Is Jordan normal form of a matrix unique? I am having doubt in matrix with repeated eigen values. For example a matrix A with eigen values $2,2,3$. What will be its Jordan form representation ? We can take A to be $\begin{bmatrix}2&4&5 \\ 0&2&1 \\ 0&0&3 \end{bmatrix}.$
 A: A Jordan canonical form is unique up to order of permutation.
For example the matrix $$\begin{pmatrix} 4&1&2\\0&4&3\\0&0&0\end{pmatrix}$$ is upper triangle (and it's Jordan blocks are $\color{blue}0$ of size $1$ and $\color{green}4$ of size $2$)
The matrix's Jordan form(s) are $\begin{pmatrix} \color{green}4&\color{green}1&0\\\color{green}0&\color{green}4&0\\0&0&\color{blue}0\end{pmatrix}$ and $\begin{pmatrix} \color{blue}0&0&0\\0&\color{green}4&\color{green}1\\0&\color{green}0&\color{green}4\end{pmatrix}$
As you can see, there are two blocks; it's just the order of which the blocks exist in the diagonal are different, we say they have the same Jordan form.
A: You may always by a change of basis reduce into blocks, i.e. $\begin{bmatrix}2&4&5 \\ 0&2&1 \\ 0&0&3 \end{bmatrix}$
becomes:
$\begin{bmatrix}2&4&0 \\ 0&2&0 \\ 0&0&3 \end{bmatrix}$.
The number 4, however, is part of the block with identical diagonal elements and can not be eliminated. You may, again by change of basis, transform to any non-zero number you want, so typically you choose 1. You then have
$\begin{bmatrix}2&1&0 \\ 0&2&0 \\ 0&0&3 \end{bmatrix}$
which is what is called the Jordan normal form for your given matrix. Up to permutations of blocks it is unique.
