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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.