Associated matrix with respect of a basis Given the following linear transformation:
$$f : \mathbb{R}^2 \to \mathbb{R}^3 | f(1; 0) = (1; 1; 0), f(0; 1)=(0; 1; 1)$$
find the associated matrix of $f$ with respect of the following basis:
$R = ((1; 0); (0; 1))$ of $\mathbb{R}^2$
and
$R^1 =
((1; 0; 0); (1; 1; 0); (0; 1; 1))$ of $\mathbb{R}^3$
--
I've found the associated matrix of $f$ with respect of the basis $R$, and it appears to be the following:
$$
\begin{bmatrix}1&0\\1&1\\0&1\end{bmatrix}
$$
Is it correct? How can I calculate the associated matrix of $f$ with respect of the basis $R^1$?
 A: Your matrix
$$
\begin{bmatrix}1&0\\1&1\\0&1\end{bmatrix}
$$
represent the function $f$ if we use the standard basis $R_3=((1,0,0);(0,1,0);(0,0,1))$ of $\mathbb{R}^3$.
If you use the basis $R_3^1=((1,0,0);(1,1,0);(0,1,1))$ note that $(1,0)$ become the second element of this new basis, and $(0,1)$ become the third element, so the matrix become:
$$
\begin{bmatrix}0&0\\1&0\\0&1\end{bmatrix}
$$ 
A: I suggest and proceed to use different letters for the bases, let's call them $\mathcal B=(b_1,b_2)$ the one given in $\Bbb R^2$ and $\mathcal C=(c_1,c_2,c_3)$ the other one given in $\Bbb R^3$.

The $i$th column of the matrix we are looking for is the image of $b_i$ under the given map, coordinated in basis $\mathcal C$.

Now, $f(b_1)=\pmatrix{1\\1\\0}=c_2$, this has coordinates $\pmatrix{0\\1\\0}$ in $\mathcal C$ as $c_2={\bf 0}c_1+{\bf 1}c_2+{\bf 0}c_3$.
A: No. The column vectors of the matrix are the coordinates, in basis $\mathcal R_1$, of the images of vectors in $\mathcal R$. Now the basis in $\mathbf R^2$ didn't change, it is still the canonical basis. The image of the first vector is $\begin{pmatrix}1\\1\\0\end{pmatrix}$, i.e. the second vector of  $\mathcal R_1$. Similarly the image of the second vector is the third vector of $\mathcal R_1$. Hence matrix of $f$ relative to $\mathcal R$ and $\mathcal R_1$ is
$$ \begin{pmatrix}0&0\\1&0\\0&1\end{pmatrix}.$$
A: Assuming your associated Matrix is with respect to $R$ and the standart basis of $\mathbb{R}^3$, it is correct. But your associated Matrix is always depending on both bases of your vector spaces.
As $f(1;0)=0\cdot(1;0;0)+1\cdot(1;1;0)+0\cdot (0;1;1)$  
and $f(0;1)=0\cdot(1;0;0)+0\cdot(1;1;0)+1\cdot(0;1;1)$  
your Matrix with respect to $R,R^1$ is     $\begin{pmatrix}0&0\\1&0\\0&1\end{pmatrix}$
