Let \(T\) be the linear operator on \(\RR^3\), the matrix of which in the standard ordered basis is \[ A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ -1 & 3 & 4 \end{bmatrix}. \] Find a basis for the range of \(T\) and a basis for the null space of \(T\).
Solution
The matrix is \[ A=\begin{bmatrix}1&2&1\\0&1&1\\-1&3&4\end{bmatrix}. \] Row reduction gives \[ \operatorname{rref}(A)= \begin{bmatrix}1&0&-1\\0&1&1\\0&0&0\end{bmatrix}. \] Thus the pivot columns are the first two columns of the original matrix, so a basis for the range is \[ \left\{(1,0,-1),(2,1,3)\right\}. \] The null equations are \(x_1-x_3=0\) and \(x_2+x_3=0\). Hence \[ (x_1,x_2,x_3)=t(1,-1,1), \] so a basis for the null space is \[ \{(1,-1,1)\}. \]