find matrix of linear operator in basis

Question

Linear operator $\mathcal{A}$ transforms vectors $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$ into $\mathbf{f}_1,\mathbf{f}_2,\mathbf{f}_3$ so:

$$\mathcal{A}(\mathbf{e}_1)= \mathbf{f}_1\\\mathcal{A}(\mathbf{e}_2)= \mathbf{f}_2\\\mathcal{A}(\mathbf{e}_3)= \mathbf{f}_3$$

Find matrix of the linear operator in $\langle \mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3 \rangle$

where:

$$\mathbf{e}_1=(1,-2,2), \quad \mathbf{e}_2=(-1,2,0),\quad \mathbf{e}_3=(1,3,1) \\ \mathbf{f}_1=(-2,9,-1), \quad \mathbf{f}_2=(2,1,3), \quad \mathbf{f}_3=(-1,7,1)$$

I more confused about task question, I do not know how to approach it, only one though I have is to find transformation matrix $\mathbf{e} \rightarrow \mathbf{f}$

But I am not sure if it is right understanding, give me right direction, please

Answer 1

$A:\mathbb{R}^{3}\to \mathbb{R}^{3}$ is a linear transformation sending $e_{1}\to f_{1}, e_{2}\to f_{2}, e_{3}\to f_{3}$. Now you are given to find the matrix of the map w.r.t $\{\ e_{1}, e_{2}, e_{3} \}\ $. So express each $f_{i}$ as a linear combinations of $e_{i}$'s. And then you can easily write the matrix by writing the coefficient column-wise.

So, $A(e_{1})=(-2,9,-1)=c_{1}(1,-2,2)+c_{2}(-1,2,0)+c_{3}(1,3,1)$

So, you have $$c_{1}-c_{2}+c_{3}=-2$$

$$-2c_{1}+2c_{2}+3c_{3}=9$$

$$2c_{1}+c_{3}=-1$$

Solve to find the unknowns.Similarly you have to do for the other two images.

Answer 2

HINT: the matrix $A$ of $T$ under the basis $e_1,e_2,e_3$ in the domain and $f_1,f_2,f_3$ in the codomain is defined by

$$Te_k=\sum_{j=1}^3A_{j,k}\,f_j,\quad k\in[3]$$

where $A_{j,k}$ are the coefficients of $A$.