Min max theorem

Matrices

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

R A ( x ) = ( A x , 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

f ( x ) = ( A x , 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 × n Hermitian matrix with eigenvalues λ₁ ≤ ... ≤ λ_k ≤ ... ≤ λ_n then

λ k = min U { max x { R A ( x ) ∣ x ∈ U  and  x ≠ 0 } ∣ dim ⁡ ( U ) = k }

and

λ k = max U { min x { R A ( x ) ∣ x ∈ U  and  x ≠ 0 } ∣ dim ⁡ ( U ) = n − k + 1 }

in particular,

λ 1 ≤ R A ( x ) ≤ λ n ∀ x ∈ C n ∖ { 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:

λ n = max { R A ( x ) : x ≠ 0 } .

Similarly, the minimal eigenvalue λ₁ is given by:

λ 1 = min { R A ( x ) : x ≠ 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 ≠ 0 in this intersection that we can write as

v = ∑ i = k n α i u i

and whose Rayleigh quotient is

R A ( v ) = ∑ i = k n λ i α i 2 ∑ i = k n α i 2 ≥ λ k

(as all λ i ≥ λ k for i=k,..,n) and hence

max { R A ( x ) ∣ x ∈ U } ≥ λ k

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

min { max { R A ( x ) ∣ x ∈ U  and  x ≠ 0 } ∣ dim ⁡ ( U ) = k } ≥ λ k

This is one inequality. To establish the other inequality, chose the specific k-dimensional space V = span{u₁, ..., u_k} , for which

max { R A ( x ) ∣ x ∈ V  and  x ≠ 0 } ≤ λ k

because λ k is the largest eigenvalue in V. Therefore, also

min { max { R A ( x ) ∣ x ∈ U  and  x ≠ 0 } ∣ dim ⁡ ( U ) = k } ≤ λ 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{u₁, ..., 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 = ∑ i = 1 k α i u i

and whose Rayleigh quotient is

R A ( v ) = ∑ i = 1 k λ i α i 2 ∑ i = 1 k α i 2 ≤ λ k

and hence

min { R A ( x ) ∣ x ∈ U } ≤ λ k

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

max { min { R A ( x ) ∣ x ∈ U  and  x ≠ 0 } ∣ dim ⁡ ( U ) = n − k + 1 } ≤ λ k

Again, this is one part of the equation. To get the other inequality, note again that the eigenvector u of λ 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

[ 0 1 0 0 ] .

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.

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

σ k ↑ = min S : dim ⁡ ( S ) = k max x ∈ S , ∥ x ∥ = 1 ( M ∗ M x , x ) 1 2 = min S : dim ⁡ ( S ) = k max x ∈ S , ∥ x ∥ = 1 ∥ M x ∥ .

Similarly,

σ k ↓ = max S : dim ⁡ ( S ) = n − k + 1 min x ∈ S , ∥ x ∥ = 1 ∥ M x ∥ .

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 α₁ ≤ ... ≤ α_n, and those of B are β₁ ≤ ... ≤ β_j ≤ ... ≤ β_m, then for all j < m + 1, α j ≤ β j ≤ α 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{b₁, ..., b_j}, then

β j = max x ∈ S j , ∥ x ∥ = 1 ( B x , x ) = max x ∈ S j , ∥ x ∥ = 1 ( P ∗ A P x , x ) ≥ min S j max x ∈ S j , ∥ x ∥ = 1 ( A x , x ) = α j .

According to first part of min-max, α_j ≤ β_j. On the other hand, if we define S_m−j+1 = span{b_j, ..., b_m}, then

β j = min x ∈ S m − j + 1 , ∥ x ∥ = 1 ( B x , x ) = min x ∈ S m − j + 1 , ∥ x ∥ = 1 ( P ∗ A P x , x ) = min x ∈ S m − j + 1 , ∥ x ∥ = 1 ( A x , x ) ≤ α 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 ≤ β_j ≤ 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

⋯ ≤ λ k ≤ ⋯ ≤ λ 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 ... ≤ λ_k ≤ ... ≤ λ₁. Then: max S k min x ∈ S k , ∥ x ∥ = 1 ( A x , x ) = λ k ↓ , min S k − 1 max x ∈ S k − 1 ⊥ , ∥ x ∥ = 1 ( A x , x ) = λ k ↓ .

A similar pair of equalities hold for negative eigenvalues.

Proof:

Self-adjoint operators

The min-max theorem also applies to (possibly unbounded) self-adjoint operators. 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 ≤ E 2 ≤ E 3 ≤ ⋯ be the eigenvalues of A below the essential spectrum. Then

E n = min ψ 1 , … , ψ n − 1 max { ⟨ ψ , A ψ ⟩ : ψ ∈ span ⁡ ( ψ 1 , … , ψ n − 1 ) } .

If we only have N eigenvalues and hence run out of eigenvalues, then we let E n := inf σ e s s ( 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 ≤ E 2 ≤ E 3 ≤ ⋯ be the eigenvalues of A below the essential spectrum. Then

E n = max ψ 1 , … , ψ n − 1 min { ⟨ ψ , A ψ ⟩ : ψ ⊥ ψ 1 , … , ψ n − 1 } .

If we only have N eigenvalues and hence run out of eigenvalues, then we let E n := inf σ e s s ( 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 ) ≥ 0 for E ∈ R if and only if σ ( A ) ⊆ [ E , ∞ ) .

Theorem. If A is self-adjoint, then

inf σ ( A ) = inf ψ ∈ D ( A ) , ∥ ψ ∥ = 1 ⟨ ψ , A ψ ⟩

and

sup σ ( A ) = sup ψ ∈ D ( A ) , ∥ ψ ∥ = 1 ⟨ ψ , A ψ ⟩ .