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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 $C_1$ $$J_1 = \left(\begin{array}{cccc} 4 & 1\\ 0 & 4 \end{array}\right) $$

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    $\begingroup$ Did you check to make sure there are actually two eigenvectors for the $C_1$? $\endgroup$ – jgon May 16 '15 at 17:21
  • $\begingroup$ I made a mistake. $\endgroup$ – user3382078 May 16 '15 at 17:37
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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 $\begin{bmatrix} 4 & 1 \\ 0 & 4 \end{bmatrix}$.

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