min-max theorem

In linear algebra and functional analysis, the min-max theorem, or variational theorem, or Courant–Fischer–Weyl min-max principle, is a result that gives a variational characterization of eigenvalues of compact Hermitian operators on Hilbert spaces. It can be viewed as the starting point of many results of similar nature.

This article first discusses the finite-dimensional case and its applications before considering compact operators on infinite-dimensional Hilbert spaces.
We will see that for compact operators, the proof of the main theorem uses essentially the same idea from the finite-dimensional argument.

In the case that the operator is non-Hermitian, the theorem provides an equivalent characterization of the associated singular values.
The min-max theorem can be extended to self-adjoint operators that are bounded below.

Matrices

Let be a Hermitian matrix. As with many other variational results on eigenvalues, one considers the Rayleigh–Ritz quotient defined by

:$R_A(x) = \frac$

where denotes the Euclidean inner product on .
Clearly, the Rayleigh quotient of an eigenvector is its associated eigenvalue. Equivalently, the Rayleigh–Ritz quotient can be replaced by

:$f(x) = (Ax, x), \; \, x\, = 1.$

For Hermitian matrices ''A'', the range of the continuous function ''R_A''(''x''), or ''f''(''x''), is a compact subset 'a'', ''b''of the real line. The maximum ''b'' and the minimum ''a'' are the largest and smallest eigenvalue of ''A'', respectively. The min-max theorem is a refinement of this fact.

Min-max theorem

Let be an Hermitian matrix with eigenvalues then

:$\lambda_k = \min_U \$
and
:$\lambda_k = \max_U \$
in particular,
:$\lambda_1 \leq R_A(x) \leq \lambda_n \quad\forall x \in \mathbf^n\backslash\$
and these bounds are attained when is an eigenvector of the appropriate eigenvalues.

Also the simpler formulation for the maximal eigenvalue ''λ''_n is given by:
:$\lambda_n = \max \.$
Similarly, the minimal eigenvalue ''λ''₁ is given by:
:$\lambda_1 = \min \.$

Counterexample in the non-Hermitian case

Let ''N'' be the nilpotent matrix

:$\begin 0 & 1 \\ 0 & 0 \end.$

Define the Rayleigh quotient $R_N(x)$ exactly as above in the Hermitian case. Then it is easy to see that the only eigenvalue of ''N'' is zero, while the maximum value of the Rayleigh quotient is . That is, the maximum value of the Rayleigh quotient is larger than the maximum eigenvalue.

Applications

Min-max principle for singular values

The singular values of a square matrix ''M'' are the square roots of the eigenvalues of ''M''*''M'' (equivalently ''MM*''). An immediate consequence of the first equality in the min-max theorem is:

:$\sigma_k^ = \min_ \max_ (M^* Mx, x)^=\min_ \max_ \, Mx \, .$

Similarly,

:$\sigma_k^ = \max_ \min_ \, Mx \, .$

Here $\sigma_k=\sigma_k^\uparrow$ denotes the ''k''^th entry in the increasing sequence of σ's, so that $\sigma_1\leq\sigma_2\leq\cdots$.

Cauchy interlacing theorem

Let be a symmetric ''n'' × ''n'' matrix. The ''m'' × ''m'' matrix ''B'', where ''m'' ≤ ''n'', is called a compression of if there exists an orthogonal projection ''P'' onto a subspace of dimension ''m'' such that ''PAP*'' = ''B''. The Cauchy interlacing theorem states:

:Theorem. If the eigenvalues of are , and those of ''B'' are , then for all ,
::$\alpha_j \leq \beta_j \leq \alpha_.$

This can be proven using the min-max principle. Let ''β_i'' have corresponding eigenvector ''b_i'' and ''S_j'' be the ''j'' dimensional subspace then

:$\beta_j = \max_ (Bx, x) = \max_ (PAP^*x, x) \geq \min_ \max_ (A(P^*x), P^*x) = \alpha_j.$

According to first part of min-max, On the other hand, if we define then

:$\beta_j = \min_ (Bx, x) = \min_ (PAP^*x, x)= \min_ (A(P^*x), P^*x) \leq \alpha_,$

where the last inequality is given by the second part of min-max.

When , we have , hence the name ''interlacing'' theorem.

Compact operators

Let be a compact, Hermitian operator on a Hilbert space ''H''. Recall that the spectrum of such an operator (the set of eigenvalues) is a set of real numbers whose only possible cluster point is zero.
It is thus convenient to list the positive eigenvalues of as

:$\cdots \le \lambda_k \le \cdots \le \lambda_1,$

where entries are repeated with multiplicity, as in the matrix case. (To emphasize that the sequence is decreasing, we may write $\lambda_k = \lambda_k^\downarrow$.)
When ''H'' is infinite-dimensional, the above sequence of eigenvalues is necessarily infinite.
We now apply the same reasoning as in the matrix case. Letting ''S_k'' ⊂ ''H'' be a ''k'' dimensional subspace, we can obtain the following theorem.

:Theorem (Min-Max). Let be a compact, self-adjoint operator on a Hilbert space , whose positive eigenvalues are listed in decreasing order . Then:
::$\begin \max_ \min_ (Ax,x) &= \lambda_k ^, \\ \min_ \max_ (Ax, x) &= \lambda_k^. \end$

A similar pair of equalities hold for negative eigenvalues.

Self-adjoint operators

The min-max theorem also applies to (possibly unbounded) self-adjoint operators.G. Teschl, Mathematical Methods in Quantum Mechanics (GSM 99) https://www.mat.univie.ac.at/~gerald/ftp/book-schroe/schroe.pdf Recall the essential spectrum is the spectrum without isolated eigenvalues of finite multiplicity.
Sometimes we have some eigenvalues below the essential spectrum, and we would like to approximate the eigenvalues and eigenfunctions.

:Theorem (Min-Max). Let ''A'' be self-adjoint, and let $E_1\le E_2\le E_3\le\cdots$ be the eigenvalues of ''A'' below the essential spectrum. Then

$E_n=\min_\max\$.

If we only have ''N'' eigenvalues and hence run out of eigenvalues, then we let $E_n:=\inf\sigma_(A)$ (the bottom of the essential spectrum) for ''n>N'', and the above statement holds after replacing min-max with inf-sup.

:Theorem (Max-Min). Let ''A'' be self-adjoint, and let $E_1\le E_2\le E_3\le\cdots$ be the eigenvalues of ''A'' below the essential spectrum. Then

$E_n=\max_\min\$.

If we only have ''N'' eigenvalues and hence run out of eigenvalues, then we let $E_n:=\inf\sigma_(A)$ (the bottom of the essential spectrum) for ''n > N'', and the above statement holds after replacing max-min with sup-inf.

The proofs use the following results about self-adjoint operators:

:Theorem. Let ''A'' be self-adjoint. Then $(A-E)\ge0$ for $E\in\mathbb$ if and only if $\sigma(A)\subseteq[E,\infty)$.

:Theorem. If ''A'' is self-adjoint, then

$\inf\sigma(A)=\inf_\langle\psi,A\psi\rangle$

and

$\sup\sigma(A)=\sup_\langle\psi,A\psi\rangle$.

See also

* Courant minimax principle
* Max–min inequality

References

External links and citations to related work

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