※ 引述《linshihhua (linshihhua)》之銘言：
: let A be a linear transformation definde on a finite dimension vector space V
: prove that dim(ker(A+I))+dim(ker(A-I))=dim(V) if and only if A^2=I
(i) A^2=I
then its minimal polynomial is x-1, x+1 or (x-1)(x+1)
Hence, A can be diagonalized, and its eigenvalues are 1 or -1
done.
(ii) dim(ker(A+I))+dim(ker(A-I))=dim(V)
assume ker(A+I)=Span{u1,u2,...,uk}
and V=Span{u1,u2,...,uk,u_{k+1},...,u_n}
if (A+I)u=0, (A-I)u=0, then u=0
hence ker(A+I)∩ker(A-I)={0}
but dim(ker(A-I))=n-k
hence, ker(A-I)=Span{u_{k+1},...,u_n}
then ker(A+I)+ker(A-I)=V
then (A+I)(A-I)=0, done.
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