# Complete an orthogonal basis of $\mathbb{R}^4$

Consider $$\mathbb{R}^4$$ defined by canonical dot product. Given two vectors $$u=(1,1,-2,0)$$ and $$v=(1,1,1,1)$$ I need to find an orthogonal basis of $$\mathbb{R}^4$$ using the vectors $$u$$ and $$v$$.

I know that generally vectors $$x=(x,y,z,w)$$ orthogonal to $$u$$ and $$v$$ need to satisfy the following condition:

$$\begin{cases}\langle x,u \rangle=0 \\ \langle x,v \rangle=0\end{cases} \rightarrow \begin{cases}x=-t_1-\frac{2}{3}t_2\\ y=t_1 \\ z=-\frac{1}{3}t_2\\w=t_2\end{cases}$$

So a basis of $$\{u,v\}^\perp$$ would be $$\{w_1=(1,-1,0,0), w_2=(2,0,1,-3)\}$$. Then why in order to find an orthogonal basis to $$\mathbb{R}^4$$ we need to compute Gram-Schmidt algorithm to the vectors $$w_1$$ and $$w_2$$?

• Because they are not orthogonal. Commented May 23, 2019 at 14:06
• $w_1$ and $w_2$ are both normal to $u$ and $v$, which is good. But if you want an othogonal basis, $w_1$ and $w_2$ need to be orthogonal to each other. Commented May 23, 2019 at 14:07
• Oh right, now I see that vectors $w_1$ and $w_2$ are not orthogonal to each other. Thanks! Commented May 23, 2019 at 14:08
• Technically you don't have to use Gram-Schmidt, but the higher the dimension, the harder it will be to find an orthogonal base without Gram-Schmidt. Commented May 23, 2019 at 14:10

Vectors $$w_1$$ and $$w_2$$ are not orthogonal to each other.