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I do know that $(n \times n)$ nilpotent matrices have the minimal polynomial $x^k$ for some positive integer $k ≤ n$.

I also do know that having the same Jordan normal form means they have the same minimal polynomial.

Any hints are welcome, I strongly suspect this statement to be false.

B.Swan
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  • what exactly does it mean that they have the same jordan normal form? – Asinomás Jun 02 '17 at 00:50
  • It means that there exist bases of generalised eigenvectors such that the representation of the matrix in that base has the eigenvalues on the diagonal and ones or zeros on the superdiagonal (one line above the diagonal), the rest of the entries are zero. Here's the Wikipedia link – B.Swan Jun 02 '17 at 00:59

2 Answers2

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False: consider $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$.

Daniel Schepler
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Following on Daniel's answer this would imply every symmetric matrix with $0$ on the diagonal is nilpotent.

This is known to be false to anyone who does spectral graph theory.

Asinomás
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