Let $J_5$ be the $5x5$ Jordan block, with eigenvalue zero, and define $B=J^2$. Find the Jordan form of $B$.

I found that the $J_5$ is a matrix with $0s$ on the diagonal and $1s$ on top. I squared it and tried to find the Jordan canonical form.

However, is there an easier way of doing this?



Note that the kernel (nullspace) of $B$ has dimension $2$, so $B$ contains two Jordan blocks.

Note that since $J^5 = 0$ but $J^4 \neq 0$, we have $B^3 = 0$ but $B^2 \neq 0$. Conclude that the largest Jordan block of $B$ has size $3$.

Conclude that $B$ has Jordan form $J_3 \oplus J_2$.

  • $\begingroup$ I am having a doubt; How did you get $\ker B$ has $\dim =2$ $\endgroup$ – Learnmore Nov 9 '16 at 5:53
  • $\begingroup$ Suppose that $Bx=0\implies J^2(x)=0\implies?$ $\endgroup$ – Learnmore Nov 9 '16 at 5:54
  • $\begingroup$ Do you mean to say that $x,Jx$ will be a basis of $\ker B$?Is so why? $\endgroup$ – Learnmore Nov 9 '16 at 5:55
  • $\begingroup$ @learnmore $J^2$ is in reduced row echelon form. Its rank is clearly $3$. $\endgroup$ – Omnomnomnom Nov 9 '16 at 11:16
  • $\begingroup$ Can you please do the problem in some more detail.I don't get it $\endgroup$ – Learnmore Nov 10 '16 at 2:11

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