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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 $w_2$?

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    $\begingroup$ Because they are not orthogonal. $\endgroup$
    – Berci
    Commented May 23, 2019 at 14:06
  • $\begingroup$ $w_1$ 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. $\endgroup$
    – bob.sacamento
    Commented May 23, 2019 at 14:07
  • $\begingroup$ Oh right, now I see that vectors $w_1$ and $w_2$ are not orthogonal to each other. Thanks! $\endgroup$
    – Kevin
    Commented May 23, 2019 at 14:08
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    $\begingroup$ 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. $\endgroup$
    – Floris Claassens
    Commented May 23, 2019 at 14:10

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

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