I would like to show that the covariance matrix is in fact a tensor, so its transformation must be linear and homogenous. Let $m_k=M(\xi_k)$, so $D_{ij}=M((\xi_i-m_i)(\xi_j-m_j)).$
To show that $D$ is a tensor, I have to rotate the coordinate system and see what happens to $D$, how its transformed. Since during a rotation the distance cannot change from the origo and the determinant of a matrix which represents a rotation must be $1$, $C\in SO(n)$
Let $\xi_k^,=\sum_{j=1}^nC_{kj}(\xi_j-m_j)$, then $M(\xi_k^,)=0$ for every $k$.
Now I just have to derive $D_{ij}^,$, but I can't. From the definition of $D_{ij}$:
$D_{ij}^,=M((\xi_i^,-m_i)(\xi_j^,-m_j))=M(\xi_i^,\xi_j^,)-M(\xi_i^,)M(\xi_j^,)$
but $M(\xi_{i,j}^,)=0$, so we have:
$D_{ij}^,=M(\xi_i^, \xi_j^,)=M[(\sum_{q=1}^nC_{iq}(\xi_q-m_q))(\sum_{l=1}^nC_{jl}(\xi_l-m_l))]$
I'm stuck here and don't know how to proceed. Any help would be much appreciated!