How to find the basis of a kernel of a linear transformation in polynomials I understand transformations very well in vectors, but I am clueless when it comes to polynomials. I am trying to solve a question where the transformation equation for $T:P_2\to P_3$ is
$$T (a + bx + cx^2) = -(a + b + c) +(a - c)x + (b - c)x^2 + cx^3$$
I need to find the kernel of this transformation but I have no idea how to go about this. 
 A: Consider the bases $\mathscr{B}=\{1,x,x^2\}$ of $P_2$ and $\mathscr{C}=\{1,x,x^2,x^3\}$ of $P_3$. Since
\begin{align}
T(1)&=-1+x\\
T(x)&=-1+x^2\\
T(x^2)&=-1-x-x^2+x^3
\end{align}
the matrix of $T$ relative to these bases is
$$
\begin{bmatrix}
-1 & -1 & -1\\
1 & 0 & -1\\
0 & 1 & -1\\
0 & 0 & 1
\end{bmatrix}
$$
A standard Gaussian elimination gives the RREF
$$
\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{bmatrix}
$$
which shows the matrix has rank $3$, so the kernel is $\{0\}$.
A: What is the kernel? It's the set of things which get sent to $0$. So suppose a polynomial $a+bx+cx^2$ is in the kernel, so we have $T(a+bx+cx^2)=0$. Then by the way the linear transformation is defined,
$$-(a+b+c)+(a-c)x+(b-c)x^2+cx^3=0$$
and because $\{1,x,x^2,x^3\}$ is a basis of $P_3$ (in particular, it's linearly independent) this implies
$$-(a+b+c)=(a-c)=(b-c)=c=0$$
since $c=0$, this reduces to
$$-(a+b)=a=b=0$$
and therefore $a+bx+cx^2$ is the zero polynomial. Because our element of the kernel was arbitrary, every element of the kernel must be the zero polynomial, so $\ker(T)=\{0\}$.
