[proofplan] We prove the min-max characterisation $\mu_k = \inf_{\dim V = k-1} \sup_{x \perp V, \|x\|=1} (Tx, x)_H$ by establishing matching upper and lower bounds. The spectral expansion $Tx = \sum_j \mu_j (x, e_j)_H e_j$ from the [Spectral Theorem for Compact Self-Adjoint Operators](/theorems/538) reduces the Rayleigh quotient to a weighted average of eigenvalues. A dimension-counting argument shows that any $(k-1)$-dimensional subspace $V$ must leave room for a unit vector in $\operatorname{span}\{e_1, \ldots, e_k\} \cap V^\perp$, yielding the lower bound. The upper bound is achieved at $V_0 = \operatorname{span}\{e_1, \ldots, e_{k-1}\}$. [/proofplan] [step:Establish the lower bound: $\sup_{x \perp V, \|x\|=1} (Tx, x)_H \geq \mu_k$ for every $(k-1)$-dimensional $V$] Let $V \subseteq H$ be any subspace with $\dim V = k - 1$. Define $W := \operatorname{span}\{e_1, \ldots, e_k\}$, which has dimension $k$. By the dimension formula: \begin{align*} \dim(W \cap V^\perp) \geq \dim W - \dim V = k - (k-1) = 1, \end{align*} so there exists a unit vector $x \in W \cap V^\perp$. Write $x = \sum_{j=1}^k c_j e_j$ with $\sum_{j=1}^k c_j^2 = 1$. Using the spectral expansion and orthonormality of the eigenvectors: \begin{align*} (Tx, x)_H = \left(\sum_{j=1}^k c_j \mu_j e_j,\, \sum_{j=1}^k c_j e_j\right)_H = \sum_{j=1}^k \mu_j c_j^2 \geq \mu_k \sum_{j=1}^k c_j^2 = \mu_k, \end{align*} where the inequality uses $\mu_j \geq \mu_k$ for $j \leq k$ (the positive eigenvalues are arranged in decreasing order). Since $x \in V^\perp$ with $\|x\|_H = 1$, $\sup_{x \perp V, \|x\|=1} (Tx, x)_H \geq \mu_k$. Since $V$ was an arbitrary $(k-1)$-dimensional subspace: \begin{align*} \inf_{\dim V = k-1} \sup_{\substack{x \in V^\perp \\ \|x\|_H = 1}} (Tx, x)_H \geq \mu_k. \end{align*} [guided] The dimension-counting step is the heart of the lower bound. We need to find a unit vector that is simultaneously in the span of the first $k$ eigenvectors and orthogonal to $V$. The space $W = \operatorname{span}\{e_1, \ldots, e_k\}$ is $k$-dimensional, and $V^\perp$ has codimension $k - 1$ in $H$. The intersection $W \cap V^\perp$ is therefore at least $1$-dimensional: in finite dimensions this is a direct consequence of $\dim(W \cap V^\perp) = \dim W + \dim V^\perp - \dim(W + V^\perp) \geq k + (\dim H - (k-1)) - \dim H = 1$; the same argument works in infinite dimensions because $V$ is finite-dimensional and we are intersecting within the finite-dimensional space $W$. Once we have $x \in W \cap V^\perp$, the Rayleigh quotient $\sum_{j=1}^k \mu_j c_j^2$ is a convex combination (since $\sum c_j^2 = 1$) of $\mu_1, \ldots, \mu_k$. Since all these eigenvalues are at least $\mu_k$, the convex combination is at least $\mu_k$. [/guided] [/step] [step:Show attainment at $V_0 = \operatorname{span}\{e_1, \ldots, e_{k-1}\}$] Set $V_0 := \operatorname{span}\{e_1, \ldots, e_{k-1}\}$. Any unit vector $x \in V_0^\perp$ can be decomposed as $x = \sum_{j=k}^\infty c_j e_j + z$, where $z \in \ker(T)$ and $\sum_{j \geq k} c_j^2 + \|z\|_H^2 = 1$. Since $Tz = 0$: \begin{align*} (Tx, x)_H = \sum_{j=k}^\infty \mu_j c_j^2 \leq \mu_k \sum_{j=k}^\infty c_j^2 \leq \mu_k, \end{align*} where the first inequality uses $\mu_j \leq \mu_k$ for $j \geq k$ (the eigenvalues are decreasing). Equality is attained at $x = e_k$ (which lies in $V_0^\perp$ since $e_k \perp e_1, \ldots, e_{k-1}$), giving: \begin{align*} \sup_{\substack{x \in V_0^\perp \\ \|x\|_H = 1}} (Tx, x)_H = \mu_k. \end{align*} [/step] [step:Combine the bounds to obtain the min-max identity] The lower bound from the first step gives $\inf_{\dim V = k-1} \sup_{x \perp V, \|x\|=1} (Tx, x)_H \geq \mu_k$. The attainment at $V_0$ gives $\inf_{\dim V = k-1} \sup_{x \perp V, \|x\|=1} (Tx, x)_H \leq \mu_k$. Therefore: \begin{align*} \mu_k = \inf_{\substack{V \subseteq H \\ \dim V = k-1}} \sup_{\substack{x \in V^\perp \\ \|x\|_H = 1}} (Tx, x)_H, \end{align*} and the infimum is attained at $V_0 = \operatorname{span}\{e_1, \ldots, e_{k-1}\}$. [/step]