linear transformation and standard basis

Question

We have a linear transformation $R^3 \rightarrow R^3$ where in a standard basis there is a matrix:

$$A =\begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 2 & 0 & 1 \end{bmatrix}$$

1.)What matrix does belong to this linear transformation in a basis:

$$B = \left\{\quad\begin{bmatrix} 0 \\ 1\\ 2 \end{bmatrix} ,\ \begin{bmatrix} 1 \\ 0\\ -1 \end{bmatrix} ,\ \begin{bmatrix} 1 \\ 1\\ 0 \end{bmatrix}\quad \right\}$$

My understanding of this:

1.The standard basis consists of vectors that are independent of each other. How can then a matrix be in standard basis. Does the matrix A consist of these vectors?

2.What does the $R^3\rightarrow R^3$ have to do with this problem , what if the question would be $R^3\rightarrow R^2$ or $R^2\rightarrow R^3$.

3. So a linear transformation is a linear mapping that maps the zero vector to a zero vector. I read all the theory behind it, but still cant figure out the initial problem

Answer 1

To find the matrix representing a linear transformation in a given basis, apply the linear transformation to each basis vector in turn and write the result as a linear combination of the basis vectors. The coefficients in that linear combination form a column of the matr8ix.

Here, the first basis vector is $\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}$. A applied to that is $\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{bmatrix}$$\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}= \begin{bmatrix}1 \\ 3 \\ 2 \end{bmatrix}$$= 0\begin{bmatrix}0 \\ 1 \\ 2 \end{bmatrix}- 2\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}+ 3\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}$.

So the first column of the new matrix is $\begin{bmatrix} 0 \\ -2 \\ 3\end{bmatrix}$.