Check if the identity matrix is an extreme point of the set $S$ I'm trying to check that $I$ is an extreme point of $S=\{A \in M_{2\times2}:\|A\|_1 \leq 1\}$?
I have done this by writing out $I=\lambda B + (1-\lambda)C$ with $B,C \in S$ and $\lambda \in (0,1).$ Then I have written out the system of equations for this e.g. $1=\lambda b_{11}+(1-\lambda) c_{11}$ etc.
Then I used a lemma that said that this implies that $b_{11}=c_{11}=1$ then used the fact that the norm of $I$ is $1$ so the other entries in $B$ and $C$ must be $0.$
Not sure if I can extrapolate so easily from the straightforward $\mathbb{R}$ case??
Is this correct?
 A: If $I$ the $2\times 2$ were not an extreme point of  $S$, then there would be $A,B\in S$ and $\lambda(0,1)$, such that 
$$
I=\lambda A+(1-\lambda)B.
$$
In particular,
$$
1=\lambda a_{11}+(1-\lambda)b_{11},
$$
and hence
$$
1\in [a_{11},b_{11}]\quad \text{or}\quad1\in [b_{11},a_{11}].
$$
If $a_{11}>1$, then for $(x,y)=(1,0)$, we would have
$$
A\binom{1}{0}=\binom{a_{11}}{a_{21}},\quad \left\|A\binom{1}{0}\right\|_1=\left\|\binom{a_{11}}{a_{21}}\right\|_1=\lvert a_{11}\rvert+
\lvert a_{21}\rvert>1. 
$$
Contradiction. Hence $a_{11}=1$, which implies that $b_{11}=1$. We would arrive to the 
same even if we had assumed that $b_{11}>1$. Thus in both cases
$$
a_{11}=b_{11}=1\quad\text{and similarly}\quad a_{22}=b_{22}=1.
$$
If $A\ne I$, then $a_{12}\ne 0$ or $a_{21}\ne 0$. Assume for example that $a_{12}\ne 0$.
Then
$$
\left\|A\binom{0}{1}\right\|=
\left\|\left(\begin{matrix} 1 & a_{12} \\ a_{21}& 1\end{matrix}\right)\binom{0}{1}\right\|=
\left\|\binom{a_{12}}{1}\right\|=1+\lvert a_{12}\rvert>1.
$$
A contradiction. Thus $a_{12}=0$ and similarly $a_{21}=1$.
Thus $A=B=I$.
