inequality by the integral given in the hint, and then how can one get or have an rearrangement of in front of ? Thanks for your help!

22 December, 2010 at 9:08 pm

Outliers in the spectrum of iid matrices with bounded rank permutations « What’s new[…] were no interlacing inequalities in this case to control bounded rank perturbations (as discussed in this post). However, as it turns out I had arrived at the wrong conclusion, especially in the exterior of the […]

19 January, 2011 at 6:55 pm

SanIn proposition3, how do you use induction hypothesis, as the hypothesis would be true for the sum of first k eigenvalues.

Thanks,

San

20 January, 2011 at 12:13 pm

Terence TaoWhen one restricts A to the span of to obtain an n-1-dimensional linear operator, the eigenvalues are now , thanks to the hypothesis that A has the standard eigenbasis (so that it is given by the diagonal matrix with entries .

20 January, 2011 at 5:41 pm

SanThanks for the reply. Can I ask where can I find the solution of exercise3.

Sangeeta

22 January, 2011 at 9:26 am

SandorA direct link to “my survey with Allen Knutson” is http://www.ams.org/notices/200102/fea-knutson.pdf

22 August, 2011 at 4:17 am

Interlaced eigenvalues « Alasdair's musings[…] Another proof uses the min-max theorem, of which a full account is given by Terry Tao. […]

13 February, 2012 at 8:52 am

kushalHi,

1.Is there any upper bound for the norm of the inverse of a matrix (i.e. ||inv(A)||), in terms of the norm of the matrix , i.e. ||A|| ?

NB. There is no restriction on A such as det(A)= ???

Is there any lower bound for problem-2 that involves only A and not inv(A)?

11 July, 2013 at 11:03 am

AnonymousHello Prof. Tao,

Can these results be applied to matrices of differnt dimensions? For example, I have a m*m matrix (A) of rank n<m, and add another matrix of dimension q*q (B), where q< n. All matrices are Hermitian. Can I find a bound on the absolute eigenvalues of the system? The eigenvalues are lambda_1, – lambda_1, lambda_2, -lambda_2, and so on.

16 April, 2014 at 3:42 pm

AnonymousIs there a relation for the eigenvalues of the sum of an anti-Hermitian A and a Hermitian matrix B? For example, if I want to know the eigenvalues of A + B, can I simply calculate the eigenvalues of A and B seperately and then get the eigenvalues of A + B from them?

16 April, 2014 at 4:03 pm

Terence TaoIf A and B commute, then the eigenvalues of A+B are the sum of eigenvalues of A and some permutation of the eigenvalues of B.

In general, though, the situation is quite complicated, since A+B will usually not be a normal matrix, and the eigenvalues will be quite badly behaved complex numbers. One can still control the singular values, and hence the eigenvalues to some extent, by the Weyl inequalities, and one also controls the trace of A+B, but other than that things look quite messy, even for the 2×2 matrix case.

26 August, 2015 at 4:38 am

HamedI do not know about the application you have in mind. But in case you need to know the bounds of eigenvalues (for example if you want to analysis the stability of some dynamical system) this structure can help you. You can use Bendixon’s theorem which says that the eigenvalues of A+B are bounded in a box, when the real parts are bounded by eigenvalues of B (Hermitian) and the imaginary parts are bounded by eigenvalues of A (anti-Hermitian).

27 June, 2014 at 2:18 am

Felix V.Dear Professor Tao,

in Exercise 10, you ask the reader to proof that

“the space of Hermitian matrices with at least one repeated eigenvalue has codimension {3} in the space of all Hermitian matrices.”

It seemed to me that “space” should be interpreted as “(real) vector space” in this context.

But then the problem is that the set of matrices with at least one repeated eigenvalue is NOT a vector space (not closed under addition), as can be seen by considering e.g. diagonal matrices:

{

\left(\begin{array}{ccc}

1\\

& 1\\

& & 2

\end{array}\right)+\left(\begin{array}{ccc}

0\\

& 1\\

& & 1

\end{array}\right)=\left(\begin{array}{ccc}

1\\

& 2\\

& & 3

\end{array}\right)

}

Best regards,

Felix V.

27 June, 2014 at 8:10 am

Terence TaoHere I am not interpreting “space” in the linear algebra sense; you may substitute “set” or “algebraic set” instead, if you prefer.

17 March, 2015 at 5:47 am

Lecture #22: Random Sampling for Fast SVDs | CMU Advanced Algos: S15[…] pointed out that it follows from the Weyl inequalities. These say that for all integers such that […]

28 March, 2015 at 7:57 pm

kumar vishwajeetDear Dr Tao,

I am currently working on the distribution of eigenvalues of non central wishart matrices. I found the following interesting thing.

Let be a random complex matrix of size . has circular Gaussian distribution with mean 0 and variance

Consider another case where, where has Gaussian distribution with mean 0 and variance . Using monte Carlo, I found that the probaility density function of the and that of only differs by a multiplying constant. Is there any mathematical proof or logical explanation for this?

29 March, 2015 at 8:08 am

Terence TaoThe eigenvalues of and are the squares of the eigenvalues of the Hermitian matrices and respectively. So my guess is that the matrices and have essentially the same spectrum (semicircle law, I think) and are behaving as if they were free of . (I’m not sure what you mean by the multiplying constant difference though – the probability density function has to integrate to 1.)

28 March, 2015 at 8:12 pm

kumar vishwajeetJust a modification of my last post: The probability density function of eigenvalues of and differ by a multiplying constant.

29 March, 2015 at 12:46 pm

kumar vishwajeetThank you for the prompt reply.

By “Multiplying constant difference” I mean that if be the pdf of eigenvalues of and be the pdf of eigenvalues of , then

, where is a constant.

29 March, 2015 at 3:12 pm

kumar vishwajeetI got it. has to be equal to one.

24 September, 2015 at 4:31 pm

AbbasWhat does $|Av|$ denote in Exercise 21?

[The magnitude of the vector formed by multiplying the matrix with the vector . -T.]30 March, 2016 at 4:39 pm

DavidProf. Tao,

I am a novice going through a few of these exercises and ran into trouble on #6. Here you suggest using Holder’s inequality to prove the p-Wielandt Hoffman inequality. This works very straightforwardly to yield an Lp norm for the right hand side of the inequality (in (14)), but the Lp norm also must be obtained on the left hand side, which is more difficult. Can you comment on how to proceed for the left hand side of (14)?

30 March, 2016 at 9:24 pm

Terence TaoUse the converse form of Holder’s inequality.

31 March, 2016 at 11:22 am

DavidVery nice, thanks.

24 May, 2016 at 4:44 am

AnonymousDear Dr. Tao,

I was thinking about a way to show that the partial trace is independent of the choice an orthonormal basis. But I kept failing to find a good proof. Is there any hint you can give me ?

24 May, 2016 at 8:35 am

Terence TaoGiven two orthonormal bases for , write the elements of one basis as a linear combination of the other basis; the coefficients will then form a unitary matrix since both bases are orthonormal. Insert these expansions of the basis elements to write the partial trace in one basis in terms of the other basis, and use the definition of a unitary matrix to simplify.

Alternatively, one can note that the partial trace is also equal to the full trace of , where is the orthogonal projection onto , and use the standard fact that the full trace is basis-independent.

24 May, 2016 at 10:32 am

AnonymousThank you, the second one is the one I was looking for.

27 May, 2016 at 12:01 am

DavidDear Dr. Tao,

I would be interested to see the proof of the dual Weyl inequality if you happen to have it handy. I have read the proof in “A Note on Weyl’s Inequality” by Steve Fisk, but I failed to really understand it.

Thanks,

David

27 May, 2016 at 7:56 am

Terence TaoOne can for instance derive the dual Weyl inequality from the original Weyl inequality by repeatedly using the identity .

24 June, 2016 at 8:43 am

vsk1996Tao sir please give some simpler version of (Courant-Fischer Theorem) and relate it with rayleigh quotient

25 October, 2016 at 3:39 am

ArchyProf. Tao, is there a simplification of this theory, if all the Hermitian matrices are rank 1? thank you very much!

26 November, 2016 at 11:05 am

keejThese notes are great. I think there is a typo in the definition of spectral projection in Exercise 1. It should be $u_j(A) u_j(A)^*$, not $u_j(A)^* u_j(A)$.

[Corrected, thanks – T.]2 March, 2017 at 12:08 am

keejIn the proof of the SVD, we use the fact that a critical point of the map must satisfy the identity . I have two questions about this.

First, I can see why this is true by expanding out all the entries and differentiating, but I feel like there should be a faster way. Is there?

Secondly, what does the derivative even mean in the case that has complex entries? The map in question is real-valued, so how can its gradient be a complex vector?

2 March, 2017 at 9:48 am

Terence TaoOne can first try to understand the situation when is a diagonal matrix, in which case the map basically splits as the sum of non-interacting one-dimensional maps, and the claim is then easy to understand in that setting. The existence of the SVD in fact tells us that the critical point behaviour of a general matrix will be the same as that of a diagonal matrix (with the singular values as entries), so if one believes the SVD to exist then this explains why this identity ought to be true. Of course one cannot use this observation to

provethe existence of SVD, as this would be circular, but it can certainly be used tomotivatesuch a proof.Any complex vector space can be viewed as a real vector space of twice the dimension, so one can take the gradient in that setting.