Centered Gaussian quadratic forms have dimension-free even-moment bounds

By [F1] choose an orthogonal matrix $O$ with $A=O\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)O^T$. By [F2], $Z=O^{T}U$ has independent standard-normal coordinates. Integration by parts in the density [F3], with $z{e}^{-z^2\mathrm{/}2}\to 0$ at both infinities, gives $E{Z}_i^2=1$. Every absolute moment of $V_i=Z_i^2-1$ is finite: for an integer $k\ge 1$, $\mathrm{\mid }{z}^2-1{\mathrm{\mid }}^k\le (1+z^2{)}^k$, and this polynomial is integrable against $e^{-z^2\mathrm{/}2}$. Thus $E{V}_i=0$, $E(U^{T}AU)={\sum }_i{\lambda }_i=\mathrm{tr}A$, and $U^{T}AU-\mathrm{tr}A={\sum }_i{\lambda }_i{V}_i$.

Expand the even power $({\sum }_i{\lambda }_i{V}_i{)}^{2r}$ over ordered tuples of $2r$ indices. By [F4], a term whose tuple has an index appearing exactly once has expectation zero. Group all other terms by the partition of their $2r$ positions into equal-index blocks. Such a partition has block sizes $k_1,\dots ,k_q\ge 2$ with ${\sum }_a{k}_a=2r$. For each fixed partition, taking absolute values and dropping the restriction that different blocks have distinct indices bounds its contribution by ${\prod }_{a=1}^{q}E\mathrm{\mid }{V}_1{\mathrm{\mid }}^{k_a}{\prod }_{a=1}^q{\sum }_i\mathrm{\mid }{\lambda }_i{\mathrm{\mid }}^{k_a}$. All these quantities are finite by step 1.1.

Put $L=({\sum }_i{\lambda }_i^2{)}^{1\mathrm{/}2}$. If $L=0$, the centered form is identically zero and the bound holds. Otherwise $\mathrm{\mid }{\lambda }_i\mathrm{\mid }\mathrm{/}L\le 1$ and ${\sum }_i(\mathrm{\mid }{\lambda }_i\mathrm{\mid }\mathrm{/}L{)}^2=1$, so for every integer $k\ge 2$, ${\sum }_i\mathrm{\mid }{\lambda }_i{\mathrm{\mid }}^k\le L^k$. Consequently each partition contribution in step 2.1 is bounded by $L^{2r}{\prod }_{a}E\mathrm{\mid }{V}_1{\mathrm{\mid }}^{k_a}$. There are finitely many partitions of the fixed set of $2r$ positions; the sum of these moment products over the partitions without singleton blocks is a finite constant $C_r$ depending only on $r$.

Since $2r$ is even, the expanded power is the absolute $2r$th power. Summing the bounds of step 3.1 therefore gives $E\mathrm{\mid }{U}^{T}AU-\mathrm{tr}A{\mathrm{\mid }}^{2r}\le C_r{L}^{2r}$. Finally $L^2=\mathrm{tr}(A^2)={\sum }_{i,k}{A}_{ik}^2$ by symmetry and orthogonal diagonalization. This proves the bound and the expectation assertion, including $A=0$, singular matrices and matrices of either sign.