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I am trying to show with two examples that even though two matrices look different they have the same Jordan canonical form.

So far I have the matrix $A=\begin{pmatrix} 3 & 0 & 0\\ 0 & 4 &-1\\ 0 & 1 & 2\end{pmatrix}$. The characteristic polynomial is $$(3-\lambda)^3=0.$$ Hence the eigenvalue for $A$ is $\lambda=3$ with eigenvectors $$\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix} \text{, and } \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}.$$ Note $$(A-3\cdot I)^2=0.$$ Thus the minimal polynomial is $$(\lambda-3)^2=0.$$ This implies the largest Jordan block is $2 \times 2$, which implies there are 2 Jordan blocks.

After alot of algebra work, I found the Jordan canonical form of $A$ is $$J=\begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{pmatrix}$$ and the basis matrix is $$P=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 1 & 1 \end{pmatrix}.$$

Now I need help finding another matrix $B$ with the same Jordan canonical form, $J$.

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    $\begingroup$ Why not $B=J$? Or more generally, keep the same $J$ and use a different basis matrix. $\endgroup$ – angryavian Apr 30 '17 at 23:35
  • $\begingroup$ I would like something more interesting.than $B=J$ $\endgroup$ – Username Unknown Apr 30 '17 at 23:41
  • $\begingroup$ As angryavian said, just pick a basis matrix $P$ and construct $B = PJP^{-1}$ $\endgroup$ – Sebastian Schulz Apr 30 '17 at 23:54
  • $\begingroup$ Can I just rearrange the columns in P? Or I need to find a completely new P? $\endgroup$ – Username Unknown Apr 30 '17 at 23:55
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Two matrices have the same Jordan canonical form iff they are conjugate. So just take any invertible matrix $T$ and let $B=TAT^{-1}$. For a few special choices of $T$ (such as the identity) you will get $B=A$, but for most choices of $T$ you will get $B\neq A$.

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