This article Mean Vector and Covariance Matrix explains the calculation of Covariance Matrix for this input data:
\begin{equation*}
X =
\begin{bmatrix}
4.0 & 2.0 & 0.60 \\
4.2 & 2.1 & 0.59 \\
3.9 & 2.0 & 0.58 \\
4.3 & 2.1 & 0.62 \\
4.1 & 2.2 & 0.63
\end{bmatrix}
\end{equation*}
where each row is an observation of three variables(three columns).
The mean vector is:
\begin{equation*}
\bar X =
\begin{bmatrix}
4.10 & 2.08 & 0.604
\end{bmatrix}
\end{equation*}
and the formula for covariance matrix is(n=5):
\begin{equation*}
S = {1 \over n-1} \sum_{i=1}^n(X_i - \bar X)(X_i - \bar X)^T
\end{equation*}
But the dimension seems not right:
Both $X_i$ and $\bar X$ is of shape(1,3), so $(X_i - \bar X)$ would be of shape (1,3) and $(X_i - \bar X)^T$ would be (3,1), S would be (1,1).
Shouldn't the equation be
\begin{equation*}
S = {1 \over n-1} \sum_{i=1}^n(X_i - \bar X)^T(X_i - \bar X)
\end{equation*}
and get S of shape (3,3)?
