Given the following data array :
$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline J/I&1 & 2 & 3 & 4 & 5 & 6\\ \hline x & 1 & 0 & 0 & 2 & 1 & 2\\ y & 0 & 0 & 1 & 2 & 0 & 3\\ z & 0 & 1 & 2 & 1 & 0 & 2\\ \hline \end{array}$$
I can get the following values for the centered data $Y$ along with the variance $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline &1 & 2 & 3 & 4 & 5 & 6 & v_2 & v_3\\ \hline \mbox{x }&1 & 0 & 0 & 2 & 1 & 2 & 1/\sqrt{6} & 1/\sqrt{2}\\ \mbox{y }&0 & 0 & 1 & 2 & 0 & 3 & 2/\sqrt{6} & 0\\ \mbox{z }&0 & 1 & 2 & 1 & 0 & 2 & 1/\sqrt{6} & -1/\sqrt{2}\\ \hline \end{array}$$
From there I can get the principal components value :
\begin{align} Vv_2&= \begin{pmatrix}4 & 4 & 0\\ 4 & 8 & 4\\ 0 & 4 & 4\end{pmatrix} \frac{1}{\sqrt 6}\begin{pmatrix} 1\\2\\1 \end{pmatrix}\\ &=\frac{2}{\sqrt 6} \begin{pmatrix} 1\\2\\1 \end{pmatrix} \end{align}
$PC_1$ value is therefore $2$.
\begin{align} Vv_3&= \begin{pmatrix}4 & 4 & 0\\ 4 & 8 & 4\\ 0 & 4 & 4\end{pmatrix} \frac{1}{\sqrt 2}\begin{pmatrix} 1\\0\\-1 \end{pmatrix}\\ &=\frac{2}{3\sqrt 2} \begin{pmatrix} 1\\0\\-1 \end{pmatrix} \end{align}
$PC_1$ value is therefore $\frac{2}{3}$
$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline &1 & 2 & 3 & 4 & 5 & 6 & v_2 & v_3\\ \hline \mbox{value $PC_1$ }&&&&&&&2&\\ \mbox{Value $PC_2$ }&&&&&&&&2/3\\ \mbox{value coef INR }&&&&&&&&\\ \mbox{value coef CTR }&&&&&&&&\\ \mbox{value coef COR }&&&&&&&&\\ \hline \end{array}$$
How to get the INR (I think it's a French acronym) the contribution of an individual $x_i$ to the total inertia $I_T$:
$$INR(i)=\frac{p_id(0,y_i)^2}{I_T}$$
With $d$ being usually the euclidean distance. We can deduce from the definition that $\sum INR(i)=1$.
\begin{align} INR(1) &= \frac{1}{6}\times\frac{(-1)²+1²}{2+\frac{2}{3}} = \frac{1}{4}\\ INR(2) &= \frac{1}{6}\times\frac{(-1)²+(-1)²}{2+\frac{2}{3}} = \frac{1}{4}\\ INR(3) &= \frac{1}{6}\times\frac{1²+1²}{2+\frac{2}{3}} = \frac{1}{4}\\ INR(4) &= \frac{1}{6}\times\frac{1²+2²+1²}{2+\frac{2}{3}} = \frac{2}{3}???\\ \end{align}
It would be much more than $1$ now.
