Proof of Perturbation Splitting of a Quadratic Form

Throughout, sums are finite sums of real numbers, $|t|$ is the absolute value of $t\in\mathbb{R}$, and $t^{2}$ abbreviates $t\cdot t$.

Squares are nonnegative. For every $s\in\mathbb{R}$ we have $0\le s^{2}$. Indeed, $|s|$ equals $s$ or $-s$ by claim 1 of Properties of the Absolute Value in an Ordered Field, so $|s|\,|s|$ equals $s\cdot s$ in the first case and, in the second, equals $(-s)(-s)=s\cdot s$ by claim 2 of Zero Products and Elementary Identities in a Field. Also $0\le|s|$ by claim 1 of Properties of the Absolute Value in an Ordered Field, so claim 5 of Elementary Arithmetic in an Ordered Field, applied to the inequality $0\le|s|$ with the nonnegative factor $|s|$, gives $|s|\cdot 0\le|s|\,|s|$, and $|s|\cdot 0=0$ by claim 1 of Zero Products and Elementary Identities in a Field.

Claim 1. By claim 4 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum,

Applying Interchange of a Finite Double Sum to the second double sum and then renaming the two summation indices, it equals

Since $A$ is symmetric we have $A_{ji}=A_{ij}$, so every summand of the last double sum equals $A_{ij}\,w_{i}\,z_{j}$, and the two double sums agree. This proves claim 1.

Claim 2. By claim 2 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product, applied with both matrices equal to $A$, we have $A^{2}\xi=(AA)\xi=A(A\xi)$. Hence claim 1, applied with $w=\xi$ and $z=A\xi$, gives

and $(A\xi)\cdot(A\xi)=\lVert A\xi\rVert^{2}$ by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$.

Claim 3. Put $z=x-\xi$; since $\mathbb{R}^{n}$ is a real vector space by Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, we have $\xi+z=x$. By claim 3 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, $Ax=A\xi+Az$, so by claims 2 and 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$,

Write $\beta=\xi\cdot(Az)$; by claim 1 the third summand also equals $\beta$, so the two middle summands sum to $\beta+\beta$. Put $a=\lVert A\xi\rVert$ and $b=\lVert z\rVert$, both nonnegative by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$.

Bounding the cross terms. By claim 1 above and claim 1 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, $\beta=z\cdot(A\xi)=(A\xi)\cdot z$, so Cauchy-Schwarz Inequality for the Euclidean Dot Product gives $|\beta|\le a\,b$, while $\beta\le|\beta|$ by claim 3 of Properties of the Absolute Value in an Ordered Field. Hence $\beta\le a\,b$, and adding this inequality to itself — legitimate by claims 3 and 2 of Elementary Arithmetic in an Ordered Field — gives $\beta+\beta\le a\,b+a\,b$.

Next put $s=a-\varepsilon^{-1}b$. By the paragraph on squares, $0\le s^{2}$, so $0\le\varepsilon\,s^{2}$ by claim 5 of Elementary Arithmetic in an Ordered Field, the factor $\varepsilon$ being nonnegative. Expanding $s^{2}$ with the field axioms and using $\varepsilon\,\varepsilon^{-1}=1$,

so claim 3 of Elementary Arithmetic in an Ordered Field turns $0\le\varepsilon\,s^{2}$ into

Combining with the previous display and using transitivity,

Bounding the last term. By claim 3 of Properties of the Absolute Value in an Ordered Field and claim 2 of Properties of the Norm of a Symmetric Real Matrix,

Adding the last two displayed inequalities to the identity for $x\cdot(Ax)$ — that is, applying claims 3 and 2 of Elementary Arithmetic in an Ordered Field to add them and then translating — yields

Identifying the right-hand side. By claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum,

so claims 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ and claim 2 above give

while distributivity in the field of real numbers gives

Since $z=x-\xi$, the previous display is exactly the asserted inequality, and claim 3 is proved.