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 $A$ be a $n \times n$ Hermitian matrix. As with many other variational results on eigenvalues, one considers the Rayleigh–Ritz quotient $RA : Cn \ {0} → R$ defined by

R_{A}(x)={\frac {(Ax,x)}{(x,x)}}

where $(\cdot, \cdot)$ denotes the Euclidean inner product on $C n$ . 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, 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 $A$ be a $n \times n$ Hermitian matrix with eigenvalues $λ 1 \leq ... \leq λ k \leq ... \leq λ n$ then

\lambda _{k}=\min _{U}\{\max _{x}\{R_{A}(x)\mid x\in U{\text{ and }}x\neq 0\}\mid \dim(U)=k\}

and

\lambda _{k}=\max _{U}\{\min _{x}\{R_{A}(x)\mid x\in U{\text{ and }}x\neq 0\}\mid \dim(U)=n-k+1\}

in particular,

\lambda _{1}\leq R_{A}(x)\leq \lambda _{n}\quad \forall x\in {\mathbf {C}}^{n}\backslash \{0\}

and these bounds are attained when $x$ is an eigenvector of the appropriate eigenvalues.

Also note that the simpler formulation for the maximal eigenvalue λ_n is given by:

\lambda _{n}=\max\{R_{A}(x):x\neq 0\}.

Similarly, the minimal eigenvalue λ₁ is given by:

\lambda _{1}=\min\{R_{A}(x):x\neq 0\}.

Proof

Since the matrix $A$ is Hermitian it is diagonalizable and we can choose an orthonormal basis of eigenvectors {u₁, ..., u_n} that is, u_i is an eigenvector for the eigenvalue λ_i and such that (u_i, u_i) = 1 and (u_i, u_j) = 0 for all i ≠ j.

If U is a subspace of dimension k then its intersection with the subspace $span{u k, ..., u n}$ isn't zero (by simply checking dimensions) and hence there exists a vector $v \neq 0$ in this intersection that we can write as

v=\sum _{{i=k}}^{n}\alpha _{i}u_{i}

and whose Rayleigh quotient is

R_{A}(v)={\frac {\sum _{{i=k}}^{n}\lambda _{i}\alpha _{i}^{2}}{\sum _{{i=k}}^{n}\alpha _{i}^{2}}}\geq \lambda _{k}

(as all $\lambda _{i}\geq \lambda _{k}$ for i=k,..,n) and hence

\max\{R_{A}(x)\mid x\in U\}\geq \lambda _{k}

Since this is true for all U, we can conclude that

\min\{\max\{R_{A}(x)\mid x\in U{\text{ and }}x\neq 0\}\mid \dim(U)=k\}\geq \lambda _{k}

This is one inequality. To establish the other inequality, chose the specific k-dimensional space $V = span{u 1, ..., u k}$ , for which

\max\{R_{A}(x)\mid x\in V{\text{ and }}x\neq 0\}\leq \lambda _{k}

because $\lambda _{k}$ is the largest eigenvalue in V. Therefore, also

\min\{\max\{R_{A}(x)\mid x\in U{\text{ and }}x\neq 0\}\mid \dim(U)=k\}\leq \lambda _{k}

In the case where U is a subspace of dimension n-k+1, we proceed in a similar fashion: Consider the subspace of dimension k, $span{u 1, ..., u k}.$ Its intersection with the subspace U isn't zero (by simply checking dimensions) and hence there exists a vector v in this intersection that we can write as

v=\sum _{{i=1}}^{k}\alpha _{i}u_{i}

and whose Rayleigh quotient is

R_{A}(v)={\frac {\sum _{{i=1}}^{k}\lambda _{i}\alpha _{i}^{2}}{\sum _{{i=1}}^{k}\alpha _{i}^{2}}}\leq \lambda _{k}

and hence

\min\{R_{A}(x)\mid x\in U\}\leq \lambda _{k}

Since this is true for all U, we can conclude that

\max\{\min\{R_{A}(x)\mid x\in U{\text{ and }}x\neq 0\}\mid \dim(U)=n-k+1\}\leq \lambda _{k}

Again, this is one part of the equation. To get the other inequality, note again that the eigenvector u of $\lambda _{k}$ is contained in $U = span{u k, ..., u n}$ so that we can conclude the equality.

Counterexample in the non-Hermitian case

Let N be the nilpotent matrix

{\begin{bmatrix}0&1\\0&0\end{bmatrix}}.

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 ratio is $1 / 2$ . 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 {σ_k} of a square matrix M are the square roots of eigenvalues of M*M (equivalently MM*). An immediate consequence of the first equality from min-max theorem is

\sigma _{k}^{{\uparrow }}=\min _{{S:\dim(S)=k}}\max _{{x\in S,\|x\|=1}}(M^{*}Mx,x)^{{{\frac {1}{2}}}}=\min _{{S:\dim(S)=k}}\max _{{x\in S,\|x\|=1}}\|Mx\|.

Similarly,

\sigma _{k}^{{\downarrow }}=\max _{{S:\dim(S)=n-k+1}}\min _{{x\in S,\|x\|=1}}\|Mx\|.

Cauchy interlacing theorem

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

Theorem. If the eigenvalues of

A

are

α 1 \leq ... \leq α n

, and those of B are

β 1 \leq ... \leq β j \leq ... \leq β m

, then for all

j \leq m

\alpha _{j}\leq \beta _{j}\leq \alpha _{{n-m+j}}.

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 $S j = span{b 1, ..., b j},$ then

\beta _{j}=\max _{x\in S_{j},\|x\|=1}(Bx,x)=\max _{x\in S_{j},\|x\|=1}(P^{*}APx,x)\geq \min _{S_{j}}\max _{x\in S_{j},\|x\|=1}(A(Px),Px)=\alpha _{j}.

According to first part of min-max, $α j \leq β j .$ On the other hand, if we define $S m - j +1 = span{b j, ..., b m},$ then

\beta _{j}=\min _{x\in S_{m-j+1},\|x\|=1}(Bx,x)=\min _{x\in S_{m-j+1},\|x\|=1}(P^{*}APx,x)=\min _{x\in S_{m-j+1},\|x\|=1}(A(Px),Px)\leq \alpha _{n-m+j},

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

Notice that, when $n - m = 1$ , we have $α j \leq β j \leq α j +1$ , hence the name interlacing theorem.

Compact operators

Let $A$ be a compact, Hermitian operator on a Hilbert space H. Recall that the spectrum of such an operator form a sequence of real numbers whose only possible cluster point is zero. Every nonzero number in the spectrum is an eigenvalue. It no longer makes sense here to list the positive eigenvalues in increasing order. Let the positive eigenvalues of $A$ be

\cdots \leq \lambda _{k}\leq \cdots \leq \lambda _{1},

where multiplicity is taken into account as in the matrix case. 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

A

be a compact, self-adjoint operator on a Hilbert space

H

, whose positive eigenvalues are listed in decreasing order

... \leq λ k \leq ... \leq λ 1

. Then:

{\begin{aligned}\max _{{S_{k}}}\min _{{x\in S_{k},\|x\|=1}}(Ax,x)&=\lambda _{k}^{{\downarrow }},\\\min _{{S_{{k-1}}}}\max _{{x\in S_{{k-1}}^{{\perp }},\|x\|=1}}(Ax,x)&=\lambda _{k}^{{\downarrow }}.\end{aligned}}

A similar pair of equalities hold for negative eigenvalues.

Proof:

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

Let S' be the closure of the linear span $S'=\operatorname {span}\{u_{k},u_{{k+1}},\ldots \}$ . The subspace S' has codimension k − 1. By the same dimension count argument as in the matrix case, S' ∩ S_k is non empty. So there exists x ∈ S' ∩ S_k with $\|x\|=1$ . Since it is an element of S' , such an x necessarily satisfy

(Ax,x)\leq \lambda _{k}.

Therefore, for all S_k

\inf _{{x\in S_{k},\|x\|=1}}(Ax,x)\leq \lambda _{k}

But $A$ is compact, therefore the function f(x) = (Ax, x) is weakly continuous. Furthermore, any bounded set in H is weakly compact. This lets us replace the infimum by minimum:

\min _{{x\in S_{k},\|x\|=1}}(Ax,x)\leq \lambda _{k}.

\sup _{{S_{k}}}\min _{{x\in S_{k},\|x\|=1}}(Ax,x)\leq \lambda _{k}.

Because equality is achieved when $S_{k}=\operatorname {span}\{u_{1},\ldots ,u_{k}\}$ ,

\max _{{S_{k}}}\min _{{x\in S_{k},\|x\|=1}}(Ax,x)=\lambda _{k}.

This is the first part of min-max theorem for compact self-adjoint operators.

Analogously, consider now a $(k - 1)$ -dimensional subspace S_k−1, whose the orthogonal complement is denoted by S_k−1^⊥. If S' = span{u₁...u_k},

S'\cap S_{{k-1}}^{{\perp }}\neq {0}.

\exists x\in S_{{k-1}}^{{\perp }}\,\|x\|=1,(Ax,x)\geq \lambda _{k}.

This implies

\max _{{x\in S_{{k-1}}^{{\perp }},\|x\|=1}}(Ax,x)\geq \lambda _{k}

where the compactness of A was applied. Index the above by the collection of k-1-dimensional subspaces gives

\inf _{{S_{{k-1}}}}\max _{{x\in S_{{k-1}}^{{\perp }},\|x\|=1}}(Ax,x)\geq \lambda _{k}.

Pick S_k−1 = span{u₁, ..., u_k−1} and we deduce

\min _{{S_{{k-1}}}}\max _{{x\in S_{{k-1}}^{{\perp }},\|x\|=1}}(Ax,x)=\lambda _{k}.

Self-adjoint operators

The min-max theorem also applies to (possibly unbounded) self-adjoint operators.^[1] ^[2] Recall the essential spectrum is the spectrum without isolated eigenvalues of finite multiplicity. Sometimes we have some eigenvalues below the bottom of the essential spectrum, and we would like to approximate the eigenvalues and eigenfunctions.

Theorem (Min-Max). Let A be self-adjoint, and let $E_{1}\leq E_{2}\leq E_{3}\leq \cdots$ be the eigenvalues of A below the essential spectrum. Then

$E_{n}=\min _{{\psi _{1},\ldots ,\psi _{{n-1}}}}\max\{\langle \psi ,A\psi \rangle :\psi \in \operatorname {span}(\psi _{1},\ldots ,\psi _{{n-1}})\}$ .

If we only have N eigenvalues and hence run out of eigenvalues, then we let $E_{n}:=\inf \sigma _{{ess}}(A)$ (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 $E_{1}\leq E_{2}\leq E_{3}\leq \cdots$ be the eigenvalues of A below the essential spectrum. Then

$E_{n}=\max _{{\psi _{1},\ldots ,\psi _{{n-1}}}}\min\{\langle \psi ,A\psi \rangle :\psi \perp \psi _{1},\ldots ,\psi _{{n-1}}\}$ .

The proofs^[1]^[2] use the following results about self-adjoint operators:

Theorem. Let A be self-adjoint. Then $(A-E)\geq 0$ for $E\in {\mathbb {R}}$ if and only if $\sigma (A)\subseteq [E,\infty )$ .

Theorem. If A is self-adjoint, then

$\inf \sigma (A)=\inf _{{\psi \in {\mathfrak {D}}(A),\|\psi \|=1}}\langle \psi ,A\psi \rangle$

and

$\sup \sigma (A)=\sup _{{\psi \in {\mathfrak {D}}(A),\|\psi \|=1}}\langle \psi ,A\psi \rangle$ .

References

1 2 G. Teschl, Mathematical Methods in Quantum Mechanics (GSM 99) http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/schroe.pdf
1 2 Lieb-Loss, Analysis 2nd ed. (GSM 14)

M. Reed and B. Simon, Methods of Modern Mathematical Physics IV: Analysis of Operators, Academic Press, 1978.

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Min-max theorem

Matrices

Min-max theorem

Proof

Counterexample in the non-Hermitian case

Applications

Min-max principle for singular values

Cauchy interlacing theorem

Compact operators

Self-adjoint operators

See also

References