Consider the kernel of the linear transformation, $\ltdefn{T}{M_{22}}{P_2}$, given by \begin{equation*} \lt{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}= (2a-b+c-5d)+(a+4b+5c+2d)x+(3a-2b+c-8d)x^2 \end{equation*}
We will begin with a matrix representation of $T$ relative to the bases for $M_{22}$ and $P_2$ (respectively), \begin{align*} B&=\set{ \begin{bmatrix}1 & 2 \\ -1 & -1\end{bmatrix},\, \begin{bmatrix}1 & 3 \\ -1 & -4\end{bmatrix},\, \begin{bmatrix}1 & 2 \\ 0 & -2\end{bmatrix},\, \begin{bmatrix}2 & 5 \\ -2 & -4\end{bmatrix} }\\ C&=\set{1+x+x^2,\,2+3x,\,-1-2x^2} \end{align*}
Then, \begin{align*} \vectrep{C}{\lt{T}{\begin{bmatrix}1 & 2 \\ -1 & -1\end{bmatrix}}} &=\vectrep{C}{4+2x+6x^2}\\ &=\vectrep{C}{2(1+x+x^2)+0(2+3x)+(-2)(-1-2x^2)}\\ &=\colvector{2\\0\\-2}\\ \vectrep{C}{\lt{T}{\begin{bmatrix}1 & 3 \\ -1 & -4\end{bmatrix}}} &=\vectrep{C}{18+28x^2}\\ &=\vectrep{C}{(-24)(1+x+x^2)+8(2+3x)+(-26)(-1-2x^2)}\\ &=\colvector{-24\\8\\-26}\\ \vectrep{C}{\lt{T}{\begin{bmatrix}1 & 2 \\ 0 & -2\end{bmatrix}}} &=\vectrep{C}{10+5x+15x^2}\\ &=\vectrep{C}{5(1+x+x^2)+0(2+3x)+(-5)(-1-2x^2)}\\ &=\colvector{5\\0\\-5}\\ \vectrep{C}{\lt{T}{\begin{bmatrix}2 & 5 \\ -2 & -4\end{bmatrix}}} &=\vectrep{C}{17+4x+26x^2}\\ &=\vectrep{C}{(-8)(1+x+x^2)+(4)(2+3x)+(-17)(-1-2x^2)}\\ &=\colvector{-8\\4\\-17} \end{align*}
So the matrix representation of $T$ (relative to $B$ and $C$) is \begin{equation*} \matrixrep{T}{B}{C} = \begin{bmatrix} 2 & -24 & 5 & -8 \\ 0 & 8 & 0 & 4 \\ -2 & -26 & -5 & -17 \end{bmatrix} \end{equation*}
We know from Theorem KNSI that the kernel of the linear transformation $T$ is isomorphic to the null space of the matrix representation $\matrixrep{T}{B}{C}$ and by studying the proof of Theorem KNSI we learn that $\vectrepname{B}$ is an isomorphism between these null spaces. Rather than trying to compute the kernel of $T$ using definitions and techniques from Chapter LT we will instead analyze the null space of $\matrixrep{T}{B}{C}$ using techniques from way back in Chapter V. First row-reduce $\matrixrep{T}{B}{C}$, \begin{equation*} \begin{bmatrix} 2 & -24 & 5 & -8 \\ 0 & 8 & 0 & 4 \\ -2 & -26 & -5 & -17 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & \frac{5}{2} & 2 \\ 0 & \leading{1} & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
So, by Theorem BNS, a basis for $\nsp{\matrixrep{T}{B}{C}}$ is \begin{equation*} \set{\colvector{-\frac{5}{2}\\0\\1\\0},\,\colvector{-2\\-\frac{1}{2}\\0\\1}} \end{equation*}
We can now convert this basis of $\nsp{\matrixrep{T}{B}{C}}$ into a basis of $\krn{T}$ by applying $\vectrepinvname{B}$ to each element of the basis, \begin{align*} \vectrepinv{B}{\colvector{-\frac{5}{2}\\0\\1\\0}}&= (-\frac{5}{2})\begin{bmatrix}1 & 2 \\ -1 & -1\end{bmatrix}+ 0\begin{bmatrix}1 & 3 \\ -1 & -4\end{bmatrix}+ 1\begin{bmatrix}1 & 2 \\ 0 & -2\end{bmatrix}+ 0\begin{bmatrix}2 & 5 \\ -2 & -4\end{bmatrix}\\ &=\begin{bmatrix}-\frac{3}{2} & -3 \\ \frac{5}{2} & \frac{1}{2}\end{bmatrix}\\ \vectrepinv{B}{\colvector{-2\\-\frac{1}{2}\\0\\1}}&= (-2)\begin{bmatrix}1 & 2 \\ -1 & -1\end{bmatrix}+ (-\frac{1}{2})\begin{bmatrix}1 & 3 \\ -1 & -4\end{bmatrix}+ 0\begin{bmatrix}1 & 2 \\ 0 & -2\end{bmatrix}+ 1\begin{bmatrix}2 & 5 \\ -2 & -4\end{bmatrix}\\ &=\begin{bmatrix}-\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} &0\end{bmatrix} \end{align*}
So the set \begin{equation*} \set{ \begin{bmatrix}-\frac{3}{2} & -3 \\ \frac{5}{2} & \frac{1}{2}\end{bmatrix},\, \begin{bmatrix}-\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} &0\end{bmatrix} } \end{equation*} is a basis for $\krn{T}$. Just for fun, you might evaluate $T$ with each of these two basis vectors and verify that the output is the zero polynomial (Exercise MR.C10).