Jordan Canonical Form (example)

Let $$A=\begin{pmatrix} 3&0&1\\ 2&2&2\\ -1&0&1\\ \end{pmatrix}$$ Determine a change of basis matrix $Q$ and a matrix $J$ such that $J=Q^{-1}AQ$ where $J$ is the Jordan canonical form of $A$.

$\textit{Solution:}$ We are typically interested in $\lambda$, $v$ such that $Av=\lambda v$. Choosing vectors in our space $V$ that are purely scaled under the linear transformation represented by the matrix $A$ can give us a subspace that is dilated by $A$. We also insist that $v \neq 0$, since 0 adds nothing to a basis.
$Av= \lambda v$
$Av-\lambda v = 0$
$(A-\lambda I)v=0$
So, we wish to look at $N(A-\lambda I) = \{v\in{V} | (A-\lambda I)v=0 \}$. With the assumption that $v \neq 0$, then this means that $N(A-\lambda I)$ is non-trivial $\Longleftrightarrow (A-\lambda I)$ is linearly dependent $\Longleftrightarrow det(A- \lambda I)=0$ by the invertible matrix theorem. To compute $det(A- \lambda I)$ we choose a cofactor expansion down the second column: $$det(A- \lambda I)=det \begin{pmatrix} 3-\lambda &0&1\\ 2&2-\lambda &2\\ -1&0&1-\lambda \end{pmatrix} =(2-\lambda)((3-\lambda)(1-\lambda)-(-1)(1)) $$ $$=(2-\lambda)(\lambda^2-4\lambda+4)=(2-\lambda)(\lambda-2)(\lambda-2)=p(\lambda)$$ So, the characteristic polynomial $p$ of $A$ has eigenvalue $\lambda=2$ with a multiplicity of 3. To check $A$ for diagonalizability, we can look at the dimension of the eigenspace corresponding to $\lambda=2$ and/or we can look at the minimal polynomial of $A$. The minimal polynomial $m$ of $A$ has the following characteristics: monic, irreducible, minimal degree, contains all the roots/eigenvalues, divides the characteristic polynomial, $m(A)=0$. With these constraints, the minimal polynomial is either $m(t)=t-2,(t-2)^2,(t-2)^3=p(t)$. $A-2I \neq 0$ so we now check $(A-2I)^2$: $$\begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix} \begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix}= \begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&0&0 \end{pmatrix}$$ So, $m(t)=(t-2)^2$, which is not separable, i.e. has repeated roots, thus $A$ does not diagonalize. We may also note that the second and third rows of $A-2I$ are multiples of the first, so $rank(A-2I)=1$. So, the dimension of the eigenspace corresponding to $\lambda=2$ is $$dim(E_{\lambda=2})=3-rank(A-2I)=3-1=2$$ which is consistent with the minimal polynomial having a repeated root. However, the dimension of the generalized eigenspace, $dim(K_{\lambda=2})=3$. So, the Jordan canonical form for $A$ must have a 2-cycle and a 1-cycle. $$J=\begin{pmatrix} 2&0&0\\ 0&2&1\\ 0&0&2 \end{pmatrix}$$ The desired cycle structure can be diagrammed as follows:
$A-2I:v' \mapsto v \mapsto 0$
$A-2I: w \mapsto 0$
Our change of basis matrix $Q$ then lines up these cycles (which go right to left) with the Jordan blocks of $J$ $$Q=\begin{pmatrix} |&|&|\\ w&v&v'\\ |&|&| \end{pmatrix}$$ Now to find $w,v$ and $v'$:
$$(A-2I) =\begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix} \sim \begin{pmatrix} 1&0&1\\ 0&0&0\\ 0&0&0 \end{pmatrix}$$ So with two rows of 0's we can have 2 free variables, so we can thus make 2 "regular" (non-generalized) eigenvectors. This is consistent with the previous observation that $dim(E_{\lambda=2})=3-1=2$. In a matrix-vector form, we have $a+c=0$ and 2 free variables. Letting $a=c=0$ and $b=1$ gives us $(0,1,0)=e_2$ and we can check that $(A-2I)e_2=0$: $$\begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix} \begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$$ So our eigenvector $w=e_2$ gives us a 1-cycle. To obtain $v'$ (our starting 2-cycle vector) we need to pick it so that it is outside our eigenspace $E_{\lambda=2}$. Even if we don't know exactly what this space is, we can often get lucky just by trying vectors in the standard basis $\{e_1,e_2,e_3\}$.
$(A-2I)e_1=(1,2,-1)$, so $$(A-2I)^2e_1=\begin{pmatrix} 1&0&1\\ 2&0&2\\ -1&0&-1 \end{pmatrix} \begin{pmatrix} 1\\ 2\\ -1\\ \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$$ This gives us our desired 2-cycle. If we are not lucky enough to find a starting cycle vector in the standard basis, we can always force a starting cycle vector by looking at $N((A-\lambda I)^2)$.
Our change of basis matrix is then $$Q=\begin{pmatrix} 0&1&1\\ 1&2&0\\ 0&-1&0 \end{pmatrix}$$ To check that indeed $J=Q^{-1}AQ$ we could find $Q^{-1}$. $Q^{-1}$ exists since a generalized eigenbasis always consists of linearly independent vectors that make up the columns of $Q$, so $Q$ is invertible. However, we don't need to actually do this. If we simply multiply the above equation through by $Q$ on the left, then we get $QJ=AQ$. $$QJ=\begin{pmatrix} 0&1&1\\ 1&2&0\\ 0&-1&0 \end{pmatrix} \begin{pmatrix} 2&0&0\\ 0&2&1\\ 0&0&2 \end{pmatrix} = \begin{pmatrix} 0&2&3\\ 2&4&2\\ 0&-2&-1 \end{pmatrix}$$ $$AQ = \begin{pmatrix} 3&0&1\\ 2&2&2\\ -1&0&1 \end{pmatrix} \begin{pmatrix} 0&1&1\\ 1&2&0\\ 0&-1&0 \end{pmatrix} = \begin{pmatrix} 0&2&3\\ 2&4&2\\ 0&-2&-1 \end{pmatrix}$$