A nilpotent matrix is similar to a strictly upper triangular matrix

Asked Nov 23 '17 at 13:20

Active Nov 23 '17 at 13:20

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Let $n>1$ and $A\in M_n(F).$ Suppose $A$ is nilpotent. Prove that $A$ is similar to a strictly upper triangular matrix by induction on $n$.

How should I do this? I don't know how to use the assumption to relate it to the definition of similarity of matrices that's why I'm stuck.

Please help. Thanks.

asked Nov 23 '17 at 13:20

Mashed Potato

Is there a way to get $S$? – Mashed Potato Nov 23 '17 at 13:30
Cayley-Hamilton is not proved to us yet so we can't use it. My teacher hinted that this involves eigenvalues and can be shown that 0 is an eigenvalue of $A$ – Mashed Potato Nov 23 '17 at 13:48
Show that every nilpotent matrix has an eigenvector (with eigenvalue zero). – Qiaochu Yuan Nov 23 '17 at 18:11

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