Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
  0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
 \end{pmatrix}$$
Then a Jordan canonical form of  A is 
Choose the correct option
$a) \begin{pmatrix}
  -1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
 \end{pmatrix}$
$b) \begin{pmatrix}
  -1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
 \end{pmatrix}$
$c) \begin{pmatrix}
  1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
 \end{pmatrix}$
$d) \begin{pmatrix}
  -1&1&0&0 \\ 0&-1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
 \end{pmatrix}$
My attempt : I know that  Determinant   of  A = product  of eigenvalues of A,  as  option c and d  is  not correct because  Here Determinant of A = 4
that is $ \det A =  -(-4) \begin{pmatrix}1 & 0 &0\\0& 1 & 0\\ 0&0&1\end{pmatrix}$
I'm  in  confusion  about  option  a) and  b).......how  can I find the  Jordan canonical form of A ?
PLiz  help  me.
Any hints/solution will be appreciated.
Thanks in advance
 A: Another answer that requires only a very minimal amount of computations.
The trace of $A$ is preserved under similarity transformations. $\operatorname{Tr}(A) = 0+0+0+0 = \lambda_1+\lambda_2+\lambda_3+\lambda_4$ is enough to exclude (c) and (d). As others have noted, (b) isn't even a Jordan canonical form (and by the way the matrix (b) is similar to (a) anyway). 
So (a) is the only remaining option and must be correct.
A: The matrix is diagonalizable, since it is a $4\times4$ matrix with $4$ distinct eigenvalues. Therefore, the correct option is a).
A: HINT
Since all options are compatible with the check on det(A)=4, we need to determine the eigenvalues by $det(A-\lambda I)=0$ and the evaluate again the given options.
Note also that b is not a Jordan normal form.
A: I'm pretty sure there is a result that a 


*

*triangular matrix is always diagonalizable with eigenvalues along the diagonal. 

*Combine this with the fact that permuting last column first makes $A$ triangular



Edit as mentioned in comments this answer is wrong. For 2. to work we will need not only permute rows or columns, but a permutation similarity, $P$ and $P^{-1}$ multiplying, one from each side. However part of 1. is true, the eigenvalues are always on the diagonal for a triangular matrix, so counting multiplicity can in that case give us hint of possible Jordan configurations.
A: All the other answers to this question so far overlook the fact that the matrix in option (b) is not in Jordan canonical form in the first place, so you can eliminate that option without doing any work at all. After eliminating (c) and (d) as you’ve done, that leaves only (a).
