Is $(x_1,x_2,x_3)^T\mapsto(x_1\cdot x_2,x_3)^T$ a linear map?

Question

Is $f:\mathbb{R}^4\rightarrow\mathbb{R}^3,(x_1,x_2,x_3)^T\mapsto(x_1\cdot x_2,x_3)^T$ a linear map ($K=\mathbb{R}$)?

---

I say no, since

\begin{align*} f([x_1,x_2,x_3]^T+[a_1,a_2,a_3]^T)&=\left[ \begin{array}\\ &(x_1+a_1)(x_2+a_2)\\ &x_3+a_3 \end{array}\right]\\ &=\left[ \begin{array}\\ &x_1x_2+a_1x_2+a_2x_1+a_1a_2\\ &x_3+a_3 \end{array}\right]\\ &=(x_1x_2+a_1x_2,x_3)^T+(a_2x_1+a_1a_2,a_3)^T\\ &\neq(x_1\cdot x_2,x_3)^T+(a_1\cdot a_2,a_3)^T. \end{align*}

---

Have I understood this correctly?

Answer 1

You're right that those expressions are not equal in general (and you want it to hold for all possible values). If you suspect it's not a linear map, try to find an easy counterexample and you can avoid the symbolic work with variables: $$\color{blue}{f\left(\left[ \begin{array}\\ 2 \\ 2 \\ 0 \end{array} \right]\right)}=\left[ \begin{array}\\ 4 \\ 0 \end{array} \right] \color{red}{\ne} \left[ \begin{array}\\ 2 \\ 0 \end{array} \right] = \color{blue}{2\,f\left(\left[ \begin{array}\\ 1 \\ 1 \\ 0 \end{array} \right]\right)}$$