Basis of a linear transformation If $V$ and $W$ are vectorial spaces of finite dimension, we denote $\mathscr{L}(V,W)$ like the set of all linear maps of $V$  in $W$. Let a linear map $L\in \mathscr{L}(\mathbb{R}^{2},\mathbb{R}^{2})$ defined like $L(x,y)=(-x+y,2x+y)$ for each $(x,y)\in \mathbb{R}^{2}$. Find a basis $S=\{u_{1},u_{2}\}$  of $\mathbb{R}^{2}$ such that
$$[L]_{S} := M_{L}^{S \rightarrow S}= \begin{pmatrix} 0 &amp; -1 \\ -3 &amp; 0 \end{pmatrix}$$
Where $M_{L}^{S \rightarrow T}$ denote the matricial representation of the linear map $L\in \mathscr{L}(V,W)$ respective to the basis $S$ of $V$ and $T$ of $W$.
Let $u_{1}=(v,w)$,$u_{2}=(a,b)\in \mathbb{R}^{2}$. We have that the columns in the matricial representation are $L(u_{1})$ and $L(u_{2})$, so
$$L_{u_{1}}=(-v+w,2v+w)= (0,-3)^{T}$$
and
$$L_{u_{2}}=(-a+b,2a+b)=(-1,0)^{T}$$
Then, $v=w=-1$,$a=\frac{1}{3}$ and $b=-\frac{2}{3}$. Am I right?
In your attempt you are forgetting that the columns of $[L]_S=M^{S\to S}_L$ are not the images of $u_1$ and $u_2$ but their coordinates with respect to $S$. Hence, the conditions on the basis $S=\{u_1, u_2\}$ given by $[L]_S$ should have been
\begin{align}
L(u_1) &amp;= \phantom{-}0 u_1 - 3 u_2, \\
L(u_2) &amp;= -1 u_1 + 0 u_2.
\end{align}
Setting $u_1=(v,w)^T$ and $u_2=(a,b)^T$ as you did, this translates to
\begin{align}
(-v+w,2v+w)^T &amp;= (-3a, -3b)^T, \\
(-a+b,2a+b)^T &amp;= (-v, -w)^T.
\end{align}
This is a homogeneous linear system of four equations and four variables:
\begin{align}
-v+w+3a\phantom{{}+3b} &amp;= 0,\\
2v+w\phantom{{}+3a}+3b &amp;= 0,\\
v\phantom{{}+w}-\phantom{3}a+\phantom{3}b &amp;= 0,\\
w+2a+\phantom{3}b &amp;= 0.
\end{align}
Row reduction yields that this is equivalent to
\begin{align}
v&amp;=a-b,\\
w&amp;=-2a-b.
\end{align}
In addition, $u_1$ and $u_2$ need to be linearly independent, which is equivalent to $vb-wa\neq 0$ and hence to $ab\neq 0$ (by substituting $v$ and $w$).
Picking $a=b=1$ we get $v=0$, $w=-3$ and indeed $u_1=(0,-3)^T$ and $u_2=(1,1)^T$ are linearly independent and yield
\begin{align}
L(u_1) &amp;= L((\phantom{-}0,-3)^T) = (-3,-3)^T = -3 u_2, \\
L(u_2) &amp;= L((\phantom{-}1,\phantom{-}1)^T) = (\phantom{-}0, \phantom{-}3)^T = -1 u_1.
\end{align}
Note that any other choice of non-zero $a,b$ also yields a valid solution.
Notice that the linear map L can be represented by the matrix $A=\left(\begin{matrix}-1&amp;1\\2&amp;1\end{matrix}\right)$ such that $L(x)=Ax$. Using your representation of $u_1=\left(\begin{matrix}v\\w\end{matrix}\right), u_2=\left(\begin{matrix}a\\b\end{matrix}\right)$ to be the new basis, the following equations are gotten:
$\left(\begin{matrix}[Au_1]_\beta&amp;[Au_2]_\beta\end{matrix}\right)= \left(\begin{matrix}0&amp;-1\\-3&amp;0\end{matrix}\right)$, where $\beta$ represents the new basis
This means
$\begin{cases}Au_1=0u_1+-3u_2\\ Au_2=-1u_1+0u_2\end{cases}$
Further elaborating,
$\begin{cases}w-v=-3a\\2v+w=-3b\\b-a=-v\\2a+b=-w\end{cases}$
$v=a-b, w=-2a-b$
Further, as mentioned in Cristoph's post, the basis vectors must be linearly independent. $u_1\ne cu_2\implies \frac v a\ne \frac w b \text{ or } bv-wa\ne0$
Picking $a=b=1,v=0,w=-3$, is a valid basis. Note that this basis is neither orthogonal nor are the vectors normalized. If you attempt to normalize the vectors, the transformation will not work due to the contraints in the equations above.
It is possible, however, to find an orthogonal basis. $\left(\begin{matrix}\frac 1 2 (13+5\sqrt {13})\\\frac 1 2 (13-\sqrt {13})\end{matrix}\right), \left(\begin{matrix}\sqrt {13}\\\frac 1 2 (-13-3\sqrt {13})\end{matrix}\right)$ is one such possible orthogonal basis. It is not possible to find an orthonormal basis. In fact, it is not possible to find a normalized basis i.e. where both vectors have euclidean norm 1.

Test transformation in standard basis and new basis:
$\left(\begin{matrix}-1&amp;1\\2&amp;1\end{matrix}\right)\left(\begin{matrix}1\\-2\end{matrix}\right)=\left(\begin{matrix}-3\\0\end{matrix}\right)$
$\left(\begin{matrix}0&amp;-1\\-3&amp;0\end{matrix}\right)\left(\begin{matrix}1\\1\end{matrix}\right)_\beta=\left(\begin{matrix}-1\\-3\end{matrix}\right)_\beta$
$\left(\begin{matrix}-3\\0\end{matrix}\right)=\left[\left(\begin{matrix}-1\\-3\end{matrix}\right)_\beta\right]_{\{e_1,e_2\}}$