TITLE: Help in finding the Jordan canonical form of a matrix
QUESTION [5 upvotes]: Determine the Jordan Canonical Form of the following matrix: 
$$A=\begin{bmatrix}
 1 & 2 & 3\\
 0 & 4 & 5\\
 0 & 0 & 4\\ \end{bmatrix}$$
I am trying to determine the Jordan Basis first. For that purpose I am trying to find out the generalized Eigenvectors of this matrix. 
Corresponding to $1$, Let $U_1$ be the generalized eigenspace. My calculations show that $$U_1=span\{(1,0,0)^t\}$$ and $U_2$ be the corresponding generalized eigenspace for $4$. I found out $$U_2=span\{(1,0,-9)^t,(0,1,6)^t\}$$All I need to do now is find the Jordan basis. Since $(A-\lambda_i I)|_{U_i  }$ is nilpotent, all I need to do is find the basis for each such $i$. 
I am confused from here on what to take as the jordan basis. I am sure that $(1,0,0)^t$ will feature as the first column. I am not sure about the other two.
Thanks for the help!!

REPLY [5 votes]: To find $E_4$ we have
$$A-4I=\pmatrix{-3&2&3\cr 0&0&5\cr0&0&0\cr}\ ,$$
which is already in echelon form; the solution is
$$x_3=0\ ,\quad x_2=t\ ,\quad x_1=\tfrac23t\ ,$$
so the eigenspace
$$E_4=\left\{t\pmatrix{\tfrac23\cr1\cr0\cr}\ \bigg|\ t\in{\Bbb R}\right\}$$
is one-dimensional.  So the Jordan basis vectors will form one chain for $\lambda=4$, and (obviously) one chain for $\lambda=1$.  So there is one Jordan block for each eigenvalue, and a Jordan form is
$$J=\pmatrix{1&0&0\cr0&4&1\cr0&0&4\cr}\ .$$