find $A ^\perp$

Question

for every $u,w \in \mathbb{R^4}$ define inner product :$\langle u,w\rangle =\sum_{k=1}^{4} u_k w_k$ and $A= \{x \in \mathbb{R^4} | x=(a,a,b,b),a,b \in \mathbb{R} \}$ now find $A ^\perp$.for non zero a,b we have : $ (\frac{1}{a},\frac{1}{a},\frac{1}{b},\frac{-3}{b} ) \in A^\perp$ and $A \oplus A^\perp = \mathbb{R^4}$

Answer 1

We have for $x=(x_1,x_2,x_3,x_4) \in \mathbb R^4$:

$x \in A^{\perp} \iff \langle x,(a,a,b,b)\rangle=0$ for all $a,b \iff a(x_1+x_2)+b(x_3+x_4)=0$ for all $a,b$.

Then it is easy to see that

$x \in A^{\perp} \iff x_2=-x_1$ and $x_4=-x_3 \iff x \in span\{(1,-1,0,0), (0,0,1,-1)\}$ .