Q:

Find the Jordan form of a 4 x 4 matrix

Find the Jordan Form of 
  $$
A=\left[\begin{array}{cccc}
0 & -16 & 0 & 0\\
1 & 8 & 0 & 0\\
0 & 0 & 0 & -6\\
0 & 0 & 1 &5
\end{array}\right]
$$

First, write the matrix $A$ in block format as:
$$
A=\left(\begin{array}{cccc}
C_1 & 0\\
0 & C_2
\end{array}\right)
$$
Then we find the jordan form of $C_1$ and $C_2$
the eigenvalues of $C_1$ is $4$ with multiplicity $2$
the rank of $[C_1-4I]$ is $2$. hence, $C_1$ is diagnolizable and its jordan form $$J_1 = \left(\begin{array}{cccc}
4 & 0\\
0 & 4
\end{array}\right)
$$
the eigenvalues of $C_2$ are 2, 3.  Hence $C_1$ is also diagonalizable and its jordan form $$J_2 = \left(\begin{array}{cccc}
3 & 0\\
0 & 2
\end{array}\right)
$$  
Finally, the jordan form of $A$ is $J_1\oplus J_2$
Is my solution correct? 
Correction:
the rank of $[C_1-4I]$ is $1$. hence, the jordan form of $C_1$ $$J_1 = \left(\begin{array}{cccc}
4 & 1\\
0 & 4
\end{array}\right)
$$

A:

The computation for $J_1$ is incorrect.  If a matrix $C$ is diagonalisable and
all eigenvalues are the same, say $\lambda$, then we must have $C = \lambda I$.
Hence the Jordan form of $C_1$ cannot be diagonal.  You can check that
$(C_1-4I) \neq 0$ and $(C_1-4I)^2 =0$, so the Jordan form of $C_1$ must be
$\begin{bmatrix} 4 & 1 \\ 0 & 4 \end{bmatrix}$.

