(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!


1 Answer 1


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, $z\in S^⊥$

  • $\begingroup$ Thanks though! I wonder why S⊥ can be written as αx+βy+z=0 without using its definition: x^Ty=0? $\endgroup$
    – Shadow Z
    Commented Dec 4, 2018 at 3:59
  • $\begingroup$ @ShadowZ Since, $S$ is the row space of $A$ $\endgroup$
    – Key Flex
    Commented Dec 4, 2018 at 4:49

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