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Let $V$ be a vector space over a field $F$ and $T:V\to V$ be a linear transformation such that $T$ has zero as eigenvalue, then, is $T$ $(a)$ diagonalizable $(b)$ nilpotent $(c)$ multiplicity of each eigenvalue of $T$ is $1$?
Could you please you give me some hints?

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  • $\begingroup$ This should give you some insight into what it means to have eigenvalues of zero. $\endgroup$ – Thomas Bladt Nov 5 '17 at 7:37

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