Solution ILT.C29 Chris Black

Since $T$ is given by matrix multiplication, $\krn{T} = \nsp{A}$. We have \begin{align*} \begin{bmatrix} 1 & 2 & 1 & 1\\ 2 & 1 & 1 & 0\\ 1 & 2 & 1 & 2 \\ 1 & 2 & 1 & 1 \end{bmatrix} &\rref \begin{bmatrix} \leading{1} & 0 & 1/3 & 0\\ 0 & \leading{1} & 1/3 & 0\\ 0 & 0 & 0 & \leading{1}\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{align*} Thus, a basis for the nullspace of $A$ is $\set{\colvector{-1\\-1\\3\\0}}$, and the kernel is $\krn{T} = \spn{\set{\colvector{-1\\-1\\3\\0}}}$. Since the kernel is nontrivial, this linear transformation is not injective.