How to get the equiangular vector in p dimension linear space? Used in Least angle regression Here I am reading the "Least angle regression" by Efron, 2004. https://projecteuclid.org/download/pdfview_1/euclid.aos/1083178935
, in page 413, he gives formula of equiangular vector of the linear space spanned by design matrix X.
This is the equation (2.4) to equation (2.7)
$X_A=(...s_jX_j...)_{j\subset A},s_j$is sign equal to $\pm1$. is the design matrix of active set A(selected covariates).
Then author gave the formula as follows,
$G_A=X^T_AX_A$ and $A^*_A=(1^TG_A^{-1}1_A)^{-\frac{1}{2}}$.
Then, the equiangular vector gives
$u_A=X_Aw_A$ where $w_A=A_AG_A^{-1}1_A$
is the unit vector give 
$X^T_Au_A=A^*_A1_A$ and $||u_A||^2=1$ .
I have no idea about how to derive the $w_A$.
As the intuition of the calculation, the triangular vector in two dimension space spanned by $x_1$ and $x_2$ is $u=\frac{\frac{x_1}{|x_1|}+\frac{x_2}{|x_2|}}{|\frac{x_1}{|x_1|}+\frac{x_2}{|x_2|}|}$, which is $\frac{1}{{|\frac{x_1}{|x_1|}+\frac{x_2}{|x_2|}|}|x_1|}x_1+\frac{1}{{|\frac{x_1}{|x_1|}+\frac{x_2}{|x_2|}|}|x_2|}x_2$
Or, as the equiangular condition, then $<u,x_1>=<u,x_2>$ or $u^Tx_1=u^Tx_2$.
But how to derive the general form for $w_A$ from the naive thinking about the equiangular vector?
The related questions: 
Least angle regression keeps the correlations monotonically decreasing and tied?
 A: For ease of notation I'm simply going to consider the following problem: Given a matrix $X$ whose columns  $x^{(j)}$ are linearly independent, each with mean zero and unit variance, we want to find a vector $u$ of unit length such that each column makes the same angle with $u,$ that is $x^{(j)}{'} u$ is the same number for all $j.$
Another way of stating the latter condition is that $X'u$ is some multiple $a\in\mathbb R_{>0}$ of the vector of ones $\mathbb 1,$ i.e. 
$$X'u=a\mathbb 1.\tag 1$$ 
If we can find any vector satisfying $(1)$ for some $a$ we can divide it by its length to get a unit vector satisfying $(1)$ with a different $a$. Now note that 
$$\mathbb 1 = X'X(X'X)^{-1}\mathbb 1.$$
Hence $X(X'X)^{-1}\mathbb 1$ is a vector satisfying $(1)$ for $a=1.$ Our vector has length 
$$((X(X'X)^{-1}\mathbb 1)'X(X'X)^{-1}\mathbb 1)^{1/2}
=(\mathbb 1'(X'X)^{-1}X'X(X'X)^{-1}\mathbb 1)^{1/2}
=(\mathbb 1'(X'X)^{-1}\mathbb 1)^{1/2}.$$
Dividing our vector by its length we get the unit-length vector
\begin{align}
u&=X(X'X)^{-1}\mathbb 1 / (\mathbb 1'(X'X)^{-1}\mathbb 1)^{1/2}\\
&=X(X'X)^{-1}\mathbb 1 \times (\mathbb 1'(X'X)^{-1}\mathbb 1)^{-1/2}\\
&=:XG^{-1}\mathbb 1 \times (\mathbb 1'G^{-1}\mathbb 1)^{-1/2}\\
&=:XG^{-1}\mathbb 1 \times A^*\\
&=XA^*G^{-1}\mathbb 1\\
&=:Xw.\\
\end{align}
