Let $\mathbf{X}=(X_1,\ldots,X_n)^T$ be a random vector. Then the covariance matrix of $\mathbf{X}$ denoted by $\mathbf{Cov(X)}$ is $\lbrace Cov(X_i,X_j) \rbrace$ The diagonals of $\mathbf{Cov(X)}$ are $Cov(X_i,X_i)=Var[X_i]$ In matrix notation, $$\mathbf{Cov(X)}=\begin{pmatrix} Var[X_1] & \cdots & Cov(X_1,X_n) \\ \vdots & & \vdots \\ Cov(X_n,X_1) & \cdots & Var[X_n] \end{pmatrix}.$$

It is easily seen that $\mathbf{Cov(X)}=\mathbf{Var[X]}$ via $$\begin{pmatrix} E[{X_1}^2]-E[X_1]^2 & \cdots & E[X_1X_n]-E[X_1]E[X_n] \\ \vdots & & \vdots \\ E[X_nX_1]-E[X_n]E[X_1] & \cdots & E[{X_n}^2]-E[X_n]^2 \end{pmatrix} = \mathbf{E\Big[\big(X-E[X]\big)\big(X-E[X]\big)^T\Big]}.$$

The covariance matrix is symmetric and if the $X_i$ s are independent, identically distributed (iid) with variance $\boldsymbol{\sigma}^2$ then $$\mathbf{Cov(X)}=\boldsymbol{\sigma}^2\mathbf{I}.$$