# Find a basis for $W^\perp$ for $W=\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_1-x_2-x_3=0\}$

Suppose $W=\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_1-x_2-x_3=0\}$. Find a basis for $W^\perp$.

I started by finding a basis for $W$. So since all elements of $W$ are expressible as $(x_2+x_3,x_2,x_3)=x_2(1,1,0)+x_3(1,0,1)$, then a basis for $W$ is $\{(1,1,0),(1,0,1)\}$.

I am not sure, however, how to find the basis for $W^\perp$. I know that all elements in $W^\perp$ must have an inner product with $(x_2+x_3,x_2,x_3)$ equal to zero.

Notice from the definition of $W$,

all elemenets of $W$ must satisfies

$$x_1-x_2-x_3=0$$

$$\langle (1,-1,-1), x\rangle = 0, \forall x \in W$$

$\{(1,-1,-1)\}$ is a basis for $W^\perp$.

• Shouldn’t it be $(1,-1,-1)$ in the last line? May 2 '18 at 18:39
• ya, thank you so much. May 2 '18 at 18:40

It is sufficient to find $(x_1,x_2,x_3) \neq (0,0,0)$ such that

$(x_1,x_2,x_3) \perp (1,1,0)$

$(x_1,x_2,x_3) \perp (1,0,1)$

i.e. $x_1+x_2=0$ and $x_1+x_3=0$

Hence $(a, -a, -a)$ for some $a\neq 0$ is basis of $W^\perp$ as dim $W^\perp =1$.