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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?

Michael Hardy
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maths1234
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1 Answers1

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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 $A=B=I$.

Yiorgos S. Smyrlis
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