Equation with linear mapping
Task number: 4463
Solution
The matrix of the mapping with respect to 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 with this matrix. Here \(z\) is a free variable and therefore the solution set is \(ker(f)=span((1{,}1,1)^T)\)
The range is the subspace generated by the columns corresponding to the leading variables, i.e. \(f(\mathbb{R}^3)=span((1{,}0,-1)^T,(-1{,}1,0)^T)\).
The preimage of the vector \((1, 1, -2)^T\) is the affine space obtained by adding one particular solution of the equation \(f(\mathbf{x})=(1, 1, -2)^T\) to the kernel. This is the solution of the non-homogeneous system with right-hand side \((1, 1, -2)^T\):
\(\left( \begin{array}{ccc|c} 1 & -1 & 0 & 1\\ 0 & 1 & -1 & 1\\ -1 & 0 & 1 & -2\\ \end{array} \right) \sim\sim \left( \begin{array}{ccc|c} 1 & -1 & 0 & 1\\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0\\ \end{array} \right) \sim \left( \begin{array}{ccc|c} 1 & 0 & -1 & 2 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0\\ \end{array} \right) \)This yields \(f^{-1}((1, 1, -2)^T)=(2{,}1,0)^T+span((1{,}1,1)^T)\).
Answer
The kernel of the mapping \(f\) is the one-dimensional subspace \(ker(f)=span((1{,}1,1)^T)\).
A image is a two-dimensional subspace \(f(\mathbb{R}^3)=span((1{,}0,-1)^T,(-1{,}1,0)^T)\).
The pre-image of the given vector is the affine subspace \(f^{-1}((1, 1, -2)^T)=(2{,}1,0)^T+span((1{,}1,1)^T)\).
