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For a nonzero positive semidefinite covariance matrix $\Sigma$, its covariance effective rank iswhere the operator norm is induced by Euclidean norm and the ${\lambda }_j$ are the nonnegative eigenvalues. Each nonzero eigenvalue contributes at most one to this ratio, soA scalar multiple of $I_p$ has covariance effective rank $p$, while a matrix $\lambda u{u}^T$ with $\lambda >0$ and a unit vector $u$ has covariance effective rank one. For $\Sigma =\mathrm{diag}(100,1,1)$, the ratio is $102/100=1.02$, although the ordinary matrix rank is three.

Thus the quantity measures total variance relative to the largest directional variance, rather than merely counting nonzero directions. Multiplying $\Sigma$ by a positive scalar leaves the ratio unchanged. Its role in the Gaussian sample-covariance operator-norm bound makes it a useful measure of the dimension that matters for covariance estimation. The zero matrix requires a separate convention because the displayed ratio is $0/0$.

Use the uncentered sample covariance matrixThen $v_{\ell }=a_{\ell }^T\Sigma a_{\ell }$ and, simultaneously for every unit vector $a_{\ell }$,Here $\Vert \Sigma {\Vert }_{\mathrm{op}}=1$ and the effective rank of a covariance matrix satisfiesThe Gaussian sample-covariance operator-norm bound therefore gives, with probability at least $1-e^{-\delta }$,Under the assumed upper bound on $\log d+1+\delta$, the second term is at most the first. Thus the stronger simultaneous estimateholds for every $\ell$. Squaring and using that the displayed ratio is at most one gives the inequality requested in the question after enlarging the universal constant $C$.