Lipschitz constant of a matrix

Question

I am studying the Lipschitz continuity and trying to solve the following question:

If a function $f(x)= Ax$ is defined for $x \in \mathbb{R}^2$ with $A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then find a constant L such that \begin{eqnarray*} ||Ax - Ay|| \le L||x - y||, x, y \in \mathbb{R}^2. \end{eqnarray*} I understand how to find the Lipschitz constant in $\mathbb{R}$, but I have no idea about how to find it in $\mathbb{R}^2$.

Answer 1

For the Frobenius norm, given by $\mid\mid A\mid\mid=\sqrt{\sum_{i=1}^n\sum_{j=1}^m\mid a_{ij}\mid^2}$.

we get $\mid\mid Ax_1-Ax_2\mid\mid\le\mid\mid A\mid\mid\cdot\mid\mid x_1-x_2\mid\mid\le2\operatorname{max}\{\mid a\mid,\mid b\mid,\mid c\mid,\mid d\mid\}\mid\mid x_1-x_2\mid\mid$.

Answer 2

The Lipschitz constant $L$ for the matrix $\boldsymbol{A}$ is given by its Spectral norm $\sigma(\boldsymbol{A})$.

Spectral norm is the maximum singular value (from SVD) of the matrix, i.e. maximum scaling of a vector.