[proofplan] We use the compact self-adjoint spectral theorem to choose an orthonormal eigenvector list $(e_j)$ corresponding to the positive eigenvalues $(\lambda_j)$, with multiplicity. The quadratic form of $T$ is diagonal in this list, and vectors orthogonal to the first $n-1$ eigenvectors have Rayleigh quotient at most $\lambda_n$. The max-min formula follows by comparing the leading span $\operatorname{span}\{e_1,\dots,e_n\}$ with an arbitrary $n$-dimensional subspace. The min-max formula is dual: compare the orthogonal complement of the first $n-1$ eigenvectors with an arbitrary closed subspace of codimension $n-1$. [/proofplan]

[step:Choose eigenvectors and write the diagonal quadratic form] For every index $j$ for which $\lambda_j$ exists, choose an eigenvector $e_j\in H$ such that \begin{align*} \|e_j\|_H=1 \end{align*} and \begin{align*} Te_j=\lambda_j e_j. \end{align*} The list is chosen orthonormally inside each eigenspace, so repeated eigenvalues are represented according to their multiplicity. By the compact self-adjoint spectral theorem for non-negative compact operators (citing a result not yet in the wiki: [Spectral theorem for compact self-adjoint operators](/theorems/538)), the positive spectral subspace of $T$ is the closed linear span of these eigenvectors, and $T$ vanishes on its orthogonal complement. Thus every $x\in H$ admits an [orthogonal decomposition](/theorems/436) \begin{align*} x=\sum_j (x,e_j)_H e_j + z, \end{align*} where $z\in \ker T$, with convergence in $H$. Moreover, \begin{align*} (Tx,x)_H=\sum_j \lambda_j |(x,e_j)_H|^2. \end{align*} The series is absolutely convergent because $\lambda_j\le \lambda_1$ and [Bessel's inequality](/theorems/540) gives \begin{align*} \sum_j |(x,e_j)_H|^2\le \|x\|_H^2. \end{align*} Define the quadratic form \begin{align*} Q:H&\to \mathbb R \end{align*} by \begin{align*} Q(x)=(Tx,x)_H. \end{align*} Then the displayed expansion says \begin{align*} Q(x)=\sum_j \lambda_j |(x,e_j)_H|^2. \end{align*} [/step]

[step:Show the leading eigenspace gives the lower bound in the max-min formula]Define the $n$-dimensional subspace \begin{align*} L_n=\operatorname{span}\{e_1,\dots,e_n\}. \end{align*} If $x\in L_n$ and $\|x\|_H=1$, then \begin{align*} x=\sum_{j=1}^n (x,e_j)_H e_j. \end{align*} Using the diagonal quadratic form and the ordering $\lambda_j\ge \lambda_n$ for $1\le j\le n$, we get \begin{align*} Q(x)=\sum_{j=1}^n \lambda_j |(x,e_j)_H|^2. \end{align*} Therefore \begin{align*} Q(x)\ge \lambda_n \sum_{j=1}^n |(x,e_j)_H|^2. \end{align*} Since $(e_1,\dots,e_n)$ is an [orthonormal basis](/page/Orthonormal%20Basis) of $L_n$ and $\|x\|_H=1$, \begin{align*} \sum_{j=1}^n |(x,e_j)_H|^2=1. \end{align*} Hence $Q(x)\ge \lambda_n$ for every unit vector $x\in L_n$, so \begin{align*} \min_{\substack{x\in L_n, \|x\|_H=1}} Q(x)\ge \lambda_n. \end{align*} Consequently, \begin{align*} \max_{\substack{L\subset H, \dim L=n}} \ \min_{\substack{x\in L, \|x\|_H=1}} Q(x)\ge \lambda_n. \end{align*}[/step]

[guided]We want a subspace on which every unit vector has Rayleigh quotient at least $\lambda_n$. The natural candidate is the span of the eigenvectors corresponding to the first $n$ eigenvalues, so define \begin{align*} L_n=\operatorname{span}\{e_1,\dots,e_n\}. \end{align*} This subspace has dimension $n$ because the vectors $e_1,\dots,e_n$ are orthonormal. Let $x\in L_n$ satisfy $\|x\|_H=1$. Since $(e_1,\dots,e_n)$ is an orthonormal basis of $L_n$, we have \begin{align*} x=\sum_{j=1}^n (x,e_j)_H e_j. \end{align*} Substituting this expansion into the diagonal formula for the quadratic form gives \begin{align*} Q(x)=\sum_{j=1}^n \lambda_j |(x,e_j)_H|^2. \end{align*} For each index $j$ with $1\le j\le n$, the eigenvalue ordering gives $\lambda_j\ge \lambda_n$. Therefore each summand is bounded below by replacing $\lambda_j$ with $\lambda_n$: \begin{align*} Q(x)\ge \lambda_n \sum_{j=1}^n |(x,e_j)_H|^2. \end{align*} The last sum is exactly $\|x\|_H^2$ by [Parseval's identity](/theorems/434) in the finite-dimensional [Hilbert space](/page/Hilbert%20Space) $L_n$. Since $\|x\|_H=1$, this becomes \begin{align*} Q(x)\ge \lambda_n. \end{align*} Thus every unit vector in $L_n$ has quadratic form at least $\lambda_n$, and so \begin{align*} \min_{\substack{x\in L_n, \|x\|_H=1}} Q(x)\ge \lambda_n. \end{align*} Because the outer maximum ranges over all $n$-dimensional subspaces, it is at least the value obtained from this particular subspace $L_n$: \begin{align*} \max_{\substack{L\subset H, \dim L=n}} \ \min_{\substack{x\in L, \|x\|_H=1}} Q(x)\ge \lambda_n. \end{align*}[/guided]

[step:Use dimension counting to bound every $n$-dimensional subspace from above] Let $L\subset H$ be an arbitrary subspace with $\dim L=n$. Define \begin{align*} E_{n-1}=\operatorname{span}\{e_1,\dots,e_{n-1}\}, \end{align*} with the convention $E_0=\{0\}$ when $n=1$. We claim that \begin{align*} L\cap E_{n-1}^{\perp}\ne \{0\}. \end{align*} Indeed, the [orthogonal projection](/theorems/437) \begin{align*} P_{E_{n-1}}:H\to E_{n-1} \end{align*} restricts to a [linear map](/page/Linear%20Map) \begin{align*} P_{E_{n-1}}|_L:L\to E_{n-1}. \end{align*} Since $\dim L=n$ and $\dim E_{n-1}=n-1$, the [rank-nullity theorem](/theorems/916) gives \begin{align*} \dim \ker(P_{E_{n-1}}|_L)\ge 1. \end{align*} But \begin{align*} \ker(P_{E_{n-1}}|_L)=L\cap E_{n-1}^{\perp}. \end{align*} Choose $0\ne y\in L\cap E_{n-1}^{\perp}$ and define \begin{align*} x=\frac{y}{\|y\|_H}. \end{align*} Then $x\in L$, $\|x\|_H=1$, and $(x,e_j)_H=0$ for $1\le j\le n-1$. Using the diagonal expansion, non-negativity of the eigenvalues, and the ordering $\lambda_j\le \lambda_n$ for every $j\ge n$, we obtain \begin{align*} Q(x)=\sum_{j\ge n} \lambda_j |(x,e_j)_H|^2. \end{align*} Hence \begin{align*} Q(x)\le \lambda_n\sum_{j\ge n}|(x,e_j)_H|^2. \end{align*} By Bessel's inequality, \begin{align*} \sum_{j\ge n}|(x,e_j)_H|^2\le \|x\|_H^2=1. \end{align*} Therefore $Q(x)\le \lambda_n$. Since this unit vector lies in $L$, \begin{align*} \min_{\substack{u\in L, \|u\|_H=1}} Q(u)\le \lambda_n. \end{align*} Because $L$ was arbitrary, this proves \begin{align*} \max_{\substack{L\subset H, \dim L=n}} \ \min_{\substack{u\in L, \|u\|_H=1}} Q(u)\le \lambda_n. \end{align*} Combining this with the previous step gives \begin{align*} \lambda_n = \max_{\substack{L\subset H, \dim L=n}} \ \min_{\substack{x\in L, \|x\|_H=1}} (Tx,x)_H. \end{align*} [/step]

[step:Use the orthogonal complement of the first eigenspaces for the upper bound in the min-max formula] Define the closed subspace \begin{align*} M_n=E_{n-1}^{\perp}. \end{align*} Since $E_{n-1}$ is finite-dimensional, $M_n$ is closed and \begin{align*} \operatorname{codim} M_n=n-1. \end{align*} If $x\in M_n$ and $\|x\|_H=1$, then $(x,e_j)_H=0$ for $1\le j\le n-1$. Thus the same diagonal estimate as above gives \begin{align*} Q(x)\le \lambda_n. \end{align*} Therefore \begin{align*} \max_{\substack{x\in M_n, \|x\|_H=1}} Q(x)\le \lambda_n. \end{align*} It follows that \begin{align*} \min_{\substack{M\subset H\ \mathrm{closed}, \operatorname{codim} M=n-1}} \ \max_{\substack{x\in M, \|x\|_H=1}} Q(x)\le \lambda_n. \end{align*} [/step]

[step:Use codimension counting to bound every finite-codimensional subspace from below] Let $M\subset H$ be a closed subspace with \begin{align*} \operatorname{codim} M=n-1. \end{align*} Define \begin{align*} F_n=\operatorname{span}\{e_1,\dots,e_n\}. \end{align*} We claim that \begin{align*} M\cap F_n\ne \{0\}. \end{align*} Let \begin{align*} \pi:H\to H/M \end{align*} be the quotient map. The restriction \begin{align*} \pi|_{F_n}:F_n\to H/M \end{align*} is linear. Since $\dim F_n=n$ and $\dim(H/M)=n-1$, rank-nullity gives \begin{align*} \dim \ker(\pi|_{F_n})\ge 1. \end{align*} But \begin{align*} \ker(\pi|_{F_n})=M\cap F_n. \end{align*} Choose $0\ne y\in M\cap F_n$ and define \begin{align*} x=\frac{y}{\|y\|_H}. \end{align*} Then $x\in M$, $\|x\|_H=1$, and \begin{align*} x=\sum_{j=1}^n (x,e_j)_H e_j. \end{align*} Using $\lambda_j\ge \lambda_n$ for $1\le j\le n$, we get \begin{align*} Q(x)=\sum_{j=1}^n \lambda_j |(x,e_j)_H|^2. \end{align*} Therefore \begin{align*} Q(x)\ge \lambda_n \sum_{j=1}^n |(x,e_j)_H|^2. \end{align*} Since $x\in F_n$ and $\|x\|_H=1$, \begin{align*} \sum_{j=1}^n |(x,e_j)_H|^2=1. \end{align*} Thus $Q(x)\ge \lambda_n$. Hence \begin{align*} \max_{\substack{u\in M, \|u\|_H=1}} Q(u)\ge \lambda_n. \end{align*} Because $M$ was arbitrary, \begin{align*} \min_{\substack{M\subset H\ \mathrm{closed}, \operatorname{codim} M=n-1}} \ \max_{\substack{u\in M, \|u\|_H=1}} Q(u)\ge \lambda_n. \end{align*} Combining this inequality with the previous step gives \begin{align*} \lambda_n = \min_{\substack{M\subset H\ \mathrm{closed}, \operatorname{codim} M=n-1}} \ \max_{\substack{x\in M, \|x\|_H=1}} (Tx,x)_H. \end{align*} [/step]

[step:Justify attainment of the inner extrema] For each finite-dimensional subspace $L\subset H$, the unit sphere \begin{align*} S_L=\{x\in L:\|x\|_H=1\} \end{align*} is compact, and the map $Q|_{S_L}:S_L\to\mathbb R$ is continuous because $T\in\mathcal L(H)$. Hence the minimum over $S_L$ is attained. For a closed subspace $M\subset H$, let \begin{align*} P_M:H\to M \end{align*} be the orthogonal projection onto $M$, and define \begin{align*} T_M:M&\to M \end{align*} by \begin{align*} T_Mx=P_MTx. \end{align*} The operator $T_M$ is compact, self-adjoint, and non-negative on $M$, and \begin{align*} (T_Mx,x)_H=(Tx,x)_H \end{align*} for every $x\in M$. By the compact self-adjoint spectral theorem for non-negative compact operators (citing a result not yet in the wiki: Spectral theorem for compact [self-adjoint operators](/page/Self-Adjoint%20Operators)), either $T_M=0$, in which case the maximum of $Q$ on the unit sphere of $M$ is $0$ and is attained at every unit vector of $M$, or the maximum equals the largest positive eigenvalue of $T_M$ and is attained at a corresponding unit eigenvector. This verifies that the maxima appearing in the min-max formula are genuine maxima. The two variational identities proved above are therefore identities with the extrema exactly as stated. [/step]