> [!theorem] > > Let $X$ be a [[Compactness|compact]] [[Riemannian Manifold|Riemannian n-manifold]] without boundary. Suppose that the [[Closed Eigenvalue Problem]] has distinct solutions > $ > 0 \le \lambda_1 \le \lambda_2 \le \cdots > $ > Let $f \in \sch(X)$ be in the [[Sobolev Space (Manifold)|Sobolev space]], then > $ > f \perp \bigoplus_{j < k}\eig{(\lambda_j)} > $ > implies that > $ > \lambda_k \le \frac{D[f, f]}{\norm{f}_{L^2}^2} > $ > with equality if and only if $f$ is an eigenvector corresponding to $\lambda_k$. > > *Proof*. Let $\seq{\proj_j}$ be the [[Orthogonal Projection|orthogonal projections]] corresponding to the eigenspaces of $\seq{\lambda_j}$. Let $N \in \nat$, then > $ > \begin{align*} > 0 &\le D\braks{f - \sum_{j = 1}^N\proj_j(f), f - \sum_{j = 1}^N\proj_j(f)}\\ > &= D\braks{f - \sum_{j = k}^N\proj_j(f), f - \sum_{j = k}^N\proj_j(f)}\\ > &= D[f, f] - 2\sum_{j = k}^nD[f, \proj_j(f)] +\sum_{j, l = k}^ND[\proj_j(f), \proj_l(f)] \\ > &= D[f, f] + 2\sum_{j = k}^N\angles{f, \Delta \proj_j(f)}_{L^2} - \sum_{j, l = k}^N \angles{\proj_j(f), \Delta \proj_l(f)}_{L^2}\\ > &= D[f, f] - 2\sum_{j = k}^N\lambda_j\angles{f, \proj_j(f)}_{L^2} + \sum_{j = k}^N \lambda_j\angles{\proj_j(f), \proj_j(f)}_{L^2}\\ > &= D[f, f] - \sum_{j = k}^N\lambda_j\norm{\proj_j(f)}_{L^2}^2 > \end{align*} > $ > so > $ > \begin{align*} > D[f, f] &\ge \sum_{j = k}^\infty \lambda_j \norm{\proj_j(f)}_{L^2}^2 \\ > &\ge \lambda_k \sum_{j = k}^\infty \norm{\proj_j(f)}_{L^2}^2 \ge \lambda_k \norm{f}_{L^2}^2 > \end{align*} > $ > where all inequalities becomes equalities if $f$ is an eigenvector corresponding to $\lambda_k$. > > [!theorem] > > Let $T: \real^d \to \real^d$ be a symmetric operator, and $\mu_1$ be its largest eigenvalue, then > $ > \mu_1 = \sup_{x \ne 0}\frac{\angles{x, Tx}}{\norm{x}^2} = \sup_{\norm{x} = 1}\angles{x, Tx} > $ > where the maximum is achieved on the eigenvectors of $\mu_1$.