Answer
Add the two vectors $\overrightarrow{u} = \begin{bmatrix} 2\\ 2 \end{bmatrix}$ and $\overrightarrow{v} = \begin{bmatrix} -3\\ -1 \end{bmatrix}$
Work Step by Step
Vector summation in $\mathbb{R}^2$ is given by $\overrightarrow{u+v} = (u_1 + v_1, u_2 +v_2)$. From this we can learn that the relation between these two summations must result in either of these types of vectors:
$\overrightarrow{u+v} = \begin{bmatrix} u_1+v_1<0\\ u_2+v_2>0 \end{bmatrix}$
or
$\overrightarrow{u+v} = \begin{bmatrix} u_1+v_1>0\\ u_2+v_2<0 \end{bmatrix}$
Since these are the areas, in which $H$ is not defined. I will make use of the vectors $\overrightarrow{u} = \begin{bmatrix} 2\\ 2 \end{bmatrix}$ and $\overrightarrow{v} = \begin{bmatrix} -3\\ -1 \end{bmatrix}$ to achive this.
$$\overrightarrow{u+v} = \begin{bmatrix} u_1+v_1\\ u_2+v_2 \end{bmatrix},$$
$$\overrightarrow{u+v} = \begin{bmatrix} 2+(-3)\\ 2+(-1) \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}.$$
Which is a vector not in $H$
