Min-max theorem

Let be a Hermitian matrix. As with many other variational results on eigenvalues, one considers the Rayleigh–Ritz quotient {{math|RA : Cn \ {0} → R}} defined by :R_A(x) = \frac{(Ax, x)}{(x,x)} where denotes the Euclidean inner product on . The Rayleigh quotient of an eigenvector v is its associated eigenvalue \lambda because R_A(v) = (\lambda x, x)/(x, x) = \lambda. For a Hermitian matrix A, the range of the continuous functions RA(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 A be Hermitian on an inner product space V with dimension n, with spectrum ordered in descending order \lambda_1 \geq ... \geq \lambda_n. Let v_1, ..., v_n be the corresponding unit-length orthogonal eigenvectors. Reverse the spectrum ordering, so that \xi_1 = \lambda_n, ..., \xi_n = \lambda_1. {{Math proof|title=Proof|proof= Part 2 is a corollary, using -A. M is a k dimensional subspace, so if we pick any list of n-k+1 vectors, their span N := span(v_k, ... v_n) must intersect M on at least a single line. Take unit x \in M\cap N. That’s what we need. : x = \sum_{i=k}^n a_i v_i, since x\in N. : Since \sum_{i=k}^n |a_i|^2 = 1, we find \langle x,Ax \rangle = \sum_{i=k}^n |a_i|^2\lambda_i \leq \lambda_k. }} {{Math theorem \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, A x\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, A x\rangle \text{. } \end{aligned} }} Define the partial trace tr_V(A) to be the trace of projection of A to V. It is equal to \sum_i v_i^*Av_i given an orthonormal basis of V. {{Math theorem|name=Wielandt minimax formula|note=|math_statement= Let 1 \leq i_1 be integers. Define a partial flag to be a nested collection V_1 \subset \cdots \subset V_k of subspaces of \mathbb{C}^n such that \operatorname{dim}\left(V_j\right)=i_j for all 1 \leq j \leq k. Define the associated Schubert variety X\left(V_1, \ldots, V_k\right) to be the collection of all k dimensional subspaces W such that \operatorname{dim}\left(W \cap V_j\right) \geq j. \lambda_{i_1}(A)+\cdots+\lambda_{i_k}(A)=\sup _{V_1, \ldots, V_k} \inf_{W \in X\left(V_1, \ldots, V_k\right)} tr_W(A) }} {{Math proof|title=Proof|proof= The \leq case. Let V_{j} = span(e_1, \dots, e_{i_j}), and any W \in X\left(V_1, \ldots, V_k\right), it remains to show that \lambda_{i_1}(A)+\cdots+\lambda_{i_k}(A) \leq tr_W(A) To show this, we construct an orthonormal set of vectors v_1, \dots, v_k such that v_j \in V_j \cap W. Then tr_W(A) \geq \sum_j \langle v_j, Av_j\rangle \geq \lambda_{i_j}(A) Since dim(V_1 \cap W) \geq 1, we pick any unit v_1 \in V_1 \cap W. Next, since dim(V_2 \cap W) \geq 2, we pick any unit v_2 \in (V_2 \cap W) that is perpendicular to v_1, and so on. The \geq case. For any such sequence of subspaces V_i, we must find some W \in X\left(V_1, \ldots, V_k\right) such that \lambda_{i_1}(A)+\cdots+\lambda_{i_k}(A) \geq tr_W(A) Now we prove this by induction. The n=1 case is the Courant-Fischer theorem. Assume now n \geq 2. If i_1 \geq 2, then we can apply induction. Let E = span(e_{i_1}, \dots, e_n). We construct a partial flag within E from the intersection of E with V_1, \dots, V_k. We begin by picking a (i_k-(i_1-1))-dimensional subspace W_k' \subset E \cap V_{i_k}, which exists by counting dimensions. This has codimension (i_1-1) within V_{i_k}. Then we go down by one space, to pick a (i_{k-1} - (i_1 - 1))-dimensional subspace W_{k-1}' \subset W_k \cap V_{i_{k-1}}. This still exists. Etc. Now since dim(E) \leq n-1, apply the induction hypothesis, there exists some W \in X(W_1, \dots, W_k) such that \lambda_{i_1 - (i_1-1)}(A|E)+\cdots+\lambda_{i_k- (i_1-1)}(A|E) \geq tr_W(A) Now \lambda_{i_j - (i_1-1)}(A|E) is the (i_j-(i_1-1))-th eigenvalue of A orthogonally projected down to E. By Cauchy interlacing theorem, \lambda_{i_j - (i_1-1)}(A|E) \leq \lambda_{i_j}(A). Since X(W_1, \dots, W_k)\subset X(V_1, \dots, V_k), we’re done. If i_1 = 1, then we perform a similar construction. Let E = span(e_{2}, \dots, e_n). If V_k \subset E, then we can induct. Otherwise, we construct a partial flag sequence W_2, \dots, W_k By induction, there exists some W' \in X(W_2, \dots, W_k)\subset X(V_2, \dots, V_k), such that \lambda_{i_2-1}(A|E)+\cdots+\lambda_{i_k-1}(A|E) \geq tr_{W'}(A) thus \lambda_{i_2}(A)+\cdots+\lambda_{i_k}(A) \geq tr_{W'}(A) And it remains to find some v such that W' \oplus v \in X(V_1, \dots, V_k). If V_1 \not\subset W', then any v \in V_1 \setminus W' would work. Otherwise, if V_2 \not\subset W', then any v \in V_2 \setminus W' would work, and so on. If none of these work, then it means V_k \subset E, contradiction. }} This has some corollaries: {{Math theorem|name=Extremal partial trace|note=|math_statement= \lambda_1(A)+\dots+\lambda_k(A)=\sup_{\operatorname{dim}(V)=k }tr_V(A) \xi_1(A)+\dots+\xi_k(A)=\inf_{\operatorname{dim}(V)=k }tr_V(A) }} {{Math theorem|name=Corollary|note=|math_statement= The sum \lambda_1(A)+\dots+\lambda_k(A) is a convex function, and \xi_1(A)+\dots+\xi_k(A) is concave. (Schur-Horn inequality) \xi_1(A)+\dots+\xi_k(A) \leq a_{i_1,i_1} + \dots + a_{i_k,i_k} \leq \lambda_1(A)+\dots+\lambda_k(A) for any subset of indices. Equivalently, this states that the diagonal vector of A is majorized by its eigenspectrum. }} {{Math theorem|name=Schatten-norm Hölder inequality|note=|math_statement= Given Hermitian A, B and Hölder pair 1/p + 1/q = 1, |\operatorname{tr}(A B)| \leq\|A\|_{S^p}\|B\|_{S^q} }} {{Math proof|title=Proof|proof= WLOG, B is diagonalized, then we need to show |\sum_i B_{ii} A_{ii} | \leq \|A \|_{S^p} \|(B_{ii})\|_{l^q} By the standard Hölder inequality, it suffices to show \|(A_{ii})\|_{l^p}\leq \|A \|_{S^p} By the Schur-Horn inequality, the diagonals of A are majorized by the eigenspectrum of A, and since the map f(x_1, \dots, x_n) = \|x\|_p is symmetric and convex, it is Schur-convex. }} 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 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 {σk} 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^{\downarrow} = \max_{S:\dim(S)=k} \min_{x \in S, \|x\| = 1} (M^* Mx, x)^{\frac{1}{2}}=\max_{S:\dim(S)=k} \min_{x \in S, \|x\| = 1} \| Mx \|. Similarly, :\sigma_k^{\downarrow} = \min_{S:\dim(S)=n-k+1} \max_{x \in S, \|x\| = 1} \| Mx \|. Here \sigma_k^{\downarrow} denotes the kth entry in the decreasing sequence of the singular values, so that \sigma_1^{\downarrow} \geq \sigma_2^{\downarrow} \geq \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_{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 {{math|Sj span{b1, ..., bj},}} then :\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, On the other hand, if we define {{math|Sm−j+1 span{bj, ..., bm},}} then :\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 , we have , hence the name interlacing theorem. Lidskii's inequality {{Math theorem & \lambda_{i_1}(A+B)+\cdots+\lambda_{i_k}(A+B) \\ & \quad \leq \lambda_{i_1}(A)+\cdots+\lambda_{i_k}(A)+\lambda_1(B)+\cdots+\lambda_k(B) \end{aligned} \begin{aligned} & \lambda_{i_1}(A+B)+\cdots+\lambda_{i_k}(A+B) \\ & \quad \geq \lambda_{i_1}(A)+\cdots+\lambda_{i_k}(A)+\xi_1(B)+\cdots+\xi_k(B) \end{aligned} }} {{Math proof|title=Proof|proof= The second is the negative of the first. The first is by Wielandt minimax. \begin{aligned} & \lambda_{i_1}(A+B)+\cdots+\lambda_{i_k}(A+B) \\ =& \sup_{V_1, \dots, V_k} \inf_{W\in X(V_1, \dots, V_k)}(tr_W(A) + tr_W(B)) \\ =& \sup_{V_1, \dots, V_k} ( \inf_{W\in X(V_1, \dots, V_k)} tr_W(A) + tr_W(B)) \\ \leq& \sup_{V_1, \dots, V_k} ( \inf_{W\in X(V_1, \dots, V_k)} tr_W(A) + (\lambda_1(B)+\cdots+\lambda_k(B))) \\ =& \lambda_{i_1}(A)+\cdots+\lambda_{i_k}(A)+\lambda_1(B)+\cdots+\lambda_k(B) \end{aligned} }} Note that \sum_i \lambda_i(A+B) = tr(A+B) = \sum_i \lambda_i(A) + \lambda_i(B) . In other words, \lambda(A+B) - \lambda(A) \preceq \lambda(B) where \preceq means majorization. By the Schur convexity theorem, we then have == Compact operators ==

Let be a compact, Hermitian operator on a Hilbert space H. Recall that the non-zero spectrum of such an operator consists of real eigenvalues with finite multiplicities whose only possible cluster point is zero. If has infinitely many positive eigenvalues, they accumulate at zero. In this case, we 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.) We now apply the same reasoning as in the matrix case. Letting Sk ⊂ 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{align} \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{align} A similar pair of equalities hold for negative eigenvalues. {{Math proof|drop=hidden|proof= 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' ∩ Sk has positive dimension. So there exists x ∈ S'  ∩ Sk with \|x\|=1. Since it is an element of S' , such an x'' necessarily satisfy :(Ax, x) \le \lambda_k. Therefore, for all Sk :\inf_{x \in S_k, \|x\| = 1}(Ax,x) \le \lambda_k But 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) \le \lambda_k. So :\sup_{S_k} \min_{x \in S_k, \|x\| = 1}(Ax,x) \le \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 -dimensional subspace Sk−1, whose the orthogonal complement is denoted by Sk−1⊥. If ''S' = span{u1...uk''}, :S' \cap S_{k-1}^{\perp} \ne {0}. So :\exists x \in S_{k-1}^{\perp} \, \|x\| = 1, (Ax, x) \ge \lambda_k. This implies :\max_{x \in S_{k-1}^{\perp}, \|x\| = 1} (Ax, x) \ge \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) \ge \lambda_k. Pick Sk−1 = span{u1, ..., uk−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. 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_{\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 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\le E_2\le E_3\le\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}, \, \| \psi \| = 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 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{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. == See also ==