Show that H=(x,y,z)inR^(3)∣x+2y=0 is a subspace of R^(3).

Check zero vector: To determine if H is a subspace of R^3, we need to check if it satisfies three properties: it must contain the zero vector, it must be closed under vector addition, and it must be closed under scalar multiplication.Verify vector addition closure: First, we check if H contains the zero vector. The zero vector in R^3 is (0,0,0). If we substitute x=0 and y=0 into the equation x+2y=0, we get 0+2(0)=0, which is true.Confirm scalar multiplication closure: Second, we check if H is closed under vector addition. Take any two vectors (x1,y1,z1) and (x2,y2,z2) in H. According to the definition of H, we have x1+2y1=0 and x2+2y2=0. We need to check if their sum (x1+x2, y1+y2, z1+z2) is also in H. We calculate (x1+x2)+2(y1+y2) = (x1+2y1)+(x2+2y2) = 0+0 = 0, which means the sum is also in H.Third, we check if H is closed under scalar multiplication. Take any vector (x,y,z) in H and any scalar k. According to the definition of H, we have x+2y=0. We need to check if k(x,y,z) = (kx,ky,kz) is also in H. We calculate kx+2(ky) = k(x+2y) = k(0) = 0, which means the scalar multiple is also in H.

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Show that H=(x,y,z)inR^(3)∣x+2y=0 is a subspace of R^(3).

Show that $\mathrm{H}=(x, y, z) \in \mathbb{R}^{3} \mid x+2 y=0$ is a subspace of $\mathbb{R}^{3}$ .

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Q. Show that

\mathrm{H}=(x, y, z) \in \mathbb{R}^{3} \mid x+2 y=0

is a subspace of

\mathbb{R}^{3}

Check zero vector: To determine if $H$ is a subspace of $\mathbb{R}^3$ , we need to check if it satisfies three properties: it must contain the zero vector, it must be closed under vector addition, and it must be closed under scalar multiplication.
Verify vector addition closure: First, we check if $H$ contains the zero vector. The zero vector in $\mathbb{R}^3$ is $(0,0,0)$ . If we substitute $x=0$ and $y=0$ into the equation $x+2y=0$ , we get $0+2(0)=0$ , which is true.
Confirm scalar multiplication closure: Second, we check if $H$ is closed under vector addition. Take any two vectors $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in $H$ . According to the definition of $H$ , we have $x_1+2y_1=0$ and $x_2+2y_2=0$ . We need to check if their sum $(x_1+x_2, y_1+y_2, z_1+z_2)$ is also in $H$ . We calculate $(x_1+x_2)+2(y_1+y_2) = (x_1+2y_1)+(x_2+2y_2) = 0+0 = 0$ , which means the sum is also in $H$ .
Confirm scalar multiplication closure: Second, we check if $H$ is closed under vector addition. Take any two vectors $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in $H$ . According to the definition of $H$ , we have $x_1+2y_1=0$ and $x_2+2y_2=0$ . We need to check if their sum $(x_1+x_2, y_1+y_2, z_1+z_2)$ is also in $H$ . We calculate $(x_1+x_2)+2(y_1+y_2) = (x_1+2y_1)+(x_2+2y_2) = 0+0 = 0$ , which means the sum is also in $H$ .Third, we check if $H$ is closed under scalar multiplication. Take any vector $(x_1,y_1,z_1)$ $2$ in $H$ and any scalar $(x_1,y_1,z_1)$ $4$ . According to the definition of $H$ , we have $(x_1,y_1,z_1)$ $6$ . We need to check if $(x_1,y_1,z_1)$ $7$ is also in $H$ . We calculate $(x_1,y_1,z_1)$ $9$ , which means the scalar multiple is also in $H$ .