Answer to Question #106261 in Linear Algebra for Sourav Mondal

Question

Let W {(x, y, z) R3: x + y + z = 0}. Check
if W is a subspace of R3 . Find a non-zero
subspace U of R3 so that W(intersection)U = (0).

Answer 1

$W=\{(x,y,z)\in\mathbb{R}^3: x+y+z=0\}$

1) $u=(0,0,0): \ u\in W?$

$0+0+0=0\ \Rightarrow u\in W$

2) $u=(x_1,y_1,z_1), \ v=(x_2,y_2,z_2), u,v\in W: \ \ u+v\in W?$

$u+v=(x_1+x_2,y_1+y_2,z_1+z_2)$

$(x_1+x_2)+(y_1+y_2)+(z_1+z_2)=(x_1+y_1+z_1)+(x_2+y_2+z_2)=0+0=0$

$\Rightarrow \ u+v\in W$

3) $u=(x,y,z)\in W, \lambda \in \mathbb{R}: \ \lambda u\in W?$

$\lambda x+\lambda y+\lambda z=\lambda(x+y+z)=\lambda \cdot 0=0\Rightarrow \ \lambda u\in W$

Therefore, $W$ is a subspace of $\mathbb{R}^3$

$U=\{(x,y,z)\in \mathbb{R}^3: x=y=z\}$ , $U$ is non-zero because $(1,1,1)\in U$

1) $u=(0,0,0): \ u\in U?$

$0=0=0 \ \Rightarrow u\in U$

2) $u=(x_1,y_1,z_1), \ v=(x_2,y_2,z_2), u,v\in U: \ \ u+v\in U?$

$x_1=y_1=z_1, \ x_2=y_2=z_2\ \Rightarrow x_1+x_2=y_1+y_2=z_1+z_2\ \Rightarrow u+v\in U$

3) $u=(x,y,z)\in U, \lambda \in \mathbb{R}: \ \lambda u\in U?$

$x=y=z\ \Rightarrow \lambda x =\lambda y=\lambda z\ \Rightarrow \lambda u\in U$

Therefore, $U$ is a (non-zero) subspace of $\mathbb{R}^3$

Suppose that $u\in W \cap U$

$\begin{cases} x+y+z=0 \\ x=y=z \end{cases} \quad \begin{cases} 3x=0 \\ x=y=z \end{cases} \Rightarrow x=y=z=0, \ \ u=(0,0,0)$

Therefore, $W\cap U=\{0\}$