Proof of Properties of the Norm of a Symmetric Real Matrix

For a symmetric real $n\times n$ matrix $P$ write $S_P=\{|\xi\cdot(P\xi)|:\xi\in\mathbb{R}^n,\ \lVert\xi\rVert\le1\}$, so that $\lVert P\rVert$ is the least upper bound of $S_P$ by Norm of a Symmetric Real Matrix. Write $0_{\mathbb{R}^n}$ for the origin of $\mathbb{R}^n$. Record that $0\,t=0$ for every $t\in\mathbb{R}$, since $0\,t=(0+0)\,t=0\,t+0\,t$ by distributivity and claim 2 of Additive Cancellation and Elementary Additive Identities in a Field applies; consequently $(-1)\,t=-t$ by claim 1 of that lemma.

Two computations are used repeatedly. First, for $\beta\in\mathbb{R}$ and $\xi\in\mathbb{R}^n$, claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ and claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ give

$$\xi\cdot\bigl((\beta I_n)\xi\bigr)=\beta\,(\xi\cdot\xi)=\beta\,\lVert\xi\rVert^{2},\qquad \xi\cdot\bigl((\beta P)\xi\bigr)=\beta\,\bigl(\xi\cdot(P\xi)\bigr).$$

Second, for $\beta,\gamma\in\mathbb{R}$ and $\xi,\zeta\in\mathbb{R}^n$, claim 3 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claims 4 and 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ give $(\beta\xi)\cdot\bigl(P(\gamma\zeta)\bigr)=\beta\,\gamma\,\bigl(\xi\cdot(P\zeta)\bigr)$.

Claim 1. Since $0_{\mathbb{R}^n}=0\,0_{\mathbb{R}^n}$ coordinatewise, the second computation gives $0_{\mathbb{R}^n}\cdot(A\,0_{\mathbb{R}^n})=0$, and $|0|=0$ by claim 1 of Properties of the Absolute Value in an Ordered Field. Moreover $\lVert0_{\mathbb{R}^n}\rVert=0$ by claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and $0\le1$ by claim 1 of Elementary Arithmetic in an Ordered Field. Hence $0\in S_A$, and since $\lVert A\rVert$ is an upper bound for $S_A$, $0\le\lVert A\rVert$.

Claim 2. If $\xi=0_{\mathbb{R}^n}$ then, as in claim 1, both sides are $0$. Otherwise claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ gives $\lVert\xi\rVert\ne0$ and claim 1 of that lemma gives $0\le\lVert\xi\rVert$, so $0<r$ where $r=\lVert\xi\rVert$; by claim 7 of Elementary Order Arithmetic in an Ordered Field, $r^{-1}$ exists and $0<r^{-1}$. Put $\zeta=r^{-1}\xi$. By claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and Absolute Value in an Ordered Field, $\lVert\zeta\rVert=|r^{-1}|\,r=r^{-1}r=1$, so $|\zeta\cdot(A\zeta)|\le\lVert A\rVert$. By the second computation above, $\zeta\cdot(A\zeta)=r^{-1}r^{-1}\bigl(\xi\cdot(A\xi)\bigr)$, hence $\xi\cdot(A\xi)=r^{2}\,\bigl(\zeta\cdot(A\zeta)\bigr)$ with $r^{2}=\lVert\xi\rVert^{2}$. Now $0\le r^{2}$ by claim 5 of Elementary Arithmetic in an Ordered Field, so $|r^{2}|=r^{2}$ by Absolute Value in an Ordered Field, and claim 4 of Properties of the Absolute Value in an Ordered Field gives $|\xi\cdot(A\xi)|=r^{2}\,|\zeta\cdot(A\zeta)|$. Applying claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative factor $r^{2}$ to $|\zeta\cdot(A\zeta)|\le\lVert A\rVert$ gives $|\xi\cdot(A\xi)|\le\lVert A\rVert\,\lVert\xi\rVert^{2}$.

Claim 3. Suppose $0\le\lambda$ and both $-\lambda I_n\preceq A$ and $A\preceq\lambda I_n$ hold, and let $\xi\in\mathbb{R}^n$ with $\lVert\xi\rVert\le1$. By the first computation above, together with $(-\lambda)\lVert\xi\rVert^{2}=-\bigl(\lambda\lVert\xi\rVert^{2}\bigr)$, these two relations read

$$-\lambda\,\lVert\xi\rVert^{2}\le\xi\cdot(A\xi)\le\lambda\,\lVert\xi\rVert^{2}.$$

Since $0\le\lVert\xi\rVert\le1$, claim 5 of Elementary Arithmetic in an Ordered Field gives $\lVert\xi\rVert^{2}\le\lVert\xi\rVert$, hence $\lVert\xi\rVert^{2}\le1$ by transitivity, and then $\lambda\,\lVert\xi\rVert^{2}\le\lambda$; claim 4 of Elementary Order Arithmetic in an Ordered Field gives $-\lambda\le-\lambda\,\lVert\xi\rVert^{2}$. So $-\lambda\le\xi\cdot(A\xi)\le\lambda$, and claim 6 of Properties of the Absolute Value in an Ordered Field gives $|\xi\cdot(A\xi)|\le\lambda$. Thus $\lambda$ is an upper bound for $S_A$, and $\lVert A\rVert\le\lambda$ because $\lVert A\rVert$ is the least upper bound.

Conversely suppose $0\le\lambda$ and $\lVert A\rVert\le\lambda$, and let $\xi\in\mathbb{R}^n$ be arbitrary. Since $0\le\lVert\xi\rVert^{2}$, claim 5 of Elementary Arithmetic in an Ordered Field and claim 2 above give

$$|\xi\cdot(A\xi)|\le\lVert A\rVert\,\lVert\xi\rVert^{2}\le\lambda\,\lVert\xi\rVert^{2},$$

so $-\lambda\,\lVert\xi\rVert^{2}\le\xi\cdot(A\xi)\le\lambda\,\lVert\xi\rVert^{2}$ by claim 6 of Properties of the Absolute Value in an Ordered Field. By the first computation above this says $\xi\cdot\bigl((-\lambda I_n)\xi\bigr)\le\xi\cdot(A\xi)\le\xi\cdot\bigl((\lambda I_n)\xi\bigr)$; since $\xi$ was arbitrary, $-\lambda I_n\preceq A$ and $A\preceq\lambda I_n$ by The Positive Semidefinite Ordering on Symmetric Matrices. The final assertion of claim 3 is this equivalence applied with $\lambda=\lVert A\rVert$, which is nonnegative by claim 1.

Claim 4. If $A=0_n$, then $A=0\,I_n$ by Scalar Multiple of a Real Matrix and Identity Matrix, so the first computation gives $\xi\cdot(A\xi)=0\,\lVert\xi\rVert^{2}=0$ for every $\xi$; hence $S_A=\{0\}$ and $\lVert A\rVert=0$. Conversely, if $\lVert A\rVert=0$, then claim 3 with $\lambda=0$ gives $-0\,I_n\preceq A$ and $A\preceq0\,I_n$, that is $0_n\preceq A$ and $A\preceq0_n$, using $-0=0$ from claim 4 of Additive Cancellation and Elementary Additive Identities in a Field. Claim 6 of The Positive Semidefinite Ordering is a Partial Order Compatible with the Linear Structure then gives $A=0_n$.

Claim 5. Let $P$ be symmetric, let $\nu\in\mathbb{R}$, and let $\xi\in\mathbb{R}^n$ with $\lVert\xi\rVert\le1$. By the first computation and claim 4 of Properties of the Absolute Value in an Ordered Field, $|\xi\cdot((\nu P)\xi)|=|\nu|\,|\xi\cdot(P\xi)|$, and claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative factor $|\nu|$ gives $|\xi\cdot((\nu P)\xi)|\le|\nu|\,\lVert P\rVert$. Hence $|\nu|\,\lVert P\rVert$ is an upper bound for $S_{\nu P}$, and

$$\lVert\nu P\rVert\le|\nu|\,\lVert P\rVert . \tag{$\dagger$}$$

If $\mu=0$, then $\mu A=0_n$ and both sides of the asserted equality are $0$, by claim 4 and $|0|=0$. If $\mu\ne0$, then $\mu^{-1}$ exists, $\mu^{-1}(\mu A)=A$ entrywise, and $|\mu|\,|\mu^{-1}|=|\mu\,\mu^{-1}|=|1|=1$ by claim 4 of Properties of the Absolute Value in an Ordered Field, so $|\mu^{-1}|=|\mu|^{-1}$. Applying $(\dagger)$ with $\nu=\mu^{-1}$ and $P=\mu A$,

$$\lVert A\rVert=\lVert\mu^{-1}(\mu A)\rVert\le|\mu|^{-1}\,\lVert\mu A\rVert ,$$

and multiplying by $|\mu|$, nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field, using claim 5 of Elementary Arithmetic in an Ordered Field, gives $|\mu|\,\lVert A\rVert\le\lVert\mu A\rVert$. Together with $(\dagger)$ for $\nu=\mu$, $P=A$, and antisymmetry of the total order $\le$, this gives $\lVert\mu A\rVert=|\mu|\,\lVert A\rVert$.

For the triangle inequality, let $\xi\in\mathbb{R}^n$ with $\lVert\xi\rVert\le1$. By claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ and claim 5 of Properties of the Absolute Value in an Ordered Field,

$$\bigl|\xi\cdot\bigl((A+B)\xi\bigr)\bigr|=\bigl|\xi\cdot(A\xi)+\xi\cdot(B\xi)\bigr|\le|\xi\cdot(A\xi)|+|\xi\cdot(B\xi)|\le\lVert A\rVert+\lVert B\rVert ,$$

the last step adding the two defining bounds by claims 2 and 3 of Elementary Arithmetic in an Ordered Field. Hence $\lVert A\rVert+\lVert B\rVert$ is an upper bound for $S_{A+B}$, and $\lVert A+B\rVert\le\lVert A\rVert+\lVert B\rVert$.