已知n维随机变量(vec{X}=(X_{1},X_{2},...,X_{n}))的协方差矩阵为(C = egin{bmatrix}c_{11} & c_{12} & ... & c_{1n} \c_{21} & c_{22} & ... & c_{2n} \. & .& &.\. & .& &.\. & .& &.\c_{n1} & c_{n2} &...&c_{nn}end{bmatrix} )，其中(c_{ij} = Eig{[X_{i}-E(X_{i})][X_{j}-E(X_{j})]ig})。那么，能否将协方差矩阵写成向量形式呢？即写成和一维随机向量的方差公式类似的形式：(D(X) = E(X-E(X))^{2} )。
我们来证明一下，设(X_{i})的样本量为m，由协方差矩阵的公式，可知：
(C = frac{1}{m}egin{bmatrix} (X_{1}-E(X_{1}))^{T}(X_{1}-E(X_{1})) & ... & (X_{1}-E(X_{1}))^{T}(X_{n}-E(X_{n})) \ (X_{2}-E(X_{2}))^{T}(X_{1}-E(X_{1})) & ... & (X_{2}-E(X_{2}))^{T}(X_{n}-E(X_{n})) \. &. & .\. &. & .\. &. & .\ (X_{n}-E(X_{n}))^{T}(X_{1}-E(X_{1})) & ... & (X_{n}-E(X_{n}))^{T}(X_{n}-E(X_{n})) end{bmatrix} )
( = frac{1}{m}egin{bmatrix} (X_{1}-E(X_{1}))^{T} \ (X_{2}-E(X_{2}))^{T} \ .\. \. \ (X_{n}-E(X_{n}))^{T} end{bmatrix} egin{bmatrix} X_{1}-E(X_{1}), & X_{2}-E(X_{2}), &...,& X_{n}-E(X_{n}) end{bmatrix} )
( = frac{1}{m}(vec{X}-E(vec{X}))^{T}(vec{X}-E(vec{X})) )
( = E((vec{X}-E(vec{X}))^{T}(vec{X}-E(vec{X}))) )
推导结果与一维随机向量的方差公式是一致的，所以，我们的猜测是正确的，证毕。