Let \(T\) be the (unique) linear operator on \(C^3\) for which \[ T\epsilon_1 = (1, 0, i), \quad T\epsilon_2 = (0, 1, 1), \quad T\epsilon_3 = (i, 1, 0). \] Is \(T\) invertible?
Solution
Relative to the standard basis, \(T\) has matrix \[ A=\begin{bmatrix}1&0&i\\0&1&1\\ i&1&0\end{bmatrix}. \] The vector \[ X=\begin{bmatrix}-i\\-1\\1\end{bmatrix} e0 \] satisfies \[ AX=\begin{bmatrix}-i+i\\-1+1\\i(-i)-1\end{bmatrix}=0. \] Thus \(T\) has nonzero nullspace, so it is not injective and therefore not invertible.