# Finding two matrices with the same Jordan canonical form

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, $J$.

• Why not $B=J$? Or more generally, keep the same $J$ and use a different basis matrix. Apr 30, 2017 at 23:35
• I would like something more interesting.than $B=J$ Apr 30, 2017 at 23:41
• As angryavian said, just pick a basis matrix $P$ and construct $B = PJP^{-1}$ Apr 30, 2017 at 23:54
• Can I just rearrange the columns in P? Or I need to find a completely new P? Apr 30, 2017 at 23:55

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 $B\neq A$.