# Linear Transformation of polynomials

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?

• 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. Feb 11, 2013 at 18:35