Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 21 September 2014

Lecture 12: Jordan Normal form

Let $A$ be an $n\times n$ matrix. If $I=\{i_1,\dots ,i_m\}$ is a subset of $\{1,2,\dots ,n\}$ and $J=\{j_1,\dots ,j_m\}$ another subset of $\{1,2,\dots ,n\}$ the $(I,J)$ minor of $t-A$ is $$\text{det}\left({(t-A)}_{IJ}\right)=\text{det}\left(\begin{array}{cccc}{(t-A)}_{i_1{j}_1}& {(t-A)}_{i_1{j}_2}& \cdots & {(t-A)}_{i_1{j}_m}\\ ⋮& \ddots & & ⋮\\ & & & \\ {(t-A)}_{i_m{j}_1}& {(t-A)}_{i_m{j}_2}& \cdots & {(t-A)}_{i_m{j}_m}\end{array}\right)\text{.}$$ If $I$ and $J$ have $m$ elements $\text{det}\left({(t-A)}_{IJ}\right)$ is an $m^{\text{th}}$ order minor of $t-A\text{.}$

The gcd of the $m^{\text{th}}$ order minors of $t-A$ is $$d_m\left(t\right)=\text{gcd}\left\{\text{det}{(t-A)}_{I_J}\hspace{0.17em}\right|\hspace{0.17em}\text{det}{(t-A)}_{IJ}\hspace{0.17em}\text{is an}\hspace{0.17em}{m}^{\text{th}}\hspace{0.17em}\text{order minor of}\hspace{0.17em}t-A\}\text{.}$$ The similarity invariants of $A,$ or invariant factors of $t-A,$ are monic polynomials $q_1\left(t\right),\dots ,q_n\left(t\right)$ such that $$d_m\left(t\right)=q_1\left(t\right)q_2\left(t\right)\cdots q_m\left(t\right),$$ for $m=1,2,\dots ,n$ (i.e. $q_m\left(t\right)=\frac{d_m\left(t\right)}{d_{m-1}\left(t\right)}\text{).}$

The characteristic polynomial of $A$ is $d_n\left(t\right)\text{.}$

The minimal polynomial of $A$ is $q_n\left(t\right)\text{.}$

A Jordan block of type ${(t-\lambda )}^e$ is the $e\times e$ matrix $$\left(\begin{array}{ccccc}\lambda & & & & \\ 1& \lambda & & & 0& \\ & 1& \lambda & & & \\ & & & \ddots & & \\ & 0& & & \lambda & \\ & & & & 1& \lambda \end{array}\right)\text{.}$$ Factor the $q_1\left(t\right),q_2\left(t\right),\dots ,q_n\left(t\right)\text{:}$ $$\begin{array}{ccc}{\displaystyle q_1\left(t\right)}& =& {\displaystyle {(t-{\lambda }_1)}^{e_{11}}{(t-{\lambda }_2)}^{e_{12}}\cdots {(t-{\lambda }_k)}^{e_{1k}}}\\ {\displaystyle q_2\left(t\right)}& =& {\displaystyle {(t-{\lambda }_1)}^{e_{21}}{(t-{\lambda }_2)}^{e_{22}}\cdots {(t-{\lambda }_k)}^{e_{2k}}}\\ {\displaystyle ⋮}& =& {\displaystyle }\\ {\displaystyle q_n\left(t\right)}& =& {\displaystyle {(t-{\lambda }_1)}^{e_{n1}}{(t-{\lambda }_2)}^{e_{n2}}\cdots {(t-{\lambda }_k)}^{e_{nk}}}\end{array}$$

There exists an invertible matrix $P$ such that $$PA{P}^{-1}=\left(\begin{array}{c}{J}_{11}\\ & J_{12}\\ & & \ddots & & & & & & & & 0\\ & & & J_{1k}\\ & & & & J_{21}\\ & & & & & J_{22}\\ & & & & & & \ddots \\ & & & & & & & \phantom{\ddots }\\ & & & & & & & & & \phantom{\ddots }\\ & & & & & & & & & \phantom{\ddots }\\ & & & 0& & & & & & & \phantom{\ddots }\\ & & & & & & & & & & & J_{n,k-1}\\ & & & & & & & & & & & & J_{n,k}\end{array}\right)$$ where $J_{m\ell }$ is a Jordan block of type ${(t-{\lambda }_{\ell })}^{e_{m\ell }}\text{.}$

If $$A=\left(\begin{array}{ccc}2& 1& 3\\ 5& 0& 1\\ 2& 6& 3\end{array}\right)$$ then $$t-A=\left(\begin{array}{ccc}t-2& 1& 3\\ 5& t& 1\\ 2& 6& t-3\end{array}\right)\text{.}$$

Subsets of $\{1,2,3\}$

If $I$ and $J$ have $3$ elements then $I=\{1,2,3\}=J$ and ${(t-A)}_{IJ}=t-A$ and $d_3=\text{det}(t-A)\text{.}$ So $$\begin{array}{ccc}{\displaystyle d_3\left(t\right)}& =& {\displaystyle \text{det}(t-A)=(t-2)(t^2-3t-6)-(5t-15-2)+3(30-2t)}\\ & =& {\displaystyle t^3-3t^2-6t-2t^2+6t+12-5t+17+90-6t}\\ & =& {\displaystyle t^3-5t^2-11t+119\text{.}}\end{array}$$ If $I$ and $J$ have $1$ element then ${(t-A)}_{IJ}$ is a single entry of $t-A\text{.}$ So $$d_1\left(t\right)=\text{gcd}\{t-2,1,3,5,t,1,2,6,t-3\}=1\text{.}$$ If $I$ and $J$ have $2$ elements then $$I\text{is}\{1,2\}\text{or}\{1,3\}\text{or}\{2,3\}\text{and}$$ $$J\text{is}\{1,2\}\text{or}\{1,3\}\text{or}\{2,3\}\text{and}$$ $$\begin{array}{ccc}{\displaystyle d_2\left(t\right)}& =& {\displaystyle \text{gcd}\left\{\begin{array}{ccc}\text{det}\left(\begin{array}{cc}t-2& 1\\ 5& t\end{array}\right),& \text{det}\left(\begin{array}{cc}t-2& 3\\ 5& 1\end{array}\right),& \text{det}\left(\begin{array}{cc}1& t\\ 3& 1\end{array}\right),\\ \text{det}\left(\begin{array}{cc}t-2& 1\\ 2& 6\end{array}\right),& \text{det}\left(\begin{array}{cc}t-2& 3\\ 2& t-3\end{array}\right),& \text{det}\left(\begin{array}{cc}1& 3\\ 6& t-3\end{array}\right),\\ \text{det}\left(\begin{array}{cc}5& t\\ 2& 6\end{array}\right),& \text{det}\left(\begin{array}{cc}5& 1\\ 2& t-3\end{array}\right),& \text{det}\left(\begin{array}{cc}t& 1\\ 6& t-3\end{array}\right)\end{array}\right\}}\\ & =& {\displaystyle \text{gcd}\{t^2-2t-5,t-10,1-3t,\dots ,5t-17,t^2-3t-6\}=1\text{.}}\end{array}$$ Wolfram alpha tells us that $$t^3-5t^2-11t+119=(t+4.23445)(t-(4.61723+2.60462i))(t-(4.61723-2.60462i))\text{.}$$ Since $d_1\left(t\right)=q_{1}t,$ $d_2\left(t\right)=q_1\left(t\right)q_2\left(t\right),$ $d_3\left(t\right)=q_1\left(t\right)q_2\left(t\right)q_3\left(t\right)$ and $$\begin{array}{ccc}{\displaystyle d_1\left(t\right)}& =& {\displaystyle 1,}\\ {\displaystyle d_2\left(t\right)}& =& {\displaystyle 1,}\\ {\displaystyle d_3\left(t\right)}& =& {\displaystyle t^3-5t^2-11t+119,}\end{array}$$ then $$\begin{array}{ccc}{\displaystyle q_1\left(t\right)}& =& {\displaystyle 1,}\\ {\displaystyle q_2\left(t\right)}& =& {\displaystyle 1,}\\ {\displaystyle q_1\left(t\right)}& =& {\displaystyle t^3-5t^2-11t+119}\end{array}$$ and the factorization of $q_1\left(t\right),$ $q_2\left(t\right)$ and $q_3\left(t\right)$ are $$\begin{array}{ccc}{\displaystyle q_1\left(t\right)}& =& {\displaystyle {(t-{\alpha }_1)}^0{(t-{\alpha }_2)}^0{(t-{\alpha }_3)}^0,}\\ {\displaystyle q_2\left(t\right)}& =& {\displaystyle {(t-{\alpha }_1)}^0{(t-{\alpha }_2)}^0{(t-{\alpha }_3)}^0,}\\ {\displaystyle q_3\left(t\right)}& =& {\displaystyle {(t-{\alpha }_1)}^1{(t-{\alpha }_2)}^1{(t-{\alpha }_3)}^1}\end{array}$$ where $$\begin{array}{ccc}{\displaystyle {\alpha }_1=4.23445,{\alpha }_2}& =& {\displaystyle 4.61723+i2.60462,}\\ {\displaystyle {\alpha }_3}& =& {\displaystyle 4.61723-i2.60462\text{.}}\end{array}$$ The Jordan normal form theorem then says that there exists $P$ such that $$PA{P}^{-1}=P\left(\begin{array}{ccc}2& 1& 3\\ 5& 0& 1\\ 2& 6& 3\end{array}\right)P^{-1}=\left(\begin{array}{ccc}{\alpha }_1& 0& 0\\ 0& {\alpha }_2& 0\\ 0& 0& a3\end{array}\right)$$ with ${\alpha }_1,{\alpha }_2,{\alpha }_3$ as above.

Notes and References

These are a typed copy of Lecture 12 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 19, 2011.

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