Solution ILT.C28 Chris Black

Since $T$ is given by matrix multiplication, $\krn{T} = \nsp{A}$. We have \begin{align*} \begin{bmatrix} 1 & 2 & 3 & 1\\ 2 & -1 & 1 & 0\\ 1 & 2 & -1 & -2 \\ 1 & 3 & 2 & 1 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 0 & 0\\ 0 & \leading{1} & 0 & 0\\ 0 & 0 & \leading{1} & 0\\ 0 & 0 & 0 & \leading{1} \end{bmatrix} \end{align*} The nullspace of $A$ is $\set{\zerovector}$, so the kernel of $T$ is also trivial: $\krn{T} = \set{\zerovector}$.