Kerne, image, preimage
Task number: 4530
Solution
The matrix of the mapping with respect to the standard bases is \[ [f]_{E,E}= \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ -1 & 0 & 1 \\ \end{pmatrix} \sim\sim \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{pmatrix} \]
The kernel of the mapping is the set of solutions of the homogeneous system of equations with this matrix. Here \(z\) is a free variable, and therefore the solution set is \(\ker(f)=\operatorname{span}((1{,}1,1)^\mathsf T)\).
The range is the space generated by the columns corresponding to the basic variables, i.e. \(f(\mathbb R^3)=\operatorname{span}((1{,}0,-1)^\mathsf T,(-1{,}1,0)^\mathsf T)\).
The preimage of the vector \((1, 1, -2)^\mathsf T\) is obtained as the affine space formed by adding one particular solution of the equation \(f(\boldsymbol x)=(1, 1, -2)^\mathsf T\) to the kernel of the mapping. This is the solution of the non-homogeneous system with the right-hand side \((1, 1, -2)^\mathsf T\):
\[ \left( \begin{array}{ccc|c} 1 & -1 & 0 & 1\\ 0 & 1 & -1 & 1\\ -1 & 0 & 1 & -2\\ \end{array} \right) \sim\sim \left( \begin{array}{ccc|c} 1 & -1 & 0 & 1\\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0\\ \end{array} \right) \sim \left( \begin{array}{ccc|c} 1 & 0 & -1 & 2 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0\\ \end{array} \right) \]Therefore \(f^{-1}((1, 1, -2)^\mathsf T)=(2{,}1,0)^\mathsf T+\operatorname{span}((1{,}1,1)^\mathsf T)\).
