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Let $V$ be a vector space of finite dimension. Find all linear transformations $T:V\rightarrow V$ that are both diagonalizable and nilpotent.

I was thinking $T=0$. But are there other such transformations?

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    $\begingroup$ What does a diagonal nilpotent matrix have to be? $\endgroup$ – Rankeya Apr 30 '15 at 19:32
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If $T$ is diagonalizable, then with respect to some basis of $V$, it look like diag($a_1,...,a_n$). Then $T^k$ looks like diag($a_1^k,...,a_n^k$). So $T^k = 0$ implies that each $a_r^k = 0$. Since we're over a field, $T = 0$.

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Hint: If $T$ is nilpotent, then its only eigenvalue can be $0$.

Now, suppose that $T$ has a basis of eigenvectors.

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