# Let $S$ be the subspace of $\Bbb R^3$ spanned by the vector $x= (x_1,x_2,x_3)^T$ and $y= (y_1,y_2,y_3)^T$

(a) Let $$S$$ be the subspace of $$\Bbb R^3$$ Spanned by the vector $$x= (x_1,x_2,x_3)^T$$ and $$y= (y_1,y_2,y_3)^T$$, let $$A =\begin{bmatrix}x_1 &x_2 & x_3 \\ y_1 &y_2 &y_3 \end{bmatrix}$$, show that $$S^⊥=N(A)$$.

(b) Find the orthogonal complement of the subspace of $$\Bbb R^3$$ spanned by $$(1,2,1)^T$$ and $$(1,-1,2)^T$$.

I'm still so confused by the concept of $$S^⊥$$, my book defines $$Y^⊥= \{x∈\Bbb R^n\mid x^T y=0 \text{ for every } y∈Y\}$$. So to solve (a), I think I have to find $$S^T$$, I guess it should be $$X^T y=0$$. Then to find $$N(A)$$, let $$Ax=0$$, but I don't know what to do next, so I really need help to solve this kind of question, thanks!

The rows of $$A$$ are $$x$$ and $$y$$, and so $$S$$ is the row space of A:$$S=\{\alpha x+\beta y:\ \alpha,\beta\in\mathbb{R}\}$$ Therefore, $$S^⊥=\{z\in\mathbb{R^3:\ (\alpha x+\beta y+z)=0}\}$$ for all $$\alpha,\beta$$ ; in particular, $$(x, z) = 0$$ and $$(y, z) = 0$$. But the definition of matrix multiplication says that$$Az=\begin{bmatrix} (x,z) \\ (y,z)\end{bmatrix}=0$$Hence, $$z\in N(A)$$
For the reverse inclusion, suppose that $$z \in N(A)$$. Now $$Az = 0$$. As above, the definition of matrix multiplication gives $$(x, z) = 0 = (y, z)$$. Hence, for all scalars $$\alpha,\beta$$ we have$$(\alpha x+\beta y,z)=\alpha(x,z)+\beta(y,z)=0$$ So, $$z\in S^⊥$$
• @ShadowZ Since, $S$ is the row space of $A$ – Key Flex Dec 4 '18 at 4:49