The Operator Norm of a Symmetric Matrix: Quadratic Forms, Eigenvalues and the Semidefinite Ordering

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let $n\in\mathbb{N}$, let $A\in\mathcal{S}(n)$, the set of symmetric real $n\times n$ matrices. Juxtaposition is the matrix product or the matrix-vector product; $\cdot$ is the dot product and $\lVert\cdot\rVert$ the Euclidean norm on $\mathbb{R}^{n}$, and also the operator norm of matrices; and $|\cdot|$ is the absolute value.

¶ The set $\{|u\cdot(Au)|:u\in\mathbb{R}^{n},\ \lVert u\rVert\le1\}$ of real numbers has supremum $\lVert A\rVert$.

¶ If $b\in\mathbb{R}$ and $|u\cdot(Au)|\le b$ for every $u\in\mathbb{R}^{n}$ with $\lVert u\rVert\le1$, then $\lVert A\rVert\le b$.

¶ Let $x\in\mathbb{R}^{n}$. Then $|x\cdot(Ax)|\le\lVert A\rVert\,\lVert x\rVert^{2}$.

¶ Let $\mathrm{O}(n)$ be the set of orthogonal matrices, with $Q^{\top}$ the transpose of $Q\in\mathrm{O}(n)$, and for $d\in\mathbb{R}^{n}$, with components $d_{k}$ (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §membership), let $\operatorname{diag}(d)$ be the diagonal matrix with diagonal $d$ and $\max_{k\in[n]}|d_{k}|$ the greatest element of $\{|d_{k}|:k\in[n]\}$, which exists by The Operator Norm of Entries, Transposes, Multiples of the Identity, Orthogonal Factors and Diagonal Matrices §diagonal. If $Q\in\mathrm{O}(n)$ and $d\in\mathbb{R}^{n}$ satisfy $A=(Q\operatorname{diag}(d))Q^{\top}$, then $\lVert A\rVert=\max_{k\in[n]}|d_{k}|$.

¶ Let $\preceq$ be the positive semidefinite ordering on $\mathcal{S}(n)$, and for $c\in\mathbb{R}$ let $cI_{n}$ be the scalar multiple of the identity matrix $I_{n}$. Let $\lambda\in\mathbb{R}$ with $0\le\lambda$. Then $\lVert A\rVert\le\lambda$ if and only if both $(-\lambda)I_{n}\preceq A$ and $A\preceq\lambda I_{n}$, the two multiples of $I_{n}$ lying in $\mathcal{S}(n)$ by the clause identity.

¶ With $\preceq$ and the multiples $cI_{n}$ as in the clause ordering, $(-\lVert A\rVert)I_{n}\preceq A\preceq\lVert A\rVert I_{n}$.

¶ Let $z\in\mathbb{R}^{n}$ and $A^{2}=AA$. Then $A^{2}\in\mathcal{S}(n)$ and $\lVert Az\rVert^{2}=z\cdot(A^{2}z)$.

¶ $\lVert A^{2}\rVert=\lVert A\rVert^{2}$, with $A^{2}=AA$.

¶ Let $t\in\mathbb{R}$ and $z\in\mathbb{R}^{n}$, with $tI_{n}$ the scalar multiple of the identity matrix $I_{n}$. Then $tI_{n}\in\mathcal{S}(n)$, $z\cdot((tI_{n})z)=t\,\lVert z\rVert^{2}$ and $\lVert tI_{n}\rVert=|t|$.

¶ Let $x,y\in\mathbb{R}^{n}$, with $x+y$ and $x-y$ the sum and difference of points. Then $|x\cdot(Ax)-y\cdot(Ay)|\le\lVert A\rVert\,\lVert x-y\rVert\,\lVert x+y\rVert$.

¶ Let $B\in\mathcal{S}(n)$, with $\preceq$ as in the clause ordering, and let $0$ be the zero matrix, which lies in $\mathcal{S}(n)$ by Rules for Symmetric and Positive Semidefinite Matrices: Closure, Self-Adjointness, Polarization, Congruence and Cones §closure. If $0\preceq A$ and $A\preceq B$, then $\lVert A\rVert\le\lVert B\rVert$.