# Jordan Canonical form with zero eigenvalue?

Can anyone tell me how to find the Jordan Canonical form of the matrix below?

$$A=\begin{pmatrix} 0 & 1 & 2\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}$$

Obviously this matrix cannot be transformed into a diagonal. Since

$$ch_A=|(λ-AI)|=λ^3 \Rightarrow λ=0$$ with multiplicity of 3.

Can anyone help me find the Jordan canonical form of it?

The matrix is obviously nilpotent, and in fact $$\;A^3=0\,,\,\,A^2\neq0\;$$ , and from here its JCF is
$$J_A=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}$$
Or easier, as you began to argue: a nilpotent matrix is diagonalizable iff it is the zero matrix, so our matrix is non-diagonalizable...so now choose the size of the biggest Jordan block according to the rank of $$\;A\;$$ ...