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 10: Eigenvectors and annihilators

Let $V$ be a vector space over $ℂ\text{.}$ Let $f:V\to V$ be a linear transformation.

The $ℂ\left[t\right]\text{-module}$ defined by $f$ is the vector space $V$ with $ℂ\left[t\right]\text{-action}$ given by $$p·v=(a_0+a_{1}f+\cdots +a_N{f}^N)v$$ if $v\in V$ and $p=a_0+a_{1}t+a_2{t}^2+\cdots +a_N{t}^N\in ℂ\left[t\right]\text{.}$

$$V=\left\{\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\\ a_4\\ a_5\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{a}_i\in ℂ\}$$ has basis $$B=\{\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right),\}$$ and the matrix $$B_f=\left(\begin{array}{ccccc}2& 1& 0& 0& 0\\ 0& 2& 1& 0& 0\\ 0& 0& 2& 0& 0\\ 0& 0& 0& 5& 1\\ 0& 0& 0& 0& 5\end{array}\right)$$ defines $f:V\to V\text{.}$ Then $$\begin{array}{ccc}{\displaystyle (3t+6t^3)·\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\\ a_4\\ a_5\end{array}\right)}& =& {\displaystyle (3\left(\begin{array}{ccccc}2& 1& & & \\ & 2& 1& & \\ & & 2& & \\ & & & 5& 1\\ & & & & 5\end{array}\right)+6{\left(\begin{array}{ccccc}2& 1& & & \\ & 2& 1& & \\ & & 2& & \\ & & & 5& 1\\ & & & & 5\end{array}\right)}^3)\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\\ a_4\\ a_5\end{array}\right)}\\ & =& {\displaystyle \dots }\end{array}$$

An $f\text{-invariant}$ subspace, or $ℂ\left[t\right]\text{submodule,}$ of $V$ is a subspace $W\subseteq V$ such that if $w\in W$ then $fw\in W\text{.}$

Let $\lambda \in ℂ\text{.}$ The $\lambda \text{-eigenspace}$ of $f$ is $$V_{\lambda }=\{v\in V\hspace{0.17em}|\hspace{0.17em}fv=\lambda v\}\text{.}$$ An eigenvector with eigenvalue $\lambda$ is a vector $v\in V_{\lambda }\text{.}$

In our previous example, $$\left(\begin{array}{ccccc}2& 1& & & \\ & 2& 1& & \\ & & 2& & \\ & & & 5& 1\\ & & & & 5\end{array}\right)\left(\begin{array}{c}7\\ 0\\ 0\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}14\\ 0\\ 0\\ 0\\ 0\end{array}\right)=2·\left(\begin{array}{c}7\\ 0\\ 0\\ 0\\ 0\end{array}\right)\text{.}$$ So $$\left(\begin{array}{c}7\\ 0\\ 0\\ 0\\ 0\end{array}\right)\in V_2\text{.}$$ $$\left(\begin{array}{ccccc}2& 1& & & \\ & 2& 1& & \\ & & 2& & \\ & & & 5& 1\\ & & & & 5\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 5\\ 0\end{array}\right)=5·\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)\text{.}$$ So $$\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)\in V_5\text{.}$$ $$\left(\begin{array}{ccccc}2& 1& & & \\ & 2& 1& & \\ & & 2& & \\ & & & 5& 1\\ & & & & 5\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ 2\\ 0\\ 0\end{array}\right)\ne 2·\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right)\text{.}$$ So $$\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right)\notin V_2\text{.}$$

The $\lambda \text{-generalized}$ eigenspace of $V$ is $$V_{\lambda }^{\text{gen}}=\{v\in V\hspace{0.17em}|\hspace{0.17em}\text{there exists}\hspace{0.17em}k\in {\mathbb{Z}}_{>0}\hspace{0.17em}\text{with}\hspace{0.17em}{(f-\lambda )}^{k}v=0\}\text{.}$$ In our previous example, $$\begin{array}{ccc}{\displaystyle f-2}& =& {\displaystyle \left(\begin{array}{ccccc}2& 1& & & \\ & 2& 1& & \\ & & 2& & \\ & & & 5& 1\\ & & & & 5\end{array}\right)-2\left(\begin{array}{ccccc}1& & & & \\ & 1& & & \\ & & 1& & \\ & & & 1& \\ & & & & 1\end{array}\right)=\left(\begin{array}{ccccc}0& 1& & & \\ & 0& 1& & \\ & & 0& & \\ & & & 3& 1\\ & & & & 3\end{array}\right),}\\ {\displaystyle {(f-2)}^2}& =& {\displaystyle \left(\begin{array}{ccccc}0& 0& 1& & \\ & 0& 0& & \\ & & 0& & \\ & & & 9& 6\\ & & & & 9\end{array}\right),}\\ {\displaystyle {(f-2)}^3}& =& {\displaystyle \left(\begin{array}{ccccc}0& 0& 0& & \\ & 0& 0& & \\ & & 0& & \\ & & & 81& 108\\ & & & & 81\end{array}\right)}\end{array}$$ and $${(f-2)}^3\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}\right)$$ if $a_1,a_2,a_3\in ℂ\text{.}$ So $$V_2^{\text{gen}}\supseteq \left\{\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\\ 0\\ 0\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{a}_1,a_2,a_3\in ℂ\}\text{.}$$

The annihilator of $V$ is $$\text{ann}\hspace{0.17em}V=\{p\in ℂ\left[t\right]\hspace{0.17em}|\hspace{0.17em}\text{if}\hspace{0.17em}v\in V\hspace{0.17em}\text{then}\hspace{0.17em}pv=0\}\text{.}$$ The minimal polynomial of $f$ is $m\in ℂ\left[t\right]$ such that $$\text{ann}\hspace{0.17em}V=mℂ\left[t\right]$$ where $mℂ\left[t\right]=\left\{mq\hspace{0.17em}\right|\hspace{0.17em}q\in ℂ\left[t\right]\}=\left\{\text{multiples of}\hspace{0.17em}m\right\}\text{.}$

In our previous example, $$\begin{array}{ccc}{\displaystyle {(f-2)}^3}& =& {\displaystyle \left(\begin{array}{ccccc}0& 0& 0& & \\ & 0& 0& & \\ & & 0& & \\ & & & 81& 108\\ & & & & 81\end{array}\right),}\\ {\displaystyle {(f-5)}^2}& =& {\displaystyle \left(\begin{array}{ccccc}9& -6& 1& & \\ & 9& -6& & \\ & & 9& & \\ & & & 0& 0\\ & & & 0& 0\end{array}\right)}\end{array}$$ and $${(f-2)}^3{(f-5)}^2=\left(\begin{array}{ccccc}0& 0& 0& & \\ 0& 0& 0& & \\ 0& 0& 0& & \\ & & & 0& 0\\ & & & 0& 0\end{array}\right)\text{.}$$ So ${(t-2)}^3{(t-5)}^2\in \text{ann}\hspace{0.17em}V\text{.}$

Let $f:V\to V$ be a linear transformation.

(a)	Let $\lambda \in ℂ\text{.}$ Then $V_{\lambda }$ is an $f\text{-invariant}$ subspace of $V\text{.}$
(b)	Let $\lambda \in ℂ\text{.}$ Then $V_{\lambda }^{\text{gen}}$ is an $f\text{-invariant}$ subspace of $V\text{.}$
(c)	If $p_1,p_2\in \text{ann}\hspace{0.17em}V$ then $p_1+p_2\in \text{ann}\hspace{0.17em}V\text{.}$
(d)	If $p\in \text{ann}\hspace{0.17em}V$ and $q\in ℂ\left[t\right]$ then $pq\in \text{ann}\hspace{0.17em}V\text{.}$

		Proof.
	(a) To show: $V_{\lambda }$ is an $f\text{-invariant}$ subspace of $V\text{.}$ To show: (aa) If $v_1,v_2\in V_{\lambda }$ then $v_1+v_2\in V_{\lambda }\text{.}$ (ab) If $c\in ℂ$ and $c\in V_{\lambda }$ then $cv\in V_{\lambda }\text{.}$ (ac) If $v\in V_{\lambda }$ then $fv\in V_{\lambda }\text{.}$ (aa) Assume $v_1,v_2\in V_{\lambda }\text{.}$ To show: $v_1+v_2\in V_{\lambda }\text{.}$ We know: $V_{\lambda }=\{v\in V\hspace{0.17em}|\hspace{0.17em}fv=\lambda v\}\text{.}$ So we know: $f(v_1+v_2)=\lambda (v_1+v_2)\text{.}$ $$f(v_1+v_2)=f{v}_1+f{v}_2=\lambda v_1+\lambda v_2=\lambda (v_1+v_2)\text{.}$$ (ab) Assume $c\in ℂ$ and $v\in V_{\lambda }\text{.}$ To show: $cv\in V_{\lambda }\text{.}$ To show: $f\left(cv\right)=\lambda cv\text{.}$ We know: $fv=\lambda v\text{.}$ $$\begin{array}{ccc}{\displaystyle f\left(cv\right)}& =& {\displaystyle cf\left(v\right),\text{since}\hspace{0.17em}f\hspace{0.17em}\text{is a linear transformation,}}\\ & =& {\displaystyle c\lambda v}\\ & =& {\displaystyle \lambda cv,\text{since}\hspace{0.17em}ℂ\hspace{0.17em}\text{is commutative.}}\end{array}$$ (ac) Assume $v\in V_{\lambda }\text{.}$ To show: $fv\in V_{\lambda }\text{.}$ We know: $fv=\lambda v\text{.}$ To show: $f\left(fv\right)=\lambda fv\text{.}$ $$ffv=f\left(\lambda v\right)=\lambda f\left(v\right),\text{since}\hspace{0.17em}f\hspace{0.17em}\text{is a linear transformation.}$$ So $V_{\lambda }$ is an $f\text{-invariant}$ subspace of $V\text{.}$ (d) Assume $p\in \text{ann}\hspace{0.17em}V$ and $q\in ℂ\left[t\right]\text{.}$ To show: $pq\in \text{ann}\hspace{0.17em}V\text{.}$ We know: $\text{ann}\hspace{0.17em}V=\{p\in ℂ\left[t\right]\hspace{0.17em}|\hspace{0.17em}\text{if}\hspace{0.17em}v\in V\hspace{0.17em}\text{then}\hspace{0.17em}pv=0\}\text{.}$ To show: If $v\in V$ then $pqv=0\text{.}$ Assume $v\in V\text{.}$ To show: $pqv=0\text{.}$ $$\begin{array}{ccc}{\displaystyle pqv}& =& {\displaystyle qpv,\text{since}\hspace{0.17em}ℂ\left[t\right]\hspace{0.17em}\text{is commutative,}}\\ & =& {\displaystyle q·0=0\text{.}}\end{array}$$ (c) Assume $p_1,p_2\in \text{ann}\hspace{0.17em}V\text{.}$ To show: $p_1+p_2\in \text{ann}\hspace{0.17em}V\text{.}$ To show: If $v\in V$ then $(p_1+p_2)v=0\text{.}$ Assume $v\in V\text{.}$ To show: $(p_1+p_2)v=0\text{.}$ $$\begin{array}{ccc}{\displaystyle (p_1+p_2)v}& =& {\displaystyle p_{1}v+p_{2}v}\\ & =& {\displaystyle 0+p_{2}v,\text{since}\hspace{0.17em}{p}_1\in \text{ann}\hspace{0.17em}V}\\ & =& {\displaystyle 0+0,\text{since}\hspace{0.17em}{p}_2\in \text{ann}\hspace{0.17em}V}\\ & =& {\displaystyle 0\text{.}}\end{array}$$ $\square$

Notes and References

These are a typed copy of Lecture 10 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 16, 2011.

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