We are given data $x_i=(x_{i1},x_{i2},...,x_{iq})^T$, $i=1,...,n$ with mean $$m_x:=m=\frac{1}{n}\sum_{i=1}^n x_i= \frac{1}{n}\sum_{i=1}^n\left(\begin{array}{c} x_{i,1} \\\vdots \\ x_{i,q} \end{array}\right)= \left(\begin{array}{c}\frac{1}{n}\sum_{i=1}^n x_{i,1} \\\vdots \\ \frac{1}{n}\sum_{i=1}^nx_{i,q} \end{array}\right)=\left(\begin{array}{c} m_{1} \\\vdots \\ m_{q} \end{array}\right)\in R^q$$ and a constant vector $a\in R^q$.
I need to show that:
i)$$\frac{1}{n}\sum_{i=1}^n(x_i-m)(x_i-m)^T=\frac{1}{n}\sum_{i=1}^nx_ix_i^T-mm^T$$
ii)$$\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n(x_i-a)(x_j-a)^T=(m-a)(m-a)^T$$
I might use the fact that for vectors $a,b,c,d$ with appropriate dimension it holds that $$(a+b)(c+d)^T=ac^T+ad^T+bc^T+bd^T$$
I tried i) for a long long time now but I always get a $1/n^2$ in the $mm^T$ part of the expression which destroys the whole thing... just to let you know if this notation confused as much as it did confuse me: $\frac{1}{n}\sum_{i=1}^nx_ix_i^T-mm^T$ is a $qxq$-Matrix.
//Edit: Solution for i \begin{align*} &\frac{1}{n}\sum_{i=1}^n(x_i-m)(x_i-m)^T \\ =& \frac{1}{n}\sum_{i=1}^n (x_ix_i^T -x_im^T-mx_i^T+mm^T) \\ =&\frac{1}{n}\sum_{i=1}^n x_ix_i^T - \frac{1}{n}\sum_{i=1}^n x_im^T- \frac{1}{n}\sum_{i=1}^n mx_i^T+\frac{1}{n}\sum_{i=1}^n mm^T \\ =&\frac{1}{n}\sum_{i=1}^n x_ix_i^T - \frac{1}{n}\sum_{i=1}^n x_im^T- m \frac{1}{n}\sum_{i=1}^nx_i^T+\frac{1}{n}\sum_{i=1}^n mm^T \\ =& \frac{1}{n}\sum_{i=1}^n x_ix_i^T - mm^T- m m^T+\frac{1}{n}n mm^T \\ =&\frac{1}{n}\sum_{i=1}^nx_ix_i^T-mm^T \end{align*} ii) similar to i)