To find jordan canonical form

Question

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?

Answer 1

Finding the eigenvalues of any of these matrices is not particularly difficult; they are all upper-triangular.

Hint: Suppose that $A$ is a $3 \times 3$ matrix with zero as its only (complex) eigenvalue. Note that $A$ has the desired J-C form if and only if $A \neq 0$ but $A^2 = 0$ (why?).

Answer 2

The given Jordan canonical form implies, minimal polynomial of corresponding matrix should be $x^2=0.$ Hence if matrix $A$ is having the property that $A \neq 0$ and $A^2=0 ,$ it will have desired Jordan canonical form.