Prove that any nilpotent matrix is similar to a block diagonal matrix whose blocks are matrices with 1's along the first super diagonal and 0's elsewhere.

I'm not sure where to start exactly. Any guidance would be helpful!

  • $\begingroup$ I am assuming over $\mathbb C$? $\endgroup$ – Couchy311 Mar 6 '17 at 4:58
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    $\begingroup$ Can you use Jordan canonical form? Combine with the fact that the eigenvalues of a nilpotent matrix are all $0$. $\endgroup$ – D_S Mar 6 '17 at 5:08
  • $\begingroup$ Do you know about quotient spaces? $\endgroup$ – Omnomnomnom Mar 6 '17 at 5:26
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    $\begingroup$ I think this is sometimes a stepping-stone to the full Jordan canonical form theorem. It doesn't matter what the base field is. You can build up a basis of the vector space with the required properties. Take $x\not=0$; there is some minimal $k$ such that $N^k x=0$. It's easy to prove that $\{x,Nx,N^2x, \dots,N^{k-1}x\}$ is LI, and action of $N$ on these is "right". Now take $x_2$ outside the span of these and repeat. $\endgroup$ – ancientmathematician Mar 6 '17 at 9:43
  • $\begingroup$ Check Daniel's answer here: math.stackexchange.com/questions/809473/… $\endgroup$ – Daniel Mar 6 '17 at 21:44

My answer is based on Wikipedia articles "Nilpotent matrix" and "Jordan normal form."

Every square matrix A over an algebraically-closed field has a Jordan canonical form J, which has zeros everywhere except A's eigenvalues as diagonal entries and possibly ones on the super diagonal. A is similar to J because there exists an invertible matrix P such that A = P J P ⁻¹. Because zero is the only eigenvalue of a nilpotent matrix, J has the form you want.


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