Please help me to answer this question: If A is a skew symmetric matrix such that $A^TA=I$, then $A^{4n-1}=$
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Since $A$ is skew-symmetric, that is , $A’=-A$. Also, $A’A=-I$ implies $-A^2=I$ and hence $A^4=I, A^{4n}=I$, then $A^{4n-1}=A^{4n}A^{-1}=IA^{-1}=-A$.
Jiabin Du
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