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: 27 September 2014

Lecture 36: Revision: $ℂ\left[t\right]\text{-modules}$

(1) Let $$A=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 9& 8& 7\end{array}\right)\text{.}$$ $A$ acts on $${ℂ}^3=\left\{\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{a}_1,a_2,a_3\in ℂ\}\text{.}$$ ${ℂ}^3$ has basis $\{e_1,e_2,e_3\}$ where $e_1=\left(1\\ 0\\ 0\\ \right),$ $e_2=\left(0\\ 1\\ 0\\ \right),$ $e_3=\left(0\\ 0\\ 1\\ \right)\text{.}$ Define an action of $ℂ\left[t\right]$ on ${ℂ}^3$ by, for example, $$(3+4t+5t^2)\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)=(3+4A+5A^2)\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)\text{.}$$ This is the $ℂ\left[t\right]\text{-module}$ defined by $A\text{.}$

(2) Let $$ℂ{\left[t\right]}^{\oplus 3}=\left\{\left(\begin{array}{c}{g}_1\\ g_2\\ g_3\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{g}_1,g_2,g_3\in ℂ\left[t\right]\}$$ so that, for example, $$\left(\begin{array}{c}3t+7\\ t^3-2t+1\\ 5t^2+t^6\end{array}\right)\in ℂ{\left[t\right]}^{\oplus 3}\text{and}\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)\in ℂ{\left[t\right]}^{\oplus 3}\text{.}$$ $ℂ\left[t\right]$ acts on $ℂ{\left[t\right]}^{\oplus 3}$ by, for example, $$(3+4t+5t^2)\left(\begin{array}{c}3\\ 1\\ 2\end{array}\right)=\left(\begin{array}{c}9+12t+15t^2\\ 3+4t+5t^2\\ 6+8t+10t^2\end{array}\right)\text{.}$$ $ℂ{\left[t\right]}^{\oplus 3}$ has basis: $$\{\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right),\left(\begin{array}{c}t\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ t\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ t\end{array}\right),\left(\begin{array}{c}{t}^2\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ t^2\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ t^2\end{array}\right),\dots \}\text{.}$$ $ℂ{\left[t\right]}^{\oplus 3}$ is infinite dimensional. $ℂ{\left[t\right]}^{\oplus 3}$ is the free $ℂ\left[t\right]\text{-module}$ generated by $\{\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\}\text{.}$ A $ℂ\left[t\right]\text{-module}$ is a vector space $V$ over $ℂ$ with a function $$\begin{array}{ccc}ℂ\left[t\right]\times V& ⟶& V\\ (g,v)& ⟼& gv\end{array}$$ such that

(a)	If $g_1,g_2\in ℂ\left[t\right]$ and $v\in V$ then $\left(g_1{g}_2\right)v=g_1\left(g_{2}v\right),$
(b)	If $v\in V$ then $1·v=v,$
(c)	If $c_1,v_2\in ℂ,$ $g_1,g_2\in ℂ\left[t\right]$ and $v\in V$ then $$(c_1{g}_1+c_2{g}_2)v=c_1{g}_{1}v+c_2{g}_{2}v,$$
(d)	If $c_1,c_2\in ℂ,$ $g\in ℂ\left[t\right]$ and $v_1,v_2\in V$ then $$g\left(c_1{v}_1+c_2{v}_2\right)=c_{1}g{v}_1+c_{2}g{v}_2\text{.}$$

A $ℂ\left[t\right]\text{-module}$ homomorphism from $V$ to $W$ is a linear transformation $f:V\to W$ such that if $g\in ℂ\left[t\right]$ and $v\in V$ then $$f\left(gv\right)=gf\left(v\right)\text{.}$$

The kernel of $f$ is $$\text{ker}\hspace{0.17em}f=\{v\in V\hspace{0.17em}|\hspace{0.17em}f\left(v\right)=0\}$$ and the image of $f$ is $$\text{im}\hspace{0.17em}f=\left\{f\left(v\right)\hspace{0.17em}\right|\hspace{0.17em}v\in V\}\text{.}$$

(3) The map $$\begin{array}{cccc}\mathrm{\Phi }:& ℂ{\left[t\right]}^{\oplus 3}& ⟶& {ℂ}^3\\ & \left(\begin{array}{c}{g}_1\\ g_2\\ g_3\end{array}\right)& ⟼& g_1\left(A\right)\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+g_2\left(A\right)\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)+g_3\left(A\right)\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\end{array}$$ is a $ℂ\left[t\right]\text{-module}$ homomorphism.

$\text{ker}\hspace{0.17em}\mathrm{\Phi }=(t-A)ℂ{\left[t\right]}^{\oplus 3}\text{.}$

More specifically, $$\begin{array}{rcl}{\displaystyle t-A}& =& {\displaystyle \left(\begin{array}{ccc}t-1& -2& -3\\ -4& t-5& -6\\ -9& -8& t-7\end{array}\right)\text{and}}\\ {\displaystyle \text{ker}\hspace{0.17em}\mathrm{\Phi }}& =& {\displaystyle \left\{\left(\begin{array}{ccc}t-1& -2& -3\\ -4& t-5& -6\\ -9& -8& t-7\end{array}\right)\left(\begin{array}{c}{g}_1\left(t\right)\\ g_2\left(t\right)\\ g_3\left(t\right)\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{g}_1\left(t\right),g_2\left(t\right),g_3\left(t\right)\}\text{.}}\end{array}$$

Consequence (1st homomorphism theorem): $${ℂ}^3\simeq \frac{ℂ{\left[t\right]}^{\oplus 3}}{(t-A)ℂ{\left[t\right]}^{\oplus 3}}$$ as $ℂ\left[t\right]\text{-modules.}$

POINT: $A$ acting in ${ℂ}^3$ is very similar to a "clock" $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$12\mathbb{Z}$}\right.\text{.}$

Use row reduction to find invertible matrices $L$ and $R$ (with entries in $ℂ\left[t\right]\text{)}$ such that $L(t-A)R$ is diagonal.

$A=\left(\begin{array}{cc}3& -7\\ 1& 5\end{array}\right)\text{.}$ Then $$(t-A)=\left(\begin{array}{cc}t-3& 7\\ -1& t-5\end{array}\right)$$ and $$\begin{array}{rcl}{\displaystyle \left(\begin{array}{cc}t-3& 7\\ -1& t-5\end{array}\right)}& ⟶& {\displaystyle \left(\begin{array}{cc}-1& t-5\\ t-3& 7\end{array}\right)}\\ & ⟶& {\displaystyle \left(\begin{array}{cc}1& -(t-5)\\ t-3& 7\end{array}\right)}\\ & ⟶& {\displaystyle \left(\begin{array}{cc}1& t-5\\ 0& 7+(t-5)(t-3)\end{array}\right)}\\ & ⟶& {\displaystyle \left(\begin{array}{cc}1& t-5\\ 0& t^2-8t+22\end{array}\right)}\\ & ⟶& {\displaystyle \left(\begin{array}{cc}1& 0\\ 0& t^2-8t+22\end{array}\right)\text{.}}\end{array}$$ $t^2-8t+22=(t-\lambda )(t-\mu )$ where $$\{\mu ,\lambda \}=\left\{\frac{8\pm \sqrt{64-44}}{2}\right\}=\left\{\frac{8\pm \sqrt{20}}{2}\right\}=\{4\pm \sqrt{5}\}$$ and $$\begin{array}{rcl}{\displaystyle L}& =& {\displaystyle \left(\begin{array}{cc}1& -(t-3)\\ 0& 1\end{array}\right)\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}-(t-3)& -1\\ 1& 0\end{array}\right),}\\ {\displaystyle R}& =& {\displaystyle \left(\begin{array}{cc}1& -(t-5)\\ 0& 1\end{array}\right)\text{.}}\end{array}$$ So $$L(t-A)R=L\left(\begin{array}{cc}t-3& 7\\ -1& t-5\end{array}\right)R=\left(\begin{array}{cc}1& 0\\ 0& t^2-8t+22\end{array}\right)$$ and $$\begin{array}{ccccc}{\displaystyle \frac{ℂ{\left[t\right]}^{\oplus 2}}{(t-A)ℂ{\left[t\right]}^{\oplus 2}}}& ⥲& {\displaystyle \frac{Lℂ{\left[t\right]}^{\oplus 2}}{L(t-A)ℂ{\left[t\right]}^{\oplus 2}}}& =& {\displaystyle \frac{ℂ{\left[t\right]}^{\oplus 2}}{L(t-A)Rℂ{\left[t\right]}^{\oplus 2}}}\\ \left(\begin{array}{c}{g}_1\\ g_2\end{array}\right)& ⟼& L\left(\begin{array}{c}{g}_1\\ g_2\end{array}\right)\end{array}$$ and $$\frac{ℂ{\left[t\right]}^{\oplus 2}}{\left(\begin{array}{cc}1& 0\\ 0& t^2-8t+22\end{array}\right)ℂ{\left[t\right]}^{\oplus 2}}=\frac{ℂ\left[t\right]}{ℂ\left[t\right]}\oplus \frac{ℂ\left[t\right]}{(t^2-8t+22)ℂ\left[t\right]}=\frac{ℂ\left[t\right]}{(t^2-8t+22)ℂ\left[t\right]}\text{.}$$ $\frac{ℂ\left[t\right]}{(t^2-8t+22)ℂ\left[t\right]}$ has basis $\{1,t\}$ with $t\text{-action}$ on this basis given by the matrix $\left(\begin{array}{cc}0& -22\\ 1& 8\end{array}\right)\text{.}$

Notes and References

These are a typed copy of Lecture 36 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 28, 2011.

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