1. **State the problem:** Determine if the subset $H$ of $V=\mathbb{R}^2$, consisting of all points in the first and third quadrants between the lines $y=2x$ and $y=\frac{x}{2}$, is a subspace of $V$. 2. **Recall subspace criteria:** A subset $H$ of a vector space $V$ is a subspace if and only if: - It contains the zero vector. - It is closed under vector addition. - It is closed under scalar multiplication. 3. **Check if $H$ contains the zero vector:** The zero vector in $\mathbb{R}^2$ is $\mathbf{0} = <0,0>$. Since $0$ lies between $y=2x$ and $y=\frac{x}{2}$ (both lines pass through the origin), and the origin is in both first and third quadrants boundary, $\mathbf{0} \in H$. 4. **Check closure under addition:** Consider vectors $\mathbf{u} = <2,1>$ and $\mathbf{v} = <-1,-3>$, both in $H$. Their sum is: $$\mathbf{u} + \mathbf{v} = <2 + (-1), 1 + (-3)> = <1, -2>$$ The vector $<1,-2>$ lies in the fourth quadrant, which is not part of $H$. Therefore, $H$ is **not closed** under addition. 5. **Check closure under scalar multiplication:** Consider scalar $-1$ and vector $\mathbf{w} = <1, 0.6>$ in $H$. Multiply: $$-1 \times <1, 0.6> = <-1, -0.6>$$ The vector $<-1, -0.6>$ lies in the third quadrant but we must check if it lies between the lines $y=2x$ and $y=\frac{x}{2}$. For $x=-1$, the lines give: $$y=2(-1) = -2$$ $$y=\frac{-1}{2} = -0.5$$ The vector's $y$-coordinate $-0.6$ lies between $-2$ and $-0.5$, so $<-1,-0.6> \in H$. However, the problem states $H$ is not closed under scalar multiplication because multiplying by $-1$ can move vectors out of $H$ if the vector is on the boundary or in the first quadrant. But since $<-1,-0.6>$ is in $H$, the example given in the problem shows a vector $<1,0.6>$ in $H$ and scalar $-1$ producing a vector in $H$, so scalar multiplication is closed for this example. However, the problem states otherwise, so the key is that scalar multiplication by negative scalars can move vectors from first quadrant to third quadrant or vice versa, but since $H$ includes both first and third quadrants between the lines, scalar multiplication by $-1$ keeps vectors in $H$. But the problem's answer is that $H$ is not closed under scalar multiplication, so the example given is $-1, <1,0.6>$, which produces $<-1,-0.6>$, which is in $H$. This suggests the problem's conclusion is that $H$ is not closed under scalar multiplication because the scalar $-1$ can produce vectors outside $H$ for some vectors, but the example given is actually inside $H$. To clarify, scalar multiplication by negative scalars flips vectors between first and third quadrants, but since $H$ includes both, scalar multiplication by $-1$ preserves membership in $H$. Therefore, $H$ is closed under scalar multiplication. 6. **Conclusion:** Since $H$ contains the zero vector and is closed under scalar multiplication but is not closed under addition, $H$ is **not a subspace** of $V$. **Final answer:** $H$ is not a subspace of $V$ because it is not closed under addition.