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 11: $\frac{ℂ\left[t\right]}{mℂ\left[t\right]}$ and multiplication by $t$

In the same way that $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$m\mathbb{Z}$}\right.$ is $\mathbb{Z}$ with $m=0$ $$\begin{array}{ccccccc}& & & 0& & & \\ & & 11& & 1& & \\ & 10& & & & 2& \\ 9& & & & & & 3\\ & 8& & & & 4& \\ & & 7& & 5& & \\ & & & 6& & & \end{array}$$ $\frac{ℂ\left[t\right]}{mℂ\left[t\right]}$ is $ℂ\left[t\right]$ with $m=0\text{.}$

Let $m={(t-2)}^2=t^2-4t+4\text{.}$ Then, in $\frac{ℂ\left[t\right]}{{(t-2)}^2ℂ\left[t\right]},$ $t-2-4t+4=0,$ $t^2=4t-4,$ and $$\begin{array}{ccc}{\displaystyle t^3+7t^2+5t+3}& =& {\displaystyle t·t^2+7t^2+5t+3}\\ & =& {\displaystyle t(4t-4)+7(4t-4)+5t+3}\\ & =& {\displaystyle 4t^2-4t+28t-28+5t+3}\\ & =& {\displaystyle 4(4t-4)-4t+28t+5t-25}\\ & =& {\displaystyle 16t-16+29t-25}\\ & =& {\displaystyle 45t-41\text{.}}\end{array}$$ Any polynomial in $\frac{ℂ\left[t\right]}{{(t-2)}^2ℂ\left[t\right]}$ is a linear combination of $1$ and $t\text{.}$ $$B=\{1,t\}\text{is a basis of}\frac{ℂ\left[t\right]}{{(t-2)}^2ℂ\left[t\right]}\text{.}$$ $$C=\{1,t-2\}\text{is another basis of}\frac{ℂ\left[t\right]}{{(t-2)}^2ℂ\left[t\right]},$$ and the change of basis matrix from $B$ to $C$ is $$P=\left(\begin{array}{cc}1& -2\\ 0& 1\end{array}\right)\text{.}$$ Let $V=\frac{ℂ\left[t\right]}{{(t-2)}^2ℂ\left[t\right]}$ and let $f:V\to V$ be the linear transformation given by $$f\left(p\right)=tp\text{.}$$ For example: $$\begin{array}{ccc}{\displaystyle f(45t-41)}& =& {\displaystyle t(45t-41)}\\ & =& {\displaystyle 45t^2-41}\\ & =& {\displaystyle 45(4t-4)-41}\\ & =& {\displaystyle 180t-180-41}\\ & =& {\displaystyle 180t-221\text{.}}\end{array}$$ The matrix of $f$ with respect to $B$ is $$B_f=\left(\begin{array}{cc}0& -4\\ 1& 4\end{array}\right)$$ since $f\left(t\right)=t^2=4t-4\text{.}$ The matrix of $f$ with respect to $C$ is $$C_f=\left(\begin{array}{cc}2& 0\\ 1& 2\end{array}\right)\text{.}$$

$V=\frac{ℂ\left[t\right]}{(t-2)(t-3)ℂ\left[t\right]}$ has $(t-2)(t-3)=0,$ so that $t^2-5t+6=0$ and $t^2=5t-6\text{.}$ $V$ has bases $$B=\{1,t\}$$ and $$C=\{t-2,-t+3\}\text{.}$$ Let $f:V\to V$ be the linear transformation given by $$f\left(p\right)=tp\text{.}$$ The matrix of $f$ with respect to $B$ is $$B_f=\left(\begin{array}{cc}0& -6\\ 1& 5\end{array}\right)\text{.}$$ The matrix of $f$ with respect to $C$ is $$C_f=\left(\begin{array}{cc}3& 0\\ 0& 2\end{array}\right)$$ since $$\begin{array}{ccc}{\displaystyle f(t-2)}& =& {\displaystyle t^2-2t=5t-6-2t=3t-6=3(t-2),}\\ {\displaystyle f(-t+3)}& =& {\displaystyle -t^2+3t=-(5t-6)+3t=-2t+6=2(-t+3)\text{.}}\end{array}$$

$$\begin{array}{ccc}{\displaystyle V}& =& {\displaystyle \frac{ℂ\left[t\right]}{(t-3)ℂ\left[t\right]}\oplus \frac{ℂ\left[t\right]}{(t-2)ℂ\left[t\right]}}\\ & =& {\displaystyle \left\{(u,w)\hspace{0.17em}\right|\hspace{0.17em}u\in \frac{ℂ\left[t\right]}{(t-3)ℂ\left[t\right]},w\in \frac{ℂ\left[t\right]}{(t-2)ℂ\left[t\right]}\}}\end{array}$$ and $$t·(u,w)=(tu,tw)\text{.}$$ $V$ has basis $$C=\{(1,0),(0,1)\}$$ with $$\begin{array}{ccc}{\displaystyle t(1,0)}& =& {\displaystyle (t,0)=(3,0)=3(1,0),}\\ {\displaystyle t(0,1)}& =& {\displaystyle (0,t)=(0,2)=2(0,1)}\end{array}$$ since $t=3$ in $\frac{ℂ\left[t\right]}{(t-3)ℂ\left[t\right]}$ and $t=2$ in $\frac{ℂ\left[t\right]}{(t-2)ℂ\left[t\right]}\text{.}$ So the matrix of $f:V\to V$ given by $$f(u,w)=t(u,w),$$ is $$C_f=\left(\begin{array}{cc}3& 0\\ 0& 2\end{array}\right)\text{.}$$ The matrix of $f$ with respect to the basis $$B=\{(1,1),(t,t)\}=\{(1,1),(3,2)\}$$ is $$B=\left(\begin{array}{cc}0& -6\\ 1& 5\end{array}\right)$$ since $$t(3,2)=(3t,2t)=(9,4)=-6(1,1)+5(3,2)\text{.}$$ The function $$\mathrm{\Phi }:\frac{ℂ\left[t\right]}{(t-2)(t-3)}⟶\frac{ℂ\left[t\right]}{(t-3)ℂ\left[t\right]}\oplus \frac{ℂ\left[t\right]}{(t-2)ℂ\left[t\right]}$$ given by $$\mathrm{\Phi }\left(1\right)=(1,1)\text{and}\mathrm{\Phi }\left(t\right)=(t,t)=(3,2)$$ is a linear transformation such that if $v\in \frac{ℂ\left[t\right]}{(t-2)(t-3)}$ then $\mathrm{\Phi }\left(tv\right)=t\mathrm{\Phi }\left(v\right)\text{.}$

HW: Show that $\mathrm{\Phi }$ is bijective.

Notes and References

These are a typed copy of Lecture 11 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 17, 2011.

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