Let $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ be linear, whose corresponding matrix with respect to the standard base $\hat{e}$ is: $$(f:\hat{e},\hat{e})=\left(\begin{array}{ccc} 1 & 1 & 0\\ 1 & 1 & 2\\ 0 & 0 & 1 \\ \end{array} \right).$$ Find bases $\hat{u},\hat{w}$ of $\mathbb{R}^3$ such that:$$(f:\hat{u},\hat{w})=\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \\ \end{array} \right).$$
Attempt. If $\hat{u}=\{u_1,u_2,u_3\},~\hat{w}=\{w_1,w_2,w_3\}$ are the bases then: $$w_1=f(u_1)=f(ae_1+be_2+ce_3)=af(e_1)+bf(e_2)+cf(e_3)=a(1,1,0)+b(1,1,0)+c(0,2,1)$$ and we get a sysetm of 3 equations and $6$ unknowns, $a,b,c,$ and the coordinates of $w_1$. Is there anothet way to approach the problem, or am I on the right path?
Thanks in advance for the help!