Solution ILT.C30 Chris Black

We can see without computing that $T$ is not injective, since the dimension of $M_{22}$ is larger than the dimension of $P_2$. However, that does not address the question of the kernel of $T$. We need to find all matrices $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ so that $(a + b) + (a + c)x + (a + d)x^2 = 0$. This means $a + b = 0$, $a + c = 0$, and $a + d = 0$, or equivalently, $b = d = c = -a$. Thus, the kernel is a one-dimensional subspace of $M_{22}$ spanned by $\begin{bmatrix} 1 & -1\\-1&-1 \end{bmatrix}$. Symbolically, we have $\krn{T} = \spn{\set{\begin{bmatrix} 1 & -1\\-1&-1 \end{bmatrix}}}$.