# Power of a Jordan Normal Form

In my notes I have that the Jordan normal form of $$B^2$$ is $$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 &1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ and the notes say that because of this, the only possibility for $$B$$ is $$\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Can anybody explain why this is the case? How can we conclude the jordan form of B from the jordan form of B^2? Thanks for your help!

If $$\lambda$$ is an eigenvalue of $$B$$, then $$\lambda^2$$ is an eigenvalue of $$B^2$$. In the Jordan Normal Form, the eigenvalues correspond to entries on the main diagonal. Since all such entries in $$B^2$$ are $$0$$, this means that all eigenvalues are $$0$$ for both $$B$$ and $$B^2$$. That takes care of the diagonal entries for the Jordan form of $$B$$.
We know that the entries above a Jordan block are all equal to $$1$$. Since we have only one block of a single eigenvalue, namely $$0$$, then the above-diagonal entries for the Jordan form of $$B$$ will be all $$1$$ and that gives the desired form.