Geometric Interpretation of Orthogonal Contrasts in ANOVA I was trying to understand orthogonal contrasts in ANOVA following the article by Harry M. Schey "A Geometric Description of Orthogonal Contrasts in One-Way Analysis of Variance" in The American Statistician, May 1985. 
The issue has to do with the description in the paper of these vectors, $\Phi_0$, defined as $[(n_1/n)^{1/2}\,(n_2/n)^{1/2} \,...(n_r/n)^{1/2}]'$, in which every $n_i$ stands for the number of observations in one factor (treatment), and $r$ is the overall number of factors.
$ W$ is defined as $I_r - \Phi_0\Phi_0'$ with $I_r$ representing the identity matrix with $r$ rows.
The statement that I have problems with is that the $SS_T$ or sum of squares between groups, defined as $\sum_{i} n_i(\bar{y}_i -\bar{y})^2$ with $\bar{y}$ representing the overall mean, can be alternatively be expressed as $z'Wz$ with $z = [n_1^{1/2}\bar{y}_1\,\,n_2^{1/2}\bar{y}_2\,\,...n_r^{1/2}\bar{y}_r]'$ according to the author.
I went through the linear algebra, and I probably made a mistake that I can't detect, but the $z'Wz$ ends up as $\bar{y}_1n_1(\bar{y}_1\,-\bar{y})\,+\,\bar{y}_2\,n_2\,(\bar{y}_2-\bar{y})\,...\,\bar{y}_rn_r\,(\bar{y}_r-\bar{y})$. Not entirely dissimilar to the original definition, but lacking the square, and with an added $\bar{y}_i$.
What am I missing?
 A: Looks like $W$ is idempotent, hence $zWz' = zWWz'$, using your notation. That will give you the correct expression. Just guessing, I did not check.
A: ... And after going over the linear algebra a bunch of times, and incorporating the idempotent property as mentioned in the article, and reminded by F. Tusell - thank you! - I got it to work... So just in case someone out there gets stuck with the same "trick" of using idempotent matrices to solve sum of squares, here it goes:
First off, $W$ is the result of:
$\small W = I_r - \Phi_0 \Phi_0'= 
\begin{bmatrix}
1&0&0&...&0_r\\
0&1&0&...&0_r\\
0&0&1&...&0_r\\
0&0&0&...&1_r
\end{bmatrix}
-\begin{bmatrix}
\sqrt{\frac{n_1}{n}}\\
\sqrt{\frac{n_2}{n}}\\
...\\
\sqrt{\frac{n_r}{n}}\\
\end{bmatrix}
\begin{bmatrix}
\sqrt{\frac{n_1}{n}}&
\sqrt{\frac{n_2}{n}}&
...&
\sqrt{\frac{n_r}{n}}\\
\end{bmatrix}=$
$\small=\begin{bmatrix}
1&0&0&...&0_r\\
0&1&0&...&0_r\\
0&0&1&...&0_r\\
0&0&0&...&1_r
\end{bmatrix}
-\begin{bmatrix}
\sqrt{\frac{n_1}{n}}\sqrt{\frac{n_1}{n}}&\sqrt{\frac{n_1}{n}}\sqrt{\frac{n_2}{n}}&...&\sqrt{\frac{n_1}{n}}\sqrt{\frac{n_r}{n}} \\
\sqrt{\frac{n_2}{n}}\sqrt{\frac{n_1}{n}}&\sqrt{\frac{n_2}{n}}\sqrt{\frac{n_2}{n}}&...&\sqrt{\frac{n_2}{n}}\sqrt{\frac{n_r}{n}}\\
...&...&...&...\\
\sqrt{\frac{n_r}{n}}\sqrt{\frac{n_1}{n}}&\sqrt{\frac{n_r}{n}}\sqrt{\frac{n_2}{n}}&...&\sqrt{\frac{n_r}{n}}\sqrt{\frac{n_r}{n}}\\
\end{bmatrix}=$
$\small=\frac{1}{n}\begin{bmatrix}
n -\sqrt{n_1}\sqrt{n_1}&-\sqrt{n_1}\sqrt{n_2}&...&-\sqrt{n_1}\sqrt{n_r}
\\
\\-\sqrt{n_2}\sqrt{n_1}&n-\sqrt{n_2}\sqrt{n_2}&...&-\sqrt{n_2}\sqrt{n_r}\\
...&...&...&...\\
-\sqrt{n_r}\sqrt{n_1}&-\sqrt{n_r}\sqrt{n_2}&...&n-\sqrt{n_r}\sqrt{n_r}\\
\end{bmatrix}$
This can be proven to be symmetrical and idempotent - just taking the first dot product of $WW$...
$\small\begin{bmatrix}1- \frac{n_1}{n}&-\sqrt{\frac{n_1}{n}}\sqrt{\frac{n_2}{n}}&...&-\sqrt{\frac{n_1}{n}}\sqrt{\frac{n_r}{n}}\end{bmatrix}\begin{bmatrix}1- \frac{n_1}{n}&-\sqrt{\frac{n_1}{n}}\sqrt{\frac{n_2}{n}}&...&-\sqrt{\frac{n_1}{n}}\sqrt{\frac{n_r}{n}}\end{bmatrix}^{T}=\\
1- 2\frac{n_1}{n}+\frac{n_1^2}{n^2} +
\frac{n_1 n_2}{n^2} +...+
\frac{n_1 n_r}{n^2}= 1 - 2\frac{n_1}{n} +\frac{n_1}{n}(\frac{n_1}{n}+\frac{n_2}{n}+...+\frac{n_r}{n})
= 1 - 2\frac{n_1}{n} +\frac{n_1}{n}(\frac{n}{n})=1 - \frac{n_1}{n}$.
And justifying,
$SS_{treatment\, =\, between\, groups} = z'Wz = z'WWz$
On the other hand,
$\small z = \begin{bmatrix}
\sqrt{n_1}&\bar{y}_1\\
\sqrt{n_2}&\bar{y}_2\\
...\\
\sqrt{n_r}&\bar{y}_r\\
\end{bmatrix}
$
Therefore, 
$z'WWz=$
$\small\begin{bmatrix}
\sqrt{n_1}\bar{y}_1\\
\sqrt{n_2}\bar{y}_2\\
...\\
\sqrt{n_r}\bar{y}_r
\end{bmatrix}^{T}
\small\frac{1}{n^2}\begin{bmatrix}
n -\sqrt{n_1}\sqrt{n_1}&-\sqrt{n_1}\sqrt{n_2}&...&-\sqrt{n_1}\sqrt{n_r}
\\
\\-\sqrt{n_2}\sqrt{n_1}&n-\sqrt{n_2}\sqrt{n_2}&...&-\sqrt{n_2}\sqrt{n_r}\\
...&...&...&...\\
-\sqrt{n_r}\sqrt{n_1}&-\sqrt{n_r}\sqrt{n_2}&...&n-\sqrt{n_r}\sqrt{n_r}\\
\end{bmatrix}^2\begin{bmatrix}
\sqrt{n_1}&\bar{y}_1\\
\sqrt{n_2}&\bar{y}_2\\
...\\
\sqrt{n_r}&\bar{y}_r\\
\end{bmatrix}$
Using the associative property, and separating the two apposed $W$ matrices:
$\small\frac{1}{n}\begin{bmatrix}
\sqrt{n_1}(n\,\bar{y}_1\,-[n_1 \bar{y}_1\,
+n_2\bar{y}_2+...+n_r\bar{y}_r])\\
\sqrt{n_2}(n\,\bar{y}_2\,-[n_1 \bar{y}_1\,
+n_2\bar{y}_2+...+n_r\bar{y}_r])\\...\\
\sqrt{n_r}(n\bar{y}_r\,-[n_1 \bar{y}_1\,
+n_2\bar{y}_2+...+n_r\bar{y}_r])
\end{bmatrix}^{T}\frac{1}{n}\begin{bmatrix}
\sqrt{n_1}(n\,\bar{y}_1\,-[n_1 \bar{y}_1\,
+n_2\bar{y}_2+...+n_r\bar{y}_r])\\
\sqrt{n_2}(n\,\bar{y}_2\,-[n_1 \bar{y}_1\,
+n_2\bar{y}_2+...+n_r\bar{y}_r])\\...\\
\sqrt{n_r}(n\bar{y}_r\,-[n_1 \bar{y}_1\,
+n_2\bar{y}_2+...+n_r\bar{y}_r])
\end{bmatrix}=$
$\small
\begin{bmatrix}\sqrt{n_1}(\bar{y}_1-\bar{y})&\sqrt{n_2}(\bar{y}_2-\bar{y}),...,\sqrt{n_r}(\bar{y}_r-\bar{y})
\end{bmatrix}\begin{bmatrix}\sqrt{n_1}(\bar{y}_1-\bar{y})\\\sqrt{n_2}(\bar{y}_2-\bar{y})\\\vdots\\\sqrt{n_r}(\bar{y}_r-\bar{y})
\end{bmatrix} = \sum_{i}n_i(\bar{y}_i-\bar{y})^2$.
