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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?

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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 $\;A\;$ ...

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