Answer:
S is not the subspace of

Explanation:
Let us suppose two vectors u and v belong to the S such that the property of xy≥0 is verified than
![v=\left[\begin{array}{c}0 \\1\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/sfu7oajfao8vt0mprac9fppuxhakgeng52.png)
Both the vectors satisfy the given condition as follows and belong to the S

Now S will be termed as subspace of R2 if
- u+v also satisfy the condition
- ku also satisfy the condition
Taking u+v
![u+v=\left[\begin{array}{c}-1 \\0\end{array}\right]+\left[\begin{array}{c}0 \\1\end{array}\right]\\u+v=\left[\begin{array}{c}-1+0 \\0+1\end{array}\right]\\u+v=\left[\begin{array}{c}-1 \\1\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/7pj1odxt5h5lgecfi23xydn6srs7om6rxq.png)
Now the condition is tested as

This indicates that the condition is not satisfied so S is not the subspace of
