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Put $H=A-B$. This is a nonzero Hermitian matrix. By the finite-dimensional spectral theorem, it has a unit eigenvector $v$ with a nonzero real eigenvalue $lambda$. Define the rank-one orthogonal projection matrixThe cyclic property of the matrix trace givesHence this $P$ distinguishes $A$ and $B$ through the required traces.

Suppose $v_i=\pm 1/\sqrt{n}$ encodes the true partition and $\hat{v}$ is a unit estimated eigenvector. Every coordinate of the wrong sign contributes at least $1/n$ to ${\min }_{\sigma =\pm 1}\Vert \hat{v}-\sigma v{\Vert }_2^2=2(1-|{\hat{v}}^{\top }v|)$. Since this is at most $2(1-|{\hat{v}}^{\top }v{|}^2)$, which is the squared Frobenius distance between the rank-one orthogonal projection matrices, the bound follows. Coordinates estimated as zero may be assigned consistently to either group.

For an orthogonal projection matrix $P$, a sum of squares in ANOVA is $Y^{T}PY=\Vert PY{\Vert }^2$. Mutually orthogonal projections yield an additive decomposition of total squared response magnitude. Removing the grand mean gives the corrected total.

For $Y\sim N_n(\mu ,{\sigma }^{2}I)$ and a fixed orthogonal projection matrix $P$ of rank $k$, the displayed expression is an unbiased estimator of the mean-vector prediction risk whenever ${\hat{\sigma }}^2$ is an unbiased estimator of ${\sigma }^2$. Independence of its two terms is unnecessary. Comparing fixed models yields the Mallows Cp penalty, but minimizing unbiased estimates does not preserve unbiasedness after selection.

For a normal distribution $N_n(\mu ,{\sigma }^{2}I)$ with unrestricted $\mu \in {\mathbb{R}}^n$ and unknown ${\sigma }^2$, an integrable data-only unbiased estimator of the mean-vector prediction risk of a fixed rank-$k$ orthogonal projection matrix exists exactly when $2k=n$. To prove necessity, randomize the mean by independent Gaussian variance $v$: conditioning adds $(n-k)v$ to the squared bias term, whereas the marginal variance identity would add $kv$. When $2k=n$, the residual sum of squares itself is unbiased.