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Let V be $ C[x]/(x^3+5x^2+6x+2) $ and let $ T:V→V $ be defined by $T(p(x))=(x+1)p(x) $. I need to find the bases for the kernel (null space) and image (range) of T.

I have to write T in matrix form right? After that we can easily find the bases. How can we write it in matrix form? Also one more puzzling thing is that do we take $p(x)=x^3+5x^2+6x+2$.

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I have the beginnings of the answer you are looking for below. But it is worth noting that your polynomial factors as $(x+1)(x^2+4x+2)$. This observation makes the answer to the kernel question apparent, if you have the right theorems available. –  alex.jordan Feb 24 '12 at 23:06

2 Answers 2

up vote 1 down vote accepted

A basis for your $V$ is $\left\{\vec{b}_1,\vec{b}_2,\vec{b}_3\right\}=\left\{1,x+1,(x+1)^2\right\}$.

$T$ takes $\vec{b}_1$ to $\vec{b}_2$, and $\vec{b}_2$ to $\vec{b}_3$.

$T$ takes $\vec{b}_3$ to $(x+1)^3=x^3+3x^2+3x+1\equiv-2x^2-3x-1=-2(x+1)^2+(x+1)$.

So with respect to the given basis, $T$ has the matrix $$\begin{bmatrix}0&0&0\\1&0&1\\0&1&-2\end{bmatrix}$$ That should be a good start.

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thanks for your answer. Don't we need to use $ x^3+5x^2+6x+2 $ –  rose Feb 24 '12 at 23:10
    
Hi rose, I used that when I reduced $x^3+3x^2+3x+1≡−2x^2−3x−1$. –  alex.jordan Feb 24 '12 at 23:22
    
yeah i misread it. now i totally get it. thanks –  rose Feb 24 '12 at 23:23
    
This shows up in the step $x^3+3x^2+3x+1 \equiv -2x^2-3x-1$ –  Brett Frankel Feb 24 '12 at 23:23

Here's a more naive approach, for when you don't see how the kernel factors: Using $\{1,x,x^2\}$ as an ordered basis, $1$ maps to $1+x$, $x$ maps to $x+x^2$, and $x^2$ maps to $x^2+x^3=x^2+x^3-x^3-5x^2-6x-2=-4x^2-6x-2$, so the matrix is: $$\begin{bmatrix} 1 & 0 & -2 \\ 1 & 1 & -6 \\ 0 & 1 & -4 \end{bmatrix}$$ The image of a matrix is always the span of the columns, in this case those polynomials $a+bx+cx^2$ with $b=c+a$ and the kernel is found by solving $$\begin{bmatrix} 1 & 0 & -2 \\ 1 & 1 & -6 \\ 0 & 1 & -4 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ So $a+bx+cx^2$ maps to $0$ if and only if $a=2c$ and $b=4c$

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Of course, you should note that the kernel is the span of $(2+4x+x^2)$, and as alex.jordan pointed out, $(2+4x+x^2)(1+x)\equiv0$. –  Brett Frankel Feb 24 '12 at 23:15
    
i got it. Thank you so much Alex and Brett –  rose Feb 24 '12 at 23:21

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