For A $\in L(\mathbb {R}^n)$ define ${\wr \hspace {-3pt}\wr \hspace {-1pt}A\hspace {-1pt}\wr \hspace {-3pt}\wr } := \inf \{\vert Ax\vert \,\big \vert \, \vert x\vert = 1\}$. Which of the following properties of a norm are true for ${\wr \hspace {-3pt}\wr \hspace {-1pt}\cdot \hspace {-1pt}\wr \hspace {-3pt}\wr }$?
a) $\forall \lambda \in \mathbb {R},A\in L(\mathbb {R}^n)\colon {\wr \hspace {-3pt}\wr \hspace {-1pt}\lambda A\hspace {-1pt}\wr \hspace {-3pt}\wr } = |\lambda |{\wr \hspace {-3pt}\wr \hspace {-1pt}A\hspace {-1pt}\wr \hspace {-3pt}\wr }$
b) $\forall A\colon {\wr \hspace {-3pt}\wr \hspace {-1pt}A\hspace {-1pt}\wr \hspace {-3pt}\wr } \geq 0$ and $({\wr \hspace {-3pt}\wr \hspace {-1pt}A\hspace {-1pt}\wr \hspace {-3pt}\wr } = 0 \Leftrightarrow A = 0)$
c) $\forall A,B\in L(\mathbb {R}^n)\colon {\wr \hspace {-3pt}\wr \hspace {-1pt}A+B\hspace {-1pt}\wr \hspace {-3pt}\wr } \leq {\wr \hspace {-3pt}\wr \hspace {-1pt}A\hspace {-1pt}\wr \hspace {-3pt}\wr } + {\wr \hspace {-3pt}\wr \hspace {-1pt}B\hspace {-1pt}\wr \hspace {-3pt}\wr }$
a) and b) are fairly easy: ${\wr \hspace {-3pt}\wr \hspace {-1pt}A\hspace {-1pt}\wr \hspace {-3pt}\wr }=0$ iff $A$ there exists a vector $x$ of norm $1$ such that $Ax=0$ iff $A$ is singular.
Here is the answer for c): Define $A,B: \mathbb R^{2} \to \mathbb R^{2}$ by $A(x_1,x_2)=(0,x_2)$ and $B (x_1,x_2)=(x_1,0)$. Then $A+B$ is the identity so ${\wr \hspace {-3pt}\wr \hspace {-1pt}A\hspace {-1pt}\wr \hspace {-3pt}\wr }=1.$ But $A$ and $B$ are singular so ${\wr \hspace {-3pt}\wr \hspace {-1pt}A\hspace {-1pt}\wr \hspace {-3pt}\wr }=0$ and ${\wr \hspace {-3pt}\wr \hspace {-1pt}B\hspace {-1pt}\wr \hspace {-3pt}\wr }=0$ so c) is false.