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
 A: 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.
A: The properties listed here can help you.
In this case you have three possibilities for the Jordan Canonical Form.These are:
$$J_1=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 0& 0\\ 0 &0& 2& 0\\ 0 &0& 0& 2\end{pmatrix} , \ J_2=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 0& 0\\ 0 &0& 2& 1\\ 0 &0& 0& 2\end{pmatrix}, \ J_3=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 1& 0\\ 0 &0& 2& 1\\ 0 &0& 0& 2 \end{pmatrix}.$$
If $(A-I)(A-2I)=0$ is $J_1$. If $(A-I)(A-2I)\neq 0$ and $(A-I)(A-2I)^2= 0$ is $J_2$. Else is $J_3.$
A: I'll answer to "What is the way of finding $A$'s jordan canonical form?".
You must first find how $A$ acts on the generalised eigenspaces. Since the characteristic polynomial of $A$ is clearly $(X-1)(X-2)^3$, the one for $\lambda=1$ is $1$-dimensional, and the one for $\lambda=2$ is $3$-dimensional. The former is just the eigenspace for $\lambda=1$, while the latter is the kernel of the linear map with matrix $(A-2I_4)^d$ for sufficiently large $d$. You can be sure that $d=3$ will suffice, but simple computation will tell you that in this case $(A-2I_4)^2$ already has rank $1$, and therefore gives a $3$-dimensional kernel which is the generalised eigenspace for $\lambda=2$. In fact it is just the space spanned by the last $3$ standard basis vectors, which subspace can be indeed seen to be stable under (the linear map with matrix) $A$, and the restriction of $A$ to it has, on the obvious basis, matrix
$$
A'=\begin{pmatrix}2& 0& 0\\0& 2& 0\\1& 0& 2\end{pmatrix}.
$$
You can see that $A'-2I_3\neq0$ and $(A'-2I_3)^2=0$ confirming the $d=2$ found above; this leaves for the Jordan canonical form of $A'$ as only possibility
$$
\begin{pmatrix}2& 1& 0\\0& 2& 0\\0& 0& 2\end{pmatrix}
$$
(indeed it suffices to permute the basis vectors, bringing the final one to the front, to obtain this form). The Jordan canonical form of $A$ then is
$$
\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 1& 0\\ 0 &0& 2& 0\\ 0 &0& 0& 2\end{pmatrix}.
$$
The basis on which this matrix is obtained is not difficult to give either: the first basis vector is an eignevector for $\lambda=1$, for instance $(-1,1,1,0)$, and for the remaing three, one can take the fourth, the second and the third standard basis vectors of $\Bbb C^4$ (assuming your field was $\Bbb C$).
This example is particularly simple. However, computing the ranks of the matrices $(A-\lambda I)^k$ for all eigenvalues $\lambda$ and sufficiently many $k$ (until the rank reaches the codimension of the generalised eigenspace) always gives you enough information to know the Jordan canonical form: when going from $k-1$ to $k$, this rank drops by the number of Jordan bocks for $\lambda$ that have size at least $k$.
