Jordan's Canonical Form of a Matrix I'm a little tied up about Jordan's canonical form of a square matrix. How would Jordan's Canonical form of the following square matrix be obtained?
$$A = \begin{pmatrix}
2 & 1  & 1 \\ 
-2 & -1 & -2\\ 
1 & 1 & -2
\end{pmatrix}$$
 A: Compute the eigenvalues, which are the roots of the characteristic polynomial. You'll get two roots: $1$ (a simple root) and $-1$ (a double one). Use this to prove that the Jordan normal form is$$\begin{pmatrix}1&0&0\\0&-1&1\\0&0&-1\end{pmatrix}.$$
A: You have to find a Jordan basis for $A$. As $-1$ is a double eigenvalue, to determine whether $A$ is diagonalisable, we have toknow whether the eigenspace for this value has dimension $2$ or not, and in the latter case determine a basis for the generalised eigenspace.
Now $\ker(A+I)$ has dimension $1$ since
$$A+I= \begin{pmatrix}
3 & 1  & 1 \\ 
-2 & 0 & -2\\ 
1 & 1 & -1
\end{pmatrix}$$
has rank $2$. The last two rows are linearly independent, and an eigenvector for the eigenvalue $-1$ satisfy the equations $x+z=0$, $\;x+y-z=0$.
$\ker (A+I)^2$ has dimension $2$ since
$$(A+I)^2= \begin{pmatrix}
8 & 4  & 0 \\ 
-8 & -4 & 0\\ 
0 & 0 & 0
\end{pmatrix}$$has rank $1$, and is defined by the equation $2x+y=0$.
To have a Jordan basis , we begin with taking $v_3\in\ker (A+I)^2 \setminus\ker (A+I)$. The simplest is to choose $v_3=(0,0,1)$.
Next set $v_2=(A+I)v_3=(1,-2,-1)$.  This is an eigenvector for the eigenvalue $-1$, and by construction we have:
$$Av_3=-v_3+v_2.$$
 Last step: find an eigenvector for the simple eigenvalue $1$, i.e. a vector in the kernel of 
$$ A-I= A = \begin{pmatrix}
1 & 1  & 1 \\ 
-2 & -2 & -2\\ 
1 & 1 & -3
\end{pmatrix}.$$
It has to satisfy the independent linear equations:
$$x+y+z=0,\enspace x+y_3z=0 \iff z=0,\enspace  x+y =0, $$
so we may choose $v_1=(1,-1,0)$.
By construction, in the basis $(v_11,v_2,v_3)$ the matrix of the endomorphism associated to $A$ is
$$J(A)= \begin{pmatrix}1&0&0\\0&-1&1\\0&0&-1\end{pmatrix}.$$
