## Equation with linear mapping

For mapping $$f: \mathbb{R}^3 \to \mathbb{R}^3$$ given by $$f(x, y, z) = (x - y, y - z, z - x)^T$$ determine the kernel $$\ker(f)$$, the image $$f(\mathbb{R}^3)$$ and the pre-image $$f^{-1}((1, 1, -2)^T)$$.
• #### 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 $$f^{-1}((1, 1, -2)^T)=(2{,}1,0)^T+span((1{,}1,1)^T)$$.

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 $$f^{-1}((1, 1, -2)^T)=(2{,}1,0)^T+span((1{,}1,1)^T)$$.