Is there a mathematical expression that shows how LASSO shrinks coefficients (including some to zero)? By using singular value decomposition (SVD), I noticed from the derivation that ridge regression shrinks the coefficients by factor $\frac{D^2}{D^2+\lambda}$, where $D$ is the diagonal matrix of the matrix $\underset{m\times n}A$. Moreover, as the penalty term $\lambda$ increases, the amount of shrinkage increases. 
But, what about LASSO regression? Unlike ridge regression, LASSO regression shrinks some of the coefficients to zero. My question: 


*

*Is there a way to show, in some mathematical fashion, that LASSO regression shrinks some of the coefficients to zero as the notation above does for ridge regression?
Using the two predictor case would make it easy to understand. Could you please provide mathematical lines?


EDIT
Knight & Fu (2000) show that $\hat{\beta}_{lasso}=0$ if and only if $−\lambda I≤2\sum\limits_{i}Y_iX_i≤\lambda I$.
How does that occur?
References:


*

*Knight, Keith, and Wenjiang Fu. "Asymptotics for lasso-type estimators." Annals of Statistics (2000): 1356-1378.

 A: Firstly,  I think it's worth noting that the description of what ridge does assumes that the data matrix is orthonormal.
Secondly, the answer to your question is yes under those circumstances.  The details may be found in "Elements of Statistical Learning" on p. 69 bis (section 3.4.3) .  The short story is that 
$ \beta \to \text{sign}(\beta)\max(\beta-\lambda,0)$ is the formula. Please see the book for the complete discussion, better formatting, and details.  
A: The question can be answered when one assumes an orthogonal matrix of predictors. Then the right singular vector matrix $\mathbf V$ equals the identity matrix $\mathbf I$, and the derivation below holds only for this case.
Consider the lasso problem
$$
\min_{\mathbf \beta} \frac12 ||\mathbf X \mathbf \beta - \mathbf y||^2_2 + \lambda ||\beta||_1
$$
Use the singular value decomposition
\begin{align}
\mathbf X &= \mathbf U \, \mathbf D\, \mathbf V^T\\
&= \sum_i \mathbf u_i \, d_i \mathbf v^T_i
\end{align}
and also expand $\beta$ and $\mathbf y$ as
\begin{align}
\beta &= \sum_i \beta_i \mathbf v_i\,\\
\mathbf y &= \sum_i y_i \mathbf u_i
\end{align}
Insert the whole stuff into the lasso functional to obtain
\begin{align}
\min_{\beta_i} \left(\sum_i \frac12 |\beta_i d_i - y_i|^2 + \lambda |\beta_i|\right)
\end{align}
In the SVD basis, the Lasso minimization problem thus decomposes into separate problems for each component. 
For a vanishing singular value, $d_i=0$, the solution is $\beta_i=0$, so let's  consider only the case $d_i >0$. For this, one can rewrite the previous equation as
$$\min_{\beta_i}  \frac12 \left|\beta_i - \frac{y_i}{d_i}\right|^2 + \frac{\lambda}{d_i^2} \left|\beta_i\right|$$
As stated in the other answer, and also derived here, the solution of this problem is
$$
\beta^{Lasso}_i \ = \ \begin{cases} 0 &,\  |d_i y_i| \le \lambda \\ \frac{y_i}{d_i} - \frac{\lambda}{d_i^2} & ,\ |d_i y_i| > \lambda \end{cases}
$$
Compare this to the corresponding solution of the ordinary least squares problem  and ridge regression (with regularization parameter $\alpha$):
\begin{align}
\beta_i^{OLS} &= \frac{y_i}{d_i}\\
\beta_i^{ridge} &= \frac{y_i}{\tilde d_i}, \quad \text{where}\ \ \tilde d_i = d_i \cdot \frac{d_i^2+\alpha}{d_i^2}
\end{align}
As is well known, ridge regression can be obtained by scaling a given singular value $d_i$ with a factor that depends on $d_i$ and $\alpha$.
On the other hand, the Lasso solution is more complex and also depends on the target $y_i$. With a good amount of enforcement, one can write the adjusted singular values as
$$
\beta_i^{Lasso} = \frac{y_i}{\bar d_i}, \quad \text{where} \ \ \bar d_i = d_i  \frac{1}{\left(1 - \frac{\lambda}{d_i y_i}\right)\theta\big(|d_i y_i|-\lambda\big)}
$$
where $\theta$ is the Heaviside step function, which is zero for $|d_i y_i|<\lambda$ and otherwise one.

EDIT: Please note that for a general predictor matrix, the above statement is wrong, for two reasons: first and informally, if it would be correct, it would be the standard approach for an easy Lasso solution. Second, the formal reason why this doesn't work is that the L1-norm is not invariant under orthogonal transformations. So when one does the expansion of $\beta$ in two orthonormal basis sets, $\beta = \sum_i \beta_i \mathbf e_i = \sum_i b_i \mathbf v_i$, one has $\sum_i |\beta_i| \neq  \sum_i |b_i|$.
So, what it does is not to solve the Lasso problem, but rather the more exotic Lasso problem
$$
\min_{\mathbf \beta} \frac12 ||\mathbf X \mathbf \beta - \mathbf y||^2_2 + \lambda ||\mathbf V^T \beta||_1
$$
