# Nilpotent Matrix is Similar to a block diagonal matrix

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!

• I am assuming over $\mathbb C$? – Couchy311 Mar 6 '17 at 4:58
• Can you use Jordan canonical form? Combine with the fact that the eigenvalues of a nilpotent matrix are all $0$. – D_S Mar 6 '17 at 5:08
• Do you know about quotient spaces? – Omnomnomnom Mar 6 '17 at 5:26
• 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. – ancientmathematician Mar 6 '17 at 9:43
• Check Daniel's answer here: math.stackexchange.com/questions/809473/… – Daniel Mar 6 '17 at 21:44