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 $QJ=AQ$. $$QJ=\begin{pmatrix} 0&1&1\\ 1&2&0\\ 0&-1&0 \end{pmatrix} \begin{pmatrix} 2&0&0\\ 0&2&1\\ 0&0&2 \end{pmatrix} = \begin{pmatrix} 0&2&3\\ 2&4&2\\ 0&-2&-1 \end{pmatrix}$$ $$AQ = \begin{pmatrix} 3&0&1\\ 2&2&2\\ -1&0&1 \end{pmatrix} \begin{pmatrix} 0&1&1\\ 1&2&0\\ 0&-1&0 \end{pmatrix} = \begin{pmatrix} 0&2&3\\ 2&4&2\\ 0&-2&-1 \end{pmatrix}$$

D+F 7.3.12

$\textbf{Exercise}[7.3.12]$ (Parts (a) and (b)) Let $D \in{\mathbb{Z}}$ that is not a perfect square and let $$S=\{\begin{pmatrix} a&b\\ Db&a \end{pmatrix} \mid a,b\in{\mathbb{Z}}\}$$
(a) Prove that $S$ is a subring of $M_{2 \times 2}(\mathbb{Z})$
(b) If $D$ is not a perfect square in $\mathbb{Z}$ prove that the map $\varphi: \mathbb{Z}[\sqrt{D}] \longrightarrow S$ defined by $$\varphi(a+b\sqrt{D})=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}$$ is a ring isomorphism.
$\textit{Proof:}$ To show that $S$ is a subring of $M_{2 \times 2}(\mathbb{Z})$, we need to show that $S \neq \emptyset $, $S$ is closed under subtraction, and $S$ is closed under multiplication. $S$ is clearly non-empty. For example, $\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} \in{S}$
Now let $A,B\in{S}$. $$A=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}, B=\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}$$ Then, $$A-B=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}-\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\begin{pmatrix} a-c&b-d\\ D(b-d)&a-c \end{pmatrix} \in{S}$$ The product, $$AB=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\begin{pmatrix} ac+bdD&ad+bc\\ cbD+adD&Dbd+ac \end{pmatrix}=\begin{pmatrix} ac+bdD&ad+bc\\ D(ad+bc)&ac+bdD \end{pmatrix}\in{S}$$ Therefore, $S$ is a subring.

We now need to show that $\varphi$ is well defined, a bijection, and a ring homomorphism. To check that $\varphi$ is well defined, we pick two equal representatives from $\mathbb{Z}[\sqrt{D}]$ and show that their images in $S$ are equal. So, let $x=a+b\sqrt{D}$ and $y=c+d\sqrt{D}$. Note that the assumption that $D$ is square-free is necessary. For example, let $D=4=2^2$. Consider $0+1 \cdot \sqrt{4}=2$ and $2+0\cdot \sqrt{4}=2$. These are 2 elements in $\mathbb{Z}[\sqrt{4}]$ that are equal, yet have a different representation. Then, $\varphi(0+1 \cdot \sqrt{4})=\begin{pmatrix} 0&1\\ 2&0 \end{pmatrix}$ however, $\varphi(1+0 \cdot \sqrt{4})=\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}$. This gives us 2 different matrices being mapped from the same element. So, if $D$ has a square in its factorization, then $\varphi$ is ill-defined.
Let $D$ be square-free. Now, $\varphi(x)=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}$ and $\varphi(y)=\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}$. Since $a,b,c,d\in{\mathbb{Z}}$ and $D$ is square-free, $x=y \Longleftrightarrow a=c$ and $b=d$ thus $\varphi(x)=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}=\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\varphi(y)$. Reversing this process, if $\varphi(x)=\varphi(y)$ then each entry in the above matrices must be equal, and we get that $a=c$ and $b=d$ so $\varphi$ is an injective map.
$\varphi$ is surjective: Let $M\in{S}$ where $M=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}$. Picking $x\in{\mathbb{Z}[\sqrt{D}]}$ where $x=a+b\sqrt{D}$, then $\varphi(x)=M$.
$\varphi$ is a ring homomorphism since it preserves both operations. Again letting $x=a+b\sqrt{D}$ and $y=c+d\sqrt{D}$, we see $\varphi$ preserves addition since: $$\varphi(x+y)=\varphi((a+b\sqrt{D})+(c+d\sqrt{D}))=\varphi((a+c)+(b+d)\sqrt{D})$$ $$=\begin{pmatrix} a+c&b+d\\ D(b+d)&a+c \end{pmatrix}=\begin{pmatrix} a&b\\ Db&a \end{pmatrix}+\begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\varphi(x)+\varphi(y)$$ $$\varphi(xy)=\varphi((a+b\sqrt{D})(c+d\sqrt{D}))=\varphi(ac+ad\sqrt{D}+bc\sqrt{D}+bdD)=\varphi(ac+(ad+bc)\sqrt{D}+bdD)$$ $$=\varphi(ac+bdD+(ad+bc)\sqrt{D})=\begin{pmatrix} ac+bdD&ad+bc\\ D(ad+bc)&ac+bdD \end{pmatrix}$$ $$\varphi(x)\varphi(y)=\begin{pmatrix} a&b\\ Db&a \end{pmatrix} \begin{pmatrix} c&d\\ Dd&c \end{pmatrix}=\begin{pmatrix} ac+bdD&ad+bc\\ D(ad+bc)&ac+bdD \end{pmatrix}=\varphi(xy)$$ Thus, $\varphi$ is a well-defined, bijective ring homomorphism i.e. a ring isomorphism.
$\therefore \quad \mathbb{Z}[\sqrt{D}] \cong S$ $\blacksquare$