[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]