Which of the following matrices have Jordan canonical form of equal to the $3\times 3$ matrix
$$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
a)$ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$
b)$ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$
c)$ \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$
d)$ \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$
Here characteristic equation of the matrix is $x^3$.Hence the 3 eigenvalues of the matrix are zero. Do we want to find the eigenvalues of all the matrices in the options?Is there any other way?
\pmatrix{a&b\\c&d}
will work; I find this to be quicker than the whole "begin/end" deal. $\endgroup$