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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!

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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.

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