Proof of The Doubling Matrix and its Elementary Properties

Throughout, $n$, $J$, $\xi$, $\eta$, $X$, $Y$, $a$ and $c$ are as in the statement, and we write $z=\iota(\xi,\eta)$.

Conventions. We use without further comment that $\le$ on $\mathbb{R}$ is transitive and compatible with addition, by the ordered field axioms and Total Order on a Set, and the following weak compatibility with multiplication: if $0\le b$ and $s\le t$, then $sb\le tb$; this is immediate when $b=0$ or $s=t$, and otherwise follows from claim 10 of Elementary Order Arithmetic in an Ordered Field. We use that $|-1|=1$ and, for positive $a$, that $|a|=a$: by claim 1 of Properties of the Absolute Value in an Ordered Field the value $|x|$ equals $x$ or $-x$ and is nonnegative, while $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field and $-1<0$, respectively $-a<0$, by claim 4 of that lemma; the case $|-1|=|1|$ being covered by claim 2 of Properties of the Absolute Value in an Ordered Field. Recall from Euclidean Space and Real Matrices: Standing Notation (iv) that $P\preceq Q$ means $w\cdot(Pw)\le w\cdot(Qw)$ for every $w\in\mathbb{R}^{2n}$, and from Euclidean Space and Real Matrices: Standing Notation (vii) that $\iota$ is a bijection, so that every $w\in\mathbb{R}^{2n}$ is of the form $\iota(\xi,\eta)$.

Claim 1. By construction $J$ is the block matrix determined by $A=I_{n}$, $B=-I_{n}$, $C=-I_{n}$ and $D=I_{n}$. By claim 3 of Elementary Properties of the Transpose of a Real Matrix we have $I_{n}^{\top}=I_{n}$, and hence $(-I_{n})^{\top}=(-1)I_{n}^{\top}=-I_{n}$ by claim 2 of that lemma. Thus $A=A^{\top}$, $D=D^{\top}$ and $C=B^{\top}$, so $J$ is symmetric by claim 3 of Action and Quadratic Form of a Block Matrix; that is, $J\in\mathcal{S}(2n)$.

By claim 1 of Action and Quadratic Form of a Block Matrix,

$$J\,\iota(\xi,\eta)=\iota\bigl(I_{n}\xi+(-I_{n})\eta,\ (-I_{n})\xi+I_{n}\eta\bigr).$$

Here $I_{n}\xi=\xi$ and $I_{n}\eta=\eta$ by claim 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, while $(-I_{n})\eta=(-1)(I_{n}\eta)=-\eta$ and $(-I_{n})\xi=-\xi$ by claim 1 of that lemma; and $\xi+(-\eta)=\xi-\eta$, $\eta+(-\xi)=\eta-\xi$. Hence $J\,\iota(\xi,\eta)=\iota(\xi-\eta,\eta-\xi)$. Taking $\eta=\xi$ and using $\xi-\xi=0_{\mathbb{R}^{n}}$ gives $J\,\iota(\xi,\xi)=\iota\bigl(0_{\mathbb{R}^{n}},0_{\mathbb{R}^{n}}\bigr)=0_{\mathbb{R}^{2n}}$, the last equality by Concatenation and the Sup-Convolution of a Sum in Separated Variables (i).

Claim 2. By claim 2 of Action and Quadratic Form of a Block Matrix, applied with $\xi'=\xi$ and $\eta'=\eta$, together with claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$z\cdot(Jz)=\xi\cdot(I_{n}\xi)+\xi\cdot\bigl((-I_{n})\eta\bigr)+\eta\cdot\bigl((-I_{n})\xi\bigr)+\eta\cdot(I_{n}\eta)=\xi\cdot\xi-\xi\cdot\eta-\eta\cdot\xi+\eta\cdot\eta .$$

On the other hand, by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and claims 3 and 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$\lVert\xi-\eta\rVert^{2}=(\xi-\eta)\cdot(\xi-\eta)=\xi\cdot\xi-\eta\cdot\xi-\xi\cdot\eta+\eta\cdot\eta .$$

The two right-hand sides coincide, which proves claim 2.

Claim 3. Let $w=\iota(\xi,\eta)\in\mathbb{R}^{2n}$. Since $0_{2n}=0\,I_{2n}$, both matrices having all entries $0$, Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (vi) gives $w\cdot(0_{2n}w)=0\,\lVert w\rVert^{2}=0$, while $w\cdot(Jw)=\lVert\xi-\eta\rVert^{2}$ is nonnegative by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Hence $0_{2n}\preceq J$.

By claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and claims 1 and 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite, the computation of claim 2 reads $\lVert\xi-\eta\rVert^{2}=\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}-2(\xi\cdot\eta)$, where $2=1+1$. Applying A Weighted Young Inequality and the Splitting of a Quadratic Form (i) with $t=1$ to the pair $\xi$ and $-\eta$, and using $\xi\cdot(-\eta)=-(\xi\cdot\eta)$ from claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite and $\lVert-\eta\rVert=\lVert\eta\rVert$ from claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, we get $-2(\xi\cdot\eta)\le\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}$. Therefore

$$w\cdot(Jw)=\lVert\xi-\eta\rVert^{2}\le 2\bigl(\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}\bigr)=2\lVert w\rVert^{2}=w\cdot\bigl((2I_{2n})w\bigr),$$

by claim 3 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space and Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (vi). Hence $J\preceq 2I_{2n}$.

Finally $w\cdot\bigl((-2I_{2n})w\bigr)=-2\lVert w\rVert^{2}\le 0=w\cdot(0_{2n}w)$, since $0\le 2\lVert w\rVert^{2}$ and hence $-2\lVert w\rVert^{2}\le 0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field; so $-2I_{2n}\preceq 0_{2n}\preceq J\preceq 2I_{2n}$, and claim 3 of Properties of the Norm of a Symmetric Real Matrix, applicable because $0\le 2$, gives $\lVert J\rVert\le 2$.

Claim 4. The matrix $J^{2}$ lies in $\mathcal{S}(2n)$ by Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (ii), and $2J\in\mathcal{S}(2n)$ by Euclidean Space and Real Matrices: Standing Notation (iii). Let $w=\iota(\xi,\eta)$. By Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (ii), claim 1 above and claim 3 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space,

$$w\cdot(J^{2}w)=\lVert Jw\rVert^{2}=\lVert\iota(\xi-\eta,\eta-\xi)\rVert^{2}=\lVert\xi-\eta\rVert^{2}+\lVert\eta-\xi\rVert^{2}=2\lVert\xi-\eta\rVert^{2},$$

since $\eta-\xi=(-1)(\xi-\eta)$ in the real vector space $\mathbb{R}^{n}$ of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space and hence $\lVert\eta-\xi\rVert=\lVert\xi-\eta\rVert$ by claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square. On the other hand, by claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite and claim 2,

$$w\cdot\bigl((2J)w\bigr)=2\bigl(w\cdot(Jw)\bigr)=2\lVert\xi-\eta\rVert^{2}.$$

The quadratic forms of $J^{2}$ and $2J$ therefore agree at every point of $\mathbb{R}^{2n}$, so $J^{2}=2J$ by A Symmetric Matrix is Determined by its Quadratic Form, and a Second-Order Expansion by its Coefficients (i).

Claim 5. That $aJ\in\mathcal{S}(2n)$ is Euclidean Space and Real Matrices: Standing Notation (iii). By claim 5 of Properties of the Norm of a Symmetric Real Matrix and claim 3, $\lVert aJ\rVert=|a|\,\lVert J\rVert=a\lVert J\rVert\le 2a$.

Since $J\preceq 2I_{2n}$ by claim 3 and $3a$ is nonnegative, the compatibility of $\preceq$ with multiplication by a nonnegative real number, recorded in Euclidean Space and Real Matrices: Standing Notation (iv), gives $3aJ\preceq 3a(2I_{2n})$; and $3a(2I_{2n})=6aI_{2n}$, the two matrices having the same entries. Hence $3aJ\preceq 6aI_{2n}$.

For the last identity, note first that $aJ+a^{-1}(aJ)^{2}$ and $3aJ$ both lie in $\mathcal{S}(2n)$, by Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (ii) and Euclidean Space and Real Matrices: Standing Notation (iii). Let $w=\iota(\xi,\eta)$ and put $q=w\cdot(Jw)=\lVert\xi-\eta\rVert^{2}$. By claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite, $w\cdot\bigl((aJ)w\bigr)=aq$. Moreover $(aJ)w=a(Jw)$ by claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, so by claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and the computation of claim 4,

$$w\cdot\bigl((aJ)^{2}w\bigr)=\lVert(aJ)w\rVert^{2}=a^{2}\lVert Jw\rVert^{2}=2a^{2}q,$$

using Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (ii) for the first equality. Consequently, again by claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$w\cdot\Bigl(\bigl(aJ+a^{-1}(aJ)^{2}\bigr)w\Bigr)=aq+a^{-1}\bigl(2a^{2}q\bigr)=aq+2aq=3aq=w\cdot\bigl((3aJ)w\bigr),$$

since $a^{-1}a^{2}=a$ in the field $\mathbb{R}$. As $w$ was arbitrary, A Symmetric Matrix is Determined by its Quadratic Form, and a Second-Order Expansion by its Coefficients (i) gives $aJ+a^{-1}(aJ)^{2}=3aJ$.

Claim 6. Assume $X\oplus(-Y)\preceq cJ$ and let $\xi\in\mathbb{R}^{n}$. Put $w=\iota(\xi,\xi)$. By Block Diagonal Symmetric Matrices and the Blocks of a Symmetric Matrix (ii), claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$w\cdot\Bigl(\bigl(X\oplus(-Y)\bigr)w\Bigr)=\xi\cdot(X\xi)+\xi\cdot\bigl((-Y)\xi\bigr)=\xi\cdot(X\xi)-\xi\cdot(Y\xi).$$

By claim 2 and claim 3 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, $w\cdot(Jw)=\lVert\xi-\xi\rVert^{2}=0$, so $w\cdot\bigl((cJ)w\bigr)=c\bigl(w\cdot(Jw)\bigr)=0$ by claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite. The assumed inequality therefore reads $\xi\cdot(X\xi)-\xi\cdot(Y\xi)\le 0$, that is $\xi\cdot(X\xi)\le\xi\cdot(Y\xi)$. As $\xi\in\mathbb{R}^{n}$ was arbitrary, $X\preceq Y$ by Euclidean Space and Real Matrices: Standing Notation (iv).