A 2 dimensional Euclidean space is represented by two different coordinate systems: the Cartesian system $(x_1,x_2)$ and an alternative system $(\xi^1,\xi^2)$ where $$x_1=\frac{1}{\sqrt2}(\xi^1+\xi^2)$$ $$x_2=\frac{1}{\sqrt2}(\xi^1-\xi^2)$$ A quantity $e(i,j)$ is defined such that in both coordinate systems
$$ e(1,1)=e(2,2)=0,$$ $$ e(1,2)=1$$
$$e(2,1)=-1$$
Do the quantities $e(i,j)$ represent the components of a tensor?
I am not sure how to start. Can anyone help me plz?
To show that this quantity is a tensor we need to show that coordinate representation of the quantity is invariant under a change of coordinates.
The question stipulates that, for the vectors with coordinates $\left(\begin{array}{c}x_1\\x_2\end{array}\right)$ and $\left(\begin{array}{c}y_1\\y_2\end{array}\right)$ in the cartesian system, the matrix representation of the quantity implies
$$ \left(x_1,\ x_2 \right)\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)\left(\begin{array}{c}y_1\\y_2\end{array}\right){}={}x_1y_2-x_2y_1\,. $$ Note, also, that this form must also hold for the alternative coordinate system as well. That is, if the vectors have the alternative coordinates $\left(\begin{array}{c}\xi_1\\\xi_2\end{array}\right)$ and $\left(\begin{array}{c}\zeta_1\\\zeta_2\end{array}\right)$, respectively, then we must also have $$ \left(\xi_1,\ \xi_2 \right)\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)\left(\begin{array}{c}\zeta_1\\\zeta_2\end{array}\right){}={}\xi_1\zeta_2-\xi_2\zeta_1\,, $$ for this to be a tensor. However, by direct substitution using the given change of coordinates transformation, we see that $$ \begin{eqnarray*} x_1y_2-x_2y_1&{}={}&\dfrac{1}{2}\bigg(\left(\xi_1+\xi_2\right)\left(\zeta_1-\zeta_2\right)-\left(\zeta_1+\zeta_2\right)\left(\xi_1-\xi_2\right)\bigg)\newline &&\newline &{}={}&\zeta_1\xi_2-\zeta_2\xi_1\newline &&\newline &{}\ne{}&\xi_1\zeta_2-\xi_2\zeta_1\,. \end{eqnarray*} $$
Therefore, the quantity is not a tensor!