Find general formula of linear operator $A \in L(\Bbb R^3, M_2(\Bbb R))$ if it's known that: $$A(1,1,1)= \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}, A(1,1,0)= \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}, A(0,-1,0)= \begin{pmatrix} 1 & 1 \\ 2 & 1 \\ \end{pmatrix}, $$ Find the range and kernel of linear transformation $A$.
For the general formula, first I wrote the canonical basis of $\Bbb R^3$ as a linear combination of the vectors $a_1=(1,1,1), a_2=(1,1,0), a_3=(0,-1,0)$.
I got that: $e_1=a_2+a_3, e_2=-a_3, e_3=a_1-a_2$.
From there, I found that:
$$A(e_1)= A(a_2+a_3)=A(a_2) + A(a_3)= \begin{pmatrix} 2 & 2 \\ 3 & 2 \\ \end{pmatrix}$$ Similarly,$$A(e_2)= \begin{pmatrix} -1 & -1 \\ -2 & -1 \\ \end{pmatrix}, A(e_3)= \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix}, $$ Is this enough to define the general formula of a linear transformation? And how do I find the range and the kernel of it? I know the definitions of these terms, but I still haven't grasped them fully and I don't really know how to find them if the transformation isn't explicitly stated.
Guide:
Just go a bit further to definte the linear transofmration.
$$A(x,y,z)= xA(e_1)+yA(e_2)+zA(e_3)$$
To find the range, find a basis for , $$\operatorname{Span}\left\{\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} ,\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix},\begin{pmatrix} 1 & 1 \\ 2 & 1 \\ \end{pmatrix}\right\}.$$
To find the kernel, solve for
$$A(x,y,z)= xA(e_1)+yA(e_2)+zA(e_3)=\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}$$
You should be solving a linear equations of $4$ equations and $3$ unknowns.