Let, \(A=\left[ {\begin{array}{cc} 1 & 2 & 1 & 1 \\ 2 & 1 & 1 & 0 \\ 1 & 2 & 1 & 2 \\ 1 & 2 & 2 & 1 \\ \end{array} } \right]\) and let T: \(\mathbb{C^4}\rightarrow\mathbb{C^4}\) be given by \(T\) (x) = Ax. Find K(T). Is T injective?
Christian’s Response:
Since T is provided by K(T) - N(A), we end up with the following:
\[ \left[ {\begin{array}{cc} 1 & 2 & 1 & 1 \\ 2 & 1 & 1 & 0 \\ 1 & 2 & 1 & 2 \\ 1 & 2 & 1 & 1 \\ \end{array} } \right] \rightarrow \left[ \begin{array}{cc} 1 & 0 & 1/3 & 0 \\ 0 & 1 & 1/3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right] \]
Below we can see both the nullspace and kernal:
Nullspace: \[\left[ {\begin{array}{cc} -1 \\ -1 \\ 3 \\ 0 \\ \end{array} } \right] \]
The Kernel is: \[\left[ {\begin{array}{cc} -1 \\ -1 \\ 3 \\ 0 \\ \end{array} } \right] \]
The kernel is nontrivial and therefore, the linear transformation is not injective.