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Let $f$ be the linear operator on finite dimensional complex vector space $V$. Prove, that operator $f$ has the only one eigenvalue $\lambda_1$ iff when operator $f_1=f-\lambda_1 Id: v \mapsto f(v)-\lambda_1 v$ is nilpotent, i.e. $f^{N}_1$ for some $N>0$. Prove, that basis $j$ in $V$ is basis of Jordan normal form of $f$ iff when it's basis of Jordan normal form for $f_1$

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For the first part, if it has the only eigenvalue $\lambda_1$, you can consider its Jordan canonical form: $J(f)-\lambda_1id$ is nilpotent; if conversely, you can consider the minimal polynomial of $f$, which must be a factor of $(\lambda-\lambda_1)^N$ and the eigen polynomial has the same roots as the minimal polynomial. For the second part, suppose $e_1,\cdots,e_n$ is a basis of $J(f)$. Then $f(e_1,\cdots,e_n)=(e_1,\cdots,e_n)J(f)$. Then $(f-\lambda_1)(e_1,\cdots,e_n)=(e_1,\cdots,e_n)(J(f)-\lambda_1id)$ and note that $J(f)-\lambda_1id$ is nothing but the Jordan canonical form of $f_1$. Similarly you can prove its converse.

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