## Group Theory and Linear Algebra

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 $ℤ}{mℤ}$ is $ℤ$ with $m=0$ $0 111 102 93 84 75 6$ $\frac{ℂ\left[t\right]}{mℂ\left[t\right]}$ is $ℂ\left[t\right]$ with $m=0\text{.}$

Let $m={\left(t-2\right)}^{2}={t}^{2}-4t+4\text{.}$ Then, in $\frac{ℂ\left[t\right]}{{\left(t-2\right)}^{2}ℂ\left[t\right]},$ $t-2-4t+4=0,$ ${t}^{2}=4t-4,$ and $t3+7t2+5t+3 = t·t2+7t2+5t+3 = t(4t-4)+7 (4t-4)+5t+3 = 4t2-4t+28t-28+5t+3 = 4(4t-4)-4t+28t +5t-25 = 16t-16+29t-25 = 45t-41.$ Any polynomial in $\frac{ℂ\left[t\right]}{{\left(t-2\right)}^{2}ℂ\left[t\right]}$ is a linear combination of $1$ and $t\text{.}$ $B={1,t}is a basis of ℂ[t](t-2)2ℂ[t].$ $C={1,t-2}is another basis of ℂ[t](t-2)2ℂ[t],$ and the change of basis matrix from $B$ to $C$ is $P= ( 1-2 01 ) .$ Let $V=\frac{ℂ\left[t\right]}{{\left(t-2\right)}^{2}ℂ\left[t\right]}$ and let $f:V\to V$ be the linear transformation given by $f(p)=tp.$ For example: $f(45t-41) = t(45t-41) = 45t2-41 = 45(4t-4)-41 = 180t-180-41 = 180t-221.$ The matrix of $f$ with respect to $B$ is $Bf= ( 0-4 14 )$ since $f\left(t\right)={t}^{2}=4t-4\text{.}$ The matrix of $f$ with respect to $C$ is $Cf= ( 20 12 ) .$

$V=\frac{ℂ\left[t\right]}{\left(t-2\right)\left(t-3\right)ℂ\left[t\right]}$ has $\left(t-2\right)\left(t-3\right)=0,$ so that ${t}^{2}-5t+6=0$ and ${t}^{2}=5t-6\text{.}$ $V$ has bases $B={1,t}$ and $C={t-2,-t+3}.$ Let $f:V\to V$ be the linear transformation given by $f(p)=tp.$ The matrix of $f$ with respect to $B$ is $Bf= ( 0-6 15 ) .$ The matrix of $f$ with respect to $C$ is $Cf= (3002)$ since $f(t-2) = t2-2t= 5t-6-2t= 3t-6=3(t-2), f(-t+3) = -t2 +3t=-(5t-6)+ 3t=-2t+6=2(-t+3).$

$V = ℂ[t](t-3)ℂ[t]⊕ ℂ[t](t-2)ℂ[t] = { (u,w) | u∈ℂ[t](t-3)ℂ[t], w∈ℂ[t](t-2)ℂ[t] }$ and $t·(u,w)=(tu,tw).$ $V$ has basis $C={(1,0),(0,1)}$ with $t(1,0) = (t,0)= (3,0)= 3(1,0), t(0,1) = (0,t)=(0,2) =2(0,1)$ since $t=3$ in $\frac{ℂ\left[t\right]}{\left(t-3\right)ℂ\left[t\right]}$ and $t=2$ in $\frac{ℂ\left[t\right]}{\left(t-2\right)ℂ\left[t\right]}\text{.}$ So the matrix of $f:V\to V$ given by $f(u,w)=t(u,w),$ is $Cf= (3002).$ The matrix of $f$ with respect to the basis $B={(1,1),(t,t)}= {(1,1),(3,2)}$ is $B=(0-615)$ since $t(3,2)= (3t,2t)= (9,4)= -6(1,1)+5(3,2).$ The function $Φ: ℂ[t](t-2)(t-3) ⟶ ℂ[t](t-3)ℂ[t]⊕ ℂ[t](t-2)ℂ[t]$ given by $Φ(1)=(1,1) andΦ(t)= (t,t)=(3,2)$ is a linear transformation such that if $v\in \frac{ℂ\left[t\right]}{\left(t-2\right)\left(t-3\right)}$ 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.