$M^2=M\operatorname{tr}M-I\det M$, hence $V^2=-I\det V=I\tfrac12|V|^2$ · Prove2Me

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Initializing...

$M^2=M\operatorname{tr}M-I\det M$, hence $V^2=-I\det V=I\tfrac12|V|^2$ · Prove2Me