Why is $X(X^TX)^{-1}X^T (y-X\beta)$ called the direction to the OLS solution? (Instead of $X(X^TX)^{-1}X^T y$) In Least Angle Regression (LAR), the coefficient $\beta_k$ is updated every $k$th step with:
$$
\beta_k=\beta_{k-1}+\lambda( X_\mathcal{A}^TX_{\mathcal{A}})^{-1}
X_{\mathcal{A}}^T(y-
X_{\mathcal{A}}\beta_{k-1})
$$
The book says that this update moves the coefficients to the direction of  $\beta$ derived from  least square (OLS).
Why?
Isn't the direction to the OLS coefficient (when multiplied to $X_{\mathcal{A}}$ )  vector given by:
$$
\lambda( X_\mathcal{A}^TX_{\mathcal{A}})^{-1}
X_{\mathcal{A}}^Ty
$$
(Notice $y$ instead of the residual $r=y-
X_{\mathcal{A}}\beta_{k-1}$)

I already know that when the residual is used instead of $y$ alone, the correlations across the predictors in $X_\mathcal{A}$ are same and decreasing as $\lambda$ gets bigger. This propoerty lead to the "Least Angle" in the method.
However, why do the paper and books say that $\lambda( X_\mathcal{A}^TX_{\mathcal{A}})^{-1}
X_{\mathcal{A}}^Tr$ is the direction to the coefficient vector derived from running OLS?
Shouldn't it be $y$ instead of $r$ as defined in OLS? I have only seen regressing on the the residual in finding orhogonal bases for $X$ however, in the end, that method still uses $y$ when obtaining the coefficient vector.

Let us assume you are currently at some coefficient $\beta_k$, and you want to move from there.  Where you will end up is at $\hat{\beta} = (X^TX)^{-1}X^TY$.  Therefore, the direction you want to move in is:
$$(X^TX)^{-1}X^TY - \beta_k$$
Premultiplying the r.h.s. by the identity matrix $(X^TX)^{-1}X^TX$ gives us:
$$(X^TX)^{-1}X^TY - (X^TX)^{-1}X^TX\beta_k$$
Rearranging terms results in:
$$(X^TX)^{-1}X^T(Y - X\beta_k)$$
... and substituting $r_k$ for $(Y - X\beta_k)$ gets you the rest of the way there.
