On $\mathbb{R}^3$, define $$A(x) = (x \cdot a)a + 2(x \cdot b)b$$ Here $$a = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, b = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$ Show that A is linear and find its matrix in the usual basis.
I did out the above equation and got $A(x) = (x*a \cdot a^2) + 2(x*b + b^2)$. This reduces to $0+2x$.
I understand that the usual basis means that $$e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, e_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, e_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$
Did I do the matrix algebra incorrectly or is the answer the actually $2x$? If so, $2x$ is linear, but you cannot reduce that to a basis in $\mathbb{R}^3$.
$$A\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=(x-z)\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)+2y\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c}x-z\\ 2y\\ -x+z\end{array}\right)=\left(\begin{array}{rrr}1 & 0 & -1\\ 0& 2& 0\\ -1& 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$$