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Work with a nonsingular multivariate normal distribution, so $V$ is positive-definite. The claimed finite maximum-likelihood estimates also require the centered sample covariance matrix to be nonsingular; the exceptional case is discussed below.

Independence gives, up to a constant, the log-likelihoodLet $\bar{Y}=n^{-1}{\sum }_i{Y}_i$ and $S=n^{-1}{\sum }_i(Y_i-\bar{Y})(Y_i-\bar{Y}{)}^T$. The centered sum is zero, and expansion therefore givesFor every fixed positive-definite matrix $V$, the extra quadratic term in the log-likelihood is $-n(\bar{Y}-\mu {)}^T{V}^{-1}(\bar{Y}-\mu )/2$, with unique maximum at $\mu =\bar{Y}$.

Assume $S$ is positive-definite. Maximizing over $V$ now means minimizing $\log \det V+\mathrm{tr}(V^{-1}S)$. Put $A=S^{1/2}{V}^{-1}{S}^{1/2}$, another positive-definite matrix. ThenIf $a_1,\dots ,a_p>0$ are its eigenvalues, the variable part is ${\sum }_j(a_j-\log a_j)$. The scalar inequality $a-\log a\ge 1$, with equality only at $a=1$, shows that this is minimized uniquely at $A=I$, or $V=S$. ThusThe divisor is $n$ because this is maximum likelihood estimation, not the unbiased covariance estimate with divisor $n-1$.

The nonsingularity condition matters. The centered scatter has rank at most $n-1$. If $S$ is singular, choose $\mu =\bar{Y}$, retain positive covariance eigenvalues on its range and let a covariance eigenvalue on its null space tend to zero. The quadratic likelihood term remains finite, while $-\log \det V\to +\infty$. This is singular covariance and nonexistence of a Gaussian maximum likelihood estimate: there is no finite maximizer over nonsingular $V$. Under a nonsingular Gaussian population, $S$ is positive definite almost surely when $n>p$; for $n\le p$ it is necessarily singular. The asserted estimate and the inverse in the next part have their usual meaning under $n>p$.