Jordan canonical form of a matrix for distinct eigenvalues How can i find the Jordan canonical form of this matrix $$A=\begin{pmatrix}1 &0 &0 &0 \\ 1& 2& 0& 0\\ 1 &0& 2& 0\\ 1 &1& 0& 2\end{pmatrix}.$$
In my book there are examples but all the matrices in these examples have only one eigenvalue repeated n times(for $n\times n$ matrices) but the matrix $A$ has eigenvalues $1$ and $2$(multiple of $3$). What is the way of finding $A$'s jordan canonical form?
Thanks

For your matrix right now, there are several Jordan structures possible. There will be a single trivial Jordan block corresponding to $1$. For the eigenvalue $2$ there will be several possibilities:


*

*We can have three trivial blocks in which case the entire sub-matrix corresponding to $2$ will look like
$$\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}$$
This is the case where the matrix is diagonaliable.

*There can be two Jordan blocks. In this case, one must be trivial and the other must be of size $2$. The sub-matrix will look like
$$\begin{pmatrix}2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}$$

*There can be a single Jordan block of size $3$
$$\begin{pmatrix}2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2\end{pmatrix}$$
The three cases are distinguished by the geometric multiplicity of the eigenvalue $2$. Remember that the number of Jordan blocks corresponding to an eigenvalue is the geometric multiplicity of that eigenvalue.
