Decomposing R squared or VIF In the context of multi-regression, I am wondering if there is a way to decompose $$VIF_i = 1/(1-R_i^2)$$ where $R_i^2$ is the r squared obtained from the regression of dependent variable = i and independent variables are all other factors.
I want to decompose $VIF_i$ or $R_i^2$ into individual factors to see how much each individual factor contributes to the $VIF_i$ or $R_i^2$
Someone recommended using the square of partial correlation coefficient and that value is linearly related to $R_i^2$. My undestanding is that partial correlation coefficient measures the correlation between two variables, holding the other variables constant.  Is this a viable option?
 A: Let's say the ith column of X is $x_i$ and the $X$ without ith column is $X_{(i)}$.
As the definition of $R_i^2$ is $SS_R/SStot$, 
$R_i^2=\frac{x_i'(X_{(i)}(X_{(i)}'X_{(i)})^{-1}X'_{(i)}-\frac{1}{n} 1_n1_n')x_i}{x_i'(I- 1_n1_n')x_i}$
The other factors are only in the term in the numerator $x_i'(X_{(i)}(X_{(i)}'X_{(i)})^{-1}X'_{(i)})x_i$.
Because $X_{(i)}(X_{(i)}'X_{(i)})^{-1}X'_{(i)}$ is idempotent matrix, it's same with $||X_{(i)}(X_{(i)}'X_{(i)})^{-1}X'_{(i)}x_i||^2$
Here, $X_{(i)}(X_{(i)}'X_{(i)})^{-1}X'_{(i)}x_i$ is actually the production of $X_i$ and the slope 'from the regression of dependent variable = i and independent variables are all other factors'.
Let's say this slope vector $a = (a_0,a_1,...,a_{i-1},a_{i+1},...,a_{k})'$ which can be calculated from $(X_{(i)}'X_{(i)})^{-1}X'_{(i)}x_i$ where $k$ is the number of regressors. ($a_0$ is the intercept though, technically we can consider it as a slope)
Then,
$\begin{align}X_{(i)}(X_{(i)}'X_{(i)})^{-1}X'_{(i)}x_i&=X_{(i)}(a_0,a_1,...,a_{i-1},a_{i+1},...,a_{k})'\\&=1_n a_0  + x_1 a_1 +x_2 a_2+...+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+...+a_{k}x_k\end{align}$.
I guess this expression shows each factor and its contribution. 
So, by putting this into the expression of $R_i^2$, I think we can get the decomposed version of $R_i^2$.
I don't know how to do with partial correlation though. 
I hope this answer is helpful.
I am not a native English speaker. Please don't mind my poor English and feel free to correct it. 
edit1
There is another way of decomposing VIF using spectral decomposition. 
When the design matrix $X$ is unit length scaled, the diagonal elements of $(X'X)^{-1}$ is the VIFs.
According to spectral decomposition theory, $(X'X)^{-1}=\sum_{i=1}^k \frac{1}{\lambda_i}v_i v_i'$ where k is the number of factors, the $\lambda_i$ is the  ith eigen value of the matrix $X'X$ and the $v_i$ is the corresponding eigen vector. The ith eigen value is related to ith factor.
So VIF of jth factor $VIF_j=\sum_{i=1}^k \frac{1}{\lambda_i}v_{ij}^2$ where $v_{ij}$ is the jth element of the eigen vector $v_i$. Each term of this summation represents the contribution of each factor. 
A: You can decomposing R2 by Shapley-Owen value in SAS, for more details You can see :
https://wernerantweiler.ca/blog.php?item=2014-10-10&fbclid=IwAR1XH5pDFFGimVAJbwR9ZUneshLx5djtZDjxoURQ2RWLIRO-Y2uStMMfTR0
my pleasure
Dr. Jawad Albakri
