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 14: The Cayley-Hamilton Theorem

Let $A$ be a matrix in Jordan normal form. Let $m_A\left(t\right)$ be the minimal polynomial of $A\text{.}$ Let $c_A\left(t\right)$ be the characteristic polynomial of $A\text{.}$ $m_A\left(t\right)$ is minimal degree such that $m\left(A\right)=0\text{.}$ $c_A\left(t\right)=\text{det}(t-A)\text{.}$

If $$J=\left(\begin{array}{c}\lambda \\ 1& \lambda \\ & 1& \lambda \\ & & 1& \lambda \end{array}\right)$$ is a Jordan block of type ${(t-\lambda )}^4$ then $$\begin{array}{ccc}{\displaystyle J-\lambda }& =& {\displaystyle \left(\begin{array}{c}0\\ 1& 0\\ & 1& 0\\ & & 1& 0\end{array}\right),}\\ {\displaystyle {(J-\lambda )}^2}& =& {\displaystyle \left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right),}\\ {\displaystyle {(J-\lambda )}^3}& =& {\displaystyle \left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 0\end{array}\right)\text{and}}\\ {\displaystyle {(J-\lambda )}^4}& =& {\displaystyle 0\text{.}}\end{array}$$ If $$\begin{array}{ccc}{\displaystyle m_A\left(t\right)}& =& {\displaystyle {(t-{\lambda }_1)}^{m_1}{(t-{\lambda }_2)}^{m_2}\cdots {(t-{\lambda }_k)}^{m_k}\text{and}}\\ {\displaystyle c_A\left(t\right)}& =& {\displaystyle {(t-{\lambda }_1)}^{c_1}{(t-{\lambda }_2)}^{c_2}\cdots {(t-{\lambda }_k)}^{c_k}}\end{array}$$ then $m_j$ is the maximal size of a Jordan block with eigenvalue ${\lambda }_j$ in $A,$ and
$c_j$ is the sum of the sizes of the Jordan blocks of eigenvalue ${\lambda }_j$ in $A\text{.}$ So, finding the minimal polynomial and characteristic polynomial of a matrix in Jordan normal form is easy.

$$A=\left(\begin{array}{c}2\\ 1& 2\\ & & 2\\ & & & 2\\ & & & 1& 2\\ & & & & 1& 2\\ & & & & & & 3\\ & & & & & & 1& 3\\ & & & & & & & & 3\\ & & & & & & & & 1& 3\\ & & & & & & & & & 1& 3\\ & & & & & & & & & & 1& 3\end{array}\right)\text{.}$$ Then $$m\left(t\right)={(t-2)}^3{(t-3)}^4\text{and}c\left(t\right)={(t-2)}^6{(t-3)}^6$$ since the sizes of Jordan blocks of eigenvalue 2 are 2, 1, 3 (with maximum 3) and the sizes of Jordan blocks of eigenvalue $3$ are 2, 4 (with maximum 4).

Let $Q$ be an $n\times n$ matrix. The Jordan normal form theorem says that: There exists $P$ such that $$PQ{P}^{-1}=A$$ is in Jordan normal form.

Let us compare $m_Q\left(t\right)$ and $m_A\left(t\right)$ and compare $c_Q\left(t\right)$ and $c_A\left(t\right)\text{.}$ $$\begin{array}{ccc}{\displaystyle c_Q\left(t\right)}& =& {\displaystyle \text{det}(t-Q),}\\ {\displaystyle c_A\left(t\right)}& =& {\displaystyle \text{det}(t-A)}\\ & =& {\displaystyle \text{det}(t-PQ{P}^{-1})}\\ & =& {\displaystyle \text{det}(tP{P}^{-1}-PQ{P}^{-1})}\\ & =& {\displaystyle \text{det}\left(P(t-Q)P^{-1}\right)}\\ & =& {\displaystyle \text{det}\left(P\right)\text{det}(t-Q)\text{det}\left(P^{-1}\right)}\\ & =& {\displaystyle \text{det}(t-Q)}\\ & =& {\displaystyle c_Q\left(t\right)\text{.}}\end{array}$$ If $\ell \left(t\right)$ is a polynomial then $$\begin{array}{ccc}{\displaystyle \ell \left(t\right)}& =& {\displaystyle {\ell }_0+{\ell }_{1}t+{\ell }_2{t}^2+\cdots +{\ell }_d{t}^d,\text{and}}\\ {\displaystyle \ell \left(Q\right)}& =& {\displaystyle {\ell }_0+{\ell }_{1}Q+{\ell }_2{Q}^2+\cdots +{\ell }_d{Q}^d,\text{and}}\\ {\displaystyle \ell \left(A\right)}& =& {\displaystyle {\ell }_0+{\ell }_{1}A+{\ell }_2{A}^2+\cdots +{\ell }_d{A}^d}\\ & =& {\displaystyle {\ell }_0+{\ell }_{1}PQ{P}^{-1}+{\ell }_2{\left(PQ{P}^{-1}\right)}^2+\cdots +{\ell }_d{\left(PQ{P}^{-1}\right)}^d\text{.}}\end{array}$$ Since $$\begin{array}{ccc}{\displaystyle {\left(PQ{P}^{-1}\right)}^2}& =& {\displaystyle PQ{P}^{-1}PQ{P}^{-1}=P{Q}^2{P}^{-1},}\\ {\displaystyle {\left(PQ{P}^{-1}\right)}^2}& =& {\displaystyle PQ{P}^{-1}PQ{P}^{-1}PQ{P}^{-1}=P{Q}^3{P}^{-1},\dots }\end{array}$$ then $$\begin{array}{ccc}{\displaystyle \ell \left(A\right)}& =& {\displaystyle {\ell }_0+{\ell }_{1}PQ{P}^{-1}+{\ell }_{2}P{Q}^2{P}^{-1}+\cdots +{\ell }_{d}P{Q}^d{P}^{-1}}\\ & =& {\displaystyle P({\ell }_0+{\ell }_{1}Q+{\ell }_2{Q}^2+\cdots +{\ell }_d{Q}^d)P^{-1}}\\ & =& {\displaystyle P\ell \left(Q\right)P^{-1}}\end{array}$$ and $$\ell \left(Q\right)=P^{-1}\ell \left(A\right)P\text{.}$$ So if $\ell \left(Q\right)=0$ then $\ell \left(A\right)=0$ and if $\ell \left(A\right)=0$ then $\ell \left(Q\right)=0\text{.}$

$m_Q\left(t\right)$ is the smallest degree monic polynomial such that $m_Q\left(Q\right)=0\text{.}$

$m_A\left(t\right)$ is the smallest degree monic polynomial such that $m_A\left(A\right)=0\text{.}$

The point: If $PQ{P}^{-1}=A,$ then $m_Q\left(t\right)=m_A\left(t\right)\text{.}$
If $PQ{P}^{-1}=A,$ then $c_Q\left(t\right)=c_A\left(t\right)\text{.}$

The Cayley-Hamilton Theorem

Let $Q$ be an $n\times n$ matrix and let $$c_Q\left(t\right)=\text{det}(t-Q)$$ be the characteristic polynomial of $Q\text{.}$ Then $$c_Q\left(Q\right)=0\text{.}$$

Notes and References

These are a typed copy of Lecture 14 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 24, 2011.

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