Finding the Jordan canonical form of A and Choose the correct option

Question

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

Answer 1

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.

Answer 2

The matrix is diagonalizable, since it is a $4\times4$ matrix with $4$ distinct eigenvalues. Therefore, the correct option is a).

Answer 3

I'm pretty sure there is a result that a

1. triangular matrix is always diagonalizable with eigenvalues along the diagonal.
2. Combine this with the fact that permuting last column first makes $A$ triangular

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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.

Answer 4

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.

Answer 5

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).