# Finding a basis for the orthhongonal complement

So I have a question that says,

Give a basis for the orthogonal complement of each of the following subspaces of $\mathbb{R}^4$

$w=\left \{ \boldsymbol{x} \in\mathbb{R}^4:x_1+3x_3+4x_4=0, x_2+2x_3-5x_4=0 \right \}$

I thought to find a basis for the orthogonal complement when the subspace is given you just solve for the nullspace of the matrix. However, the answer is suppose to be

{$(1,0,3,4),(0,1,2,-5)$}.

But if I solve for the nullspace I get

{ ${(-3,2,1,0)(-4,5,0,1)}$}

So I'm confused as to what I'm doing wrong? Could someone explain?

• Nullspace of which matrix? What it looks like is that you found basis for the original subspace, not its orthogonal complement. – Ennar Dec 7 '16 at 5:52
• Wouldn't you just create a matrix from $x_1+3x_3+4x_4=0,x_2+2x_3−5x_4=0$? – user6259845 Dec 7 '16 at 5:54
• Nullspace of that matrix is the original subspace, since it is given by homongeneous equations, and not its orthogonal complement. Hint: youw equations can be written as $\langle (1,0,3,4),x\rangle = 0$ and $\langle (0,1,2,-5),x\rangle = 0$ where $x$ is in the subspace. – Ennar Dec 7 '16 at 5:58
• Yes, I originally saw if you just take out the x from the equations you get (1,0,3,4) and (0,1,2,-5), but I was more wondering is that how your suppose to solve it? Because just taking out the x doesn't seem like a correct process to solving it. – user6259845 Dec 7 '16 at 6:05
• The two given equations can be written as $(1,0,3,4)\cdot(x_1,x_2,x_3,x_4)=0$ and $(0,1,2,-5)\cdot(x_1,x_2,x_3,x_4)=0$. What do you know about pairs of vectors whose dot product vanishes? – amd Dec 7 '16 at 8:35

Let $S =\{x\in \Bbb R^4\mid x_1+3x_3+4x_4=0, x_2+2x_3-5x_4=0\}$. Obviously, $S$ is space of solutions of linear equation $Ax = 0$ where, $$A = \begin{pmatrix}1& 0& 3& 4\\ 0& 1& 2& -5\end{pmatrix}$$
Here comes your error, you solved equation $Ax = 0$ and found vectors $(-3,2,1,0)$ and $(-4,5,0,1)$ (which is wrong, by the way, but that's not important), but you incorrectly interpret that the solution space is orthogonal to $S$. On the contrary, the solution space is equal to $S$, as I've already said.
So, how do you find $S^\perp$? Well, by definition $S^\perp = \{x\in\Bbb R^4\mid \langle x,y\rangle = 0, \forall y\in S\}$.
Let us denote $u = (1,0,3,4)$, $v = (0,1,2,-5)$, $S' =\operatorname{span}\{u,v\}$. Then we can rewrite $$S = \{x\in\Bbb R^4\mid \langle u,x\rangle = 0, \langle v, x\rangle = 0 \} \stackrel{(*)}{=}\{x\in\Bbb R^4\mid \langle x,y\rangle = 0,\forall y\in S'\} = (S')^\perp$$ and hence, $S^\perp = ((S')^\perp)^\perp = S'$.
$(*)$ This is because of bilinearity of inner product.