Mapping given by the image of the basis
Task number: 4531
Consider the linear map \(f : \mathbb R \to \mathbb R\) given by the image of the basis \(B\):
\(f((2, 1, 1)^\mathsf T) = (1, 2, 3)^\mathsf T\), \(f((1, 3, 5)^\mathsf T) = (3, 2, 1)^\mathsf T\), \(f((7, 1, 4)^\mathsf T) = (1, 1, 1)^\mathsf T\).
Determine whether this mapping is injective (if it is not, find different vectors \(\boldsymbol u, \boldsymbol v \in \mathbb R^3\) such, that \(f (\boldsymbol u) = f (\boldsymbol v)\) and also whether \(f\) is surjective (if not, find a vector that does not have preimage, i.e. \(\boldsymbol u \in \mathbb R^3\) such that \(\forall\boldsymbol v \in \mathbb R^3 : f (\boldsymbol v) \ne \boldsymbol u\)).
For both the image and the kernel of this linear mapping, determine the dimension and bases of these two subspaces.
Solution
The matrix of this mapping from basis \(B\) to the standard basis \(E\) is:
\( [f]_{B,E}= \begin{pmatrix} 1 & 3 & 1 \\ 2 & 2 & 1 \\ 3 & 1 & 1 \end{pmatrix} \sim \sim \begin{pmatrix} 1 & 3 & 1 \\ 0 & -4 & -1 \\ 0 & -8 & -2 \end{pmatrix} \sim \sim \begin{pmatrix} 1 & 3 & 1 \\ 0 & 4 & 1 \\ 0 & 0 & 0 \end{pmatrix} \)
This square matrix is singular so the mapping \(f\) is not a bijection and therefore neither injective nor surjective.
The set of solutions of the system with the matrix \([f]_{B,E}\) corresponds to the vectors of coordinates \([\boldsymbol u]_B\) of the vectors \(\boldsymbol u\) belonging to the kernel of \(\ker f\). Every two vectors from the kernel have the same image and thus prove that \(f\) is not injective.
Specifically, by the backward substitution, we can find that \(\ker f=\{\boldsymbol u\colon [\boldsymbol u]_B=p(1{,}1,-4)^\mathsf T,p\in\mathbb R\}\) and hence, for example, the vector \(\boldsymbol u=(2, 1{,}1)^\mathsf T+(1{,}3,5)^\mathsf T-4(7{,}1,4)^\mathsf T=(-25{,}0,-10)^\mathsf T\in \ker f\) and it has the same image as the zero vector: \(f(\boldsymbol u)=f(\mathbf 0)=\mathbf 0\).
The image of the mapping \(f\) is generated by the rows of the matrices:
\( [f]_{B,E}^\mathsf T= \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 1 & 1 & 1 \end{pmatrix} \sim \sim \begin{pmatrix} 1 & 2 & 3 \\ 0 & -4 & -8 \\ 0 & -1 & -2 \end{pmatrix} \sim \sim \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix} \)
For example, the vector \(\boldsymbol v=(0{,}0,1)^\mathsf T\) is one of the vectors that can be used to extend the basis of the image \(f\) to the basis of \(\mathbb R^3\), and therefore this vector does not belong to the image of the mapping \(f\).
Answer
The mapping \(f\) is not injective because, for example, for the vectors \(\boldsymbol u=(5{,}0,2)^\mathsf T\) and \(\boldsymbol v=\mathbf 0\), \(f(\boldsymbol u)=f(\boldsymbol v)\) holds.
The kernel of \(f\) has dimension 1 with possible basis \(\{(5{,}0,2)^\mathsf T\}\).
The mapping \(f\) is not surjective. The image \(\operatorname{Im} f\) has dimension 2, with a possible basis \(\{(1{,}2,3)^\mathsf T,(0{,}1,2)^\mathsf T\}\).
The vector \(\boldsymbol u=(0{,}0,1)^\mathsf T\) is one of the vectors that is not a preimage of any vector \(\boldsymbol \in\mathsf R\).
Comment
Both calculations could be done at once by transforming the block matrix \(([f]_{B,E}^\mathsf T|\mathbf I_3)\) into a row echelon form, since the coefficients of the linearly dependent vectors being sought would be on the right-hand side of the zero row.
