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Let $V= \Bbb{R}_2[x]=\{P(x)=a_0+a_1x+a_2x^2 \mid a_0,a_1,a_2 \in \Bbb{R}\}$ and let $f: V \to \Bbb{R[x]} $ defined by $f(P(x))=xP(x)- \frac 1 2x^2P'(x)$

  1. Find the matrix of $f$ in the basis $\{1,x,x^2\}$.

  2. Find $\ker(f)$ and $Im(f)$.

For part one 1 just took the linear transformation of each vector of the basis and put them in columns to get $$ \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & \frac 1 2 & 0 \end{array} \right] $$

My problem is with part 2, as this is the first time I encounter linear algebra with polynomials I have no idea what $$ \left[ \begin{array}{ccc} 1 & 0 & 0 | & 0\\ 0 & \frac 1 2 & 0| & 0\\ 0 & 0 & 0 | & 0\end{array} \right] $$ means, neither is it clear to me what $$ \left[ \begin{array}{ccc} 1 & 0 & 0 | & a\\ 0 & \frac 1 2 & 0| & b\\ 0 & 0 & 0 | & c\end{array} \right] $$ means. Can anyone explain to me clearly what the solutions of these systems represent and how do I find the kernal and image.

Edit I read the comments and found that the kernel of the matrix is $\ker(f)=\{(0,0,t) \mid t \in \Bbb{R}\}$ but in the image I have problem in the last equation where I get $0z= c$ how should I proceed? Do I take two cases?

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On 2. you are looking for $P(x)$ s.t. $2xP(x)=x^2P'(x)$. But don't forget that your space contains only degree 2 polynomials. – Sigur Feb 11 '13 at 18:35
You have calculated your first matrix correctly. Now note that the first two columns span the image, and the the third basis vector (corresponding to $x^2) is in the kernel. – Andreas Caranti Feb 11 '13 at 18:42
How do you go back from the columns of the (correct) matrix in part 1 to your original $V$? Permuting rows of the matrix without permuting the given ordered basis will give you a different linear transformation. – Barbara Osofsky Feb 11 '13 at 18:55
up vote 1 down vote accepted

First find kernel and image of your matrix $A$. Then remember that $A$ represents your linear mapping $f$ with respect to the basis $\{1,x,x^2\}$. Thus, you can translate the kernel (respectively image) of $A$ to the kernel (resp. image) of $f$.

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Please review my edit concerning the image part. – user10444 Feb 11 '13 at 18:58
Your kernel is correct. So -- translating it back using the basis $\{1,x,x^2\}$ -- what polynomials are in the kernel? Regarding the image: You should have learned some method to compute the image of a matrix. Apply it. – azimut Feb 11 '13 at 19:02
I found the image to be $(a,2b,0)$ then the kernel is the set of polynomials of the form $tx^2$ and the image is the set of the polynomials of the form $a+2bx$ , is this true? Thank you – user10444 Feb 11 '13 at 19:26
Yes, this is correct. You still can make the result a little bit nicer: By substituting $b' = 2b$ you can remove the $2$ in the expressions $2b$. BTW, if this answer was helpful, feel free to upvote. – azimut Feb 11 '13 at 19:29

The whole point of choosing a basis so you can write a matrix is so that the origins of the problem don't matter. You have algorithms for finding the image and kernel of a matrix -- i.e. the column space and the (right) nullspace -- these algorithms don't care where you got the matrix from, they simply work on the matrix. Once you have those, you can use your basis again to convert the result back to polynomials.

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