Jordan form of a matrix through intuition? Find the normal form of the matrix $A$: $$\begin{bmatrix}2 & 0 & 0 & 0\\1 & 2 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 0 & 1\end{bmatrix}$$
It looks like A's jordan decomposition should be: 
$$\begin{bmatrix}2 & 1 & 0 & 0\\0 & 2 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 0 & 1\end{bmatrix}$$
And from here it's easy to find the change of basis matrix that Jordanizes' $A$.
My question is this: What do I add to make this proof formal and acceptable?
Thanks for your time.
 A: A quick argument would be something along these lines.
You have a block matrix
$$C=\begin{pmatrix}A & 0 \\ 0 & B\end{pmatrix}$$
So the Jordan form of $C$ is just the direct sum of the Jordan forms of $A$ and $B$. Since $B$ is already in Jordan form, we only have to worry about $A$.
Now $A$ is the transpose of a Jordan block, and a matrix is similar to its transpose. Therefore you can conclude that the Jordan form of $A$ is the same as the Jordan form of $A^\mathrm{T}$, which is just $A^\mathrm{T}$ itself.
A: The matrix $P=\begin{pmatrix}
  0 & 1 & 0 & 0 \\
  1 & 0 & 0 & 0 \\
  0 & 0 & 1 & 0 \\
  0 & 0 & 0 & 1 
\end{pmatrix}$ is the permutation matrix corresponding to interchanging 1 and 2. It is clear that $P^{-1}=P$ and if you multiply a matrix by $P$ from the left, you interchange the first two rows and if you multiply if from the right, you interchange the first two columns.
From this is should be clear that $PAP^{-1}=PAP=J=\begin{pmatrix}2 & 1 & 0 & 0\\0 & 2 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 0 & 1\end{pmatrix}$ (and it can be verified by direct computation). Hence $A$ and $J$ are similar and $J$ is the Jordan canonical form for $A$.
