The [adjoint page](/page/The%20Adjoint%20of%20an%20Operator) developed the construction $T \mapsto T^*$ for bounded operators between [Hilbert spaces](/page/Hilbert%20Space) and established the defining identity $(Tx, y)_K = (x, T^*y)_H$. A natural question emerges: what happens when an operator equals its own adjoint? The condition $T = T^*$ — self-adjointness — is the infinite-dimensional analogue of the symmetry condition $A = A^\top$ for real matrices. In finite dimensions, the spectral theorem guarantees that every real symmetric matrix is diagonalisable by an orthogonal change of basis, with real eigenvalues. The central achievement of this page is the extension of that result to compact self-adjoint operators on separable Hilbert spaces: every such operator admits a complete orthonormal eigenbasis with real eigenvalues accumulating only at zero. Self-adjointness is not merely a convenient technical condition. It is the *minimal algebraic requirement* that forces the spectrum to be real, the eigenspaces to be orthogonal, and the operator norm to be captured by the quadratic form $(Tx, x)_H$. These three properties — established on [the adjoint page](/page/The%20Adjoint%20of%20an%20Operator) — are the prerequisites for all that follows here. The present page develops the deeper structural theory: the spectral localisation result that pins the spectrum to the interval $[m, M]$ determined by the quadratic form, the full spectral theorem for compact self-adjoint operators, the Courant–Fischer variational characterisation of eigenvalues, and the consequences of positivity. We also introduce the operator order on self-adjoint operators and the notion of a positive square root. ## Motivation [motivation] ### What Diagonalisability Means in Infinite Dimensions In $\mathbb{R}^n$, diagonalising a symmetric matrix $A$ means finding an orthonormal basis $\{e_1, \dots, e_n\}$ of eigenvectors: $Ae_k = \lambda_k e_k$. This decomposition is equivalent to writing \begin{align*} Ax = \sum_{k=1}^n \lambda_k (x \cdot e_k) e_k \quad \text{for all } x \in \mathbb{R}^n, \end{align*} which represents $A$ as a sum of rank-one projections scaled by eigenvalues. In an infinite-dimensional Hilbert space $H$, the analogous statement would be: a self-adjoint operator $T$ admits an orthonormal basis $\{e_k\}$ of eigenvectors such that \begin{align*} Tx = \sum_{k=1}^\infty \mu_k (x, e_k)_H e_k, \end{align*} with convergence in the norm of $H$. This is precisely what the spectral theorem for compact self-adjoint operators achieves — but two new phenomena appear that have no finite-dimensional counterpart. ### Two Obstacles Absent in Finite Dimensions First, a bounded self-adjoint operator on an infinite-dimensional space need not have *any* eigenvalues. The multiplication operator $M_t: L^2([0,1]) \to L^2([0,1])$ defined by $(M_t f)(x) = xf(x)$ is self-adjoint with $\sigma(M_t) = [0,1]$, but the equation $xf(x) = \lambda f(x)$ forces $f = 0$ a.e. for every $\lambda$ — there are no eigenvectors. The spectrum is entirely "[continuous](/page/Continuity)." Self-adjointness alone does not guarantee a diagonalisation. Second, even when eigenvalues exist and span a dense subspace, the convergence of the [series](/page/Series) $\sum \mu_k (x, e_k)_H e_k$ to $Tx$ is not automatic — it requires a norm estimate on the partial sums, which in turn requires control on how fast the eigenvalues decay. Compactness provides this control: if $T$ is compact, the eigenvalues must converge to zero, and the tail of the series is bounded by the largest remaining eigenvalue. ### The Role of Compactness Compactness resolves both obstacles simultaneously. A compact self-adjoint operator on an infinite-dimensional space always has eigenvalues (the operator norm is attained, and the extremal value is an eigenvalue), the eigenspaces are finite-dimensional (by compactness of the unit ball in finite dimensions), and the eigenvalues converge to zero (since the images of the orthonormal eigenvectors must have a convergent subsequence). These three facts combine to produce the full spectral decomposition. The spectral theorem for compact self-adjoint operators is the workhorse of elliptic PDE theory: the eigenvalue problems $-\Delta u = \lambda u$ on bounded domains, Sturm–Liouville problems, and the spectral theory of [integral](/page/Integral) operators with symmetric kernels all reduce to this result. [/motivation] ## Recap: Definition and Basic Properties The [adjoint page](/page/The%20Adjoint%20of%20an%20Operator) established the following, which we use freely throughout this page. [definition:Self-Adjoint Operator] Let $H$ be a Hilbert space with inner product $(\cdot, \cdot)_H$. A bounded linear operator $T \in \mathcal{L}(H)$ is **self-adjoint** if $T^* = T$, that is, \begin{align*} (Tx, y)_H = (x, Ty)_H \quad \text{for all } x, y \in H. \end{align*} [/definition] The basic properties proved on [the adjoint page](/page/The%20Adjoint%20of%20an%20Operator) are: (a) every eigenvalue of a self-adjoint operator is real; (b) eigenvectors for distinct eigenvalues are orthogonal; (c) the operator norm satisfies $\|T\| = \sup_{\|x\|=1} |(Tx, x)_H|$. Property (c) is the key that connects the analytic quantity $\|T\|$ to the quadratic form, and it is used repeatedly in the spectral theory below. ## Positive Operators and the Operator Order The quadratic form $x \mapsto (Tx, x)_H$ is real-valued for any self-adjoint operator $T$ (since $(Tx, x)_H = (x, Tx)_H = \overline{(Tx, x)_H}$). This opens the door to comparing self-adjoint operators through their quadratic forms, in the same way that real numbers are ordered. The simplest and most important class of self-adjoint operators are those whose quadratic form is nonnegative. These are the operators for which "all eigenvalues are nonnegative" — the infinite-dimensional analogues of positive semidefinite matrices. [definition:Positive Operator] Let $H$ be a Hilbert space. A self-adjoint operator $T \in \mathcal{L}(H)$ is **positive** (or **positive semidefinite**), written $T \ge 0$, if \begin{align*} (Tx, x)_H \ge 0 \quad \text{for all } x \in H. \end{align*} More generally, for self-adjoint $S, T \in \mathcal{L}(H)$, we write $S \le T$ if $T - S \ge 0$, i.e., if \begin{align*} (Sx, x)_H \le (Tx, x)_H \quad \text{for all } x \in H. \end{align*} This defines the **operator order** on the set of self-adjoint operators. [/definition] The operator order is a partial order (reflexive, antisymmetric, transitive), but it is *not* a total order: there exist self-adjoint operators $S$ and $T$ such that neither $S \le T$ nor $T \le S$. This is already visible in $\mathbb{R}^2$: the diagonal matrices $\operatorname{diag}(1, 0)$ and $\operatorname{diag}(0, 1)$ are both positive, but neither dominates the other. The operator order interacts well with algebraic operations: if $S \le T$, then $S + R \le T + R$ and $\lambda S \le \lambda T$ for $\lambda \ge 0$. However, the product of positive operators need not be positive (or even self-adjoint): if $S, T \ge 0$, the product $ST$ is self-adjoint only if $S$ and $T$ commute. [example:The Laplacian As A Positive Operator] Let $U \subset \mathbb{R}^n$ be a bounded [open set](/page/Open%20Set) with smooth boundary, and consider the operator $A := -\Delta$ acting on $H^1_0(U) \cap H^2(U) \subset L^2(U)$. For $u \in H^1_0(U) \cap H^2(U)$, [integration by parts](/theorems/210) gives \begin{align*} (-\Delta u, u)_{L^2} = \int_U |\nabla u|^2 \, d\mathcal{L}^n \ge 0, \end{align*} where the [boundary](/page/Boundary) term vanishes because $u \in H^1_0(U)$. So $-\Delta$ is positive (in the sense of quadratic forms on its domain). This is the prototypical example: the positivity of $-\Delta$ is the reason the Dirichlet eigenvalue problem $-\Delta u = \lambda u$, $u|_{\partial U} = 0$ has only nonnegative eigenvalues. [/example] ### Why the Operator Order Matters The operator order is not just an algebraic convenience — it gives a direct comparison tool for quadratic forms and hence for eigenvalues. If $S \le T$ are self-adjoint and compact, then the $k$-th eigenvalue of $S$ is at most the $k$-th eigenvalue of $T$ (this is a consequence of the [Courant–Fischer min-max principle](/theorems/553), which we state below). This monotonicity is the basis of eigenvalue comparison theorems in PDE theory: if one elliptic operator has "larger" coefficients than another, its eigenvalues are correspondingly larger. ## The Spectrum of a Self-Adjoint Operator The basic properties from the adjoint page tell us that eigenvalues of a self-adjoint operator are real and that $\|T\| = \sup_{\|x\|=1} |(Tx, x)_H|$. The following theorem strengthens both statements: the *entire spectrum* (not just the eigenvalues) is real and confined to the interval determined by the quadratic form. This is a much stronger assertion, because the spectrum of a non-compact self-adjoint operator can contain points that are not eigenvalues — the continuous spectrum and residual spectrum also lie in $[m, M]$. The proof exploits the fact that self-adjointness gives a lower bound on $\|(T - \lambda I)x\|_H$ whenever $\lambda$ has nonzero imaginary part or lies outside $[m, M]$. Such a lower bound shows $T - \lambda I$ is bounded below, and the [kernel-range duality](/theorems/551) then implies it is invertible. [quotetheorem:552] This theorem has several consequences worth emphasising. First, the spectral radius of a self-adjoint operator equals its norm: $r(T) = \max\{|m|, |M|\} = \|T\|$. For a general operator, one only has $r(T) \le \|T\|$, and equality can fail (the right shift on $\ell^2$ has $r(S) = 1 = \|S\|$, but a nilpotent operator has $r = 0 < \|T\|$). For self-adjoint operators, the spectral radius formula $r(T) = \lim_{n \to \infty} \|T^n\|^{1/n}$ simplifies to $r(T) = \|T\|$ because the $C^*$-identity gives $\|T^2\| = \|T\|^2$, whence $\|T^n\|^{1/n} = \|T\|$ for all $n$ that are powers of $2$. Second, a positive self-adjoint operator ($T \ge 0$) satisfies $m \ge 0$, so $\sigma(T) \subseteq [0, \|T\|]$ — the spectrum is nonnegative. Conversely, if $\sigma(T) \subseteq [0, \infty)$, then $m \ge 0$ (since $m \in \sigma(T)$), so $T \ge 0$. Positivity is therefore equivalent to nonnegativity of the spectrum. [example:Spectrum Of A Multiplication Operator] Let $H = L^2([0,1], \mathcal{L}^1)$ and define $M_\varphi: L^2([0,1]) \to L^2([0,1])$ by $(M_\varphi f)(x) = \varphi(x) f(x)$, where $\varphi \in L^\infty([0,1])$ is real-valued. Then $M_\varphi$ is self-adjoint (as established on [the adjoint page](/page/The%20Adjoint%20of%20an%20Operator)). The quadratic form is \begin{align*} (M_\varphi f, f)_{L^2} = \int_0^1 \varphi(x) |f(x)|^2 \, d\mathcal{L}^1(x), \end{align*} so $m = \operatorname{ess\,inf} \varphi$ and $M = \operatorname{ess\,sup} \varphi$. The theorem gives $\sigma(M_\varphi) \subseteq [\operatorname{ess\,inf} \varphi, \operatorname{ess\,sup} \varphi]$. In fact, the spectrum equals the essential range of $\varphi$: every value $\lambda$ such that $\{\varphi \in (\lambda - \varepsilon, \lambda + \varepsilon)\}$ has positive measure for all $\varepsilon > 0$ is in $\sigma(M_\varphi)$, and these are exactly the points of the essential range. For $\varphi(x) = x$, we get $\sigma(M_x) = [0, 1]$, which confirms the claim from the motivation: the spectrum is entirely continuous, with no eigenvalues. The operator $M_x - \lambda I$ is not bounded below for any $\lambda \in [0,1]$ (because $(x - \lambda)f(x)$ can be made arbitrarily small by concentrating $f$ near $x = \lambda$), but $M_x - \lambda I$ *is* injective for every $\lambda$ (since $(x - \lambda)f(x) = 0$ a.e. implies $f = 0$ a.e.). So $\lambda$ is in the continuous spectrum, not the point spectrum. [/example] ## The Spectral Theorem for Compact Self-Adjoint Operators When $T$ is compact — in addition to being self-adjoint — the spectral theory becomes as complete as in finite dimensions: $T$ can be diagonalised by an orthonormal eigenbasis. The compactness provides the crucial ingredient that the general theory lacks: the operator norm is *attained* at a unit vector (rather than merely approached as a supremum), and this vector is an eigenvector. Iterating this observation by restricting to the orthogonal complement of the eigenspaces already found produces the full eigensequence. The spectral expansion $Tx = \sum_k \mu_k (x, e_k)_H e_k$ is the operator-theoretic analogue of the matrix diagonalisation $A = P \Lambda P^\top$. The eigenvalues are real (by self-adjointness), accumulate only at zero (by compactness), and the eigenvectors form an orthonormal set that spans the orthogonal complement of the kernel. The convergence of the expansion is in operator norm, not merely pointwise — the "error" $\|T - T_n\|$ after $n$ eigenvalues is at most $|\mu_{n+1}|$, which tends to zero. [quotetheorem:538] Several features of this theorem deserve discussion. The expansion $Tx = \sum_k \mu_k (x, e_k)_H e_k$ says that $T$ acts diagonally in the eigenbasis: it simply scales the $k$-th Fourier coefficient by $\mu_k$. The kernel of $T$ (which may be infinite-dimensional) is the orthogonal complement of the eigenspaces for nonzero eigenvalues, so the full decomposition is $H = \ker(T) \oplus \overline{\operatorname{span}\{e_k\}}$. The extremal characterisation $\|T\| = \max_k |\mu_k|$ shows that the operator norm is *attained* — not merely approximated — at an eigenvector. This is a special feature of compact operators. For the multiplication operator $M_x$ on $L^2([0,1])$, which is self-adjoint but not compact, $\|M_x\| = 1$ is not attained at any $f \in L^2$. The finite-dimensionality of each eigenspace $\ker(T - \mu_k I)$ distinguishes the infinite-dimensional setting from the finite-dimensional one. In $\mathbb{R}^n$, the eigenspace of a symmetric matrix can have any dimension up to $n$. For a compact self-adjoint operator on an infinite-dimensional space, every *nonzero* eigenvalue has finite multiplicity, but zero (if it is an eigenvalue) can have infinite multiplicity. This is because the unit ball of an infinite-dimensional eigenspace is not compact, which would contradict the compactness of $T$ if the eigenvalue were nonzero. ### The Spectral Theorem for the Laplacian The most important application of the spectral theorem is to the eigenvalue problem for elliptic operators on bounded domains. Consider $-\Delta$ on a bounded domain $U \subset \mathbb{R}^n$ with Dirichlet boundary conditions. The resolvent $(-\Delta)^{-1}: L^2(U) \to L^2(U)$ (defined via the [Lax–Milgram theorem](/theorems/91) and the compact embedding $H^1_0(U) \subset\subset L^2(U)$) is compact and self-adjoint and positive. The spectral theorem applied to $(-\Delta)^{-1}$ produces an orthonormal basis $\{e_k\}$ of $L^2(U)$ consisting of Dirichlet eigenfunctions: \begin{align*} -\Delta e_k = \lambda_k e_k \quad \text{in } U, \qquad e_k = 0 \quad \text{on } \partial U, \end{align*} with eigenvalues $0 < \lambda_1 \le \lambda_2 \le \dots \to \infty$. The eigenvalues of $(-\Delta)^{-1}$ are $1/\lambda_k \to 0$, consistent with compactness. Every $f \in L^2(U)$ has a unique eigenfunction expansion $f = \sum_k (f, e_k)_{L^2} e_k$, which converges in $L^2$. ## The Courant–Fischer Min-Max Principle The spectral theorem tells us that the eigenvalues of a compact self-adjoint operator exist and can be ordered. But how can we characterise the $k$-th eigenvalue *without* first computing the eigenvectors? The answer is the Courant–Fischer min-max principle, which expresses each eigenvalue as an optimisation problem over subspaces of a given dimension. This variational characterisation is one of the most powerful tools in spectral theory, because it provides eigenvalue bounds without solving the eigenvalue problem. The intuition behind the min-max principle is geometric. The largest eigenvalue $\mu_1$ is the maximum of the Rayleigh quotient $(Tx, x)_H / \|x\|_H^2$. The second eigenvalue $\mu_2$ is the maximum of the Rayleigh quotient *restricted to the orthogonal complement of the first eigenvector*. More generally, $\mu_k$ is what you get by "removing" $k - 1$ directions (the first $k - 1$ eigenvectors) and maximising over what remains. The min-max formulation says that you get the same answer no matter *which* $k - 1$ directions you remove — the $k$-th eigenvalue is the best worst case over all possible removals. [quotetheorem:553] The power of this result lies in its upper and lower bounds. To prove $\mu_k \le c$, choose a specific $(k-1)$-dimensional subspace $V$ and show that $(Tx, x)_H \le c$ for all unit vectors $x$ orthogonal to $V$. To prove $\mu_k \ge c$, choose a specific $k$-dimensional subspace $W$ and show that $(Tx, x)_H \ge c$ for all unit vectors $x$ in $W$. Neither bound requires knowledge of the eigenvectors — only a clever choice of test subspace. ### Eigenvalue Monotonicity An immediate corollary of the min-max principle is the eigenvalue monotonicity theorem for the operator order: if $S \le T$ are compact and self-adjoint, then $\mu_k(S) \le \mu_k(T)$ for each $k$ (where the eigenvalues are arranged in decreasing order). The proof is one line: for any $(k-1)$-dimensional $V$, \begin{align*} \sup_{\substack{x \perp V \\ \|x\| = 1}} (Sx, x)_H \le \sup_{\substack{x \perp V \\ \|x\| = 1}} (Tx, x)_H, \end{align*} so taking the infimum over $V$ preserves the inequality. [example:Eigenvalue Comparison For Sturm-Liouville Problems] Consider the Sturm–Liouville eigenvalue problems on $[0, \pi]$ with Dirichlet boundary conditions: \begin{align*} -u'' + q_1(x) u = \lambda u, \qquad -u'' + q_2(x) u = \mu u, \qquad u(0) = u(\pi) = 0, \end{align*} where $q_1, q_2 \in L^\infty([0, \pi])$ are real-valued potentials with $q_1(x) \le q_2(x)$ for $\mathcal{L}^1$-a.e. $x \in [0, \pi]$. The associated operators $T_1 := -\partial_{xx} + q_1$ and $T_2 := -\partial_{xx} + q_2$ satisfy $T_1 \le T_2$ in the operator order (since $(T_2 u - T_1 u, u)_{L^2} = \int_0^\pi (q_2 - q_1)|u|^2 \, d\mathcal{L}^1 \ge 0$). By the eigenvalue monotonicity theorem, the $k$-th eigenvalue of $T_1$ is at most the $k$-th eigenvalue of $T_2$. For instance, if $q_1 = 0$ and $q_2 = q \ge 0$, the eigenvalues of $-u''$ on $[0, \pi]$ are $\lambda_k = k^2$ (with eigenfunctions $e_k(x) = \sqrt{2/\pi}\sin(kx)$). The comparison gives $k^2 \le \mu_k(q)$ for every nonnegative potential $q$ — the eigenvalues can only increase when a nonnegative potential is added. [/example] ## Positive Square Roots The spectral theorem for compact self-adjoint operators gives a powerful tool for constructing new operators from old ones: if $T$ is compact, self-adjoint, and positive, we can define its square root by taking the square root of each eigenvalue. The existence of a positive square root is not an obvious fact, and it has no counterpart for general bounded operators. It relies on the positivity of $T$ (so all eigenvalues are nonnegative and their square roots are real) and on the completeness of the eigenexpansion (so the square root is well-defined on all of $H$). [definition:Positive Square Root] Let $H$ be a [separable](/page/Separable) Hilbert space and let $T \in \mathcal{L}(H)$ be compact, self-adjoint, and positive. Let $\{\mu_k\}$ be the nonzero eigenvalues of $T$ (all positive) with corresponding orthonormal eigenvectors $\{e_k\}$. The **positive square root** of $T$ is the operator \begin{align*} T^{1/2}: H &\to H \\ x &\mapsto \sum_{k} \sqrt{\mu_k} \, (x, e_k)_H \, e_k. \end{align*} [/definition] The positive square root satisfies $T^{1/2} \circ T^{1/2} = T$ (immediate from the eigenexpansion), $T^{1/2}$ is self-adjoint (the eigenvalues $\sqrt{\mu_k}$ are real), positive (the eigenvalues are nonnegative), and compact (since $\sqrt{\mu_k} \to 0$). Moreover, $T^{1/2}$ is the *unique* positive self-adjoint operator whose square is $T$. The uniqueness is not difficult: if $S \ge 0$ is self-adjoint with $S^2 = T$, then $S$ commutes with $T$ (since $ST = S^3 = TS$), hence $S$ preserves each eigenspace of $T$, and on the eigenspace for $\mu_k$ the only positive square root is $\sqrt{\mu_k}$. The square root construction extends beyond the compact case: every positive self-adjoint *bounded* operator has a unique positive self-adjoint square root, but the proof requires the continuous functional calculus rather than the eigenexpansion. This is one of the motivations for developing the general spectral theory of bounded self-adjoint operators, which is beyond the scope of this page. [example:Square Root Of An Integral Operator] Let $T: L^2([0, 1]) \to L^2([0, 1])$ be the integral operator with kernel $k(x, t) = \min(x, t)$: \begin{align*} (Tf)(x) = \int_0^1 \min(x, t) f(t) \, d\mathcal{L}^1(t). \end{align*} The kernel is symmetric ($k(x,t) = k(t,x)$), so $T$ is self-adjoint. It is positive because \begin{align*} (Tf, f)_{L^2} = \int_0^1 \int_0^1 \min(x,t) f(x) f(t) \, d\mathcal{L}^1(t) \, d\mathcal{L}^1(x) = \int_0^1 \left|\int_0^x f(t) \, d\mathcal{L}^1(t)\right|^2 d\mathcal{L}^1(x) \ge 0, \end{align*} where the second equality follows from integration by parts (the kernel $\min(x,t)$ is the Green's [function](/page/Function) for $-u'' = f$ on $[0,1]$ with $u(0) = u(1) = 0$). The eigenvalues of $T$ are $\mu_k = 1/((k - 1/2)^2 \pi^2)$ for $k \in \mathbb{N}$, with eigenfunctions $e_k(x) = \sqrt{2}\sin((k - 1/2)\pi x)$. The positive square root $T^{1/2}$ has eigenvalues $1/((k - 1/2)\pi)$ on the same eigenfunctions. [/example] ## Problems [problem] Let $H$ be a separable Hilbert space and let $T \in \mathcal{L}(H)$ be compact and self-adjoint. Show that $T^2$ is compact, self-adjoint, and positive, and that its eigenvalues are exactly $\{\mu_k^2\}$ where $\{\mu_k\}$ are the eigenvalues of $T$. Use this to prove that $\|T^2\| = \|T\|^2$ directly from the spectral theorem (without the $C^*$-identity). [/problem] [solution] **Step 1: $T^2$ is compact, self-adjoint, and positive.** Compactness: $T^2 = T \circ T$, and the composition of a compact operator with a bounded operator is compact (the image of a bounded set under $T$ is precompact, and $T$ maps precompact [sets](/page/Set) to precompact sets since it is continuous). Self-adjointness: $(T^2)^* = (T \circ T)^* = T^* \circ T^* = T \circ T = T^2$, using the composition rule $(AB)^* = B^*A^*$ and $T^* = T$. Positivity: $(T^2 x, x)_H = (Tx, Tx)_H = \|Tx\|_H^2 \ge 0$ for all $x \in H$. **Step 2: Eigenvalues of $T^2$.** By the [Spectral Theorem for Compact Self-Adjoint Operators](/theorems/538), $T$ has the expansion $Tx = \sum_k \mu_k (x, e_k)_H e_k$. Applying $T$ again: \begin{align*} T^2 x = T\left(\sum_k \mu_k (x, e_k)_H e_k\right) = \sum_k \mu_k (x, e_k)_H Te_k = \sum_k \mu_k^2 (x, e_k)_H e_k. \end{align*} This is the spectral expansion of $T^2$, so the eigenvalues of $T^2$ are $\mu_k^2$ with the same eigenvectors $e_k$. **Step 3: The norm identity.** By the extremal characterisation (part 6 of the spectral theorem): \begin{align*} \|T^2\| = \max_k |\mu_k^2| = \max_k |\mu_k|^2 = \left(\max_k |\mu_k|\right)^2 = \|T\|^2. \end{align*} This recovers the $C^*$-identity $\|T^*T\| = \|T\|^2$ (since $T^*T = T^2$ when $T$ is self-adjoint) purely from the spectral theorem, without using the general algebraic proof. [/solution] [problem] Let $H$ be a separable Hilbert space and let $S, T \in \mathcal{L}(H)$ be compact, self-adjoint, and positive with $S \le T$ (i.e., $(Sx, x)_H \le (Tx, x)_H$ for all $x$). Prove that $\mu_k(S) \le \mu_k(T)$ for every $k$, where $\mu_k(S)$ and $\mu_k(T)$ denote the $k$-th eigenvalues (arranged in decreasing order). [/problem] [solution] **Step 1: Setup using Courant–Fischer.** By the [Courant–Fischer min-max principle](/theorems/553), for each $k$: \begin{align*} \mu_k(S) = \inf_{\substack{V \subseteq H \\ \dim V = k - 1}} \sup_{\substack{x \perp V \\ \|x\|_H = 1}} (Sx, x)_H, \qquad \mu_k(T) = \inf_{\substack{V \subseteq H \\ \dim V = k - 1}} \sup_{\substack{x \perp V \\ \|x\|_H = 1}} (Tx, x)_H. \end{align*} **Step 2: Pointwise comparison of the quadratic forms.** Since $S \le T$, we have $(Sx, x)_H \le (Tx, x)_H$ for every $x \in H$. Therefore, for any fixed $(k-1)$-dimensional subspace $V$: \begin{align*} \sup_{\substack{x \perp V \\ \|x\|_H = 1}} (Sx, x)_H \le \sup_{\substack{x \perp V \\ \|x\|_H = 1}} (Tx, x)_H. \end{align*} **Step 3: Taking the infimum.** Taking the infimum over all $(k-1)$-dimensional subspaces $V$ on both sides preserves the inequality (if $f(V) \le g(V)$ for all $V$, then $\inf_V f(V) \le \inf_V g(V)$): \begin{align*} \mu_k(S) = \inf_V \sup_{x \perp V, \|x\| = 1} (Sx, x)_H \le \inf_V \sup_{x \perp V, \|x\| = 1} (Tx, x)_H = \mu_k(T). \end{align*} [/solution] ## References - Brezis, *Functional Analysis, [Sobolev Spaces](/page/Sobolev%20Space) and Partial Differential Equations* (2011), Chapter 6. - Conway, *A Course in Functional Analysis* (1990), Chapters II and VII. - Evans, *Partial Differential Equations* (2010), §6.5. - Lax, *Functional Analysis* (2002), Chapters 28–31. - Reed and Simon, *Methods of Modern Mathematical Physics I: Functional Analysis* (1980), Chapter VI. - Zeidler, *Applied Functional Analysis* (1995), Chapter 4.