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

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

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

${\displaystyle f(x)=(Ax,x),\;\|x\|=1.}$

For Hermitian matrices A, the range of the continuous function RA(x), or f(x), is a compact interval [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 ${\textstyle A}$ be Hermitian on an inner product space ${\textstyle V}$ with dimension ${\textstyle n}$, with spectrum ordered in descending order ${\textstyle \lambda _{1}\geq ...\geq \lambda _{n}}$.

Let ${\textstyle v_{1},...,v_{n}}$ be the corresponding unit-length orthogonal eigenvectors.

Reverse the spectrum ordering, so that ${\textstyle \xi _{1}=\lambda _{n},...,\xi _{n}=\lambda _{1}}$.

(Poincaré’s inequality) — Let ${\textstyle M}$ be a subspace of ${\textstyle V}$ with dimension ${\textstyle k}$, then there exists unit vectors ${\textstyle x,y\in M}$, such that

${\textstyle \langle x,Ax\rangle \leq \lambda _{k}}$, and ${\textstyle \langle y,Ay\rangle \geq \xi _{k}}$.

Proof

Part 2 is a corollary, using ${\textstyle -A}$.

${\textstyle M}$ is a ${\textstyle k}$ dimensional subspace, so if we pick any list of ${\textstyle n-k+1}$ vectors, their span ${\textstyle N:=span(v_{k},...v_{n})}$ must intersect ${\textstyle M}$ on at least a single line.

Take unit ${\textstyle x\in M\cap N}$. That’s what we need.

${\textstyle x=\sum _{i=k}^{n}a_{i}v_{i}}$, since ${\textstyle x\in N}$.
Since ${\textstyle \sum _{i=k}^{n}|a_{i}|^{2}=1}$, we find ${\textstyle \langle x,Ax\rangle =\sum _{i=k}^{n}|a_{i}|^{2}\lambda _{i}\leq \lambda _{k}}$.

min-max theorem —

{\displaystyle {\begin{aligned}\lambda _{k}&=\max _{\begin{array}{c}{\mathcal {M}}\subset V\\\operatorname {dim} ({\mathcal {M}})=k\end{array}}\min _{\begin{array}{c}x\in {\mathcal {M}}\\\|x\|=1\end{array}}\langle x,Ax\rangle \\&=\min _{\begin{array}{c}{\mathcal {M}}\subset V\\\operatorname {dim} ({\mathcal {M}})=n-k+1\end{array}}\max _{\begin{array}{c}x\in {\mathcal {M}}\\\|x\|=1\end{array}}\langle x,Ax\rangle {\text{. }}\end{aligned}}}

Proof

Part 2 is a corollary of part 1, by using ${\textstyle -A}$.

By Poincare’s inequality, ${\textstyle \lambda _{k}}$ is an upper bound to the right side.

By setting ${\textstyle {\mathcal {M}}=span(v_{1},...v_{k})}$, the upper bound is achieved.

### Counterexample in the non-Hermitian case

Let N be the nilpotent matrix

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

Define the Rayleigh quotient ${\displaystyle R_{N}(x)}$ 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 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 {σk} of a square matrix M are the square roots of the eigenvalues of M*M (equivalently MM*). An immediate consequence[citation needed] of the first equality in the min-max theorem is:

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

Similarly,

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

Here ${\displaystyle \sigma _{k}=\sigma _{k}^{\uparrow }}$ denotes the kth entry in the increasing sequence of σ's, so that ${\displaystyle \sigma _{1}\leq \sigma _{2}\leq \cdots }$.

### Cauchy interlacing theorem

Let A be a symmetric n × n matrix. The m × m matrix B, where mn, is called a compression of A 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 A are α1 ≤ ... ≤ αn, and those of B are β1 ≤ ... ≤ βj ≤ ... ≤ βm, then for all jm,
${\displaystyle \alpha _{j}\leq \beta _{j}\leq \alpha _{n-m+j}.}$

This can be proven using the min-max principle. Let βi have corresponding eigenvector bi and Sj be the j dimensional subspace Sj = span{b1, ..., bj}, then

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

According to first part of min-max, αjβj. On the other hand, if we define Smj+1 = span{bj, ..., bm}, then

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

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

When nm = 1, we have αjβjαj+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 (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 A as

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

where entries are repeated with multiplicity, as in the matrix case. (To emphasize that the sequence is decreasing, we may write ${\displaystyle \lambda _{k}=\lambda _{k}^{\downarrow }}$.) 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 SkH 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 ... ≤ λk ≤ ... ≤ λ1. Then:
{\displaystyle {\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

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

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

Therefore, for all Sk

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

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:

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

So

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

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

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

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

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

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

So

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

This implies

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

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

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

Pick Sk−1 = span{u1, ..., uk−1} and we deduce

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

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 essential spectrum, and we would like to approximate the eigenvalues and eigenfunctions.

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

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

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

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

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

Theorem. Let A be self-adjoint. Then ${\displaystyle (A-E)\geq 0}$ for ${\displaystyle E\in \mathbb {R} }$ if and only if ${\displaystyle \sigma (A)\subseteq [E,\infty )}$.[1]: 77
Theorem. If A is self-adjoint, then

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

and

${\displaystyle \sup \sigma (A)=\sup _{\psi \in {\mathfrak {D}}(A),\|\psi \|=1}\langle \psi ,A\psi \rangle }$.[1]: 77