Multivariate Statistics: Show that SST = SSTR + SSE (More detailed question below) Show that
$$\sum_{j=1}^J \sum_{i=1}^{n_{j}} (y_{ij} - \bar{y}_{..})(y_{ij} - \bar{y}_{..})^T = \sum_{j=1}^J \sum_{i=1}^{n_{j}} (y_{ij} - \bar{y}_{. j})(y_{ij} - \bar{y}_{. j})^T + \sum_{j=1}^J n_{j}  (\bar{y}_{. j} - \bar{y}_{..})(\bar{y}_{. j} - \bar{y}_{..})^T $$

So far I've tried to use the following property for SSTR:
$$SSTR = \sum_{j=1}^J n_{j} \bar{y}_{. j} \bar{y}_{. j}^T - n \bar{y}_{..} \bar{y}_{..}^T$$
$$SST = \sum_{j=1}^J \sum_{i=1}^{n_{j}} {y}_{ij} {y}_{ij}^T - n \bar{y}_{..} \bar{y}_{..}^T$$
Within the equality SST = SSTR + SSE I can see that I can cancel $n \bar{y}_{..} \bar{y}_{..}^T$ but now I'm unsure of how to implement SSE. I tried expanding within the double summation in SSE but wasn't sure how to proceed from there.
Thank you in advance!
 A: $$
\left[ \begin{array}{c}
y_{11} - \overline y_{\cdot\cdot} \\ \vdots \\ y_{n_1 1} - \overline y_{\cdot\cdot} \\ \vdots \\ \vdots \\ y_{1J} - \overline y_{\cdot\cdot} \\ \vdots \\ y_{n_J J} - \overline y_{\cdot\cdot}
\end{array} \right] = \left[ \begin{array}{c} y_{11} - \overline y_{\cdot1} \\ \vdots \\ y_{n_1 1} - \overline y_{\cdot 1} \\ \vdots \\ \vdots \\ y_{1J} - \overline y_{\cdot J} \\ \vdots \\ y_{n_J J} - \overline y_{\cdot J} \end{array} \right] + \left[ \begin{array}{c} \overline y_{\cdot1} - \overline y_{\cdot\cdot} \\ \vdots \\ \overline y_{\cdot 1} - \overline y_{\cdot\cdot} \\ \vdots \\ \vdots \\ \overline y_{\cdot J} - \overline y _{\cdot\cdot} \\ \vdots \\ \overline y _{\cdot J} - \overline y_{\cdot\cdot} \end{array} \right] 
$$
Show that the dot-product of the two vectors on the right is $0.$ That can be done by showing that the sum of the first $n_1$ products is $0,$ then that the sum of the next $n_2$ products is $0,$ and so on.
For two vectors $a,b,$ you have
\begin{align}
\|a+b\|^2 & = (a+b)\cdot(a+b) = a\cdot a+a\cdot b+b\cdot a + b\cdot b \\[10pt]
& = \|a\|^2 + 0 + 0 + \|b\|^2 \text{ if } a\cdot b=0.
\end{align}
Apply that to the two vectors on the right side above.
