Finding bases such that the matrix representation is a block matrix where one submatrix is the identity matrix Problem: Let $L: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a linear map with \begin{align*} [L]_{\alpha}^{\beta} = \begin{pmatrix} 2 & 3 \\ 4 & 6 \\ 6 & 9 \end{pmatrix} \end{align*} as the matrix representation with respect to the standard bases $\alpha$ for $\mathbb{R}^2$ and $\beta$ for $\mathbb{R}^3$. Now find a basis $\mathcal{V}$ for $\mathbb{R}^2$ and a basis $\mathcal{W}$ for $\mathbb{R}^3$ such that \begin{align*} [L]_{\mathcal{V}}^{\mathcal{W}} = \begin{pmatrix} \mathbb{I}_r & O \\ O & O \end{pmatrix} \end{align*} where each $O$ represents a block matrix with all zeroes and/or does not appear.
Attempt at solution: I'm not sure if I understand what's being asked here. Since the matrix $[L]_{\alpha}^{\beta}$ is with respect to the standard bases, we have $L(1,0) = (2,4,6)$ and $L(0,1) = (3,6,9)$. From this I determined the general formula of $L$ as $L(x,y) = (2x + 3y, 4x + 6y, 6x + 9y)$.
Now, I assume the condition $\begin{pmatrix} \mathbb{I}_r & O \\ O & O \end{pmatrix}$ means we want a matrix representation of the form $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$. Let $\mathcal{V} = \left\{v_1, v_2\right\}$ and $\mathcal{W} = \left\{w_1, w_2, w_3\right\}$ be the other two bases we seek. So we want \begin{align*} L(v_1) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = 1 w_1 + 0 w_2 + 0 w_3 \qquad \text{and} \qquad L(v_2) = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = 0 w_1 + 1 w_2 + 0 w_3 \end{align*} Now I'm stuck, and I don't know how to find a concrete example of $\mathcal{V}$ and $\mathcal{W}$ that would fit with the explicit formula for $L$ I found earlier.
 A: so as I mentioned in the comment, we first choose a proper basis $B$ in $\mathbb{R}^2$, we take 
$$
B=\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1.5 \\ 1 \end{pmatrix}\right\}
$$
and first keep the standard basis $A$ in $\mathbb{R}^3$, therefore our matrix looks like
$$
L^{B}_A=\begin{pmatrix} 2 & 0 \\ 4 & 0 \\ 6 & 0 \end{pmatrix}
$$
Next thing we do is to find the proper basis $C$ in $\mathbb{R}^3$, this is rather obvious and we take 
$$
C=\left\{\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right\}
$$
and therefore our linear maps L looks with respect to the basis $B$ and $C$
$$
L^B_C=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}
$$
and that's it. By the way, the choice of 
$ \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ is arbitrary, you just need to fill up $\left\{\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}\right\}$ to a basis of $\mathbb{R}^3$.
The complete line of computation looks like 
$$
\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}=
\begin{pmatrix} 2 & 0 & 0\\ 4 & 1 & 0\\ 6 & 0 & 1 \end{pmatrix}^{-1}
\begin{pmatrix} 2 & 3 \\ 4 & 6 \\ 6 & 9 \end{pmatrix}
\begin{pmatrix} 1 & -1.5 \\ 0 & 1 \end{pmatrix}
$$
bests
