Given the Lasso problem $$ min_\beta (Y-X\beta)^\top(Y-X\beta) \quad s.t. \|\beta\|_1\leq\lambda, $$ and assuming that X is orthonormal such that $X^\top X=I$, we know that the closed form solution can be written as $$ \hat{\beta}^{Lasso} = sgn(\hat{\beta}^{LS})(|\hat{\beta}^{LS}|-\lambda)_+, $$ where $\hat{\beta}^{LS}=X^\top Y$.
However, even if X is non-orthonormal, we can use a SVD transformation of $X^\top X$ to generate transformed data $\tilde{X}$ such that $\tilde{X}^\top \tilde{X} = I$, and apply the Lasso to the transformed data. In particular, consider the SVD: $$ X^\top X= Q\Delta Q^\top $$ and the transformed matrix $\tilde{X}=XQ\Delta^{-1/2}$, so that $\tilde{X}^\top \tilde{X} = I$.
Next, using the transformed data $(Y,\tilde{X})$, we get the orthonormalized estimate: $\hat{\tilde{\beta}}^{Lasso} = sgn(\hat{\tilde{\beta}}^{LS})(|\hat{\tilde{\beta}}^{LS}|-\lambda)$, where $\hat{\tilde{\beta}}^{LS} = \tilde{X}^\top Y$.
The non-orthonormalized estimate can be recovered as $\hat{\beta}^{Lasso} = Q\Delta^{-1/2}\hat{\tilde{\beta}}^{Lasso}$.
$\textbf{Question}$: This seems like a valid method of estimating the Lasso and such orthonormalization technique is used in the Group Lasso context (for e.g. in these lecture notes and in Wei, Huang and Li (2011)). However, I am unable to find anything that discusses this as a way to exploit the closed form solution for the Lasso. I am thinking that perhaps the SVD of a large data matrix might be computationally expensive relative to numerical/iterative methods. Are there any reasons why this is not done?