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up vote 2 down vote favorite 1 What is wrong with this proof: Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$ . By property of similar matrices: $R=STS^{-1}$ Therefore: $Rv = STS^{-1}v$ $2v = STS^{-1}v$ $2S^{-1}v = TS^{-1}v$ $2S^{-1}Sv = Tv$ $2Iv = Tv$ $Tv = 2v$ Thus $v$ is also an eigenvector of $T$ with eigenvalue 2. Clearly this proof is incorrect, but where does it go wrong? linear-algebra matrices eigenvalues-eigenvectors inverse share | cite | improve this question asked 8 hours ago Justin Sanders 11 1