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Suppose $A$ is a real $n\times n$ matrix with real eigenvalues: $$ 1=\lambda_1>|\lambda_2|\ge \ldots\ge |\lambda_n|>0. $$ Suppose $B$ is an involution, for simplicity let us assume that $B$ is diagonal, has $k$ ones on the diagonal, and $n-k$ minus ones. Suppose I also know that $AB$ has an eigenvector $ABx=x$. Thus I know that $$ ABx=x,\quad Ay=y $$ for some $y$.

Can I conclude that $A(I+B)/2$ has eigenvalue $1$ as well, ie. $$ A\cdot \frac {I+B}{2} z=z $$ for some $z$?

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Is the $\leq$ after $|\lambda_2|$ a typo for $>$? –  Andreas Blass Feb 12 '13 at 12:36
    
Yes, a typo, thank you. –  user12345678 Feb 12 '13 at 13:01

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