Uniqueness of traceless symmetric operator Let $S:V^2\to V^2$ be a symmetric operator defined into a $2$-dimensional vector space $V$. Suppose that $S$ is traceless, that is, the trace of $S$ is null and for $x,y\in V$, such that $x\neq 0$, we have $Sx=y$.
My question is: There is $R:V^2\to V^2$ a traceless symmetric operator, $R\neq S$, such that $Rx=y$? Ie, the condition $x\neq 0$ and $Sx=y$ becomes $S$ unique?
 A: I am assuming $V$ is a vector space over a "real" field, that is, over some field $\Bbb F$ with $\Bbb Q \subset \Bbb F \subset \Bbb R$; this assumption is in accord with my understanding of the usual meaning of he word "symmetric" as applied to linear operators, which word is, to my knowledge, most typically used in the "real" context.
Having said this, we may resolve this question by representing $S$ and $R$ in matrix form.  Since $S$ is traceless and symmetric, we may write
$S = \begin{bmatrix} a & b \\ b & -a \end{bmatrix}, \tag 1$
where $a, b \in \Bbb F$; now if
$x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \in \Bbb F^2, \tag 2$
and 
$y = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} \in \Bbb F^2, \tag 3$
the equation
$Sx = y \tag 4$
becomes
$\begin{bmatrix} a & b \\ b & -a \end{bmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}, \tag 5$
which when written out in terms of components is
$ax_1 + bx_2 = y_1, \tag 6$
and
$bx_1 - ax_2 = y_2; \tag 7$
we observe that (6)-(7) may be written as a two-dimensional system in $a$ and $b$, setting
$X\begin{pmatrix} a \\ b \end{pmatrix} = \begin{bmatrix} x_1 & x_2 \\ -x_2 & x_1 \end{bmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}; \tag 8$
furthermore, 
$\det X = x_1^2 + x_2^2 \ne 0 \; \text{if} \; x \ne 0; \tag 9$
thus $X^{-1}$ exists; it follows that $a$ and $b$, and hence $S$, are uniquely determined by (4), so we may conclude that
$R = S. \tag {10}$
NB:  Does a similar result hold over complex fields $E$ such that $\Bbb Q(i) \subset \Bbb E \subset \Bbb C$?  Perhaps if we replace the word "symmetric" with "hermitian"? inquiring minds want to  know . . . End of Note.
Note Added in Edit, Wednesday 27 September 2017 9:23 PM PST:  This in response to a further question posed by our OP Irddo in his comment(s) below.
If instead of hypothesizing $S$ be traceless, we wish to find $S$ such that $\text{Tr} S = t \in \Bbb F$, we can set 
$T = \begin{bmatrix} 0 & 0 \\ 0 & t \end{bmatrix} \tag{11}$ and look at the equation
$(S + T) x = y \tag{12}$
for given $x$ and $y$.  Then
$Sx = y - Tx, \tag{13}$
and we now obtain a system
$X\begin{pmatrix} a \\ b \end{pmatrix} = y - Tx, \tag{14}$
which may be solved as before:
$\begin{pmatrix} a \\ b \end{pmatrix} = X^{-1}(y - Tx). \tag{15}$
One caveat:  we have to be a little careful to avoid the situation 
$\det(S + T) = ta - (a^2 + b^2) = 0, \tag{15}$
or we won't necessarily be able to choose $x$ and $y$ independently.  Also, $a$ and $b$ will now exhibit a nonlinear dependence on $x$; I leave the details to my readers to discover.  End of Note.
A: If $R,S:V\to V$ are both traceless symmetric operators such that $Rx=Sx$ for some nonzero $x$, then $R-S:V\to V$ is a traceless symmetric operator such that $(R-S)x=0$.  If we can show that $R-S=0$, then we know $R=S$.  Let $T=R-S$.
We have $Tx=0$, and since $T$ is symmetric, $x^tT=0$, too.  Let $y\in V$ be a vector linearly independent of $x$.  So far we have $x^tTx=0$, $x^tTy=0$, and $y^tTx=0$.
Trace does not depend on the basis, so $\operatorname{tr}T=\frac{1}{x^tx}x^tTx+\frac{1}{y^ty}y^tTy=\frac{1}{y^ty}y^tTy$.  Tracelessness implies the last thing we need: $y^tTy=0$. Note: this depends on these denominators not being zero, which is not a problem over $\mathbb{R}$.
Since $\{x,y\}$ is a basis of $V$, we can conclude $T=0$, and therefore $R=S$.
This argument carries over immediately to the case $\operatorname{tr}R=\operatorname{tr}S$ because we did not rely on the tracelessness of $R$ or $S$, only that $\operatorname{tr}(R-S)=0$.
